Related Rates Problems

Transcription

1 These notes closely follow the presentation of the material given in James Stewart s textbook Calculus, Concepts and Contexts (2nd edition). These notes are intended primarily for in-class presentation and should not be regarded as a substitute for thoroughly reading the textbook itself and working through the exercises therein. Related Rates Problems In solving a related rates problem, one attempts to find the rate of change of some quantity based on the rate of change of some related quantity. The basic strategy for solving related rates problems is outlined on page 267 of our textbook. Example Suppose that one leg of a right triangle remains of fixed length while the other leg grows at a constant rate k 0 and that all the while the triangle remains a right triangle. Find a formula that gives the rate at which the hypotenuse is growing. Solution Let a be the length of the fixed leg of the triangle. Let x be the length of the other leg and let y be the length of the hypotenuse. Note that a is a constant quantity but x and y are functions of time ( t ). We are given that dx/ k and we want to determine /. Since the triangle remains a right triangle at all times, then by the Pythagorean Theorem, a 2 x 2 y 2 for all t 0. (Recall that x and y are functions of t.) Differentiating both sides of this equation with respect to t, we obtain d a2 x 2 d y2 or Since dx/ k, this gives 2x dx 2y. kx y. For example, suppose that we are given that a 6 inches and k 2 inches per second and we wish to know at what rate the hypotenuse is growing at the instant when the growing leg has length 12 inches. The rate of growth of the hypotenuse at this instant is. x 12 When x 12, the Pythagorean Theorem gives us y 2 which gives y 6 5. Thus, 1

2 x We conclude that at the instant when x 12 inches, the hypotenuse is growing at a rate of 4 5/ inches per second. 2

3 Example A plane flying horizontally at an altitude of one mile and a speed of 500 miles per hour passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing at the instant when the plane is two miles away from the station. Solution Let O be the location of the radar station and let P be the position of the plane when it is directly over the radar station. Let Q be the position of the plane at any given time t after flying over the radar station. Then OPQ is a right triangle with leg OP of length one mile, leg PQ growing at a constant rate of 500 miles per hour and hypotenuse OQ also growing with time. Let x be the length of PQ and let y be the length of OQ. From the work done in the previous example, we have 500x y. We wish to know the value of / at the instant when y 2 miles. By the Pythagorean Theorem, when y 2 we have 1 2 x which gives that x 3. Thus, y Our conclusion is that the distance of the plane from the radar station is increasing at a rate of miles per hour at the instant when the plane is two miles from the radar station. 3

4 Example A water trough is 10 m long and a cross section has the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top, and has height 50 cm. If the trough is being filled with water at the rate of 0.2 cubic meters per minute, how fast is the water level rising when the water is 30 cm deep? Solution Let x be the depth and V be the volume of the water in the trough at any given time t. Referring to the figure, which shows a cross section of the trough, we reason that the volume of water in the trough at any time t is given by V x rx. By properties of similar triangles, which implies that Thus, we obtain r x r 1 2 x. V x 1 2 x x. which is equivalent to V 3x 5x 2. Differentiation with respect to t yields dv 3 10x dx. Since we are given that dv/ 0.2, then x dx. Solving for dx/ gives dx x Substitution of x 0.3 into this equation yields dx 1 x Thus, the water level is rising at a rate of 1/30 of a meter per minute at the instant when the water level is 0.3 meters. 4

5 Example Water is leaking out of an inverted conical tank at a rate of 10,000 cubic centimeters per minute at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 meters and the diameter at the top is 4 meters. If the water level is rising at a rate of 20 centimeters per minute at the instant when the water level is 2 meters, find the rate at which water is being pumped into the tank. Solution Let k be the constant rate at which water is being pumped into the tank. This is the unknown. Let x be the water level in the tank at any given time t and let r and V be, respectively, the radius of the water surface and the volume of water at any given time t. Since the water in the tank also takes the shape of a cone, then by referring to the formula for the volume of a cone on the inside front cover of Stewart, we obtain V 1 3 r2 x. Using properties of similar triangles, we note the relation r x 2 6 which implies that r 1 3 x. Thus, V x x 2 x Differentiation with respect to t yields dv x2 dx 1 9 x2 dx. Since water is being pumped into the tank at the constant rate k and simultaneously being pumped out of the tank at the constant rate of 10,000 cm 3 /min, then the net rate of change of the volume of the water in the tank is dv/ k 10,000 cm 3 /min. (Note that dv/ is constant.) We thus have k 10, x2 dx. (1) Finally, we are given that the water level is rising at a rate of 20 cm/min at the instant when the water level is 200 centimeters. This is expressed in Leibniz notation as dx 20. x 200 Setting x 200 in Equation (1) gives k 10, Solving for k then gives 5

6 k 10, , ,253 We conclude that water is being pumped into the tank at the rate of about 289,253 cm 3 /min. 6