RELATED RATES By: Kyler Hall Autrey: AP Calculus March 7, 2011 THINGS TO KNOW BEFORE STARTING A RELATED RATES PROBLEM! Chain Rule – If g is differentiable at x and f is differentiable at g(x), then the composite function F = fog defined by F(x) = f’(g(x)) is differentiable at x and F’ is given by the product: F’(x) = f’(g(x)) • g’(x) STEPS FOR SOLVING RELATED RATES PROBLEMS 1. 2. 3. Identify all given quantities and the quantities that need to be determined. Make a sketch of the problem and label the quantities. Write an equation that uses all of the variables that have rates of change that are labeled or need to be found. STEPS FOR SOLVING RELATED RATES PROBLEMS CONTINUED… 4. Using the Chain Rule, implicitly differentiate both sides of the equation with respect to time t. 5. After completing the previous step, substitute the quantities into the equation as well as their rates of change, if they are given. Then solve for the rate of change that is required to answer the problem and include any units given by the problem. All edges of a cube are expanding at a rate of 6 centimeters per second. How fast is the volume changing when each edge is 2 centimeters? - First, the quantities need to be identified, once this has been done they need to be labeled onto a picture of the cube X=2 X X dy =2 dx Now, an equation needs to be written that includes all of the variables and rates of change either labeled or that need to be found . In the case of this problem the formula for the volume of a cube needs to be written: V = (l)(w)(h) But since this since the problems labels the sides as X the problem can be written: V=X3 Now using the Chain rule, take the derivate in terms of time t of the equation, plug in the values given into the equation of the derivate, and then solve for the missing variable and include its units: V X3 dV 2 dx 3X dt dt dV (3)(2) 2 (6) dt dV 72 cm3/sec dt An airplane is flying at an altitude of 5 miles and passes directly over a radar antenna. When the plane is 10 miles away, the radar detects that the distances is changing at a rate of 240 miles per hours. What is the speed of the plane? First draw the shape and label the quantities indicated by the problem: dx x? dt x y 5 miles y dy 0 miles hour dt z ? z 10 miles dz 240 miles hour dt Then make an equation that would include any rates of change indicated by the problem: x2 y 2 z 2 Now take the derivative, using the Chain rule in terms of time t, of the formula found in the last step: x2 y 2 z 2 2x dx dy dz 2y 2z dt dt dt Which can be rewritten as: x dx dy dz y z dt dt dt Now plug in the values given by the problem: dx x 5(0) 10(240) dt Note: The rate of change of y is equal to 0 since y is a constant. With this problem another variable, x, has to b found because there are two variables that do not have numerical values: x y z x 2 (5)2 (10)2 2 2 2 x 2 75 x5 3 Now solve for the rate of change of x: dx 5 3 10(240) dt dx 2400 dt 5 3 dx 480 3 miles 160 3 hour dt 3 Example of an FRQ Related Rates Problem 1977 AB 6 A rectangle has a constant area of 200 square meters and its length is increasing at the rate of 4 meters per second. a) Find the width W at the instant the width is decreasing at the rate of 0.5 meters per second. b) At what rate is the diagonal D of the rectangle changing at the instant when the width is 10 meters. FRQ example continued… Part A: (Finding the Width of W) 1. First indentify the quantities and label them on a picture of the problem. dL 4 meters/second L dt W A LW 200 m2 dW ? dt dA 0 dt meters/second Since A is a constant FRQ example continued… 2. Then write an equation that includes all of the rates of change indicated by the problem, and use the chain rule to take the