MATH 12002 - CALCULUS I 2.3 (Part 2): Rates of Change, Revisited - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §2.3 (Part 2): Rates of Change, Revisited

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Rates of Change

Recall that if a quantity y is a function of another quantity x, say y = f (x), then for a value a of x, the derivative f ′(a) is the instantaneous rate of change of y with respect to x at x = a. For example: If x is time and y is the position of an object along a straight line, then f ′(a) is the rate of change of position with respect to time, i.e., velocity of the object, at time x = a. If x is time and y is the velocity of an object, then f ′(a) is the rate of change of velocity with respect to time, i.e., acceleration of the

• bject, at time x = a.

If x is time and y is the number of deer in a park, then f ′(a) is the rate of change of population with respect to time at time x = a. This measures how fast the population is increasing or decreasing at a given time. The main idea is simply that any rate of change is just a DERIVATIVE.

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Example

Example

Suppose the radius of a circle starts at 0 centimeters and increases at a constant rate of 10 centimeters per second. How fast is the area of the circle increasing after 5 seconds? First, “10 centimeters per second” is a rate of change. It is the rate of change of the radius with respect to time. It tells us that the radius increases by 10 centimeters each second. Since the rate of change is constant, between any given time and one second later, the radius will increase by 10 centimeters. Moreover, during any fraction of a second, the radius will change by that fraction of 10 centimeters.

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Example

Since the radius starts at 0 centimeters, we have the following: Time Radius 0 sec 0 cm 1 sec 10 cm 2 sec 20 cm 2.5 sec 25 cm 3 sec 30 cm 3.2 sec 32 cm 5 sec 50 cm Therefore, t seconds after the start, the radius is r(t) = 10t centimeters. Note that the derivative of r(t) is dr

dt = 10,

which is indeed the rate of change of r with respect to t at any time t. The area of the circle is given by A = πr2, or as a function of time, A(t) = πr(t)2 = π(10t)2 = π100t2 = 100πt2.

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Example

We have that the area of the circle at time t seconds is A(t) = 100πt2. We want to know how fast the area is increasing after 5 seconds. This is the instantaneous rate of change of area, A(t), at time t = 5, which is the value of the derivative, A′(t) = dA

dt = 200πt, when t = 5.

Since this is the rate of change of area (in square centimeters, cm2) with respect to time (in seconds, s), the units are square centimeters per second, written cm2/s. Therefore, after 5 seconds, the area of the circle is increasing at a rate of A′(5) = 200π(5) = 1000π cm2/s ≈ 3141.59 cm2/s. This answers our original question, but what does it actually mean?

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Linear Approximation

Recall that at time t = 5 seconds, the radius of the circle is r = 50 cm, and so the area is A = π(50)2 = 2500π cm2, and the area is increasing at a rate of 1000π cm2/s. This means that when t = 5, the area increases by 1000π cm2 each second. Therefore, if the rate of change were constant, then at t = 6 seconds, the area would be 2500π + 1000π = 3500π cm2. However, the rate of change at time t is A′(t) = 200πt, so is not constant. Based on this information (and not the formula we derived for A(t)), would you expect the actual area at time t = 6 to be greater or less than 3500π cm2? The rate of change of the area is A′(t) = 200πt, so the greater t is, the greater the rate of change, and so the faster the area is increasing. Hence the actual area at time t = 6 is greater than 3500π cm2.

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Linear Approximation

One final question: What is the meaning of the 3500π cm2 area? Recall that A(t) = 100πt2, and so the actual area when t = 6 seconds is A(6) = 100π(62) = 3600π cm2, as indicated on the graph of A(t) below. Recall also that A′(5) = 1000π cm2/s. This is the slope of the tangent line at the point (5, 2500π). The tangent line shows what the graph of A(t) would look like for t 5, if the rate of change remained constant at 1000π cm2/s. Therefore, 3500π cm2 is the A coordinate on the tangent line at t = 6. We arrive at this point by going to the right 1 sec and up 1000π cm2.

A(t)

✲ ✻

1 2 3 4 5 6 7 8 t A

q

2500π

q

3600π

• 1 sec

1000π cm2

q

3500π D.L. White (Kent State University) 7 / 7