RELATED RATES PRACTICE                                      NAME: SOLUTIONS
             March22,2017
                                 STEPS FOR SOLVING PROBLEMS WITH RELATED RATES
                (1) Drawapicturerepresentingtheproblem.
                (2) Write down what you knowandwhatyouneedtofindout.
                (3) Label your picture with the known and unknown variables.
                (4) Find the equation that relates the variables.
                (5) Take the derivative of the equation relating the variables with respect to t.
                   RememberyourDerivativeRules!
                (6) Substitute the given values into the equation from Step 5.
                (7) Use the equation from Step 4 to find the missing pieces of information.
                (8) Solve for the final missing piece and you’re done!
               (1) If the radius of a balloon is increasing at a constant rate of 0.03 in/min, how fast is the volume of the
                 balloon changing at the time when its radius is 5 inches?
                 SOLUTION: Wecanrepresenttheballoonbyasphere. Thevolumeofasphereis
                                                        4  3
                                                    V = 3πr .
                 Wearegiven dr = 0.03in/minandaskedtofind dV whenr=5inches. Tothatend,wedifferentiate
                            dt                          dt
                 the formula for the volume with respect to t.
                                                   dV =4πr2dr
                                                   dt       dt
                 Thenpluginthevaluesr=5and dr =0.03tofind
                                            dt
                                             dV                3
                                             dt =4(25)(0.03) = 3 in /min
             This material is based upon original Active Calculus materials produced by the University of Nebraska at Omaha.
               (2) Anoil spill expands in a circular pattern. Its radius increases at a constant rate of 1 m/s. At time t = 0,
                  the radius is 120 meters. What is the rate of change of the area of the spill at time t = 1 minute?
                  SOLUTION: Theoilspillisacircleofradiusr. Therelation of the area of this circle and the radius is
                                                           2
                                                     A=πr .
                  Wearegiven dr = 1m/sandaskedtofind dA whent=1minute.
                            dt                      dt
                  To relate the radius and time, note that the initial radius is 120 meters, and it expands at one meter
                  every second. So for t measured in seconds, r(t) = 120+t meters.
                  Todothat,wedifferentiate both sides of the area equation by t to find
                                                    dA =2πrdr.
                                                    dt      dt
                  Nowplug in dr = 1 and r(t) = 120+t. We seek the answer for t = 1 minute = 60 seconds, so
                             dt
                  r(60) = 180. Therefore,
                                             dA                   2
                                             dt =2π(180)(1) = 360πm /s
               (3) Thelengthofarectangleincreasesby3ft/minwhilethewidthdecreasesby2ft/min. Whenthelength
                  is 15 ft and the width is 40 ft, what is the rate of change of:
                   (a) the area?
                      SOLUTION: TheareaofarectangleisgivenbyA = ℓw,whereℓisthelengthandwisthewidth.
                     Wearegiven dℓ = 3ft/minand dw = −2ft/minandℓ = 15ftandw = 40ft. Weareaskedto
                                dt              dt
                     find dA. Sowedifferentiate the formula for area:
                          dt
                                                    dA = dℓw+ℓdw
                                                    dt   dt    dt
                     Nowpluginthegivenvaluestofind
                                             dA =3(40)+15(−2)=90ft2/min.
                                             dt
                  (b) the perimeter?
                      SOLUTION: Theperimeterofarectangleisgivenby
                                                      P = 2ℓ+2w,
                     whereℓisthelengthandwisthewidth. Wearegiven dℓ = 3ft/minand dw = −2ft/minand
                                                                dt              dt
                     ℓ = 15 ft and w = 40 ft. We are asked to find dP. So we differentiate the formula for perimeter:
                                                        dt
                                                    dP =2dℓ +2dw
                                                    dt    dt   dt
                     Andthenpluginthegivenvaluestofind
                                               dP =2(3)+2(−2)=2ft/min
                                               dt
             This material is based upon original Active Calculus materials produced by the University of Nebraska at Omaha.
                   (4) The volume of a tree is given by V =  1 C2h,whereCisthecircumferenceofthetreeatgroundlevel
                                                            12π
                       (in meters) and h is the height of the tree (in meters). Both C and h are functions of time (in years).
                        (a) Find a formula for dV. What does it represent?
                                               dt
                            SOLUTION: Differentiate both sides of the formula for volume with respect to t. Don’t forget the
                            productrule!                                                 
                                                             dV     1         dC     2dh
                                                             dt = 12π 2Chdt +C dt
                            This represents the rate of change of volume of the tree, or how fast the tree is growing.
                        (b) Suppose the circumference grows at a rate of 0.2 meters per year and the height grows at a rate of
                            4metersperyear. Howfastisthetreegrowingwhenthecircumferenceis5metersandtheheight
                            is 22 meters?
                            SOLUTION: Wearegiven dC = 0.2metersperyearand dh = 4metersperyear,aswellasC = 5
                                                      dt                            dt
                            metersandh=22meters. Weareaskedtofind dV,whichisgivenasintheformulafrompart
                                                                             dt
                            (a). So plug the values in
                                 dV      1                     2        1 (         )    1           12
                                 dt = 12π 2(5)(22)(0.2)+(5) (4) = 12π 44+100 = 12π(144) = π meters/year.
                 This material is based upon original Active Calculus materials produced by the University of Nebraska at Omaha.