Miscellaneous Related Rates Solutions

Miscellaneous Related Rates: Solutions To do these problems, you may need to use one or more of the following: The Pythagorean Theorem, Similar Triangles, Proportionality ( A is proportional to B means that A = kB, for some constant k). 1. The top of a 25-foot ladder, leaning against a vertical wall, is slipping down the wall at a rate of 1 foot per second. How fast is the bottom of the ladder slipping along the ground when the bottom of the ladder is 7 feet away from the base of the wall? First, make a sketch of a triangle whose hypotenuse is the ladder. Let y(t) be the height of the ladder with the vertical wall, and let x(t) be the length of the bottom of the ladder with the vertical wall. Then x2 (t) + y 2 (t) = 252 The problem can then be interpreted as: If x = 7? dy dt = −1, what is dx dt when Differentiating with respect to time: 2x(t) dx dy + 2y(t) =0 dt dt we have numbers for dy dt and x(t)- we need a number for y in order to solve for dx . Use the original equation, and dt 72 + y 2 (t) = 252 ⇒ y = 24 Now plug everything in and solve for 2·7· dx dt : dx dx 24 + 2 · 24 · (−1) = 0 ⇒ = dt dt 7 2. A 5-foot girl is walking toward a 20-foot lamppost at a rate of 6 feet per second. How fast is the tip of her shadow (cast by the lamppost) moving? Let x(t) be the distance of the girl to the base of the post, and let y(t) be the distance of the tip of the shadow to the base of the post. If you’ve drawn the right setup, you should see similar triangles... Hgt of post Dist of tip of shadow to base = Hgt of girl Dist of girl to base In our setup, this means: 20 y(t) = 5 y(t) − x(t) With a little simplification, we get: 3y(t) = 4x(t) 1 dx We can now interpret the question as asking what dy dt is when dt = −6. Differentiating, we get dx dy 3 =4 dt dt dy so that the final answer is dt = −8, which we interpret to mean that the tip of the shadow is approaching the post at a rate of 8 feet per second. 3. Under the same conditions as above, how fast is the length of the girl’s shadow changing? Let L(t) be the length of the shadow at time t. Then, by our previous setup, L(t) = y(t) − x(t) so dL dt = dy dt − dx dt = −8 − (−6) = −2. 4. A rocket is shot vertically upward with an initial velocity of 400 feet per second. Its height s after t seconds is s = 400t − 16t2 . How fast is the distance changing from the rocket to an observer on the ground 1800 feet away from the launch site, when the rocket is still rising and is 2400 feet above the ground? We can form a right triangle, where the launch site is the vertex for the right angle. The height is s(t), given in the problem, the length of the second leg is fixed at 1800 feet. Let u(t) be the length of the hypotenuse. Now we have: u2 (t) = s2 (t) + 18002 and we can interpret the question as asking what Differentiating, we get 2u(t) du dt is when s(t) = 2400. du ds du ds = 2s(t) or u(t) = s(t) dt dt dt dt ds To solve for du dt , we need to know s(t), u(t) and dt . We are given s(t) = 2400, so we can get u(t): p u(t) = 24002 − 18002 = 3000 2 Now we need ds dt . We are given that s(t) = 400t − 16t , so That means we need t. From the equation for s(t), ds dt = 400 − 32t. 2400 = 400t − 16t2 ⇒ −16t2 + 400t − 2400 = 0 Solve this to get t = 10 or t = 15. Our rocket is on the way up, so we choose t = 10. Finally we can compute ds dt = 400 − 32(10) = 80. Now, 3000 so du dt du = 2400(80) dt = 64 feet per second (at time 10). 