Math 113/114 Lecture 22

Math 113/114 Lecture 22 Xi Chen 1 1 University of Alberta October 31, 2014

Outline 1 2

(Application of Implicit Differentiation) Given a word problem about related rates, we need to do: interpret the problem correctly; establish the variables; find the relation of the variables (F(x, y) = 0); determine which derivative (dx/ or dy/) is given and which derivative we are after; apply implicit differentiation to solve the problem.

(Application of Implicit Differentiation) Given a word problem about related rates, we need to do: interpret the problem correctly; establish the variables; find the relation of the variables (F(x, y) = 0); determine which derivative (dx/ or dy/) is given and which derivative we are after; apply implicit differentiation to solve the problem.

Examples of Air is being pumped into the balloon at a rate of a cm 3 /s. How fast is the radius of the balloon increasing where the radius of the balloon is b cm? Solution. Let t be the time variable in seconds, r be the radius in cm and V be the volume in cm 3. Then V = 4πr 3 /3. Differentiating both sides with respect to t, we obtain So dv = d dr = 1 dv 4πr 2 ( ) 4π 3 r 3 dr = 4πr 2 dr. = r=b a 4πb 2 cm/s.

Examples of Air is being pumped into the balloon at a rate of a cm 3 /s. How fast is the radius of the balloon increasing where the radius of the balloon is b cm? Solution. Let t be the time variable in seconds, r be the radius in cm and V be the volume in cm 3. Then V = 4πr 3 /3. Differentiating both sides with respect to t, we obtain So dv = d dr = 1 dv 4πr 2 ( ) 4π 3 r 3 dr = 4πr 2 dr. = r=b a 4πb 2 cm/s.

Examples of Air is being pumped into the balloon at a rate of a cm 3 /s. How fast is the surface area of the balloon increasing where the radius of the balloon is b cm? Solution. Let t be the time variable in seconds, r be the radius in cm, V be the volume in cm 3 and S be the surface area in cm 2. Then V = 4πr 3 /3 and S = 4πr 2. Differentiating both identities with respect to t, we obtain So ds dv = 8πr dv 4πr 2 = 4πr 2 dr = 2 r dv and ds ds = 8πr dr. = 2a r=b b cm2 /s.

Examples of Air is being pumped into the balloon at a rate of a cm 3 /s. How fast is the surface area of the balloon increasing where the radius of the balloon is b cm? Solution. Let t be the time variable in seconds, r be the radius in cm, V be the volume in cm 3 and S be the surface area in cm 2. Then V = 4πr 3 /3 and S = 4πr 2. Differentiating both identities with respect to t, we obtain So ds dv = 8πr dv 4πr 2 = 4πr 2 dr = 2 r dv and ds ds = 8πr dr. = 2a r=b b cm2 /s.

Examples of At noon, ship A is 150km west of ship B. Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h. How fast is the distance between the two ships changing at 4pm? Solution. Suppose that ship A is at the origin at noon, the positions of A and B are (a, 0) and (150, b) and let y be the distance between A and B. Then y 2 = (150 a) 2 + b 2. Differentiating with respect to the time variable t, we obtain 2y dy = 2(a 150) da +2b db dy = a 150 da y + b db y. When t = 4, a = 4 35 = 140, b = 4 25 = 100 and y = (150 140) 2 + (100) 2 = 10 101. So 101 km/h. dy = 215

Examples of At noon, ship A is 150km west of ship B. Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h. How fast is the distance between the two ships changing at 4pm? Solution. Suppose that ship A is at the origin at noon, the positions of A and B are (a, 0) and (150, b) and let y be the distance between A and B. Then y 2 = (150 a) 2 + b 2. Differentiating with respect to the time variable t, we obtain 2y dy = 2(a 150) da +2b db dy = a 150 da y + b db y. When t = 4, a = 4 35 = 140, b = 4 25 = 100 and y = (150 140) 2 + (100) 2 = 10 101. So 101 km/h. dy = 215

Examples of A spotlight on the ground shines on a wall 12m away. If a man 2m tall walks from the spotlight toward the building at a speed of 1.6m/s. How fast is the length of his shadow on the building changing when he is 4m from the building? Solution. Let x be the distance between the man and the spotlight and y be the length of his shadow. Then x 12 = 2 y xy = 24. Differentiating with respect to t, we obtain ( ) ( ) d dx dy (xy) = 0 y + x = 0 dy = y x When x = 12 4 = 8, y = 24/8 = 3 and hence dy = 3 (1.6) = 0.6 m/s. x=8 8 ( ) dx.

