MATH 12002 - CALCULUS I § 2.3 (Part 2): Rates of Change, Revisited Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 7
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 object, 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. D.L. White (Kent State University) 2 / 7
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. D.L. White (Kent State University) 3 / 7
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 ) = 10 t 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 = π r 2 , or as a function of time , A ( t ) = π r ( t ) 2 = π (10 t ) 2 = π 100 t 2 = 100 π t 2 . D.L. White (Kent State University) 4 / 7
Example We have that the area of the circle at time t seconds is A ( t ) = 100 π t 2 . 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 = 200 π t , when t = 5. Since this is the rate of change of area (in square centimeters, cm 2 ) with respect to time (in seconds, s ), the units are square centimeters per second, written cm 2 / s . Therefore, after 5 seconds, the area of the circle is increasing at a rate of A ′ (5) = 200 π (5) = 1000 π cm 2 / s ≈ 3141 . 59 cm 2 / s . This answers our original question, but what does it actually mean ? D.L. White (Kent State University) 5 / 7
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 = 2500 π cm 2 , and the area is increasing at a rate of 1000 π cm 2 / s . This means that when t = 5, the area increases by 1000 π cm 2 each second. Therefore, if the rate of change were constant , then at t = 6 seconds, the area would be 2500 π + 1000 π = 3500 π cm 2 . However, the rate of change at time t is A ′ ( t ) = 200 π t , 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 3500 π cm 2 ? The rate of change of the area is A ′ ( t ) = 200 π t , 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 3500 π cm 2 . D.L. White (Kent State University) 6 / 7
Linear Approximation One final question: What is the meaning of the 3500 π cm 2 area? Recall that A ( t ) = 100 π t 2 , and so the actual area when t = 6 seconds is A (6) = 100 π (6 2 ) = 3600 π cm 2 , as indicated on the graph of A ( t ) below. Recall also that A ′ (5) = 1000 π cm 2 / s . This is the slope of the tangent line at the point (5 , 2500 π ). The tangent line shows what the graph of A ( t ) would look like for t � 5, if the rate of change remained constant at 1000 π cm 2 / s . Therefore, 3500 π cm 2 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 1000 π cm 2 . A ✻ � A ( t ) � � 3600 π q 3500 π q � 1000 π cm 2 � 2500 π q 1 sec � � � ✲ t 1 2 3 4 5 6 7 8 D.L. White (Kent State University) 7 / 7
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