Limits Definition 1 (Limit) . If the values f ( x ) of a function f : - - PDF document

Limits

Definition 1 (Limit). If the values f(x) of a function f : A → B get very close to a specific, unique number L when x is very close to, but not necessarily equal to, a limit point c, we say the limit of f(x), as x approaches c, is L and write limx→c f(x) = L. The number c is referred to as the limit point. Note: This is not a rigorous definition. It is really a preliminary attempt at a definition and will be replaced later by a more rigorous definition. One obvious problem is that the concept of closeness is subjective. Some limits are relatively easy to calculate. Example: limx→5(x2 + 3). It’s obvious that x2 will be close to 25 and x2 + 3 will be close to 28 when x is close to 5, so limx→5(x2 + 3) = 28. Example: limt→7 3t + 9 t − 5 . It’s obvious that 3t is close to 21, 3t + 9 is close to 30, t − 5 is close to 2 and thus 3t + 9 t − 5 is close to 30 2 = 15 when t is close to 7, so limt→7 3t + 9 t − 5 = 15. Most of the limits that come up naturally are not quite so obvious and easy to calculate, but often yield to a little algebraic manipulation and common sense. Example: limx→3 x2 + x − 12 x − 3 . If one examines the numerator and denominator of x2 + x − 12 x − 3 for values of x close to 3, it is obvious both are close to 0 and one cannot immediately estimate the quotient. Such limits are referred to as indeterminate. This particular type of indeterminate is sometimes referred to symbolically as the 0 0 case. We say symbolically because, of course, there is no quotient 0

• 0. It is a short way of referring

to the fact that we are trying to find the limit of a quotient where both the numerator and the denominator are getting very close to 0. However, one may simplify x2 + x − 12 x − 3 , finding x2 + x − 12 x − 3 = (x − 3)(x + 4) x − 3 = x + 4 when x = 3. Since x + 4 is clearly close to 7 when x is close to 3, so must x2 + x − 12 x − 3 and it follows that limx→3 x2 + x − 12 x − 3 = 7. We have implicitly made use of the following theorem, which is an almost immediate consequence of the preliminary definition of a limit. Theorem 1. If f(x) = g(x) for x = c and either f or g has a limit at c, then both must have a limit at c and their limits must be the same. In effect, this is a reiteration of the fact that the value of a function at a limit point has no effect on a limit. We use this theorem in the calculation of the limit almost every time we are faced with an indeterminate. The calculation generally looks like the following: limx→c f(x) = · · · = limx→c g(x) = L.

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The idea is that we start with an indeterminate f(x) and keep simplifying, each time getting either a new form of f(x) or another function that is equal to f(x) except at c, until we have found some function g(x) whose limit is easy to find. For example, the calculation in the previous example might look like the following: limx→3 x2 + x − 12 x − 3 = limx→3 (x − 3)(x + 4) x − 3 = limx→3(x + 4) = 7. To repeat, in a nutshell, we usually wind up calculating limits of indeterminates by sim- plifying until we’ve found a function, equal to the original except at the limit point, which is not indeterminate. Most of the time, the algebraic manipulation we need to do to simplify is in one of the following three categories. (1) Factor and cancel. (2) Rationalize and cancel. (3) Simplifying a complex fractional expression The example we saw fell in the first category, factor and cancel. Consider the following

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Technique 2: limz→4

1 z − 1 4

z − 4 = limz→4

1 z − 1 4

z − 4 · 4z 4z = limz→4 4 − z 4z(z − 4) = limz→4 −1 4z = − 1 16. Here, we observe the complications arose from the denominators of 4 and z in the numer- ator and simply multiplied by 1 in the form 4z 4z. There will be other manipulations that will come up, and sometimes the simplication involved using these techniques may not be quite as straightforward, but most of the examples we run across will fall into one of these three cases.

Different Types of Limits

Besides ordinary, two-sided limits, there are one-sided limits (left-hand limits and right- hand limits), infinite limits and limits at infinity.

One-Sided Limits

limx→5 √ x2 − 4x − 5. One might think that since x2 − 4x − 5 → 0 as x → 5, it would follow that limx→5 √ x2 − 4x − 5 = 0. But since x2 − 4x − 5 = (x − 5)(x + 1) < 0 when x is close to 5 but x < 5, √ x2 − 4x − 5 is undefined for some values of x very close to 5 and the limit as x → 5 doesn’t exist. But we would still like a way of saying √ x2 − 4x − 5 is close to 0 when x is close to 5 and x > 5, so we say the Right-Hand Limit exists, write limx→5+ √ x2 − 4x − 5 = 0 and say √ x2 − 4x − 5 approaches 0 as x approaches 5 from the right. Sometimes we have a Left-Hand Limit but not a Right-Hand Limit. Sometimes we have both Left-Hand and Right-Hand Limits and they’re not the same. Sometimes we have both Left-Hand and Right-Hand Limits and they’re equal, in which case the ordinary limit exists and is the same. f(x) =      x2 if x < 1 x3 if 1 < x < 2 x2 if x > 2. limx→1− f(x) = limx→1+ f(x) = 1, so the left and right hand limits are equal and limx→1 f(x)1. limx→2− f(x) = 8 while limx→2+ f(x) = 4, so the left and right hand limits are different and limx→2 f(x) doesn’t exist.

Limits at Infinity

Suppose we’re interested in estimating about how big 2x x + 1 is when x is very big. It’s easy to see that 2x x + 1 = 2x x(1 + 1

x ) =

2 1 + 1

x

if x = −1 and thus 2x x + 1 will be very close to 2 if x is very big. We write limx→∞ 2x x + 1 = 2 and say the limit of 2x x + 1 is 2 as x approaches ∞.

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Similarly, 2x x + 1 will be very close to 2 if x is very small and we write limx→−∞ 2x x + 1 = 2 and say the limit of 2x x + 1 is 2 as x approaches −∞. Here, small does not mean close to 0, but it means that x is a negative number with a large magnitude (absolute value). A convenient way to find a limit of a quotient at infinity (or minus infinity) is to factor

• ut the largest term in the numerator and the largest term in the denominator and cancel

what one can. limx→∞ 5x2 − 3 8x2 − 2x + 1 = limx→∞ x2(5 − 3

x2)

x2(8 − 2

x + 1 x2) =

limx→∞ 5 − 3

x2

8 − 2

x + 1 x2

= 5 8 limx→∞ 5x − 3 8x2 − 2x + 1 = limx→∞ x(5 − 3

x )

x2(8 − 2

x + 1 x2) =

limx→∞ 5 − 3

x

x(8 − 2

x + 1 x2) = 0

Infinite Limits

If x is close to 1, it’s obvious that 1 (x − 1)2 is very big. We write limx→1 1 (x − 1)2 = ∞ and say the limit of 1 (x − 1)2 is ∞ as x approaches 1. Similarly, limx→1 − 1 (x − 1)2 = −∞. Technically, a function with an infinite limit doesn’t actually have a limit. Saying a function has an infinite limit is a way of saying it doesn’t have a limit in a very specific way. Infinite limits are inferred fairly intuitively. If one has a quotient f(x) g(x), one may look at how big f(x) and g(x) are. For example, If f(x) is close to some positive number and g(x) is close to 0 and positive, then the limit will be ∞. If f(x) is close to some positive number and g(x) is close to 0 and negative, then the limit will be −∞. If f(x) is close to some negative number and g(x) is close to 0 and positive, then the limit will be −∞. If f(x) is close to some negative number and g(x) is close to 0 and negative, then the limit will be ∞.

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Variations of Limits

One can also have one-sided infinite limits, or infinite limits at infinity. limx→1+ 1 x − 1 = ∞ limx→1− 1 x − 1 = −∞

Asymptotes

If limx→∞ f(x) = L then y = L is a horizontal asymptote. If limx→−∞ f(x) = L then y = L is a horizontal asymptote. If limx→c+ f(x) = ±∞ then x = c is a vertical asymptote. If limx→c− f(x) = ±∞ then x = c is a vertical asymptote.

Properties of Limits

When calculating limits, we intuitively make use of some basic properties it’s worth noting. Each can be proven using a formal definition of a limit. We list some of them, usually both using mathematical notation and using plain language. It is the plain language that should be remembered. In each case, unless noted otherwise, we assume the limits written down actually exist. As is usually the case, x, y, z, u, v, s, t will represent variables and a, b, c, k, L will represent constants.

• limx→c k = k

The limit of a constant is that constant.

• limx→c x = c

When x gets close to c, x gets close to c.

• limx→c[kf(x)] = k limx→c f(x)

The limit of a constant times a function is equal to the constant times the limit.

• limx→c[f(x) ± g(x)] = limx→c f(x) ± limx→c g(x)

The limit of a sum or difference is the sum or difference of the limits.

• limx→c[f(x) · g(x)] = [limx→c f(x)] · [limx→c g(x)]

The limit of a product is the product of the limits.

• limx→c

f(x) g(x) = limx→c f(x) limx→c g(x) The limit of a quotient is equal to the quotient of the limits. Of course, this depends

• n the quotient existing, so the denominator on the right must not equal 0.
• limx→c f ◦ g(x) = limx→c f(g(x))

= f(limx→c g(x)) if f is continuous at limx→c g(x). The limit of a composition is the composition of the limits, provided the outside function is continuous at the limit of the inside function. Example: limx→3 √5x + 1 = √ 16 = 4.

• (Squeeze Theorem) If f(x) ≤ g(x) ≤ h(x) in some open interval containing c and

limx→c f(x) = limx→c h(x) = L, then limx→c g(x) = L. If the values of a function lie between the values of two functions which have the same limit, then that function must share that limit.

Continuity

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Definition 2 (Continuity). A function f is said to be continuous at c if limx→c f(x) = f(c). Goemetrically, this corresponds to the absence of any breaks in the graph of f at c. When we’ve calculated limits, most of the time we started with a function that was not continuous at the limit point, simplified to get another function which was equal to the

• riginal function except at the limit point but was continuous at the limit point, and then

it was easy to find the limit of the latter function. Rule of Thumb: Most functions we run across will be continuous except at points where there is an obvious reason for them to fail to be continuous. Examples of Continuous Func- tions:

• Polynomial Functions
• Rational Functions (Quotients of Polynomial Functions) – except where the denom-

inator is 0.

• The exponential function
• The natural logarithm function
• sin and cos
• tan – except at odd multiples of π/2, where it obviously isn’t since tan = sin

cos and cos

takes on the value 0 at odd multiples of π/2. When we perform most algebraic manipulations involving continuous functions, we wind up with continuous functions. Again, the exception is if there’s an obvious reason why the new function wouldn’t be continuous somewhere.

• The sum of continuous functions is a continuous function.
• The difference of continuous functions is a continuous function.
• The product of continuous functions is a continuous function.
• The quotient of continuous functions is a continuous function – except where the

denominator is 0.

• The composition of continuous functions is a continuous function.

Theorem 2 (Intermediate Value Theorem). If a function is continuous on a closed interval [a, b], then the function must take on every value between f(a) and f(b). Corollary 3 (Zero Theorem). If a function is continuous on a closed interval [a, b] and takes on values with opposite sign at a and at b, then it must take on the value 0 somewhere between a and b. The Zero Theorem leads to the Bisection Method, which is a foolproof way of estimate a zero of a continuous function to any desired precision provided we are able to find both positive and negative values of the function. The Bisection Method works as follows. We wish to estimate a zero for a continuous function f. We start by finding points a0, b0 at which the function has opposite sign. Without loss of generality (WLOG), let us assume a0 < b0. We evaluate the function at the midpoint m0 a0 + b0 2 . Unless f(m0) = 0, in which case we have found a zero for f and can stop, f(m0) must have a different sign than either f(a0) or f(b0).

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If f(m0) differs in sign from f(a0), then we know f has a zero on the interval [a0, m0];

• therwise, we know f has a zero on the interval [m0, b0].

Either way, we now have an interval we may denote by [a1, b1] which is half the width of the original interval but which also contains a zero of f. We can repeat this process until we have an interval as small as we desire.

Number of Iterations

We can also determine, in advance, how many iterations we need in order to obtain an interval of width smaller than any predetermined number ǫ > 0. Clearly, after the first iteration we will have an interval of width b0−a0

2

. After the next iteration, we will have an interval of width b0−a0

4

= b0−a0

22 .

After the third iteration, we will have an interval of width b0−a0

8

= b0−a0

23 .

Continuing, it should be clear that after n iterations we will have an interval of width

b0−a0 2n .

