1. speed = speed = 7 9 ft/sec. 3. speed = 1 ft/sec. 4. speed = 8 9 ft/sec. 5. speed = 11

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1 Version PREVIEW HW 07 hoffman 57225) This print-out should have 0 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. CalCi02s points A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at a rate of 6 ft/sec, at what speed is the area of the ripple increasing when its radius is feet?. speed = 5 sq. ft/sec 2. speed = π sq. ft/sec. speed = 5π sq. ft/sec 4. speed = 6π sq. ft/sec correct 5. speed = sq. ft/sec pole at a rate of 2 feet per second. At what speed is the length of the person s shadow growing?. speed = speed = 7 9 ft/sec. speed = ft/sec 4. speed = 8 9 ft/sec 5. speed = 9 ft/sec ft/sec correct If x denotes the length of the person s shadow and y denotes the distance of the person from the pole, then shadow and the lightpole are related in the following diagram 6. speed = 4π sq. ft/sec 7. speed = 6 sq. ft/sec 8. speed = 7 sq. ft/sec The area, A, of a circle having radius r is given by A = πr 2. Differentiating implicitly with respect to t we thus see that da = 2πr dr. When dr r =, = 6, therefore, the speed at which the area of the ripple is increasing is given by speed = 6π sq. ft/sec. 4 y By similar triangles, 5 x + y 5 x = 4 x + y, x CalCi09s points A street light is on top of a 4 foot pole. A person who is 5 feet tall walks away from the so 5y = 4 5)x. Thus, after implicit differentiation with respect to t, 5 dy = 4 5) dx.

2 Version PREVIEW HW 07 hoffman 57225) 2 When dy/ = 2, therefore, the length of the person s shadow is growing with speed = 0 9 ft/sec. CalCi5s points The height of a triangle is increasing at a rate of 5 cm/min while its area is increasing at a rate of sq. cms/min. At what speed is the base of the triangle changing when the height of the triangle is 5 cms and its area is 5 sq. cms?. speed = 2 5 cms/min 2. speed = 5 cms/min. speed = 27 5 cms/min 4. speed = 24 5 cms/min correct 5. speed = 26 5 cms/min Let b be the length of the base and h the height of the triangle. Then the triangle has area = A = 2 b h. Thus by the Product Rule, and so da = 2 b ) + hdb, db = 2 da h b ) = 2 h da A h since b = 2A/h. Thus, when ), we see that db = 2 5A h h ) cms/min. Consequently, at the moment when h = 5 and A = 5, the base length is changing at a speed = 24 5 cms/min. CalCi2s points Gravel is being dumped from a conveyor belt at a rate of 25 ft /min and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 7 ft high? ft/min ft/min ft/min ft/min ft/min correct h h/2 = 5, and da =,

3 Version PREVIEW HW 07 hoffman 57225) We are given that When h = 7 ft, dv = 25 ft /min V = πr2 h V = ) 2 h h = πh 2 2 dv = dv 25 = πh2 4 = 00 πh 2 = 00 π7) 2 = 00 49π ft/min CalCi25s points Two sides of a triangle are 4 inches and 8 inches in length and the angle θ between them is increasing at a rate of /8 radians per second. Find the rate at which the area of the triangle is increasing when θ = π/.. rate = 4 sq.ins/sec 2. rate = 7 8 sq.ins/sec. rate = sq.ins/sec correct 4. rate = 9 8 sq.ins/sec 5. rate = 5 4 sq.ins/sec For any triangle its Area = height base. 2 In the case of the triangle therefore, its area A = h 4 sin θ = 6 sinθ, since by right triangle trigonometry the triangle has height h = 4 sinθ. Differentiating now with respect to t we thus see that da Consequently, when θ = π, = 6 cosθ dθ. dθ θ = 8 rads/sec, the area of the triangle is increasing at a rate = da = CalCis points = sq. ins/sec. A weather balloon is rising vertically at 40 meters per minute. An observer is standing on the ground 24 meters from the point at which the balloon was released. Determine in meters per minute) the rate at which the distance between the feet of the observer and the balloon is changing when the balloon is 2 meters high. Hint: remember -4-5 right triangles.). rate = 2 meters/min correct 2. rate = meters/min

4 Version PREVIEW HW 07 hoffman 57225) 4. rate = 4 meters/min 4. rate = 6 meters/min 5. rate = 5 meters/min Let h be the height of the balloon t seconds after it is released and s the distance of the balloon from the observer as shown in the figure balloon A man starts at a point A and walks 8 feet north. He then turns and walks due east at 8 feet per second. If a searchlight placed at A follows him, at what rate is the light turning seconds after he started walking east?. rate = 5 rads/sec 2. rate = 7 0 rads/sec. rate = 6 rads/sec 4. rate = 2 5 rads/sec h s 5. rate = 0 rads/sec correct 24 observer Let x = xt) be the distance the man has travelled t seconds after he started walking east as shown in the figure Then by Pythagoras theorem, h = s 2. Differentiating this equation implicitly with respect to t we see that 2h = 2s ds, i.e., ds = h s. But, again by Pythagoras theorem, h = 2 = s = = 40 remember -4-5 right-angled triangles!), and so ds = 2 40 = 2 meters/min when / = 40 meters/min. 8 A N θ x By right triangle trigonometry, the angle θ satisfies the equation tanθ = x 8. Differentiating this equation implicitly with respect to t we see that CalCi4s points sec 2 θ dθ = dx 8.

5 Version PREVIEW HW 07 hoffman 57225) 5 Now sec 2 θ = + tan 2 θ = 82 + x Consequently, dθ = After seconds, therefore, x) = dx 8 dx x 2. = 54. Hence the rate at which the search light is turning after seconds is given by dθ = t= 0 rads/sec. CalCj06s points Find the linearization of at x = 0.. Lx) = fx) = x ) + x the Chain Rule ensures that Consequently, and so f x) = 2 + x) /2. f0) =, f 0) = 6, Lx) = 6 x ) CalCj09s points Use linear approximation with a = 6 to estimate the number 6.7 as a fraction Lx) = + 6 x ) correct. Lx) = + 6 x ) Lx) = x 5. Lx) = + x 6. Lx) = 6 x ) correct The linearization of f is the function But for the function fx) = Lx) = f0) + f 0)x. + x = + x) /2, For a general function f, its linear approximation at x = a is defined by Lx) = fa) + f a)x a) and for values of x near a fx) Lx) = fa) + f a)x a) provides a reasonable approximation for fx). Now set fx) = x, f x) = 2 x.

6 Version PREVIEW HW 07 hoffman 57225) 6 Then, if we can calculate a easily, the linear approximation a + h a + h 2 a provides a very simple method via calculus for computing a good estimate of the value of a + h for small values of h. In the given example we can thus set a = 6, h = 7 0. For then CalCjs points Use linear appproximation to estimate the value of 7 /4. Hint: 6) /4 = 2.). 7 / / / / correct 5. 7 /4 6 Set fx) = x /4, so that f6) = 2 as the hint indicates. Then df dx = 4x /4. By differentials, therefore, we see that fa + x) fa) df x = x dx x=a 4a /4. Thus, with a = 6 and x =, 7 /4 2 = 2. Consequently, 7 /