#### Transcript 3.4 Rates of change

3.4 RATES OF CHANGE RATES OF CHANGE Example 1: Find rate ofinchange the Area of a circle with If r is the measured inches of and A is measured in square inches, would be appropriate for dA/dr? respectwhat to itsunits radius. A r 2 dA 2r dr Evaluate in2/in These are units, so we do not cancel them!!! the rate of change of A at r = 5 and r = 10. dA 2 (5) 10 dr dA 2 (10 ) 20 dr MOTION Displacement of an object is how far an object has moved over time. s f (t t ) f (t ) Average velocity is the slope of a displacement vs. time graph. vavg d s1 s2 t t1 t2 MOTION Instantaneous Velocity is the velocity at a certain point. Instantaneous Velocity is the first derivative of the position function. ds v dt Speed is the absolute value of velocity. MOTION Acceleration is the rate at which a particle’s velocity changes. Measures how quickly the body picks up or loses speed. 2nd derivative of the position function!!! 2 d s a 2 dt EXAMPLE A particle moves along a line so that its position at any time t ≥ 0 is given by the function s(t) = 2t2 – 5t + 3 where s is measured in meters and t is measured in seconds. a.) Find the displacement of the particle during the first 2 seconds. s s(2) s(0) 1 3 2 meters b.) Find the average velocity of the particle during the first 6 seconds. vavg s (6) s (0) 45 3 42 7 meters second 60 60 6 EXAMPLE A particle moves along a line so that its position at any time t ≥ 0 is given by the function s(t) = 2t2 – 5t + 3 where s is measured in meters and t is measured in seconds. c.) Find the instantaneous velocity of the particle at 6 seconds. velocity s' (t ) 4t 5 t 6 m 19 s d.) Find the acceleration of the particle when t = 6. acceleration s' ' (t ) v' (t ) 4 t 6 m 4 2 s EXAMPLE A particle moves along a line so that its position at any time t ≥ 0 is given by the function s(t) = 2t2 – 5t + 3 where s is measured in meters and t is measured in seconds. e.) When does the particle change directions? v(t ) 0 4t 5 0 5 t seconds 4 It is important to understand the relationship between a position graph, velocity and acceleration: acc neg vel pos & decreasing acc neg vel neg & decreasing acc zero vel pos & constant distance velocity zero acc pos vel pos & increasing acc zero vel neg & constant acc pos vel neg & increasing acc zero, velocity zero time FREE-FALL Gravitational Constants: Free-fall equation: ft g 32 sec 2 1 2 s gt 2 m g 9.8 sec 2 cm g 980 sec 2 s is the position at any time t during the fall g is the acceleration due to Earth’s gravity (gravitational constant) VERTICAL MOTION Example: A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft after t seconds. a.) How high does the rock go? Find when position = 0 and divide by 2 (symmetric path) s(t ) 160t 16t 0 16t (10 t ) 0 16t 0 t 0 (when blast occurs) 2 10 t 0 t 10 (when rock hitsground) Since it takes 10 seconds for the rock to hit the ground, it takes it 5 seconds to reach it max height. s(5) 160(t ) 16(5)2 400feet VERTICAL MOTION Example: A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft after t seconds. a.) How high does the rock go? Find when velocity = 0 (this is when the rock changes direction) v(t ) 160 32t 0 t 5 s(5) 160(5) 16(5)2 400feet VERTICAL MOTION Example: A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft after t seconds. b.) What is the velocity and speed of the rock when it is 256 ft above the ground on the way up? At what time is the rock 256 ft above the ground on the way up? t 20 s(t ) 256 160t 16t 2 16t 2 160t 256 0 t2 16(t 2 10t 16) 0 v(t ) 160 32t 16(t 2)(t 8) 0 m v ( 2) 96 s t 8 0 t 8 m speed velo 96 s VERTICAL MOTION Example: A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft after t seconds. b.) What is the velocity and speed of the rock when it is 256 ft above the ground on the way down? At what time is the rock 256 ft above the ground on the way down? t 20 s(t ) 256 160t 16t 2 t2 16t 2 160t 256 0 16(t 2 10t 16) 0 16(t 2)(t 8) 0 t 8 0 t 8 v(t ) 160 32t m v(8) 96 s m speed velo 96 s VERTICAL MOTION Example: A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s = 160t – 16t2 ft after t seconds. c.) What is the acceleration of the rock at any time t at any time t during its flight? feet a (t ) 32 sec 2 from Economics: Marginal cost is the first derivative of the cost function, and represents an approximation of the cost of producing one more unit. Marginal revenue is the first derivative of the revenue function, and represents an approximation of the revenue of selling one more unit.