Math 1120 Class 1 Dan Barbasch Aug. 23, 2012 Course Website Math - - PowerPoint PPT Presentation

Math 1120 Class 1 Dan Barbasch

Math 1120 Class 1

Dan Barbasch

• Aug. 23, 2012

Math 1120 Class 1 Dan Barbasch

Course Website

http://www.math.cornell.edu/˜web1120/index.html Dan Barbasch (dmb14) Office Hours: Mo, We 1:30-2:30 in Malott 543 or by appointment Other lecture sections will announce their own office hours. Regular Office Hour Schedule will be at website. First Assignment due 8/30 or 8/31 depending on the lecture section. Frequent quizes and assignments due during the week. Evening Prelims Tuesday Spetember 25, and Thursday October 25.

Math 1120 Class 1 Dan Barbasch

Is 1120 the right course for you?

Math 1120 is a second semester course, so you need to remember material from first semester calculus. If you are interested in a more theoretical course that covers somewhat more material, you might want to take Math 1220 - Honors Calculus II. If you need to take a mathematics course but don’t need to take calculus, you might want to take Math 1105, 1300, 1340, 1350, or 1710. Math 1340: Mathematics and Politics Math 1710: Statistics Math 1105: Finite Math. for the Life and Social Sciences (Fall) Math 1350: Art of Secret Writing (Fall) If you have an AP score of 4 or 5 in BC Calculus, we do not recommend you repeat the course.

Math 1120 Class 1 Dan Barbasch

Grading

At the end of the term we will assign grades based on each prelim 20%, final 30%, 40% in fall 2015 homework 20%, and section grade in fall 2015 section grade 10%. The section grade will be based on participation, quizzes, warmup questions, interaction with your lecturer, and the quality (not just correctness) of your homework writeups. Typically half the students receive a grade of B or higher.

Math 1120 Class 1 Dan Barbasch

Math 1012

Math 1012 is a course that parallels Math 1120.

• www. lsc. sas. cornell. edu/ suppcourses/

Supplemental_ courses_ descriptions/ Math1011_ 1012. html Wednesday 2:55-4:10pm in Malott 251. Wednesday 7:30-8:45pm in Malott 251.

Math 1120 Class 1 Dan Barbasch

Review of Calculus 1

1 Derivative 2 Rules for Sums, Products, Quotients, Trigonometric

Functions

3 Chain Rule 4 Linear Approximation 5 Powers, Logs 6 Monotonicity, Critical Points, Maximum/Minimum, 2nd

Derivative Test

7 Inverse Functions 8 Riemann Sums, Antiderivative

Check review Sessions of MSC, August 28 and Auguast 30 First new topic of the course: Fundamental theorem of Calculus

Math 1120 Class 1 Dan Barbasch

Definition of Derivative

f ′(x) = lim

∆x→0

f (x + ∆x) − f (x) ∆x The interpretation of the derivative is the (instantaneous) rate

• f change of the quantity f with respect to the quantity x.

Math 1120 Class 1 Dan Barbasch

Chain Rule

Example: Compute d

dx

• (3x2 + 1)4.

For differentiating compositions of functions, many ways to write the formula: e.g.f = g ◦ h f ′(x) = g′(h(x)) · h′(x).

• r z = g(y),

y = h(x) dz dx = dz dy dy dx

• r z = g(y),

y = h(x), z = f (x) f ′(x) = g′(y)h′(x)

Math 1120 Class 1 Dan Barbasch

Powers and Logs

y = ex is the unique function which satisfies y′ = y and y(0) = 1. This function is very important in applications. The equation y′ = ky where k is a constant is interpreted as encoding the fact that the rate of change of y with respect to x is proportional to y; radioactive decay, population growth, continuously compounded interest rate. The general solution is y = Cekx.

Math 1120 Class 1 Dan Barbasch

Powers and Logs

The inverse function of ex, y = ln x (defined for x > 0 only!) is the unique function which satisfies y′(x) = 1 x and y(1) = 0. It is also the unique function satisfying eln x = x, ln (ex) = x, Arithmetic rules for logarithms and exponentials ea+b = ea · eb (ea)b = eab ln(ab) = ln a + ln b ln(ab) = b ln a ln(a + b) = ln a + ln b .

Math 1120 Class 1 Dan Barbasch

Powers and Logs

Calculus of logs and powers with other bases can be reduced to base e using eln b = b ln eb = b. Example:

d dx (10x) =?

Example:

• 10x dx =?

Example:

d dx (log2 x) =?

All logarithms are proportional: logb x = ln x ln b y = logb x ⇐ ⇒ by = x ⇐ ⇒ ln(by) = ln x ⇐ ⇒ y ln b = ln x.

Math 1120 Class 1 Dan Barbasch

Inverse functions

Inverse functions are about solving equations. If y = f (x), express x as a function of y. x is a function of y only if the equation has a UNIQUE solution x for each y. We also write y = f −1(x) ⇐ ⇒ x = f (y). Because we like the argument of a function to be called x, we write y = f (x), has inverse functiony = f −1(x), satisfying f (f −1(x)) = x, f −1(f (x)) = x.

