3-4 VELOCITY & OTHER RATES OF CHANGE. General Rate of Change The (instantaneous) rate of change of f with respect to x at a is the derivative! Ex 1a)

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3.4 Velocity and Other Rates of Change
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3-4 VELOCITY & OTHER RATES OF CHANGE

General Rate of Change The (instantaneous) rate of change of f with respect to x at a is the derivative! Ex 1a) Find the rate of change of the area A of a circle with respect to its radius r. b) Evaluate the rate of change of A at r = 5 and at r = 10. c)If r is measured in inches and A is measured in square inches, what units would be appropriate for

Velocity The (instantaneous) velocity is the derivative of the position function s = f (t) Velocity shows direction (+)  forward (–)  backward Speed Acceleration Acceleration is the derivative of velocity (aka the 2 nd derivative of the position) Note: The free fall constant for Earth is

Ex 4) A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec (about 109 mph). It reaches a height of s = 160t – 16t 2 after t seconds. a) How high does the rock go? v = s ' = 160 – 32t = = 32t t = 5 sec s(5) = 160(5) – 16(25) = 800 – 400 = 400 ft b) What is the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down? s = 256 = 160t – 16t 2 16t 2 – 160t = 0 t 2 – 10t + 16 = 0 (t – 2)(t – 8) = 0 t = 2, 8 sec v(2) = 160 – 32(2) = 96 ft/sec v(8) = 160 – 32(8) = –96 ft/sec *Speed is 96 ft/sec for both (when deriv = 0)

Ex 4c) What is the acceleration of the rock at any time t during its flight (after the blast)? Ex 4d)When does the rock hit the ground? At 10 sec a(t) = v ' (t) = – 32 ft/sec 2 160t – 16t 2 = 0 16t(10 – t) = 0 t = 0, 10 sec

Ex 5) A particle moves along a line so that its position at any time t  0 is given by the function s(t) = t 2 – 4t + 3, where s is measured in meters and t is measured in seconds. a) Find the displacement of the particle during the first 2 sec s(2) – s(0) = (4 – 8 + 3) – (0 – 0 + 3) = – 44 meters left b) Find the average velocity of the particle during the first 4 s. c) Find the instantaneous velocity of the particle when t = 4. v = s' = 2t – 4 v(4) = 8 – 4 = 4 m/sec

d) Find the acceleration of the particle when t = 4. e) Describe the motion of the particle. At what values of t does the particle change directions? a = v ' = 2 m/sec 2 Look at the graph of s & v left 0  t < t = 2 changes direction right t > 2

HOMEWORK Pg. 124 #23, 24, 26, 28, 29 Pg. 135 #2, 3, 5, 8, 9, 14, 16, 23, 32