We can easily find a value of n for which this width is less than ǫ by solving as follows.

b0−a0 2n

< ǫ

b0−a0 ǫ

< 2n 2n > b0−a0

ǫ

n ln 2 > ln b0−a0

ǫ

• n > ln

b0−a0

ǫ

• ln 2

. The number of iterations we will need is thus the smallest integer greater than or equal to ln b0−a0

ǫ

• ln 2

. One should not memorize this formula; the calculations are routine enough so that you should be able to carry them out fairly easily and quickly when you need to and doing so is good practice.

Definition of a Limit

Definition 3 (Limit). We say limx→c f(x) = L if for every ǫ > 0 there is some δ > 0 such that |f(x) − L| < ǫ whenever 0 = |x − c| < δ. This gives a precise meaning to the vague idea of being close to. Essentially, regardless of how close (ǫ) someone insists f(x) be to the limit L in order to be convinced f(x) is close to L, it must be possible to show that f(x) is at least that close provided x is close enough (within δ of) to c. There are variations of this definition that give rigorous meaning to each of the variations:

• ne-sided limits, infinite limits and limits at infinity.

Speed

Suppose an object is moving along a straight line and the distance it has traveled at a given time is given by the formula s = f(t), where t represents time and s represents distance. The average speed of the object over a time period t0 ≤ t ≤ t1 would be equal to ∆s ∆t = f(t1) − f(t0) t1 − t0 .

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If t1 was close to t0, this would likely be close to the instantaneous speed of the object when t = t0, suggesting that the instantaneous speed is equal to limt1→t0 f(t1) − f(t0) t1 − t0 , which is equal to the simpler looking expression limt→t0 f(t) − f(t0) t − t0 .

Slope of a Tangent Line

Consider the graph of a function y = f(x). The slope of a line connecting two points (x0, y0) and (x1, y1) on its graph would be equal to ∆y ∆x = y1 − y0 x1 − x0 = f(x1) − f(x0) x1 − x0 . If the two points are close together on the graph, which will be the case if x1 is close to x0, this would likely be close to the slope of the line tangent to the graph at the point (x0, y0), suggesting that the slope of the tangent is equal to limx1→x0 f(x1) − f(x0) x1 − x0 , which is equal to the simpler looking expression limx→x0 f(x) − f(x0) x − x0 .

The Derivative

The similarity between limt→t0 f(t) − f(t0) t − t0 and limx→x0 f(x) − f(x0) x − x0 suggest something more general is going on and that it would be worthwhile to study that expression, leading to the definition of a derivative. Definition (The Derivative). f ′(c) = limx→c f(x) − f(c) x − c is called the derivative of f at c if the limit exists. Alternative Form: f ′(c) = limh→0 f(c + h) − f(c) h . If f ′(c) exists, then f is said to be differentiable at c; otherwise, it is said f fails to be differentiable at c.

The Derivative Function

The correspondence c → f ′(c) which associates f ′(c) with c for every c ∈ Df effectively defines a function, which we also call the derivative and denote by f ′. We may use the alternate form of the derivative at a point to create the following definition. Definition (The Derivative Function). The derivative of a function f is the function f ′ given by the formula f ′(x) = limh→0 f(x + h) − f(x) h wherever the limit exists.

Variations

f ′(x) = lim∆x→0 f(x + ∆x) − f(x) ∆x f ′(x) = limz→x f(z) − f(x) z − x . f ′(x) = lim∆x→0 ∆y ∆x

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The process of finding a derivative is called differentiation.

Notation

The derivative function may be denoted by each of the following equivalent notations: f ′(x) = y′ = dy dx = d dx (f(x)). The latter two versions are called Leibniz’ Notation. dy dx should be read the derivative of y (with respect to x). The value of the derivative at a point c may be denoted by each of the following equivalent notations: f ′(c) = y′

• x=c = dy

dx

• x=c = d

dx (f(x))

• x=c.

Often, the part

• x=c indicating the derivative is evaluated at c will be omitted and must

be inferred from the context.

Differentiable Functions Are Continuous

Suppose f is differentiable at c. Then f ′(c) = limx→c f(x) − f(c) x − c

• exists. Intuitively, it

would appear that f(x) would have to be close to f(c) if x is close to c (otherwise, the limit wouldn’t exist), which is essentially what it means for f to be continuous at c. This can be proven as follows.

• Proof. Suppose f is differentiable at c. Then limx→c f(x) =

limx→c

• f(c) + f(x) − f(c)

x − c · (x − c)

• =

f(c) + limx→c f(x) − f(c) x − c · limx→c(x − c) = f(c)+f ′(c)·0 = f(c), where each calculation is justifiable from either the properties of limits

• r the definition of a derivative.
• Properties and Formulas

dk dx = 0 – The derivative of a constant function is identically 0. d dx (kx) = k d dx (ax + b) = a Power Rule: d dx (xn) = nxn−1 The derivative of a constant times a function is equal to the constant times the derivative. This may be written as d dx (ku) = kdu dx or d dx (kf(x)) = kf ′(x). The derivative of a sum is equal to the sum of the derivatives. This may be written as d dx (u + v) = du dx + dv dx or (f + g)′(x) = f ′(x) + g′(x). The derivative of a difference is equal to the difference of the derivatives. This may be written as

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d dx (u − v) = du dx − dv dx or (f − g)′(x) = f ′(x) − g′(x). These three properties, taken together, imply that we may differentiate term-by-term. This means that if we want to calculate the derivative of a function containing several terms, we may find the derivative of each term separately and add the derivatives together. In particular, combined with the Power Rule, this makes the differentiation of polynomials routine. Example: d dx (5x8 + 3x5 − 8x2 + 11x + 4) = 5 d dx (x8) + 3 d dx (x5) − 8 d dx (x2) + d dx (11x + 4) = 5 · 8x7 + 3 · 5x4 − 8 · 2x + 11 = 40x7 + 15x4 − 16x + 11. Each of these properties may be proven using the definition of a derivative. For example, we may prove the derivative of a sum is equal to the sum of the derivatives with the following routine calculation.

• Proof. Let φ(x) = f(x) + g(x). By definition,

φ′(x) = limz→x φ(z) − φ(x) z − x = limz→x [f(z) + g(z)] − [f(x) + g(x)] z − x = limz→x f(z) + g(z) − f(x) − g(x) z − x = limz→x f(z) − f(x) + g(z) − g(x) z − x = limz→x f(z) − f(x) z − x + g(z) − g(x) z − x

• Since the limit of a sum is equal to the sum of the limits, we get

φ′(x) = limz→x f(z) − f(x) z − x + limz→x g(z) − g(x) z − x . Once again by the definition of a derivative, the expression on the right is equal to f ′(x)+ g′(x).

• The Product Rule

The formula for the derivative of a product turns out to be more complicated than the formulas for derivatives of sums and differences. In plain language, The derivative of a product is equal to the first factor times the derivative of the second plus the second factor times the derivative of the first. Symbolically, this may be written several ways. (fg)′ = fg′ + gf ′ d dx (f(x)g(x)) = f(x)g′(x) + g(x)f ′(x) d dx (uv) = u · dv dx + v · du dx

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Example: d dx (x2(5x − 3)) = x2 d dx (5x − 3) + (5x − 3) d dx (x2) = x2(5) + (5x − 3) · 2x This may be proven with the following calculation.

• Proof. Let φ(x) = f(x)g(x). By the definition of a derivative,

φ′(x) = limz→x φ(z) − φ(x) z − x = limz→x f(z)g(z) − f(x)g(x) z − x = limz→x f(z)g(z) − f(z)g(x) + f(z)g(x) − f(x)g(x) z − x = limz→x f(z)[g(z) − g(x)] + [f(z) − f(x)]g(x) z − x = limz→x

• f(z) · g(z) − g(x)

z − x + f(z) − f(x) z − x · g(x)

• .

From the properties of limits, the expression on the right may be written as φ′(x) = limz→x f(z) · limz→x g(z) − g(x) z − x + limz→x f(z) − f(x) z − x · g(x) From the definition of a derivative, limz→x f(z) − f(x) z − x = f ′(x) and limz→x g(z) − g(x) z − x = g′(x). Since f is continuous (differentiable functions are continuous), it also follows that limz→x f(z) = f(x), so we get φ′(x) = f(x)g′(x) + f ′(x)g(x).

• The Quotient Rule

The derivative of a quotient is equal to the denominator times the derivative of the numer- ator minus the numerator times the derivative of the denominator, all divided by the square

• f the denominator.

Symbolically, this may be written several ways. f g ′ = gf ′ − fg′ g2 d dx f(x) g(x)

• = g(x)f ′(x) − f(x)g′(x)

(g(x))2 d dx u v

• =

vdu dx − udv dx v2 u v ′ = vu′ − uv′ v2

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Example: d dx sin x x2 + 3

• =

(x2 + 3) d dx (sin x) − sin x d dx (x2 + 3) (x2 + 3)2 = (x2 + 3) cos x − sin x(2x) (x2 + 3)2 = (x2 + 3) cos x − 2x sin x (x2 + 3)2 . The Quotient Rule may be proven with a calculation similar to that used to prove the Product Rule.

• Proof. Let φ(x) = f(x)

g(x). By the definition of a derivative, φ′(x) = limz→x φ(z) − φ(x) z − x = limz→x

f (z) g(z) − f (x) g(x)

z − x = limz→x

f (z)g(x)−f (x)g(z) g(x)g(z)

z − x = limz→x f(z)g(x) − f(x)g(x) − f(x)g(z) + f(x)g(x) g(x)g(z)(z − x) = limz→x g(x)[f(z) − f(x)] − f(x)[g(z) − g(x)] g(x)g(z)(z − x) = limz→x g(x)f(z) − f(x) z − x − f(x)g(z) − g(x) z − x g(x)g(z) . Using the properties of limits, we get φ′(x) = g(x) limz→x

f (z)−f (x) z−x

− f(x) limz→x f(x)g(z)−g(x)

z−x

g(x) limz→x g(z) = g(x)f ′(x) − f(x)g′(x) (g(x))2 .

• Derivatives of Trigonometric Functions

Let f(x) = sin x. From the definition of a derivative, f ′(x) = limh→0 f(x + h) − f(x) h = limh→0 sin(x + h) − sin x h . Conveniently, we have a trigonometric identity that enables us to rewrite sin(x + h) as sin x cos h + cos x sin h, so we have f ′(x) = limh→0 sin x cos h + cos x sin h − sin x h = limh→0 sin x cos h − sin x + cos x sin h h = limh→0 sin x(cos h − 1) + sin h cos x h =

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limh→0

• sin xcos h − 1

h + cos xsin h h

• =

sin x limh→0 cos h − 1 h + cos x limh→0 sin h h . We will show that limh→0 sin h h = 1 and limh→0 1 − cos h h = 0, from which it will follow that f ′(x) = (sin x) · 0 + (cos x) · 1 = cos x. We thus have the formula d dx (sin x) = cos x subject to proving the claims about the limits

• f sin h

h and 1 − cos h h . Claim 1. limh→0 sin h h = 1.

• Proof. First consider 0 < h < π/2, draw the unit circle with center at the origin, and consider

the sector with central angle h where one side lies along the x-axis and the other side lies in the first quadrant. Since the area of the circle is π and the ratio of the area of the sector to the area of the circle is h 2π, the area of the sector is h 2π · π = h 2. Now consider the right triangle where the hypotenuse coincides with the side of the sector lying in the first quadrant and the base lies along the x-axis. The vertices will be (0, 0), (cos h, 0), (cos h, sin h), so its legs will be of length cos h, sin h and its area will be 1 2·cos h sin h. Since the triangle is contained within the sector, its area will be smaller than the area of the sector. Hence 1 2 · cos h sin h < h 2. Multiplying both sides by 2 h cos h yields the inequality sin h h < 1 cos h. Now consider the right triangle with one leg coinciding with the side of the sector lying along the x-axis and the hypotenuse making an angle h with that leg. Its vertices are (0, 0), (1, 0), (1, tan h), so its legs will be of length 1, tan h and its area will be 1 2 · tan h. Since the sector is contained within this triangle, its area will be smaller than the area of the triangle. Hence h 2 < 1 2 · tan h. Multiplying both sides by 2 cos h h and making use of the identity tan h cos h = sin h yields the inequality cos h < sin h h . Combining the two inequalities we have obtained yields (1) cos h < sin h h < 1 cos h if 0 < h < π/2. Now, suppose −π/2 < h < 0. Then 0 < −h < π/2 and the double inequality (1) yields

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(2) cos(−h) < sin(−h) −h < 1 cos(−h). Since cos(−h) = cos h and sin(−h) = − sin h, it follows that sin(−h) −h = − sin h −h = sin h h and (2) becomes (3) cos h < sin h h < 1 cos h. We thus see (1) holds both for 0 < h < π/2 and for −π/2 < h < 0. Since limh→0 cos h = limh→0 1 cos h = 1, by the Squeeze Theorem it follows that limh→0 sin h h = 1

• Claim 2. limh→0

1 − cos h h = 0.