Math 1120 Class 1 Dan Barbasch

Inverse functions

Example: y = ex has inverse x = ln y; we write the new function as y = ln x, interchanging x and y. Note that ex is increasing, has domain all real x, but has range y > 0. So y = ln x is also increasing, but has domain x > 0 and range all y. Thus the inverse function ln x is only defined for x > 0. The equations eln x = x, ln ex = x encode the fact that these functions are inverse to each other.

Math 1120 Class 1 Dan Barbasch

Inverse functions

Example: y = sin x DOES NOT have an inverse because it is not 1-1. It is 1-1 and onto if x is restricted to the interval −π/2 ≤ x ≤ π/2. The inverse function is called y = Arcsin x; it has domain −1 ≤ x ≤ 1 and range −π/2 ≤ y ≤ π/2. In words: y = Arcsin x is the unique value −π/2 ≤ y ≤ π/2 which satisfies sin y = x. In equation form: sin (Arcsin x) = x, Arcsin (sin x) = x ONLY IF −π/2 ≤ x ≤ π/2.

Math 1120 Class 1 Dan Barbasch

Inverse functions

REFERENCE section 3.9. y = Arccos x, y = Arctan x, y = Arccot x, y = Arcsec x, y = Arccsc x are in section 3.9. The notation in the text is sin−1 x rather than Arcsin x, but I think this leads to the possible confusion that sin−1 x might be 1/ sin x.

Math 1120 Class 1 Dan Barbasch

Inverse functions

The derivative of y = f −1(x) is y′ = 1 f ′(f −1(x)) also written as dy dx = 1

dx dy

This follows from the chain rule: x = f (f −1(x)) ⇔ 1 = f ′(f −1(x)) ·

• f −1(x)

′ ⇔

• f −1′ (x) =

1 f ′ (f −1(x)). Exercise: d dx (Arccsc x) =?

Math 1120 Class 1 Dan Barbasch

Solution to Exercise

The function y = Arccsc x is defined as: Given x, Arccsc x is the unique value −π/2 ≤ y ≤ π/2 which satisfies csc y = x. Rewrite the equation as 1/ sin y = x, or sin y = 1/x, and differentiate both sides (using the chain rule) cos y · y′ = −1/x2 ⇔ y′ = − 1 x2 cos y We need to express y′ in terms of x alone. For this note that cos2 y + sin2 y = 1, so solving for cos y we get cos y = ±

• 1 − sin2 y = ±
• 1 − 1/x2.

However −π/2 ≤ y ≤ π/2 so cos y ≥ 0. Since square roots are always positive quantities, the final answer is y′ = − 1 x2 1 − 1/x2

Math 1120 Class 1 Dan Barbasch

Solution to Exercise continued

You can use a right triangle with one angle equal to y, the hypotenuse of size 1, and the opposite side equal to 1/x to compute cos y. You still have the problem of deciding between ±.

Math 1120 Class 1 Dan Barbasch

Montonicity 1st Derivative Test

Suppose f (x) is continuous on [a, b] and f ′(x) > 0 for x ∈ (a, b). Then f (x) is (strictly) increasing on [a, b]. This means x < y ⇒ f (x) < f (y). Suppose f (x) is continuous on [a, b] and f ′(x)< 0 for x ∈ (a, b). Then f (x) is (strictly) decreasing on [a, b]. This means x < y ⇒ f (x) > f (y). f ′(x) ≤ 0 would give f nonincreasing: x < y ⇒ f (x) ≥ f (y).

Math 1120 Class 1 Dan Barbasch

Montonicity

Exercise: Is 1 1 + x always bigger than 1 − x for x > 0? e.g. 1 1 + .25 = .8 > 1 − .25 = .75.

Math 1120 Class 1 Dan Barbasch

Solution to Exercise

The question is the same as same as Is f (x) :=

1 (1+x) − (1 − x) ≥ 0 for x ≥ 0?

We answer it by sketching the graph of f using the 1st derivative test: f ′(x) = − 1 (1 + x)2 − (−1) = 1 − 1 (1 + x)2 . So we need to decide when this is ≥ 0, and when it is ≤ 0. This is why you need to be able to solve inequalities. f ′(x) ≥ 0 for x ≥ 0. Indeed, x ≥ 0 ⇒ 1 + x ≥ 1 ⇒ (1 + x)2 ≥ 12 ≥ 1 ⇔ 1 ≥ 1 (1 + x)2 ⇔ 1 − 1 (1 + x)2 ≥ 0. Since f ′(x) ≥ 0, f (x) is increasing (nondecreasing) for x ≥ 0. But f (0) = 0, so f (x) ≥ f (0) = 0 for x ≥ 0.

Math 1120 Class 1 Dan Barbasch

MaxMin and the 2nd Derivative Test

Exercise: Find and classify all local and global maxima and minima of f (x) = x3 + ax for any value of a > 0? Exercise: Find and classify all local and global maxima and minima of f (x) = x3 + ax for a any value of a < 0? Exercise: Find and classify all local and global maxima and minima of f (x) = x3. REFERENCE: Sections 4.1 - 4.4 cover this material.

