A balloon rises at the rate of 8 ft/sec from a point on the ground 60 ft from the observer. How to find the rate of change of the angle of elevation when the balloon is 25 ft above the ground?

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To address this rate (of change) issue:Let       y   = the height of the balloon and let       θ   = the angle of elevation.We are told that                               d          y                                      d          t                      =    8   ft/sec.We are asked to find                     d        θ                              d          t                     when       y    =    25   ft.Draw a right triangle with base = 60 ft (that doesn&#039;t change), height       y   and angle opposite height       θ  Then             tan      θ        =          y      60       and       y    =    60          tan      θ      Differentiating in relation to t results in:            d                        d          t                            (      y      )        =          d                        d          t                            (      60              tan        θ            )                                    d          y                                      d          t                      =    60                  sec        2            θ                      d        θ                              d          t                    We are asked to find                     d        θ                              d          t                     when       y    =    25  We have:        8    =    60                  sec        2            θ                      d        θ                              d          t                    , so                     d        θ                              d          t                      =          8      60                      cos        2            θ        =          2      15                      cos        2            θ      We need             cos      θ       when       y    =    25  With base = 60 and height = 25, we get hypotneuse       c    =                            60          2                +                  25          2                      =                                        (            5            ⋅            12            )                    2                +                              (            5            ⋅            5            )                    2                      =    5                                        (            12            )                    2                +                              (            5            )                    2                      =    5    ⋅    13    =    65  So, when       y    =    25  , we have:              cos      θ        =          60      65        =          12      13      So                     d        θ                              d          t                      =          2      15                      cos        2            θ        =          2      15                      (                  12          13                )            2        =          96      845       radians / sec(Remember, in order to use             d              d        θ                    (              tan        θ            )        =                  sec        2            θ      , we must have       θ   either a real number or the radian (not degree) measure of an angle.)

A balloon rises at the rate of 8 ft/sec from a point on the ground 60 ft from the observer. How to find the rate of change of the angle of elevation when the balloon is 25 ft above the ground?

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