• Proof. We make use of the identity involving sin and an algebraic manipulation reminiscent
• f rationalization, enabling us to prove the claim with a fairly routine calculation.

limh→0 1 − cos h h = limh→0 1 − cos h h · 1 + cos h 1 + cos h = limh→0 1 − cos2 h h(1 + cos h) = limh→0 sin2 h h(1 + cos h) = limh→0 sin h h · sin h 1 + cos h = limh→0 sin h h · limh→0 sin h 1 + cos h = 1 · 0 = 0.

• Derivatives of Logarithmic and Exponential Functions

We will ultimately go through a far more elegant development then what we can do now. Consider first an exponential function of the form f(x) = ax for some constant a > 0. Note the difference between a power function x → xn and an exponential function x → ax. For a power function, the variable is raised to a power; for an exponential function, a constant is raised to a variable power. Let’s try to calculate the derivative for f. Using the definition of a derivative, we may write f ′(x) = limh→0 f(x + h) − f(x) h = limh→0 ax+h − ax h = limh→0 ax(ah − 1) h = ax limh→0 ah − 1 h . We may write d dx (ax) = k · ax, where k = limh→0 ah − 1 h depends on a.

15

We are assuming k exists! It does, but this is not so easy to show. Evaluating 2h − 1 h for values of h close to 0 yields values close to 0.69, while evaluating 3h − 1 h for values of h close to 0 yields values close to 1.1. k · ax is obviously simplest if k = 1. The numerical calculations suggest there is some 2 < a < 3 for which a = 1. That number is called e, yielding the formula d dx (ex) = ex. The function x → ex is called the exponential function and is often denoted by exp. The exponential function is essentially unique in having the property that it’s its own derivative! The adjective essentially is used because every constant multiple of the exponential function has the same property, but no other function has that property! If we interpret the derivative as a measure of rate of change, the fact that the exponen- tial function is its own derivative may be interpreted to mean that the rate at which the exponential function changes is equal to the magnitude of the exponential function. It turns out that all functions whose rates of change are proportional to their sizes are exponential functions. Note the omission of the definite article.

The Natural Logarithm Function

Recall the definition of a logarithm function: logbx is the power which b must be raised to in order to obtain x. In other words, l = logbx if bl = x. The logarithm with base e is known as the natural logarithm function and is denoted by

• ln. Thus, l = ln x if and only el = x.

We’ll try to figure out the derivative of the natural logarithm function ln. Our calculations will not be rigorous; we will obtain the correct formula, but a legitimate derivation will have to wait until we learn about the definite integral. Let f(x) = ln x. Let’s start calculating f ′(x). According to the definition of a derivative, f ′(x) = limz→x f(z) − f(x) z − x = limz→x ln z − ln x z − x . We need to estimate the difference quotient ln z − ln x z − x when z is close to x. We’ll do it in a rather strange way. Let Z = ln z and X = ln x. Then we know z = eZ and x = eX and we may write ln z − ln x z − x as Z − X eZ − eX . Now, let’s go back and take another look at the derivative of the exponential function, but from a different perspective and with slightly different notation. Sometimes it pays to write something a few different ways! Let g(X) = eX . By the definition of a derivative, g′(X) = limZ →X eZ − eX Z − X .

16

But we know g′(X) = eX , so this suggests that when Z is close to X, eZ − eX Z − X is close to eX . But eZ − eX Z − X is the reciprocal of Z − X eZ − eX , suggesting that Z − X eZ − eX is close to 1 eX . On the other hand, when trying to find the derivative of the natural log function we came up something suggesting Z − X eZ − eX was close to the derivative. This suggests that the derivative is equal to 1 eX . Recall eX = x, so 1 eX = 1 x, which suggests d dx (ln x) = 1 x. Indeed, this is the derivative, although we’re not ready for a rigorous derivation. We will, however, make use of this formula.

Logarithms to Other Bases

The key properties of logarithms are: logb(xy) = logbx + logby (The log of a product equals the sum of the logs.) logb(x/y) = logbx − logby (The log of a quotient equals the difference of the logs.) logb(xr ) = r logbx (The log of something to a power is the power times the log.) We can use these properties to show that, in a very real sense, natural logarithms suffice and we can always write any logarithm in terms of a natural logarithm. Suppose l = logbx. It follows that bl = x and thus ln(bl) = ln x. Using the third property of logarithms, we see l ln b = ln x and, solving for l, we get l = ln x ln b . This gives us the crucial identity logbx = ln x ln b . This enables us to calculate derivatives involving logarithms to any base, as shown in the following general example. Example: Find d dx (logbx). Solution: d dx (logbx) = d dx ln x ln b

• =

1 ln b d dx (ln x) = 1 ln b · 1 x = 1 x ln b.

Other Exponential Functions

A calculation similar to the derivation of the identity logbx = ln x ln b yields a useful identity involving exponential functions. Let ax = y. Then: ln(ax) = ln y x ln a = ln y ex ln a = eln y. Since eln y = y and y = ax, it follows that ax = ex ln a.

Caution:

17

In high school algebra, a meaning was given to rational exponents: If a > 0, m, n ∈ Z, n > 0, then am/ n =

n

√am. However, no meaning was given to ax if x is irrational. That can be done and the identity ax = ex ln a will play a key role. Let’s take another look at the derivative of ordinary exponential functions. We found d dx (ax) = k · ax, where k = limh→0 ah − 1 h . Let’s play the same sort of game we played when trying to calculate the derivative of the natural log function and let ah = H, noting that ah − 1 h will be close to k when h is close to 0, and hence when H is close to 1. ah − 1 h may be written as H − 1 loga H − loga 1, since loga 1 = 0. If we let f(x) = loga x and tried to calculate f ′(1), we might write f ′(1) = limH →1 f(H) − f(1) H − 1 = limH →1 loga H − loga 1 H − 1 . But we earlier showed that f ′(x) = 1 x ln a, so f ′(1) = 1 ln a. This suggests that loga H − loga 1 H − 1 is close to 1 ln a if H is close to 1, which suggests its reciprocal H − 1 loga H − loga 1, and hence ah − 1 h as well, is close to ln a if H is close to 1 and h is close to 0. We thus expect k = ln a and d dx (ax) = ax ln a.

The Chain Rule

Suppose y = f ◦ g(x). Assuming f and g have derivatives where appropriate, the Chain Rule says that (f ◦ g)′ = (f ′ ◦ g) · g′. In more practical language, if we write y = f(u) and u = g(x), it comes out as dy dx = dy du · du dx. To see why, go back to the definition of a derivative and write dy dx = (f ◦ g)′(x) = limh→0 f ◦ g(x + h) − f ◦ g(x) h = limh→0 f(g(x + h)) − f(g(x)) h . We’d like to write this as limh→0 f(g(x + h)) − f(g(x)) g(x + h) − g(x) · g(x + h) − g(x) h , but it’s possible that g(x+h) = g(x), and hence the denominator g(x + h) − g(x) = 0, for some h = 0, so we consider two separate cases. If there are arbitrarily small values of h for which g(x + h) = g(x), then both (f ◦ g)′(x) and g′(x) will have to equal 0 and the Chain Rule certainly holds in a trivial fashion. So we need only verify the Chain Rule in the remaining case where g(x + h) = g(x) is h is close to 0. For this case, we write

18

u = g(x), g(x + h) = g(x) + k, k = g(x + h) − g(x). We can then write dy dx = limh→0 f(g(x + h)) − f(g(x)) g(x + h) − g(x) · g(x + h) − g(x) h = limh→0 f(u + k) − f(u) k · g(x + h) − g(x) h . Since the limit of a product is equal to the product of limits, we have dy dx = limh→0 f(u + k) − f(u) k · limh→0 g(x + h) − g(x) h . Since k ≈ 0 when h ≈ 0, the limit limh→0 f(u + k) − f(u) k will equal the limit limk→0 f(u + k) − f(u) k , which is equal to f ′(u) or dy du, while limh→0 g(x + h) − g(x) h is, by definition, equal to g′(x) = du dx. This demonstrates that dy dx = dy du · du dx.

Strategy For Calculating Derivatives

The following strategy will work without fail to calculate the derivative of any function defined implicitly, provided the function is constructed from the basic elementary functions.

Strategy

Differentiate Term-By-Term. For each term, determine whether it is a basic, elementary function, a product, a quotient or a composite function and then use the appropriate formula

• r rule.

That’s all there is.

The Catch

One must, of course, correctly recognize elementary functions: the power functions, sin, cos, tan, sec, csc, cot, ln, exp. One must also remember the Product and Quotient Rules and be able to use the Chain Rule. A Common Mistake: Not recognizing that a term can be rewritten in the form of a basic elementary function, product or quotient and thus incorrectly jumping to the conclusion that the term is a composite function.

Implicit Differentiation

Functions may be defined several ways. One of the most common ways is via an explicit formula, such as f(x) = x2 or y = 5 sin x + tan x. It is also possible to define functions implicitly. One uses an equation involving an inde- pendent variable and a dependent variable, often x and y, and defines a function y = f(x) by saying y = f(x) for a particular value of x if those values for x and y satisfy the equation.

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Sometimes additional conditions must be placed on the solutions in order to ensure a function is defined, that is, so that for any value of x there is only one value of y satisfying the equation along with the conditions.

Examples:

xy = 1 (The same function may be defined explicitly via the formula f(x) = 1

• x. Its graph is an

hyperbola.) x2 + y2 = 1, y ≥ 0 (The same function may be defined explicitly via the formula f(x) = √ 1 − x2. Its graph is a unit semi-circle, centered at the origin, above the x-axis.) x2 + y2 = 1, y ≤ 0 (The same function may be defined explicitly via the formula f(x) = − √ 1 − x2. Its graph is a unit semi-circle, centered at the origin, below the x-axis.) Derivative of functions defined implicitly may be calculated using a technique called Im- plicit Differentiation. The process is diabolically simple. One starts with an equation of the form F(x, y) = G(x, y), where F(x, y) and G(x, y) are algebraic expressions involving the variables x and y, defining a function y = f(x) implicitly and observes d dx (F(x, y)) = d dx (G(x, y)). If one differentiates correctly, one obtains a linear equation in dy dx which may be solved easily by moving all the terms involving dy dx to the left side, all the other terms to the right side, and then dividing both sides by dy dx.

What’s the Catch?

One must calculate d dx (F(x, y)) and d dx (G(x, y)) correctly! One must recognize the independent variable is x and remember that y is a function of x.

Example

Consider the function y = f(x) defined implicitly by xy = 1. One calculates as follows: d dx (xy) = d dx (1) xdy dx + ydx dx = 0 Using the Product Rule on the left. xdy dx + y · 1 = 0 xdy dx + y = 0 xdy dx = −y dy dx = −y x .

Example

20

Consider the function y = f(x) defined implicitly by x2 + y2 = 1, y ≥ 0. One calculates as follows: d dx (x2 + y2) = d dx (1) d dx (x2) + d dx (y2) = 0 2x + d dx (y2) = 0 To calculate d dx (y2), one needs to use the Chain Rule. Write z = y2. We need a decomposition of the composite function z = y2. It is deceptively simple: z = y2 y = f(x). We may then apply the Chain Rule as follows: dz dx = dz dy · dy dx = 2y dy dx. Plugging this derivative into the equation 2x + d dx (y2) = 0 yields 2x + 2y dy dx = 0 2y dy dx = −2x dy dx = −2x 2y dy dx = −x y .

Related Rates

Related rates problems deal with situations in which two or more variables are changing and the rates at which they’re changing are related – hence the term Related Rates. Steps

• Recognize the basic variables involved and represent them by symbols. Remember

that the independent variable is usually time and it is common to represent time by t.

• Determine the relationships between the underlying variables. Every fact and every

relationship generally may be described by an equation, so one obtains one or more equations: A1 = B1 A2 = B2 A3 = B3 . . . An = Bn

21

• For each equation, differentiate each side with respect to t, and equate the two sides

to get one or more equations relating the rates of change: d dt (A1) = d dt (B1) d dt (A2) = d dt (B2) d dt (A3) = d dt (B3) . . . d dt (An) = d dt (Bn)

• Solve the equations algebraically to obtain the rates desired.
• State the conclusion implied by the solution to the equations.