Math 1120 Class 1 Dan Barbasch

Riemann Sums and Integrals

The definition of the symbol b

a f (x) dx called the definite

integral of y = f (x) over the interval [a, b] is as follows.

• 1. we start with a partition {x0, x1, . . . , xn−1, xn} of the interval

[a, b] a = x0 < x1 < . . . < xn−1 < xn = b with, for simplicity xi+1 − xi = ∆x = b−a

n .

• 2. We then choose a point x∗

i within each subinterval [xi−1, xi].

The Riemann sum associated to this partition is

n

• i=1

f (x∗

i ) ∆x.

If the Riemann sums settle down to a limit as → ∞, that limit is called b

a f (x) dx.

Math 1120 Class 1 Dan Barbasch

Riemann Sums and Integrals

Riemann sums are important because they represent physical quantities; areas and volumes are the most intuitive ones. REFERENCE: You can find a discussion of Riemann sums and applications in sections 5.1 to 5.3. Exercise: Compute the area under the curve y = 1 − x2 above the x−axis.

Math 1120 Class 1 Dan Barbasch

Riemann Sums and Integrals

The definition of the area of a rectangle, triangle and more general polygonal objects is defined in elementary geometry starting with the basic definition that the area of a rectangle is the product of its base with its height, and deriving the rest from the additivity property of area. The region in the exercise IS NOT of this type. Such an area is DEFINED as a limit of polygonal regions; the definite integral of a nonnegative function, A := b

a f (x) dx.

Math 1120 Class 1 Dan Barbasch

Riemann Sums and Integrals

Exercise: Graph the region and compute the area under the curve y = 1 − x2 above the x−axis. Dividing the interval [−1, 1] into n equal parts gives x0 = −1 <x1 = −1 + 2 n = −n + 2 n < · · · < < · · · <xi = −1 + i 2 n = −n + 2i n < · · · < xn = 1. Choosing the left endpoint x∗

i = −n+2i n

gives the Riemann sum Rn =

n−1

• i=0

2 n

• 1 −

−n + 2i n 2 . This is not easy to evaluate. In addition we have to take a limit n → ∞. You can see how to carry out this process in sections 5.1-5.3.

Math 1120 Class 1 Dan Barbasch

Fundamental Theorem of Calculus

The FTC (Fundamental Theorem of Calculus) provides a powerful method to compute such limits. Theorem: Let y = f (x) be continuous on the interval [a, b]. Then f (x) has an antiderivative F(x), and b

a

f (x) dx = F(b) − F(a).

Math 1120 Class 1 Dan Barbasch

Fundamental Theorem of Calculus

The graph of y = 1 − x2 in the region −1 ≤ x ≤ 1 is FTC reduces the problem of computing the area to finding the antiderivative of f (x) = 1 − x2. We know that F(x) = x − x3/3 satisfies F ′(x) = 1 − x2. So by FTC, A = 1

−1

• 1 − x2

dx =

• x − x3/3
• |1

−1=

= (1 − 1/3) − (−1 + 1/3) = 4/3. Question: How many antiderivatives can a function y = f (x) have?

Math 1120 Class 1 Dan Barbasch

Fundamental Theorem of Calculus

The mean value theorem implies that if F ′(x) = G ′(x) over some interval, then there is a constant C such that F(x) = G(x) + C. Question: How does this fit with FTC?

Math 1120 Class 1 Dan Barbasch

Fundamental Theorem of Calculus

The theorem comes from the following picture:

Math 1120 Class 1 Dan Barbasch

Fundamental Theorem of Calculus

F(x):= x

a f (t) dt is a function of x.

We use x in the limit of the integral and t for the variable, because one designates the limit of the integral, the other the variable for the function!

Math 1120 Class 1 Dan Barbasch

Fundamental Theorem of Calculus

To compute its derivative we take F(x + ∆x) = x+∆x

a

f (t) dt

Math 1120 Class 1 Dan Barbasch

Fundamental Theorem of Calculus

and subtracting F(x) = x

a f (t) dt we get

F(x + ∆x) − F(x) = x+∆x

x

f (t) dt ∼ f (x)∆x. This “proves” FTC.

Math 1120 Class 1 Dan Barbasch

Fundamental Theorem of Calculus

More rigorously, m∆x ≤ x+∆x

x

f (t) dt ≤ M∆x where m = minx≤t≤x+∆x f (t), and M = maxx≤t≤x+∆x f (t). so m ≤ F(x + ∆x) − F(x) ∆x ≤ M, and because f (t) is continuous, lim∆x→0 m = lim∆x→0 M = f (x), the squeeze theorem implies F ′(x) = lim

∆x→0

F(x + ∆x) − F(x) ∆x = f (x).

Math 1120 Class 1 Dan Barbasch

Fundamental Theorem of Calculus

x+∆x

a

f (x) dx − x

a f (x) dx =

x+∆x

x

f (x) dx