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Math Help - Related Rates #2

  1. #1
    Junior Member
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    Related Rates #2

    I was asked to make a new thread for each problem, so this is thread 2

    2) A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 feet about the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and the balloon increasing 3 sec later?

    **Answer from Book**

    11
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  2. #2
    MHF Contributor
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    We are looking for ds/dt.

    Find the relation of s to the givens so that we can have an equation involving
    the s.

    The kite is rising vertically, the bicycle is running horizontally, the distance, s, between the kite and the bicycle is moving "diagonally".
    So the figure is that of a right triangle whose parts after time t are:
    ---vertical leg = (65 +1*t) = (65 +t)
    ---horizontal leg = 17*t = 17t
    ---hypotenuse = s

    By Pythagorean theorem,
    s^2 = (65 +t)^2 +(17t)^2 ------------------------(i)
    Differentiate both sides with respect to time t,
    2s(ds/dt) = 2(65 +t)(1 dt/dt) +2(17t)(17 dt/dt) <----I purposely showed dt/dt---just for comparison.
    s(ds/dt) = (65 +t) +(289t)
    s(ds/dt) = 290t +65 -----------------------(ii)

    We were given t = 3 sec

    What about s?
    We can compute s when t = 3 sec.
    s^2 = (65 +t)^2 +(17t)^2 -------------------------(i)
    s^2 = (65 +3)^2 +(17*3)^2
    s^2 = 7225
    s = sqrt(7225) = 85 ft

    So, substituting those into (ii),
    s(ds/dt) = 290t +65
    85(ds/dt) = 290*3 +65
    85(ds/dt) = 935
    ds/dt = 935/85 = 11 ft/sec

    Therefore, at that instant, s is increasing at the rate of 11 ft/sec. ----answer.
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  3. #3
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    Hello, Super Mallow!

    I use a slightly different set-up . . .


    2) A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec.
    Just when the balloon is 65 feet about the ground,
    a bicycle moving at a constant rate of 17 ft/sec passes under it.
    How fast is the distance s(t) between the bicycle and the balloon increasing 3 sec later?
    Code:
          B *
            ↑\
            ↑ \
          t ↑  \
            ↑   \
            ↑    \  s(t)
          P *     \
            |      \
         65 |       \
            |        \
            * → → → → *
            A   17t   C

    The balloon is at P when the cycle passes under it (at A): . AP = 65

    In the next t seconds, the balloon rises t feet to point B: . BP = t

    In the same t seconds, the bicycle moves 17t feet to point C: . AC = 17t

    Using Pythagorus: . s(t) \;=\;\sqrt{(17t)^2 + (t+65)^2} \;=\;\left(290t^2 + 130t + 4225\right)^{\frac{1}{2}}

    Differentiate with respect to time:
    . . \frac{ds}{dt} \;=\;\frac{1}{2}\left(290t^2+130t+4225\right)^{-\frac{1}{2}}(580t  + 130) \;=\;\frac{290t+65}{\sqrt{290t^2+130t+4225}}

    And now let t = 3.

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  4. #4
    Junior Member
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    There's 2 parts I do not understand about this...

    Where does the (T+65) come in?
    Also, how did we get 290T+65 on the top of the Square root?
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