The Mean Value Theorem

A student at the University of Connecticut happens to be travelling to Boston. He enters the Massachussetts Turnpike at the Sturbridge Village entrance at 9:15 in the morning. Since he uses Fast Lane, he never actually picks up a ticket, but sensors record that he goes through the toll lane at Newton at 9:53, just 38 minutes later. He soon receives a traffic summons in the mail indicating that he violated the posted speed limit and decides to appeal, since there is no indication he was caught by radar. Appearing in court, the prosecutor argues that he travelled 44.8 miles in 38 minutes and therefore travelled at an average speed of 70.74 miles per hour, more than 5 miles above the highest posted speed limit of 65 miles per hour. The student argues that, although he may have averaged more than 65 miles per hour, there is no evidence that he actually ever travelled at an instantaneous speed of more than 65 miles per hour. The judge considers the arguments and quickly rejects the student’s plea, noting that the Mean Value Theorem implies that since his average speed was 70.74 miles per hour, there had to be at least one instant during which his instantaneous speed was 70.74 miles per hour.

Rolle’s Theorem

Theorem 4 (Rolle’s Theorem). Suppose a function f is continuous on the closed interval [a, b], differentiable on the open interval (a, b) and f(a) = f(b). Then there is a number c ∈ (a, b) such that f ′(c) = 0. Geometrically, this says that if there are two points on a smooth curve takes at the same height, there must be a point in between where the tangent is horizontal. Rolle’s Theorem is a special case of the Mean Value Theorem.

• Proof. Suppose f satisfies the hypotheses of Rolle’s Theorem. By the Extreme Value Theorem

for Continuous Function, there must be some point in [a, b] at which f attains a minimum and some point at which f attains a maximum. One possibility is that f is constant on the entire interval, in which case f ′ is identically 0 on (a, b) and the conclusion is clearly true.

22

So let’s consider the other possibility, that f is not constant. Then either the minimum

• r the maximum must occur at some point c ∈ (a, b), that is, at some point other than the

endpoints. We will show that f ′(c) = 0 if f has a maximum at c. Similar reasoning would show f ′(c) = 0 if f had a minimum at c, showing the conclusion is true and completing the proof. We know f ′(c) = limx→c f(x) − f(c) x − c . Since this ordinary limit exists, it follows that both the left hand limit and the right hand limit exist and are both equal to f ′(c). We’ll consider them separately. The Left Hand Limit: f ′(c) = limx→c− f(x) − f(c) x − c . Since f has a maximum at c, if x < c, then f(x) ≤ f(c), so f(x)−f(c) ≤ 0. But, if x < c, it’s also true that x−c < 0 and it follows that f(x) − f(c) x − c ≥ 0. We see limx→c− f(x) − f(c) x − c is the limit of non-negative numbers and therefore can’t be negative. It follows that f ′(c) ≥ 0. The line of reasoning for the right hand limit is similar. The Right Hand Limit: f ′(c) = limx→c+ f(x) − f(c) x − c . Since f has a maximum at c, if x > c, then f(x) ≤ f(c), so f(x)−f(c) ≤ 0. But, if x > c, it’s also true that x−c > 0 and it follows that f(x) − f(c) x − c ≤ 0. We see limx→c+ f(x) − f(c) x − c is the limit of numbers less than or equal to 0 and therefore can’t be positive. It follows that f ′(c) ≤ 0. Since f ′(c) ≥ 0 and f ′(c) ≤ 0, it follows that f ′(c) = 0

• Consequences of Rolle’s Theorem

Besides being a special case of the Mean Value Theorem and being a step in the path to proving the Mean Value Theorem, Rolle’s Theorem has some interesting applications of its

• wn.

Corollary 5. A polynomial equation of degree n has at most n solutions. This is equivalent to the statement that a polynomial of degree n has at most n zeros. Since a polynomial may be differentiated as many times as necessary, with each derivative being a polynomial of lower degree, one immediate consequence of Rolle’s Theorem is that the derivative of a polynomial has at least one zero between each pair of distinct zeros of the

• riginal polynomial.

The derivative of a linear polynomial is a non-zero constant, having no zeros, so a linear polynomial can’t have more than 1 zero. The derivative of a quadratic polynomial is linear, having no more than 1 zero, so a quadratic can’t have more than 2 zeros. The derivative of a cubic polynomial is quadratic, having no more than 2 zeros, so the cubic can’t have more than 3 zeros. This clearly goes on forever. The argument can be made rigorous through the use of Mathematical Induction.

The Mean Value Theorem

23

Theorem 6 (The Mean Value Theorem). Suppose a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then there is a number c ∈ (a, b) such that f(b) − f(a) = f ′(c)(b − a) or, equivalently, f ′(c) = f(b) − f(a) b − a . Geometrically, the Mean Value Theorem says that if there is a smooth curve connecting two points, there must be some point in between at which the tangent line is parallel to the line connecting those two points. Analytically, the Mean Value Theorem says the rate of change of a differentiable function must, at some point, take on its average, or mean value. The proof depends on the fact that the particular point at which the tangent line is parallel to the line connecting the endpoints also happens to be the point at which the curve is furthers away from that line. The proof essentially consists of applying Rolle’s Theorem to the function measuring the distance between the line and the curve. Since the line goes between the points (a, f(a)) and (b, f(b)), its slope will be f(b) − f(a) b − a and its equation may be written, in point-slope form, y − f(a) = f(b) − f(a) b − a · (x − a). Solving for y, we may write the equation in the form y = f(a) + f(b) − f(a) b − a · (x − a). Since the second coordinate of a point on the curve with first coordinate x is f(x), the vertical distance between the line and the curve will equal f(x) −

• f(a) + f(b) − f(a)

b − a · (x − a)

• = f(x) − f(a) − f(b) − f(a)

b − a · (x − a). We are now prepared to prove the Mean Value Theorem.

• Proof. Let φ(x) = f(x) − f(a) − f(b) − f(a)

b − a · (x − a). It is easy to see that φ satisfies the hypotheses of Rolle’s Theorem on the interval [a, b]. Certainly, the fact that φ is both continuous and differentiable on [a, b] follows immediately from the fact that f is. In addition, φ(a) = f(a) − f(a) − f(b) − f(a) b − a · (a − a) = 0 and φ(b) = f(b) − f(a) − f(b) − f(a) b − a · (b − a) = f(b) − f(a) − [f(b) − f(a)] = 0. Thus, there must be some c ∈ (a, b) such that φ′(c) = 0. We first obtain φ′(c) as follows. φ′(x) = f ′(x) − f(b) − f(a) b − a φ′(c) = f ′(c) − f(b) − f(a) b − a Since φ′(c) = 0, it follows that

24

f ′(c) − f(b) − f(a) b − a = 0. f ′(c) = f(b) − f(a) b − a f ′(c)(b − a) = f(b) − f(a)

• Consequences of the Mean Value Theorem

Perhaps the most important consequence of the Mean Value Theorem is that it gives precise meaning to the most important single concept in elementary Calculus,

The Derivative Measures Rate of Change.

Theorem 7.

• a. If the derivative of a function is positive at all points on an interval, then the function

is increasing on that interval.

• b. If the derivative of a function is negative at all points on an interval, then the function

is decreasing on that interval. To prove this, we need a precise definition of what it means to be increasing or decreasing. Definition 4 (Strictly Increasing). A function f is said to be strictly increasing on an open interval I if f(a) < f(b) whenever a, b ∈ I and a < b. Definition 5 (Nondecreasing). A function f is said to be nondecreasing on an open interval I if f(a) ≤ f(b) whenever a, b ∈ I and a < b. Note the subtle difference. Often, we will simply say a function is increasing. Generally, in those cases, it will not really be important whether we mean strictly increasing or simply nondecreasing. There are similar definitions of the terms strictly decreasing and nonincreasing. Here, too, we will often use the ambiguous term decreasing. Definition 6 (Strictly Decreasing). A function f is said to be strictly decreasing on an open interval I if f(a) > f(b) whenever a, b ∈ I and a < b. Definition 7 (Nonincreasing). A function f is said to be nonincreasing on an open interval I if f(a) ≥ f(b) whenever a, b ∈ I and a < b. With these definitions, we are ready to prove the derivative measures rate of change. We will prove just one of what are really four different parts, that if the derivative is strictly positive then the function is strictly increasing.

• Proof. Suppose f ′(x) > 0

∀x ∈ I and let a, b ∈ I, a < b. We need to prove that f(a) < f(b). Note: We’ve introduced the notation ∀ to mean for all or for every. Since f ′(x) > 0 ∀x ∈ I, it follows that f is continuous on [a, b] and differentiable on (a, b), so by the Mean Value Theorem there is some c ∈ (a, b) such that f(b) − f(a) = f ′(c)(b − a). Since f ′(x) > 0 ∀x ∈ I, it follows that f ′(c) is positive. Since a < b, it follows that b − a is also positive, so that f ′(c)(b − a) is also positive and hence f(b) − f(a) must be positive. Clearly, f(a) must be smaller than f(b).

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Very Different Functions Can’t Have the Same Derivative

It’s obvious that if two functions differ by only a constant term, then they will have the same derivative. Example: d dx (x2) = d dx (x2 + 5) = 2x. A natural question is whether only functions differing by a constant can share the same

• derivative. The Mean Value Theorem enables us to see this is true.

Theorem 8. If f ′(x) = g′(x) for all x in some interval, then there is some constant k such that f(x) = g(x) + k.

• Proof. Let φ = f − g and let a be some fixed point in the interval. Now let x be in the
• interval. Clearly, the φ is both continuous and differentiable on the interval with endpoints

a and x and we can apply the Mean Value Theorem. We use that language because it is possible that a < x and the interval is [a, x] but also possible that a > x and the interval is [x, a]. By the Mean Value Theorem, there is some c in the interval such that φ(x) − φ(a) = φ′(c)(x − a). Because φ = f − g, it follows that φ′(c) = 0, so φ(x) − φ(a) = 0. It follows that φ(x) = φ(a), or φ(x) = k, where we let k = φ(a). Since φ = f − g, we have f(x) − g(x) = k, or f(x) = g(x) + k.

• Corollary 9. Only constant functions have derivatives which are identically 0.

This theorem will prove useful when we try to find functions with a given derivative. One application will be to determine the acceleration due to gravity.Example An object is dropped from a height of 128 feet. How long does it take to reach the ground?

Solution

Let:

• h be the height of the object, measured in feet.
• v be the velocity of the object, measured in feet per second.
• t be the time since the object was dropped, measured in seconds.

We know:

• h = 128 when t = 0
• v = 0 when t = 0
• dh

dt = v (Since velocity is the rate at which the height changes.)

• dv

dt = −32 (Since acceleration is the rate at which the velocity changes and the acceleration due to gravity is 32 feet per second per second and is in the downward,

• r negative, direction.)

We want:

• The value of t when h = 0

Since d dt (−32t) = −32 and two functions can have the same derivative only if they differ by a constant, it follows that

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v = −32t + c for some constant c. Since v = 0 when t = 0, it follows that 0 = −32 · 0 + c c = 0 v = −32t Since d dt (−16t2) = −32t, dh dt = v = −32t and two functions can have the same derivative

• nly if they differ by a constant, it follows that

h = −16t2 + k for some constant k. Since h = 128 when t = 0, it follows that 128 = −16 · 02 + k k = 128 h = −16t2 + 128 Thus, when h = 0, 0 = −16t2 + 128 16t2 = 128 t2 = 128 16 = 8 t = √ 8 = 2 √ 2 ≈ 2.828 It thus takes 2 √ 2 seconds for the object to fall to the ground. We can also figure out how fast it was going when it hit the ground: Since v = −32t, when t = 2 √ 2 we have v = −32·2 √ 2 = −64 √ 2 ≈ −90.510, so the object is travelling at a speed of 64 √ 2 feet per second when it hits the ground.Example Determine the acceleration due to gravity.

Solution:

We may perform the following experiment: We drop a coin from the ceiling and measure the time it takes to hit the ground. Let:

• t be the time since the coin is dropped
• T be the amount of time it takes to hit the ground
• h be the height of the coin
• H be the height of the ceiling
• v be the speed of the coin
• g be the acceleration due to gravity

We know:

• dh

dt = v

• dv

dt = g

• h = H when t = 0
• v = 0 when t = 0
• h = 0 when t = T

We want:

• The value of g.

