Calculus

1

APPLICATIONS OF DERIVATIVES (PART I) I. The Derivative as a Rate of Change A. Average and Instantaneous Rates of Change Let x and y be two quantities related by y = f (x) . The average rate of change of y with respect to a change in x ( or the average rate of change of y as the value of x changes from x1 to x 2 ) is given by

∆y f ( x2 ) − f ( x1 ) f ( x1 + ∆ x) − f ( x1 ) = = . ∆x x2 − x1 ∆x f ( x1 + ∆ x) − f ( x1 ) ∆y The instantaneous rate of change of y at x = x1 is given by lim . = lim ∆ x→ 0 ∆ x ∆ x→ 0 ∆x Remark: The derivative f ' ( x1 ) is the instantaneous rate of change.

B. Exercises 1. Suppose the object moves according to the position function f (t ) = t 2 + t − 2 along a straight path, where f (t ) is the directed distance (in feet) of the object from a reference point at time t (in seconds). a) Find the average velocity and the average accelaration of the object during the time interval [2, 5]. How about when t changes from 6 to 10 seconds? b) What is the (instantaneous) velocity of the object when t = 2? When t = 5? How about its (instantaneous) accelaration at each of these t values? 2. The width of a rectangle is 3 inches less than its length, and this 3-inch difference in dimensions is maintained as the rectangle increases in size. a) Compute the average rate of change of the area of the rectangle as the length changes from 3 to 3.2 inches. b) Compute for the instantaneous rate of change of the area when the length is 3 inches.

3. The number of gallons Q of water in the tank t minutes after the tank has started to drain is given by Q(t ) = 200(30 − t ) 2

a) How fast is the water running out at the end of 10 minutes (or on the 10th minute)? b) What is the average rate at which the water flows out during the first 10 minutes?

4. A manufacturer of an industrial liquid determines that the cost of manufacturing x gallons of liquids is C ( x) = x 2 − 2 x + 5 . By increasing production from 5 to 10 gallons, find the average rate of increase...