Examples of A spotlight on the ground shines on a wall 12m away. If a man 2m tall walks from the spotlight toward the building at a speed of 1.6m/s. How fast is the length of his shadow on the building changing when he is 4m from the building? Solution. Let x be the distance between the man and the spotlight and y be the length of his shadow. Then x 12 = 2 y xy = 24. Differentiating with respect to t, we obtain ( ) ( ) d dx dy (xy) = 0 y + x = 0 dy = y x When x = 12 4 = 8, y = 24/8 = 3 and hence dy = 3 (1.6) = 0.6 m/s. x=8 8 ( ) dx.

Examples of A lighthouse is located on a small island 4km away from the nearest point P on a straight shoreline and its light makes 4 revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1km from P? Solution. Let O be the lighthouse, Q be the intersection of the beam of light with the shoreline, y be the distance PQ and x be the angle POQ. Then y = 4 tan x. Differentiating it with respect to t, we obtain dy = 4 sec 2 x dx When y = 1, tan x = 1/4 and sec 2 x = 1 + tan 2 x = 17/16. So ( ) dy 17 = 4 (8π) = 34π km/min. y=1 16

Examples of A lighthouse is located on a small island 4km away from the nearest point P on a straight shoreline and its light makes 4 revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1km from P? Solution. Let O be the lighthouse, Q be the intersection of the beam of light with the shoreline, y be the distance PQ and x be the angle POQ. Then y = 4 tan x. Differentiating it with respect to t, we obtain dy = 4 sec 2 x dx When y = 1, tan x = 1/4 and sec 2 x = 1 + tan 2 x = 17/16. So ( ) dy 17 = 4 (8π) = 34π km/min. y=1 16

Given f (x) which is differentiable at a, the tangent line of y = f (x) at (a, f (a)) is y f (a) = f (a)(x a) y = f (a) + f (a)(x a) We call f (a) + f (a)(x a) the linear (1st order) approximation of f (x) at a: f (x) f (a) + f (a)(x a) for x close to a. The linear approximation of sin x at x = 0 is sin x sin 0 + (sin x) x=0 (x 0) = 0 + 1(x 0) = x. The linear approximation of ln x at x = 1 is ln(x) ln 1 + (ln x) x=1 (x 1) = 0 + 1(x 1) = x 1.

Given f (x) which is differentiable at a, the tangent line of y = f (x) at (a, f (a)) is y f (a) = f (a)(x a) y = f (a) + f (a)(x a) We call f (a) + f (a)(x a) the linear (1st order) approximation of f (x) at a: f (x) f (a) + f (a)(x a) for x close to a. The linear approximation of sin x at x = 0 is sin x sin 0 + (sin x) x=0 (x 0) = 0 + 1(x 0) = x. The linear approximation of ln x at x = 1 is ln(x) ln 1 + (ln x) x=1 (x 1) = 0 + 1(x 1) = x 1.

Given f (x) which is differentiable at a, the tangent line of y = f (x) at (a, f (a)) is y f (a) = f (a)(x a) y = f (a) + f (a)(x a) We call f (a) + f (a)(x a) the linear (1st order) approximation of f (x) at a: f (x) f (a) + f (a)(x a) for x close to a. The linear approximation of sin x at x = 0 is sin x sin 0 + (sin x) x=0 (x 0) = 0 + 1(x 0) = x. The linear approximation of ln x at x = 1 is ln(x) ln 1 + (ln x) x=1 (x 1) = 0 + 1(x 1) = x 1.

Given f (x) which is differentiable at a, the tangent line of y = f (x) at (a, f (a)) is y f (a) = f (a)(x a) y = f (a) + f (a)(x a) We call f (a) + f (a)(x a) the linear (1st order) approximation of f (x) at a: f (x) f (a) + f (a)(x a) for x close to a. The linear approximation of sin x at x = 0 is sin x sin 0 + (sin x) x=0 (x 0) = 0 + 1(x 0) = x. The linear approximation of ln x at x = 1 is ln(x) ln 1 + (ln x) x=1 (x 1) = 0 + 1(x 1) = x 1.