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Since d dt (gt) = g, dv dt = g and two functions can have the same derivative only if they differ by a constant, it follows that v = gt + c for some constant c. Since v = 0 when t = 0, it follows that 0 = g · 0 + c c = 0 v = gt Since d dt 1

2gt2

= gt, dh dt = v = gt and two functions can have the same derivative only if they differ by a constant, it follows that h = 1

2gt2 + k for some constant k.

Since h = H when t = 0, it follows that H = 1

2g · 02 + k

k = H h = 1

2gt2 + H

Since h = 0 when t = T, it follows that 0 = 1

2gT 2 + H 1 2gT 2 = −H

g = −H

1 2T 2

g = −2H T 2 So the acceleration due to gravity is 2H T 2 feet per second per second. Note that g is negative, since the acceleration is downward.

Linear Approximations

Suppose we want to approximate the value of a function f for some value of x, say x1, close to a number x0 at which we know the value of f. By its nature, the tangent to a curve hugs the curve fairly closely near the point of tangency, so it’s natural to expect the 2nd coordinate of a point on the tangent line close to the point (x0, f(x0)) will be fairly close to the actual value of f(x1). That value is called the Linear Approximation to f(x1), or the Tangent Line Approxima- tion. Since the tangent line goes through (x0, f(x0)) and has slope f ′(x0), it will have equation y − f(x0) = f ′(x0)(x − x0), which may also be written as y = f(x0) + f ′(x0)(x − x0). Definition 8 (Linear Approximation). The linear approximation of f(x) near x = x0 is L(x) = f(x0) + f ′(x0)(x − x0).

Example

Find an approximation to

3

√ 8.01.

Solution:

Let f(x) =

3

√x. We use the linear approximation for f(x) near x = 8.

28

We need f(8) and f ′(8) and find them as follows: f(x) =

3

√x = x1/ 3 f(8) =

3

√ 8 = 2 f ′(x) = 1 3 · x−2/ 3 = 1 3 · 1 x2/ 3 = 1 3( 3 √x)2 f ′(8) = 1 3(

3

√ 8)2 = 1 3 · 22 = 1 12 Using the formula L(x) = f(x0) + f ′(x0)(x − x0), we get L(x) = 2 + 1 12 · (x − 8). Our approximation to

3

√ 8.01 is L(8.01) = 2 + 1 12 · (8.01 − 8) = 2 + 1 12 · 0.01 = 2 + 1 1200 ≈ 2.00083333. Note a calculator approximation for

3

√ 8.01 is 2.00083299.

Differentials

An equivalent method of approximating values of functions is called an approximation using differentials. Definition 9 (Increment of x). ∆x = x1 − x0 Definition 10 (Increment of y). ∆y = y1 − y0 = f(x1) − f(x0) With this notation, f(x1) = f(x0) + ∆y. The Linear Approximation assumes f(x1) ≈ f(x0) + f ′(x0)(x1 − x0) = f(x0) + f ′(x0)∆x. Definition 11 (Differential of x). dx = ∆x. The differential of x is synonymous with the increment of x. Definition 12 (Differential of y). dy = f ′(x0)∆x = f ′(x0)dx = dy dx · dx. With this definition, the Linear Approximation may be written f(x1) ≈ f(x0) + dy. Independently, dy may be thought of as an approximation to the amount y, or f(x), changes.

Example

A layer of paint 0.001 inches thick is applied to a spherical object 16 inches in diameter. Approximately how much paint is used?

Solution:

The amount of paint is equal to the amount the volume of a sphere will change by if its radius increases from 8 inches to 8.001 inches. Letting V be the volume of the sphere and r be its radius, we know V = 4 3πr3.

29

dV dr = 4πr2. Taking r0 = 8, r1 = 8.001, we find: dr = ∆r = 8.001 − 8 = 0.001 dV dr

• r =8 = 4π · 82 = 256π

dV = dV dr · dr = 256π · 0.001 = 0.256π ≈ 0.8042 So it will take ≈ 0.256π ≈ 0.8042 cubic inches of paint.

Newton’s Method

Newton’s Method is designed to estimate a zero of a differentiable function. It generally works faster than the Bisection Method when it works and it does not require one to first find two points at which the function has opposite signs. The drawback is that it doesn’t always work. Given a function f and an initial value x0, one obtains a sequence of values x0, x1, x2, x3, . . . using the formula xn = xn−1 − f(xn−1) f ′(xn−1). If one is fortunate, the sequence quickly approaches a zero of f. It is generally relatively easy to set up a spreadsheet to do the calculations, which can also be done using a calculator. Students familiar with any programming languages should find writing a program to implement Newton’s Method an easy yet educational exercise. Geometrically, xn is the first coordinate of the point at which the line tangent to the graph

• f f at the point (xn−1, f(xn−1)) crosses the x−axis. This may be seen as follows.

The equation of the tangent line is y − f(xn−1) = f ′(xn−1)(x − xn−1). If xn is the first coordinate of the point where this line crosses the x−axis, since the second coordinate of that point is 0, we may plug x = xn−1, y = 0 in that equation to get 0 − f(xn−1) = f ′(xn−1)(xn − xn−1). We may then solve for xx as follows: −f(xn−1) = f ′(xn−1)(xn − xn−1) − f(xn−1) f ′(xn−1) = xn − xn−1 xn = xn−1 − f(xn−1) f ′(xn−1)

L’Hˆ

• pital’s Rule

L’Hˆ

• pital’s Rule provides a convenient way of finding limits of indeterminate quotients.

Effectively, it states that if one wishes to find a limit of a quotient f (x)

g(x) and both f(x) and

g(x) either → 0 or both → ±∞, then the limit of the quotient f (x)

g(x) is equal to the limit of

the quotient f ′(x)

g′(x) of the derivatives if the latter limit exists.

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Theorem 10 (L’Hˆ

• pital’s Rule). Let f and g be differentiable on an open interval containing

c except possibly at some fixed point c in the interval. If limx→c f(x) = limx→c g(x) = 0 and limx→c

f ′(x) g′(x) exists, then limx→c f (x) g(x) exists and

limx→c

f (x) g(x) = limx→c f ′(x) g′(x).

This is just the most basic of literally dozens of variations. The analogous result holds if the numerator and denominator both approach ±∞, or if the limit is one-sided, or if the limit is at ∞ or at −∞. Let’s look at some examples and then we will prove L’Hˆ

• pital’s Rule. The proof will

depend on a generalization of the Mean Value Theorem, the General Mean Value Theorem. Example: limh→0

sin h h

= limh→0

cos h 1

= 1. Example: limh→0

1−cos h h

= limh→0

sin h 1

= 0. Note both of these have been calculated before. Indeed, the reasoning here is somewhat circular, since it depends on the calculation of the derivatives of the sin and cos function, which were themselves derived using these limits. Example: limx→∞

5x2+8 sin x−1 x2+5x+3

= limx→∞

10x+8 cos x 2x+5

= limx→∞

10x 2x+5 + limx→∞ 8 cos x 2x+5 =

limx→∞

10 2 + 0 = 5.

With some creative algebra, we can use L’Hˆ

• pital’s Rule to help calculate limits in cases

where it cannot be applied directly. Symbolically, we may think of the cases to which we can directly apply L’Hˆ

• pital’s Rule

as the

0 and ∞ ∞

cases. We can also use it indirectly in cases we may think of symbolically as the following: 0 · ∞ 00 1∞ ∞0 We give examples of each. The general idea, in each case, is to rewrite the expression so that one winds up needing the limit of something to which one may apply L’Hˆ

• pital’s Rule.

Example of the 0 · ∞ Case: limx→0+ x ln x. Here, x → 0 but ln x → −∞. We change it into a quotient by writing x ln x = ln x

1/ x . The

calculation becomes limx→0+ x ln x = limx→0+ ln x

1/ x = limx→0+ 1/ x −1/ x2 = limx→0+(−x) = 0.

Example of the 00 Case: limx→0+ xsin x. Here, we use the definition of an exponent to write xsin x as esin x ln x. We then need only find the limit of the exponent, which brings us to the 0 · ∞ case we already dealt with. The calculations are as follows. limx→0+ sin x ln x = limx→0+

ln x 1/ sin x =

limx→0+

1/ x − cos x

sin2 x = limx→0+ − sin2 x

x cos x .

We can apply L’Hˆ

• pital’s Rule again to evaluate the latter limit!

limx→0+ − sin2 x

x cos x = limx→0+ − 2 sin x cos x −x sin x+cos x = 0.

So, limx→0+ xsin x = e0 = 1.

31

Example of the 1∞ Case: limx→∞(1 + 5/x)x. Again, we use the definition of an exponent to write (1 + 5/x)x = ex ln(1+5/ x) and find the limit of the exponent. limx→∞ x ln(1 + 5/x) = limx→∞

ln(1+5/ x) 1/ x

= limx→∞

−5/ [x2(1+5/ x)] −1/ x2

= limx→∞

5 1+5/ x = 5.

So, limx→∞(1 + 5/x)x = e5. The ∞0 Case: This may be handled like the 00 case. Indeed, the reciprocal of the ∞0 case is the 00 case.

Summary of Special Cases

• 0 · ∞

lim f(x)g(x), where f(x) → 0 and g(x) → ∞ Write f(x)g(x) as either f(x) 1/g(x) (giving the 0 0 case) or g(x) 1/f(x), (giving the ∞ ∞ case).

• 00

lim f(x)g(x), where f(x) → 0+ and g(x) → 0 Write f(x)g(x) as eg(x) ln f (x). lim g(x) ln f(x) is then the 0 · ∞ case, since ln f(x) → −∞ as f(x) → 0+.

• 1∞

lim f(x)g(x), where f(x) → 1 and g(x) → ∞ Write f(x)g(x) as eg(x) ln f (x). lim g(x) ln f(x) is then the 0 · ∞ case, since ln f(x) → 0 as f(x) → 1.

• ∞0

lim f(x)g(x), where f(x) → ∞ and g(x) → 0. Write f(x)g(x) as eg(x) ln f (x). lim g(x) ln f(x) is then the 0 · ∞ case, since ln f(x) → ∞ as f(x) → ∞.

The Generalized Mean Value Theorem

We will prove the basic case of L’Hˆ

• pital’s Rule. The proof depends on a more general

version of the Mean Value Theorem. We will state the Generalized Mean Value Theorem, show how it may be used to prove L’Hˆ

• pital’s Rule, and then prove the Generalized Mean

Value Theorem. Theorem 11 (The Generalized Mean Value Theorem). Suppose functions f and g are continuous on the closed interval [a, b], differentiable on the open interval (a, b) and g′(x) is never 0 on (a, b). Then there is a number c ∈ (a, b) such that f (b)−f (a)

g(b)−g(a) = f ′(c) g′(c).

Note, if g is the identity function g(x) = x, the Generalized Mean Value Theorem reduces to the original Mean Value Theorem. The proof will be very similar. First, we’ll use the Generalized Mean Value Theorem to prove L’Hˆ

• pital’s Rule.
• Proof. Assume f and g satisfy the hypotheses of L’Hˆ
• pital’s Rule. Since the values of f(c)

and g(c) don’t matter, we may assume f(c) = g(c) = 0. Then, for any x near c, the hypotheses of the Generalized Mean Value Theorem will hold in the interval containing c and x and there will be some number ξ between c and x such that

32

f (x) g(x) = f (x)−f (c) g(x)−g(c) = f ′(ξ) g′(ξ).

Obviously, ξ → c as x → c, so limx→c

f (x) g(x) = limx→c f ′(ξ) g′(ξ) = limx→c f ′(x) g′(x).

• We’re now ready to prove the Generalized Mean Value Theorem.

Theorem 12 (The Generalized Mean Value Theorem). Suppose functions f and g are continuous on the closed interval [a, b], differentiable on the open interval (a, b) and g′(x) is never 0 on (a, b). Then there is a number c ∈ (a, b) such that f (b)−f (a)

g(b)−g(a) = f ′(c) g′(c).

Recall the proof of the Mean Value Theorem depended on a function φ(x) = f(x)−f(a)− f(b) − f(a) b − a · (x − a) which represented the distance between a line and a curve. To prove the Generalized Mean Value Theorem, we use a similar function.

