Unit #5 - Implicit Differentiation, Related Rates Section 4.6

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1 Unit #5 - Implicit Differentiation, Related Rates Section 4.6 Some material from Calculus, Single and MultiVariable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 2005 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc. TEST PREPARATION PROBLEMS 3. (a) s(w = 60,h = 150) = 0.01(60) 0.25 (150) 0.75 = 1.19m 2 (b) We are asked for ds dw, given that h is constant with time, and = 0.5 kg/year. Taking d of both sides of the s formula, ds d5 ( ds = w 0.75dw ) (h 0.75 ) w=62,h=150 = 0.01 (0.25 (62) 0.75 (0.5) meters 2 /year 6. (a) The rate of change of the period is give by dt dl = 2π 9.8 ( 1 2 l 1/2 ) = π l ) (150) (a) (b) As l increases, the denominator gets larger, so the rate decreases. C = a q 1 +b so dc dq = aq 2 = a q 2 SinceC istheaveragecostpercellphone,itisindollars; q is#units. Theunitsof dc dq are therefore (dollars/unit). (An argument could be made that C is in dollars/unit too, so the derivative would be (dollars/unit) per unit: I would accept that as well based on the problem wording.) (b) Careful here: C is the average cost alrea, the rate of change of the average cost is dc dq. We are given = 100. Differentiating the original equation with respect to t 1

2 now, dc = a dq q 2 = a q The rate is negative (we were told a was a positive constant), so the average cost per unit is decreasing with time. 11. (a) F = Ar 2 +Br 3 so df = 2Ar 3 3Br 4 The units are N/m (assuming the force F is in Newtons, and r is in meters). (b) We are told that = k. Differentiating the original equation with respect to time, so df F = Ar 2 +Br 3 = 2Ar 3 3Br 4 = 2Ar 3 k 3Br 4 k The units would be in Newtons per unit time ( seconds usually) 21. The general relationship between depth and volume in a cylinical tank (with r = 3 is To find, we take d If dd = 0.2 ft/s, then of both sidesḋv V = π3 2 d = 9πd = 9π dd = 9π(0.2) 5.65 feet 3 /sec The fact that d = 4 at the time in question turns out to be irrelevant. 2

3 25. This is mostly a chain rule question, with some estimation of derivatives: dh = dh H y We are given that = 3000 ft/min (upwards, so positive rate). However, in the units of y (thousands of feet), we should write this as = 3 thousand feet/min. At 4,000 ft, dh H y = = 7 degrees/thousand feet. Therefore, 6 4 dh H y Note that you could also estimate dh = ( 7)(3) = 21 degrees/min using y = 2 and y = 4: you would get -4 instead of -7 degrees/thousand feet, and a corresponding decrease in the estimated rate of dh. 26. T = k 4 M = km 1/4. We are told k = 17.4 if units are seconds and kilograms for time and mass respectively. dt = M 3/4 dm Substituting M = 45 and dm =.1 we have ( ) dt 1 = (45) 3/4 (0.1) = The circulation time is increasing at a rate of seconds per month. 31. Let V be the snowball volume, and r be its radius. Taking d of both sides, For a sphere, V = 4 3 πr3 = 4 3 π(3r2 ) = 4πr2 When r = 15 cm and = 0.2 cm/hr, = 180π r=15, = cm3 /hr 32. (a) We only need to work with the area at first, given the change in radius. A = πr 2 da = π2r When r = 150 and = 0.1, da = π2(150)(0.1) = 30π 94m 2 /min 3

4 (b) V = πr 2 h The oil volume is constant, but r and h are changing with time, so If r = 150, h = 0.02, and = 0.1, d (V) = d ( πr 2 h ) [ 0 = π 2r ] h+r2 r 2 = 2rh = 2h r = 2(0.02) 150 (0.1) The thickness of the oil slick is decreasing by meters/minute. 33. The volume of the piece is constant, but as a cylinder the volume is V = πr 2 L, and both r and L are changing with time. 34. (a) When dl d V = d ( πr 2 L ) [ 0 = π 2rL = r2 2rL = 0.1 cm/s, r = 1 and L = 5 cm, = (1) 2(5) The radius is decreasing at 0.01 cm/s. dl = r 2L +r2 dl dl ] (0.1) = 0.01 cm/s 4

5 At all points, x 2 +y 2 = h 2, so 2x dx Solving for, +2y = 2h = 1 h [ x dx ] +y At the point of interest, dx = 100 miles/hr (negative, moving towards the gas station) = 80 miles/hr, (positive, away from station) x = 3, y = 4, and so h = = 5. = 1 [(3) ( 100)+(4)(80)] = 4 miles/hr 5 The distance between the truck and police car is growing at 4 miles/hr. 1. If the truck is only going 70 miles/hr, then 35. (a) = 1 [(3) ( 100)+(4)(70)] = 4 miles/hr 5 In this case, the distance between the truck and police car is shrinking at 4 miles/hr. x = z 2 and z is positive, so z = x (b) z = distance from camera to the train. It s actually easier to compute dz from the original Pythagorean equality than it is from part a, though: x = z 2 2x dx dz = 2z dz = x dx z Since dx = 0.8, and z = 1 gives x = 1.25, dz 0.75 = (0.8) 0.69 km/min 1 5

6 (c) The rotation rate will be in radians/min, or dθ. tan(θ) = x 0.5 = 2x d (tan(θ)) = d (2x) 1 dθ cos 2 (θ) = 2 dx dθ = 2 dx cos 2 (θ) At the moment we are interested in, z = 1 so θ arccos km/min. dθ = 2(0.8)cos2( π ) = 2(0.8)(0.5) 2 = 0.4 rad/min 3 ( ) 0.5 = π dx. and 1 3 = The distance between the spot of light to O is x. We are asked for how this changes with θ, or dx dθ. tan(θ) = x 2 ( x 2) so d dθ (tan(θ)) = d dθ 1 cos 2 (θ) = 1 dx 2 dθ dx dθ = 2 cos 2 (θ) 38. Both V and r of the balloon are changing with time, so V = 4 3 πr3 = 4π 3 ( 3r 2 ) = 4πr 2 When = 2 cm/s, and r = 10, = 4π(10) 2 (2) = 800π cm 3 /s 6

7 42. The pile of sand will always have a cross-section of a triangle, similar to the 1/2/ 3 triangle shown. V = 1 3 πr2 h but r/h = 1/ 3 so r = h 3 V = π ( ) h 2 h 3 3 Taking time derivatives, V = π 9 h3 = π ( 3h 2 ) 9 = π h2 3 Since we are adding sand at 0.1 cubic meters per hour, 0.1 = π h2 3 = 0.3 π h 2 meters/hr We are also asked for the rate at which the radius increases. Since r = h/ 3, = 1 3 = 0.3 3π h 2 = 0.1, so 44. 7

8 From this sketch, r/h = 8/10 = 4/5, so r = 4h/5. V = π 3 r2 h = π ( ) Taking d, = 16π ( 3h 2 ) 75 h 3 = 16π 75 h3 = 16π h2 25 The information in this problem is a little different from others, but we can infer that = rate in - rate out = h 2. Thus h 2 = 16π h2 25 Under what conditions will the tank overflow? Well, for that to happen, there must be water flowing into the tank when h = 10 (depth of the tank). That corresponds either to (or ) being positive when h = 10. When h = 10, = (10) 2 = 0.3 m 3 /min, so the tank is emptying. h=10 Also = ( (10) 2 25 ) < 0 indicating that if the tank were ever full, h=10 16π(10) 2 the water level would op, and not increase past full. 8