Math 220 Groupwok 10/12/17
                   Related Rates Word Problems
                                       SOLUTIONS
              (1) One car leaves a given point and travels north at 30 mph. Another car leaves
                 1 HOUR LATER, and travels west at 40 mph. At what rate is the distance
                 between the cars changing at the instant the second car has been traveling
                 for 1 hour?
                           z        x
                           y
                   Set up the problem by extracting information in terms of the variables x,
                 y, and z, as pictured on the triangle:
                   • First sentence: dx = 30 and x(t) = 30t.
                                 dt
                   • Second: dy = 40 and y(t) = 40(t−1) (Start one hour late!)
                            dt
                   • Goal: Find dz at t = 2.
                              dt
                 The property that combines the sides of a triangle is Pythagorean Theorem:
                                       x2 +y2 = z2.
                 At t = 2, x(2) = 60 and y(2) = 40. Using Pythagorean Theorem: z(2) =
                 √ 2     2
                   60 +40 ≈72.111. Taking the derivative in t:
                                    2xdx +2ydy =2zdz.
                                      dt    dt    dt
                 Plug in:
                             2·60·30+2·40·40=2·72.111· dz.
                                                         dt
                 Thus, dz ≈ 47.150.
                       dt
                                            1
                 2
                    (2) A 50ft ladder is placed against a large building. The base of the ladder is
                       resting on an oil spill, and it slips at the rate of 3 ft. per minute. Find
                       the rate of change of the height of the top of the ladder above the ground at
                       the instant when the base of the ladder is 30 ft. from the base of the building.
                                50        x
                                y
                         Organizing information:
                         • dy = 3
                           dt
                         • Goal: Find dx when y = 30.
                                    dt
                       Weuse Pythagorean Theorem again:
                                        2    2    2
                                       x +30 =50 =⇒x=40.
                       And differentiating (notice how the hypotenuse is constant):
                              2xx′ +2yy′ = 0x′          =−2yy′ = −yy′
                                                            2x     x
                                  ′
                       Plugging in, x = −30·3÷40 = −2.25.
                         Note: x′ is negative, that means the distance x is decreasing—the ladder
                       is slipping down the building.
                                                                        3
              (3) A stone dropped in a pond sends out a circular ripple whose radius increases
                at a constant rate of 4 ft/sec. After 12 seconds, how rapidly is the area in-
                closed by the ripple increasing?
                  Organizing information:
                  • dr = 4
                    dt
                  • Goal: Find dA when t = 12.
                             dt
                Weuse the area formula for a circle.
                                       A=πr2
                Differentiate both sides with respect to t:
                                     dA =2πrdr
                                      dt     dt
                Plug in dr = 4. When t = 12 seconds, r = 4*12 = 48.
                       dt
                              dA                  2
                               dt = 2π(48)∗4 = 384πft /sec
                  4
                     (4) A spherical balloon is being inflated so that its diameter is increasing at a
                        rate of 2 cm/min. How quickly is the volume of the balloon increasing when
                        the diameter is 10 cm?
                          Organizing information:
                           • dd = 2
                             dt
                           • Goal: Find dV when d = 10.
                                       dt
                        Weuse the volume formula for a sphere, but rewrite it with the diameter
                                                 V = 4πr3
                                                     3
                                                V = 4π(d)3
                                                    3π 2
                                                        3
                                                 V = 6d
                        Differentiate both sides with respect to t:
                                               dV =3πd2dd
                                                dt    6  dt
                        Plug in dd = 2 and d = 10
                               dt
                                       dV    π    2           3
                                       dt = 36(10) ∗2 = 100πcm /min