• Proof. Let φ(x) = f(x) − f(a) − f(b) − f(a)

g(b) − g(a) · (g(x) − g(a)). It is easy to see that φ satisfies the hypotheses of Rolle’s Theorem on the interval [a, b]. Thus, there must be some c ∈ (a, b) such that φ′(c) = 0. We first obtain φ′(c) as follows. φ′(x) = f ′(x) − f(b) − f(a) g(b) − g(a) · g′(x) φ′(c) = f ′(c) − f(b) − f(a) g(b) − g(a) · g′(c) Since φ′(c) = 0, it follows that f ′(c) − f(b) − f(a) g(b) − g(a) · g′(c) = 0 f ′(c) = f(b) − f(a) g(b) − g(a) · g′(c)

f ′(c) g′(c) = f(b) − f(a)

g(b) − g(a)

• Newton’s Method

Newton’s Method is a way of estimating zeros of functions which is highly efficient when it works. One starts with an initial estimate x0 and, based on each estimate xn obtains another, hopefully better, estimate xn+1. One thus obtains a sequence x0, x1, x2, x3, . . . of estimates which hopefully approach an actual zero of the function. Geometrically, one obtains xn+1 by considering the line tangent to the graph of f at (xn, f(xn)) and letting xn+1 be the first coordinate of the point where that line crosses the x−axis. Since the tangent line goes through (xn, f(xn)) and has slope f ′(xn), it has equation y − f(xn) = f ′(xn)(x − xn). One finds where this crosses the x−axis by solving the equation 0 − f(xn) = f ′(xn)(x − xn). −f(xn) = f ′(xn)(x − xn) x − xn = − f (xn)

f ′(xn)

33

x = xn − f (xn)

f ′(xn)

This gives the formula for obtaining xn+1: xn+1 = xn − f (xn)

f ′(xn)

Absolute and Local Extrema

Definition 13 (Absolute Maximum). A function f has an absolute maximum at c ∈ S if f(x) ≤ f(c) ∀x ∈ S. We call f(c) the absolute maximum of f on S. Definition 14 (Local Maximum). A function f has a local maximum at c if f(x) ≤ f(c) ∀x in some open interval containing c. We call f(c) a local maximum of f. Definition 15 (Absolute Minimum). A function f has an absolute minimum at c ∈ S if f(x) ≥ f(c) ∀x ∈ S. We call f(c) the absolute minimum of f on S. Definition 16 (Local Minimum). A function f has a local minimum at c if f(x) ≥ f(c) ∀x in some open interval containing c. We call f(c) a local minimum of f. In the proof of Rolle’s Theorem, we actually demonstrated the following theorem. Theorem 13. If f has a local extremum at a point c and f ′(c) exists, then f ′(c) = 0. This leads to the following definition and theorem. Definition 17 (Critical Number). A number c in the domain of a function f is called a critical number of f if either f ′(c) = 0 or f ′(c) is undefined. Theorem 14 (Fermat’s Theorem). Suppose f has a local extremum at c. Then c must be a critical number of f. Thus, to find extrema of a function, we calculate the derivative and find where it equals 0 and where it doesn’t exist.

Monotonicity - Increasing and Decreasing

If a function is increasing on an interval, or if it is decreasing on an interval, we say it is monotonic. When we wish to determine where a function is increasing and where it is decreasing, we say we wish to determine its monotonicity. We have previously shown, using the Mean Value Theorem, that the derivative measures rate of change in the sense that if the derivative is positive then the function is increasing and if the derivative is negative then the function is decreasing.

Curve Sketching

Simply analyzing monotonicity and finding possible critical points is often enough to get a rough idea of what a graph looks like. We will soon round out the picture by looking at some additional factors.

Analysis of Extrema

We can combine Fermat’s Theorem and an analysis of monotonicity to determine the nature of local extema. For example, if a continuous function has a critical point and is increasing to the left and decreasing to the right, it obviously must have a local maximum at the critical point. A similar argument deals with minima, leading to the following test.

34

Theorem 15 (First Derivative Test for Extrema). Suppose that f is continuous on an open interval I containing a point c.

• i. If f ′(x) > 0 for x ∈ I, x < c and f ′(x) < 0 for x ∈ I, x > c, then f has a local

maximum at c.

• ii. If f ′(x) < 0 for x ∈ I, x < c and f ′(x) > 0 for x ∈ I, x > c, then f has a local

minimum at c. Note: This is not really something to memorize. Simply visualize whether the function is increasing or decreasing on either side of a critical point.

Concavity We say a curve is concave up if the slope of its tangent increases as we go left to right and say it is concave down if the slope of its tangent decreases as we go left to right. Since the derivative measures rate of change, this immediately leads to the following theorem enabling us to use the derivative of the derivative, known as the second derivative, to analyze concavity. Theorem 16 (Concavity). If f ′′(x) > 0 for all x in some interval, then the graph of f is concave up on that interval. If f ′′(x) < 0 for all x in some interval, then the graph of f is concave down on that interval. A graph is generally concave up near a minimum and concave down near a

• maximum. Knowing where a graph is concave down and where it is concave

up further helps us to sketch a graph. Summary So Far We can often sketch a graph through the following steps suggested by the analysis so far of monotonicity and concavity. (1) Calculate f ′(x) and f ′′(x). (2) Factor f ′(x) and f ′′(x) completely. (3) Find the critical points of f. (4) Analyze monotonicity. (5) Find the possible points of inflection by determining where f ′′(x) = 0

• r f ′′(x) doesn’t exist.

(6) Analyze concavity. (7) Plot the critical points and possible points of inflection. (8) Connect those points appropriately using the information about mono- tonicity and concavity. Other Things To Consider

35

Often, an examination of monotonicity and concavity will give enough infor- mation to sketch a graph. Sometimes, it is useful to look at other properties such as intercepts, symmetry and asymptotes. These are described below. As you gain experience sketching graphs, you will begin to get a feeling for when it is worthwhile to give these properties more than a cursory consideration. Intercepts A point where a curve crosses one of the axes is called an intercept. There are two types of intercepts, x−intercepts and y−intercepts. Since a curve crosses the y−axis at a point where the first coordinate equals 0, you can calculate the y−intercept by simply evaluating f(0). This is usually fairly easy to calculate. Since a curve crosses the x−axis at a point where the second coordinate equals 0, you can calculate the x−intercept by solving the equation f(x) = 0. Depending on the formula for f, this may be very difficult to solve and often is not worth the effort. It’s generally important to find the y−intercept when it’s unclear whether the curve crosses the y−axis above or below the origin, since drawing a picture with the curve crossing the y−axis on the wrong side of the origin presents a misleading picture. It’s important to find the x−intercepts when you’re not sure whether or not the curve crosses the x−axis. Symmetry There are two types of symmetry that are sometimes worth paying attention to, symmetry about the y−axis and symmetry about the origin. Even functions are symmetric about the y−axis. You can check whether a function is even by seeing if f(−x) = f(x) for all values of x. Polynomial functions that contain only even powers, such as f(x) = x8 − 5x2 + 3, are examples of even functions. The cosine function is also an example of an even function. Odd functions are symmetric about the origin. You can check whether a function is odd by seeing if f(−x) = −f(x) for all values of x. Polynomial functions that contain only odd powers, such as f(x) = 10x7 + 8x3 − x, are examples of odd functions. The sine function is also an example of an odd function. It is never actually absolutely necessary to check for symmetry. However, if you recognize from the formula for a function that its graph should exhibit

36

symmetry, then you have another check for whether you have sketched the graph correctly. Asymptotes There are two types of asymptotes, horizontal and vertical. A graph will have a horizontal asymptote y = α if limx→∞ f(x) = α. In that case, the right side of the curve will get closer and closer to the horizontal asymptote, the line y = α. For example, if f(x) = (10x + 3)/(2x − 1), then limx→∞ f(x) = 5, so that the line y = 5 is a horizontal asymptote. Similarly, a graph will have a horizontal asymptote y = α if limx→−∞ f(x) = α. In that case, the left side of the curve will get closer and closer to the horizontal asymptote, the line y = α. For example, if f(x) = (15x+3)/(5x− 1), then limx→−∞ f(x) = 3, so that the line y = 3 is a horizontal asymptote. It’s worth checking for a horizontal asymptote on the right hand side if, for large x, either f is increasing and concave down or decreasing and concave

• up. (Clearly, there can be no such asymptote if f is increasing and concave

up or decreasing and concave down. Draw a picture to see why.) A graph will have a vertical asymptote x = α if either limx→α+ f(x) = ∞, limx→α+ f(x) = −∞, limx→α− f(x) = ∞ or limx→α− f(x) = −∞. For example, if limx→α+ = ∞, it follows that if a point (x, y) is on the graph and x is just a little bigger than α, then y must be very large and hence the curve must be close to the line x = α. Similar arguments hold for each of the other cases. Since vertical asymptotes, by their very nature, can exist only at discontinu- ities, it is generally a simple matter to recognize possible asymptotes. You can look for some of the the same clues that lead you to look for discontinuities— denominators that are zero. Once you suspect that x = α is a vertical asymptote, check the two one-sided limits at α. For example, let f(x) = x/(x − 3)2. Clearly, the denominator is zero when x = 3, so x = 3 is a possible vertical asymptote. Since limx→3+ x/(x−3)2 = ∞, the line x = 3 is a vertical asymptote for the portion

• f the curve on the right. Since limx→3− x/(x − 3)2 = ∞ also, the line x = 3

is also vertical asymptote for the portion of the curve on the left. Summary of Curve Sketching Putting everything together, we have the following steps. (1) Calculate f ′(x) and f ′′(x). (2) Factor f ′(x) and f ′′(x) completely.

37

(3) Find the critical points of f. (4) Analyze monotonicity. (5) Find the possible points of inflection by determining where f ′′(x) = 0

• r f ′′(x) doesn’t exist.

(6) Analyze concavity. (7) Plot the critical points and possible points of inflection. (8) Connect those points appropriately using the information about mono- tonicity and concavity. (9) Find intercepts if necessary. (10) Check for symmetry if useful. (11) Check for horizontal and vertical asymptotes if needed. Optimization Calculus and derivatives may be used to solve problems, sometimes called word problems or verbal problems or applications, which effectively call for finding a maximum or minimum of a function. These problems should be approached in the same way that word problems that come up in algebra, geometry or trigonometry are approached. The only real difference is that whereas the solutions to problems in those areas are usually ultimately found by solving equations that arise when analyzing the information given, these problems will involve optimization, that is, finding a minimum or maximum, so besides solving equations one will generally have to calculate derivatives and usually solve an equation or equations obtained when one sets a derivative to 0. Guidelines

• Read the question!
• Read the question!
• Read the question!

Has the point been made? Everything that must be done generally becomes apparent if one reads the question! Ideas to Keep in Mind

• What do you know?
• What don’t you know?
• What can you figure out or infer?
• What do you want?

This is actually the least important!

38

What You Know Every fact translates into a mathematical statement, generally a formula, equation or statement that a particular variable takes on a certain value when another variable takes on a certain value. The key to writing down an appropriate formula corresponding to a given fact is to write down that fact in plain language and then rewrite the fact using the descriptions of variables already defined and a verb such as is or equals which indicates that two quantities are equal. What You Don’t Know This is often the key. Any unknown quantity can often be profitably rep- resented by a variable. Related to unknown quantities are variable quantities. These almost always need to be represented by variables. What You Want This needs to be kept in the back of your mind. One common mistake is to concentrate too hard on what you want; it’s generally more advantageous to concentrate on what you know, what you don’t know and what you can figure out from what you know. Just remember to notice when you’ve actually figured out what you ulti- mately want. Suggestions

• Draw pictures, charts, graphs or anything visual that may help you

understand the problem. The key is understanding and translating facts to mathematics.

• Look for variables and unknowns. Represent them by symbols. Write

down what each stands for and make sure you don’t use the same symbol to represent two different quantities.

• Write down known facts in terms of the variables and unknowns you’ve
• defined. These will generally be in the form of equations and formulas.
• Solve equations where possible.
• If there’s a variable that needs to be optimized, solve for it in terms of
• ther variables, find its derivative and see when its derivative is equal

to 0 or undefined. Remember: A derivative is always undefined at the endpoints! Antiderivatives

39

Definition 18 (Antiderivative). If F ′(x) = f(x) we call F an antiderivative

• f f.

Definition 19 (Indefinite Integral). If F is an antiderivative of f, then

• f(x) dx = F(x) + c is called the (general) Indefinite Integral of f, where c

is an arbitrary constant. The indefinite integral of a function represents every possible antiderivative, since it has been shown that if two functions have the same derivative on an interval then they differ by a constant on that interval. Terminology: When we write

• f(x) dx, f(x) is referred to as the integrand.

Basic Integration Formulas As with differentiation, there are two types of formulas, formulas for the integrals of specific functions and structural type formulas. Each formula for the derivative of a specific function corresponds to a formula for the derivative

• f an elementary function. The following table lists integration formulas side

by side with the corresponding differentiation formulas.