2 5. A small funnel in the shape of a cone is being emptied of fluid at the rate of 12 cubic centimeters per second (the tip of the cone is downward). The height of the cone is 20 cm and the radius of the top is 4 cm. How fast is the fluid level dropping when the level stands 5 cm above the vertex of the cone [The volume of a cone is V = 13 πr2 h]. Draw a picture of an inverted cone. The radius at the top is 4, and the overall height is 20. Inside the cone, draw some water at a height of h(t), with radius r(t). We are given information about the rate of change of volume of water, so we are given that dV dt = −12. Note that the formula for volume is given in terms of r and h, but we only want dh dt . We need a relationship between r and h... You should see similar triangles (Draw a line right through the center of the cone. This, and the line forming the top radius are the two legs. The outer edge of the cone forms the hypotenuse). radius of top overall height 4 20 = ⇒ = radius of water level height of water r h so that r = h5 . Substituting this into the formula for the volume will give the volume in terms of h alone: V = 1 2 1 h 1 πr h = π( )2 h = πh3 3 3 5 75 Now, and we know dV dt dV 3π 2 dh = h dt 75 dt = −12, h = 5, so −12 dh = dt π 6. A balloon is being inflated by a pump at the rate of 2 cubic inches per second. How fast is the diameter changing when the radius is 12 inch? The volume is V = 34 πr3 (this formula would be given to you on an exam/quiz). If we let h be the diameter, then we know that 2r = h, so we can make V depend on diameter instead of radius: V = 4 h 3 π π( ) = h3 3 2 6 Now, the question is asking for dV dt = 2. Differentiate, and dh dt when h = 1, and we are given that dV π dh π dh = 3h2 = h2 dt 6 dt 2 dt so that dh dt = 4 π. 3 7. A particle moves on the hyperbola x2 − 18y 2 = 9 in such a way that its y coordinate increases at a constant rate of 9 units per second. How fast is the x−coordinate changing when x = 9? In this example, we don’t need any labels. The question is to find dy dt = 9. Differentiate to get: 2x dx dt when dy dx − 36y =0 dt dt We’re going to need the y value when x = 9, so go back to the original equation: 92 − 18y 2 = 9 ⇒ y = ±2 so we need to consider 2 y−values. Putting these into our derivative, we get: dx 2·9· − 36 · 2 · 9 = 0 dt so that dx dt = 36. Put in y = −2 to get the second value of dx dt = −36 8. An object moves along the graph of y = f (x). At a certain point, the slope of the curve is 12 and the x−coordinate is decreasing at 3 units per second. At that point, how fast is the y−coordinate changing? The key point here is that we want to think about both x and y as functions of t, so that when we differentiate, we get (use chain rule): dy dx = f 0 (x) dt dt The slope of the curve at a certain point is f 0 (x) = 12 , and dx dt = −3, so we can plug these values in to get the change in y (with respect to time): dy 1 −3 = · (−3) = dt 2 2 9. A rectangular trough is 8 feet long, 2 feet across the top, and 4 feet deep. If water flows in at a rate of 2 cubic feet per minute, how fast is the surface rising when the water is 1 foot deep? The trough is a rectangular box. Let x(t) be the height of the water at time t. Then the volume of the water is: V = 16x ⇒ Put in dV dt = 2 to get that dx dt dV dx = 16 dt dt = 18 . 10. If a mothball (sphere) evaporates at a rate proportional to its surface area 4πr2 , show that its radius decreases at a constant rate. 4 Let V (t) be the volume at time t. We are told that dV = kA(t) = k4πr2 dt We want to show that dr dt is constant. By the formula for V (t) = 43 πr3 , we know that: dr dV = 4πr2 dt dt Now compare the two formulas for the constant of proportionality! dV dt , and we see that dr dt = k, which was 11. If an object is moving along the curve y = x3 , at what point(s) is the y−coordinate changing 3 times more rapidly than the x−coordinate? Let’s differentiate: dy dx = 3x2 dt dt dx From this, we see that if we want dy dt = 3 dt , then we must have x = ±1. We also could have x = 0, y = 0, since 0 is 3 times 0. All the points on the curve are therefore: (0, 0), (−1, 1), (1, 1) 5