derivative. A LW dA dW L dt dt dL W dt Since A equals 200 and A = LW then LW = 200. Therefore: 200 L W dA 200 dW dt W dt dL W dt FRQ example continued… 3. Now insert any known values into the equation and solve for W. 200 1 0 4 W W 2 100 4W W 4W 2 100 W 25 W 5 meters 2 FRQ example continued… Part B (Find the rate of change of diagonal D. 1. Identify the quantities and label them onto a picture of the problem. dL 4 meters/second L dt D W dW ? dt W = 10 meters A LW 200 m2 dA 0 meters/second dt Since A is a constant FRQ example continued… 2. Write an equation that include all the variables that have rates of change indicated by the problem, and use the chain rule to take its derivate. W L D dW dL dD 2W 2L 2D dt dt dt dW dL dD W L D dt dt dt 2 2 2 FRQ example continued… 3. Insert any known values and find values that are missing that aren’t needed to answer the problem. dD dW dL dD W L D dt dt dt 200 LW 10W W 20 W 2 L2 D 2 (20) 2 (10) 2 D 2 D 2 500 D 500 10 5 10(.05) 20(4) 10 5 dt dD 5 80 10 5 dt 75 dD 10 5 dt 75 5 dD 10 5 5 dt dD 75 5 3 5 dt 50 2 meters/second TRY IT YOURSELF!! A ladder 15 feet long is leaning against a building so that the end X is on level ground and end Y is on the wall. X is moved away from the building at the constant rate of .5 feet per second. a) Find the rate in feet per second at which the length OY is changing when X is 9 feet from the building. b) Find the rate of change in square feet per second of the area of the triangle XOY when X is 9 feet from the building. A NSWERS a) b) dOY 3 feet/second dt 8 dA 33 feet2/second dt 16 H ERE ’ S HOW TO DO THE PROBLEM !! Y dXY 0 dt O X feet/second dOX 1 dt 2 XY 15 feet feet/second Part A (Find the rate of change of OY) OX 9 OX 2 OY 2 XY 2 OX 2 OY 2 XY 2 dOX dOY dXY OX OY XY dt dt dt 1 dOY 9 12 15(0) 2 dt 9 OY 2 152 2 OY 2 144 OY 12 dOY 9 12 dt 2 dOY 9 3 dt 24 8 PART B (F INDING AREA ) 1 A OX OY feet/second THE RATE OF CHANGE OF THE 2 dA 1 dOY 1 dOX OX OY dt 2 dt 2 dt dA 1 3 1 1 9 12 dt 2 8 2 2 dA 27 15 27 60 dt 16 4 16 16 dA 33 feet2/second dt 16 If a snowball melts so that its surface area decreases at a rate of 1 cm2/min, find the rate at which the diameter decreases when the diameter is 10 cm. Answer: dD 1 dt 20 cm/min dSA 1 cm2/min dt R 2 SA 4 R 2 D 2R D R 2 D SA 4 SA D 2 2 dSA dD 2 D dt dt dD 1 20 dt dD 1 dt 20 The volume V of a cone is increasing at the rate of 28π cubic inches per second. At the instant when the radius R on the cone is 3 units, its volume is 12π cubic inches, and the radius is increasing at ½ inches per second. a) b) At the instant when the radius of the cone is 3 units, what is the rate of change of the area of its base? At the instant when the radius of the cone is 3 units, what is the rate of change of its height H? a) 3π in2/sec b) 8 in/sec dV 28 dt cubic inches per second When: H r 3 R inches V 12 dR 1 dt 2 cubic inches Inches per second Part A (Finding the rate of change of the area of the base.) 2 Abase R dA dR 2 R dt dt dA 1 2 3 3 dt 2 inches2/second Part B (Finding the rate of change of the height H) 3V H R2 312 H 2 3 H 4 inches 1 V R2 H 3 dV 1 2 dH R dt 3 dt 2 dR HR 3 dt 1 dH 2 1 28 3 4 3 3 dt 3 2 dH 28 3 4 dt dH 24 3 dt dH 8 inches/second dt WORKS CITED Stewart, James. Calculus: Early Transcendentals. Sixth Edition. United States: Thomson Brooks/Cole, 2008. 241-246. Print. Leithold, First. The Calculus with Analytic Geometry. Third edition. New York, NY: Harper & Row, 1976. 156-157. Print. Larson, First, First Hostetler, and First Edwards. Calculus I with Precalculus: A One-Year Course. Boston: Houghton Mifflin Company, 2002. 300-305. Print. Larson, First, and First Edwards. Calculus. Ninth edition. Belmont, CA: Brooks/Cole, Cengage Learning, 2010. 150-156. Print.