• xn dx = xn+1

n + 1 if n = −1 d dx (xn) = nxn−1

• sin x dx = − cos x + c

d dx (cos x) = − sin x

• cos x dx = sin x + c

d dx (sin x) = cos x

• sec2 x dx = tan x + c

d dx (tan x) = sec2 x

• ex dx = ex + c

d dx (ex) = ex 1 x dx = ln x + c d dx (ln x) = 1 x

• k dx = kx + c

d dx (kx) = k Structural Type Formulas We may integrate term-by-term:

• kf(x) dx = k
• f(x) dx
• f(x) ± g(x) dx =
• f(x) dx ±
• g(x) dx

40

In plain language, the integral of a constant times a function equals the constant times the derivative of the function and the derivative of a sum or difference is equal to the sum or difference of the derivatives. These formulas come straight from the corresponding formulas for calculat- ing derivatives and are used the same way.Integrating Individual Terms When calculating derivatives of individual terms, one needs to recognize whether the term is an elementary function, a product, a quotient or a com- posite function. There is a little bit more art to integration, at least if the term is not the derivative of an elementary function. Integration is essentially the reverse of differentiation, so one might expect formulas for reversing the effects of the Product Rule, Quotient Rule and Chain Rule. This is almost the case. There is a formula, called the Inte- gration By Parts Formula, for reversing the effect of the Product Rule and there is a technique, called Substitution, for reversing the effect of the Chain

• Rule. There is no specific formula or technique for reversing the effect of

the Quotient Rule, but one is not really necessary since the Quotient Rule is redundant. Integration also becomes an art because not only isn’t it always obvious whether one should resort to Integration By Parts or the Substitution Tech- nique but the use of the Integration By Parts Formula and the Substitution Technique is not as straightforward as the use of the Product, Quotient or Chain Rule. The Substitution Technique The substitution technique may be divided into the following steps. Every step but the first is purely mechanical. With a little bit of practice (in other words, make sure you do the homework problems assigned), you should have no more difficulty carrying out a substitution than you should be having by now when you differentiate. Note: In the following, we will assume that you are trying to calculate an integral

• f(x)dx. If the dummy variable is called something other than x,

then some of the names you would use for variables might be different. (1) Choose a substitution u = g(x). Some suggestions on how to choose a substitution will be made later. (2) Calculate the derivative du dx = g′(x). (3) Treating the derivative as if it were a fraction, solve for dx:

41

du dx = g′(x), du = g′(x)dx, dx = du g′(x). (4) Go back to the original integral and replace g(x) by u and replace dx by du g′(x). (5) Simplify. Every incidence of x should cancel out at this step. If not, you will need to try another substitution. Make sure that you simplify properly—this is the easiest step to make mistakes during. (6) Integrate. (7) Replace u by g(x) in your result. (8) Check your answer (of course). Choosing an Appropriate Substitution This is the only non-routine part of carrying out a substitution, but should not be at all difficult for any student who has mastered the art of differen-

• tiation. There are two basic tactics for choosing a substitution. Each will

work in the vast majority of cases where a routine substitution is needed. Since neither will work in all cases, you need to be comfortable with both. Therefore, you should try using both methods on the same problem wherever

• possible. (There are quite a few non-routine substitutions that are used in

special situations. They need to be learned separately.) The First Method The method most students probably find easiest to use relies on familiarity with the chain rule for differentiation. In order to decide on a useful sub- stitution, look at the integrand and pretend that you are going to calculate its derivative rather than its integral. (You usually don’t actually have to write anything down—you can usually just visualize the steps.) Look to see if there is any step during which you would have to use the chain rule. If so, think of the decomposition you would have to make, i.e. the step where you would write down something like y = f(u), u = g(x). Try the substitution u = g(x). The Second Method This method involves looking at parts of the integrand and observing whether the derivative of part of the integrand equals some other factor of

42

the integrand. If so, u may be substituted for that part. (In deciding, you may ignore constant factors, since they are easy to manipulate around.) Σ Sigma Notation Sigma notation is a mathematical shorthand for expressing sums where every term is of the same form. For example, suppose we want to write out the sum of all the integers from 1 to 100, inclusively. One might write 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+ 19 + 20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 + 53 + 54 + 55 + 56 + 57 + 58 + 59 + 60 + 61 + 62 + 63 + 64 + 65 + 66 + 67+68+69+70+71+72+73+74+75+76+77+78+79+80+81+82+83+ 84+85+86+87+88+89+90+91+92+93+94+95+96+97+98+99+100 This might strike one as being somewhat tedious. We might try writing something like 1 + 2 + 3 + · · · + 98 + 99 + 100, leaving the middle terms to the imagination, or we might use Sigma Notation and simply write 100

k=1 k,

which we may read as the sum, for k taking on every integer value starting with 1 and going up to 100, of all numbers of the form k. Suppose we want to add together the squares of all the integers from 1 to

• 100. Again, we might write

12+22+32+42+52+62+72+82+92+102+112+122+132+142+152+162+ 172+182+192+202+212+222+232+242+252+262+272+282+292+302+ 312+322+332+342+352+362+372+382+392+402+412+422+432+442+ 452+462+472+482+492+502+512+522+532+542+552+562+572+582+ 592+602+612+622+632+642+652+662+672+682+692+702+712+722+ 732+742+752+762+772+782+792+802+812+822+832+842+852+862+ 872+882+892+902+912+922+932+942+952+962+972+982+992+1002, but it would be more convenient to write 12 + 22 + 32 + · · · + 982 + 992 + 1002

• r

100

k=1 k2.

We might write 5

k=1(2k + 1) rather than writing 3 + 5 + 7 + 9 + 11.

In each of these examples, k is the index. The index does not have to be

• k. Other frequently used symbols are i, j, m and n.

43

Also, the index doesn’t have to start with the value 1. As an example, we could have 6

k=3(2k + 1)

rather than writing 7 + 9 + 11 + 13

• r

−2

k=−5 5k

rather than writing (−25) + (−20) + (−15) + (−10). In general, we may interpret β

k=α ak as the sum of all terms of the form

ak for all integer values of k between α and β. Some Useful Formulas Involving Sums β

k=α cak = c β k=α ak

β

k=α(ak + bk) = β k=α ak + β k=α bk

These two formulas are generalizations of the associative, commutative and distributive laws. n

k=1 1 = n

n

k=1 k = n(n + 1)

2 n

k=1 k2 = n(n + 1)(2n + 1)

6 These three formulas are useful in a number of calculations. We will demon- strate proofs of the second and third of these. The first should be obvious. We will prove the second formula two different ways. The first comes from simply writing down the sum two different ways, frontwards and backwards.

• Proof. Let S = n

k=1 k. We may write S = 1+2+3+· · ·+(n−2)+(n−1)+n =

n + (n − 1) + (n − 2) + · · · + 3 + 2 + 1. Writing 2S = S + S = 1 + 2 + · · · + n − 1 + n + n + n − 1 + · · · + 2 + 1 Adding all the terms by adding together the pairs which lie above and below each other, we get 2S = (n + 1) + (n + 1) + · · · + (n + 1) + (n + 1) = n(n + 1), from which it follows that S = n(n + 1) 2 .

• The second proof will use a very useful method called Mathematical Induc-

tion.

44

Mathematical Induction is often used to prove that statements involving an arbitrary integer n are true for all positive integral values of that integer. It works as follows. We may have a proposition we may denote by P(n), where n represents an arbitrary integer. Using Mathematical Induction, we need to prove two statements: (1) P(1) is true. (2) If P(n) is true, then P(n + 1) is also true. If we can prove both statements, then P(n) must be true for all positive integers n.

• Proof. In this instance, the assertion P(n) corresponds to the assertion n

k=1 k =

n(n + 1) 2 and P(n+1) corresponds to the assertion n+1

k=1 k = (n + 1)(n + 2)

2 . Thus, P(1) corresponds to the assertion 1

k=1 k = 1(1 + 1)

2 . But 1

k=1 k = 1, while 1(1 + 1)

2 = 2 2 = 1, so P(1) is certainly true. Now assume P(n) is true. This means n

k=1 k = n(n + 1)

2 . It follows that n+1

k=1 k = n k=1 k + (n + 1) = n(n + 1)

2 + (n + 1) = (n + 1) n 2 + 1

• = (n + 1) · n + 2

2 = (n + 1)(n + 2) 2 , so P(n + 1) is true.

• We will also prove the third formula two different ways. The first proof

will involve Mathematical Induction, while the second proof will involve an interesting algebraic trick that can be generalized to prove similar formulas for summing higher powers.

• Proof. We wish to prove

n

k=1 k2 = n(n + 1)(2n + 1)

6 is true for all positive integer values of n. If n = 1, this corresponds to the statement 1

k=1 k2 = 1 · 2 · 3

6 , which is obviously true. Now suppose the assertion is true for a given value of n. It then follows that n+1

k=1 k2 = n k=1 k2 + (n + 1)2 =

n(n + 1)(2n + 1) 6 + (n + 1)2 =

45

n + 1 6 · (n(2n + 1) + 6(n + 1)) = n + 1 6 · (2n2 + n + 6n + 6) = n + 1 6 · (2n2 + 7n + 6) = n + 1 6 · (n + 2)(2n + 3) = [n + 1]([n + 1] + 1)(2[n + 1] + 1) 6 .

• The next proof involves the interesting algebraic trick.
• Proof. We start with the following calculation:

Let S = [1 · 2 · 3 − 0 · 1 · 2] + [2 · 3 · 4 − 1 · 2 · 3] + [3 · 4 · 5 − 2 · 3 · 4] + · · · + [n · (n + 1) · (n + 2) − (n − 1) · n · (n + 1)]. On the one hand, this is a telescoping sum, with a lot of cancellation, leaving the conclusion S = n(n + 1)(n + 2) after all the cancellation. On the other hand, we can write S = 3(1 · 2 + 2 · 3 + 3 · 4 + . . . n · (n + 1)). Looking at the two representations of S, we can conclude 1 · 2 + 2 · 3 + 3 · 4 + . . . n · (n + 1) = n(n + 1)(n + 2) 3 . In Sigma Notation, this may be written n

k=1 k(k + 1) = n(n + 1)(n + 2)

3 . We may then observe n

k=1 k2 = n k=1 k · k =

n

k=1[k(k + 1) − k] =

n

k=1 k(k + 1) − n k=1 k =

n(n + 1)(n + 2) 3 − n(n + 1) 2 = n(n + 1) 6 · (2(n + 2) − 3) = n(n + 1) 6 · (2n + 1) = n(n + 1)(2n + 1) 6 .

• Riemann Sums

46

Definition 20 (Partition). A partition of an interval [a, b] with a ≤ b is a set {x0, x1, x2, . . . , xn} where a = x0 ≤ x1 ≤ x2 ≤ · · · ≤ xn = b. Definition 21 (Riemann Sum). Let f be a function defined on [a, b] and let P = {x0, x1, . . . , xn} be a partition of [a, b]. Let ∆xk = xk − xk−1 for k = 1, 2, 3, . . . , xn and let ck ∈ [xk−1, xk] for k = 1, 2, 3, . . . , xn. We define the Riemann Sum for f for the partition P on the interval [a, b] as R(f, P, a, b) = n

k=1 f(ck)∆xk.

If f is a nice, positive, continuous function then R(f, P, a, b) may be inter- preted as a sum of areas of rectangles, giving an approximation to the area bounded by the graph of f, the x−axis and the lines x = a, x = b. We will sometimes write R(f, P) or even just R(f) instead of R(f, P, a, b). The Definite Integral As the widths ∆xk of the subintervals approach 0, the Riemann Sums hopefully approach a limit. If that happens, we call the limit the definite integral of f from a to b and denote it by b

a f(x) dx.

We immediately get the application that if f is positive and continuous, then b

a f(x) dx is equal to the area of the region bounded by the graph of f,

the x−axis and the lines x = a, x = b. Calculating Definite Integrals The Fundamental Theorem of Calculus Theorem 17 (FTC-Part I). If f is continuous on [a, b], then F(x) = x

a f(t) dt

is defined on [a, b] and F ′(x) = f(x). Theorem 18 (FTC-Part II). If f is continuous on [a, b] and F(x) =

• f(x) dx
• n [a, b], then

b

a f(x) dx = F(x)

• b

a = F(b) − F(a).

Note the introduction of a notation: F(x)

• b

a = F(b) − F(a)

Example 1

0 x2 dx = x3

3

• 1

0 = 13

3 − 03 3 = 1 3. Example π

0 sin x dx = (− cos x)

• π

0 − cos π − (− cos 0) = −(−1) − (−1) = 2.

Example Calculate 1 x x2 + 1 dx.

47

This will require slightly more work, since it’s a little harder to find an antiderivative in this case. We’ll use the method of substitution. Let I =

• x

x2 + 1 dx and let u = x2 + 1. du dx = 2x du = 2xdx dx = du 2x I = x u du 2x = 1

2

1 u du = 1

2 ln |u| = 1 2 ln(x2 + 1) + c

Thus 1 x x2 + 1 dx = 1

2 ln(x2 + 1)

• 1

0 = 1 2 ln(12 + 1) − 1 2 ln(02 + 1) = 1 2 ln 2 − 1 2 ln 1 = 1 2 ln 2.

Trapezoid Rule The Trapezoid Rule is used to estimate an integral b

a f(x) dx.

Let: h = ∆x = b − a n xk = a + kh yk = f(xk) b

a f(x) dx

≈ h 2(y0 + 2y1 + 2y2 + · · · + 2yn−1 + yn) = b − a 2n (y0 + 2y1 + 2y2 + · · · + 2yn−1 + yn) Area Under a Parabola It will be shown that the integral of a quadratic function depends only

• n the width of the interval over which it’s integrated and the values of the

function at the midpoint and endpoints. To simplify the calculations, assume that the interval is of the form [−h, h] and that the quadratic function is of the form f(x) = ax2 + bx + c. Let I = h

−h f(x) dx. This may be integrated easily using the Fundamental

Theorem of Calculus.

48

I = h

−h

f(x) dx = h

−h

ax2 + bx + c dx = ax3/3 + bx2/2 + cx|h

−h

= ah3/3 + bh2/2 + ch − {a(−h)3/3 + b(−h)2/2 + c(−h)} = ah3/3 + bh2/2 + ch + ah3/3 − bh2/2 + ch = 2ah3/3 + 2ch = h 3 · (2ah2 + 6c) Let y−h = f(−h) = ah2 − bh + c y0 = f(0) = c yh = f(h) = ah2 + bh + c Since y−h +yh = 2ah2 +2c, it is easily seen that 2ah2 +6c = y−h +4y0 +yh, and thus I = h 3 · (y−h + 4y0 + yh). Simpson’s Rule The Parabola Rule Simpson’s Rule may be used to approximate b

a f(x) dx. It takes the idea

• f the Trapezoid Rule one degree higher.

Rationale Partition the interval [a, b] evenly into n subintervals, where n is even, so that each subinterval has width h = b − a n and let yk = f(xk). Estimate the integral over adjacent pairs of integrals by the integral of a quadratic function agreeing with f at the midpoint and endpoints of the interval. x2

x0 f(x) dx ≈ h

3 · (y0 + 4y1 + y2) x4

x2 f(x) dx ≈ h

3 · (y2 + 4y3 + y4)

49

x6

x4 f(x) dx ≈ h

3 · (y4 + 4y5 + y6) . . . xn

xn−2 f(x) dx ≈ h

3 · (yn−2 + 4yn−1 + yn) If everything is added together, we obtain the estimate b

a f(x) dx ≈ h

3 · (y0 + 4y1 + 2y2 + 4y3 + 2y4 + · · · + 2yn−2 + 4yn−1 + yn). This is known as Simpson’s Rule. Midpoint Rule b

a f(x) dx

≈ h ·

• f

x0+x1

2

• + f

x1+x2

2

• + · · · + f

xn−1+xn

2

• Trapezoid Rule

b

a f(x) dx

≈ h 2(y0 + 2y1 + 2y2 + · · · + 2yn−1 + yn) = b − a 2n (y0 + 2y1 + 2y2 + · · · + 2yn−1 + yn) Simpson’s Rule b

a f(x) dx

≈ h 3 · (y0 + 4y1 + 2y2 + 4y3 + 2y4 + · · · + 2yn−2 + 4yn−1 + yn) = b − a 3n · (y0 + 4y1 + 2y2 + 4y3 + 2y4 + · · · + 2yn−2 + 4yn−1 + yn) Error Estimates Let ET be the error in the Trapezoid Rule. Let EM be the error in the Midpoint Rule. Let ES be the error in Simpson’s Rule. Let K be a bound on the second derivative. Let K∗ be a bound on the fourth derivative. |ET| ≤ K(b − a)3 12n2 |EM| ≤ K(b − a)3 24n2 |ES| ≤ K∗(b − a)5 180n4 Applications of Integration

50

Many applications of definite integrals arise from one of the following two situations.

• A quantity may be viewed as a function Q(x) of a variable x and the

derivative Q′(x) = q(x) is known. A value Q(a) is known–usually, the value is 0–and the value of Q(b) is needed. From the Fundamental Theorem of Calculus, one observes Q(b) = Q(a) + b

a q(x) dx.

• One needs to obtain some quantity Q and is not able to evaluate it
• directly. However, one is able to estimate Q by dividing an interval [a, b]

into subintervals and obtaining an approximation Q ≈ n

k=1 f(ck)∆xk,

where the ck are points in the subintervals and the ∆xk are the lengths

• f the subintervals. One observes the approximation is a Riemann Sum

R(f, P, a, b) for f and concludes the quantity Q desired is actually equal to the integral b

a f(x) dx.

Most of our applications we be obtained through the latter route. Areas The definite integral eminated out of the problem of estimating the area

• f a region of the form {(x, y) : a ≤ x ≤ b, 0 ≤ y ≤ f(x)}. With a little

imagination, the areas of most plane regions one comes across can be obtained by breaking the region into pieces, each of which may be described as a set

• f the form {(x, y) : a ≤ x ≤ b, f(x) ≤ y ≤ g(x)}. The area of such a region

is given by b

a g(x) − f(x) dx.

This is apparent if f(x) ≥ 0, since then the area of the region desired is clearly the difference between the area of the region between the x−axis and the graph of f and the area of the region between the graph of the x−axis and the graph of g. The former area is b

a f(x) dx and the area of

the latter is b

a g(x) dx, so the difference is equal to

b

a g(x) dx −

b

a f(x) dx =

b

a g(x) − f(x) dx.

If one interprets g(x) − f(x) as the height of the region, it’s clear that the requirement f(x) ≥ 0 is unnecessary. Volume Consider a solid for which one wants the volume. Suppose an x−axis is drawn and it is found that for a particular value of x the cross-sectional area is given by a function A(x).

51

This means that if one slices the solid by making a cut through the solid perpendicular to the x−axis, the cut goes through a portion of the solid along a plane (the cross-section) and the area of that portion is A(x). Now, suppose one cuts the solid into thin slices by cutting perpendicular to the points {x0, x1, x2, . . . , xn} along the x−axis. This determines a partition P of [a, b], where a = x0 and b = xn. The volume ∆Vk of the kth slice may be ap- proximated by taking the area A(ck) of the cross-section anywhere in the interval [xk−1, xk] and multiplying by the thickness ∆xk of the slice. So ∆Vk ≈ A(ck)∆xk. The total volume is thus n

k=1 ∆Vk ≈ n k=1 A(ck)∆xk.

This is the Riemann Sum R(A, P, a, b) and we may conclude the volume is equal to the integral b

a A(x) dx.

Volume of a Solid of Revolution Suppose a solid is obtained by taking a region of the form {(x, y) : a ≤ x ≤ b, 0 ≤ y ≤ f(x)} and rotating it about the x−axis. The solid obtained is called a Solid of Revolution. Each cross section is a circle of radius f(x), so the cross-sectional area is given by πf(x)2 and the volume will be b

a πf(x)2 dx = π

b

a f(x)2 dx.

This is sometimes referred to as the Disk Method, since the individual slices look like disks. Variations

• One may also take a region bounded by a curve of the form x = g(y),

two horizontal lines and the y−axis and rotate it about the y−axis. One gets a similar formula, with the roles of x and y interchanged.

• One may rotate around a line parallel to one of the coordinate axes.

In this case, the key point to remember is the cross-section is a circle so one must figure out what the radius is. Cylindrical Shells Suppose one creates a solid by taking a region of the form {(x, y) : a ≤ x ≤ b, 0 ≤ y ≤ f(x)} but rotating it about the y−axis rather than around the x−axis. One may still be able to find its volume by employing the Disk Method in ingenious ways, but it will usually be easier to use another method called the Method of Cylindrical Shells.

52

To make the analysis simpler, assume a ≥ 0, so the region is in the first quadrant, and create a partition P = {x0, x1, x2, . . . , xn} of [a, b]. Let Sk be the solid obtained by rotating the vertical strip {(x, y) : xk−1 ≤ x ≤ xk} about the y−axis and let ∆Vk be the volume of Sk. The volume

• f the entire solid will be the sum n

k=1 ∆Vk. We will estimate the entire

volume by finding an estimate for each ∆Vk and adding them up. This will give us a Riemann Sum. Sk may be visualized as almost being a cylindrical shell. It’s relatively simple to visualize the volume of a cylindrical shell of a given height, radius and thickness as follows. Visualize making a vertical cut through the height of the shell. (Many tin cans have seams; one may imagine cutting along the seam.) Now separate the edges along the cut and flatten the shell. You’ll obtain a very thin rectangular solid whose volume, which is obviously equal to the volume of the original shell, will be equal to the product of its length, width and height. The height of the rectangular solid is clearly the same as the height of the shell and the width is clearly the same as the thickness of the shell. The length of the rectangular solid is clearly equal to the circumference of the shell, which is 2π× the radius of the shell. Thus the volume of the solid, and hence the shell, will equal the product

• f the height, thickness and 2π× its radius.

Sk is not quite a cylindrical shell, but this calculation may be used to estimate its volume. The height of Sk varies, if we choose some ck in the interval [xk−1, xk] the height will not vary much from f(ck). The thickness of Sk is clearly ∆xk. The radius of Sk is not unambiguously defined, but it certainly isn’t much different from ck. We thus get ∆Vk ≈ f(ck) · ∆xk · 2πck = 2πckf(ck)∆xk. We may thus approximate the volume by n

k=1 2πckf(ck)∆xk = 2π n k=1 ckf(ck)∆xk.

This is 2πR(xf(x), P, a, b) and we conclude the volume is the integral 2π b

a xf(x) dx.

53

As with the Disk Method, there will be variations. The key point to re- member is to visualize a cylindrical shell and pick out the radius, height and thickness. It is the thickness which determines whether one integrates with respect to x or with respect to y. If one obtains the shells by partitioning an interval along the y−axis, then one will need to integrate with respect to y. Arc Length Consider a curve y = f(x), a ≤ x ≤ b. We may estimate its length by dividing it into small sub-arcs, estimating the length of each sub-arc by the length of a straight line segment and adding those lengths together. With a little simplification and the judicious use of the Mean Value Theorem, we’ll

• btain a Riemann Sum and conclude the length of the curve is equal to a

specific definite integral. As usual, create a partition P = {x0, x1, x2, . . . , xn} of the interval [a, b]. Let ∆sk be the length of the portion of the curve for which xk−1 ≤ x ≤ xk. Since the length s of the curve is equal to n

k=1 ∆sk, we may obtain an

approximation to s by adding together our approximations to each of the ∆sk. Since the portion of the curve for which xk−1 ≤ x ≤ xk has endpoints (xk−1, f(xk−1)) and (xk, f(xk)), we may approximate ∆sk by the length of the line segment joining (xk−1, f(xk−1)) and (xk, f(xk)). Using the distance formula, we obtain ∆sk ≈

• (xk − xk−1)2 + (f(xk) − f(xk−1))2.

If f is differentiable on [a, b], it will satisfy the hypotheses of the Mean Value Theorem and we may infer the existence of some point ck ∈ (xk−1, xk) such that f(xk) − f(xk−1) = f ′(ck)(xk − xk−1). Hence, ∆sk ≈

• (xk − xk−1)2 + (f ′(ck)(xk − xk−1))2 =
• (1 + f ′(ck)2)(xk − xk−1)2 =
• 1 + f ′(ck)2(xk − xk−1) =
• 1 + f ′(ck)2∆xk.

We may then approximate the length of the curve by s = n

k=1 ∆sk ≈ n k=1

• 1 + f ′(ck)2∆xk.

This is a Riemann Sum R(

• 1 + f ′(x)2, P, a, b) for the function
• 1 + f ′(x)2

and we conclude the arc length is equal to the integral s = b

a

• 1 + f ′(x)2 dx.