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  1. #1
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    word problem...

    A sphere shrinks at a rate of 1 cm^3 per second. At what rate does the diameter decrease when the diameter is 10 cm? And yes, this is all the info I was given.
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    Quote Originally Posted by driver327 View Post
    A sphere shrinks at a rate of 1 cm^3 per second. At what rate does the diameter decrease when the diameter is 10 cm? And yes, this is all the info I was given.
    Well, for a sphere
    V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left ( \frac{d}{2} \right ) ^3

    V = \frac{1}{6} \pi d^3

    So
    \frac{dV}{dt} = \frac{1}{2} \pi d^2 \frac{dd}{dt}

    The rest you can do yourself.

    -Dan
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  3. #3
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    Quote Originally Posted by driver327 View Post
    A sphere shrinks at a rate of 1 cm^3 per second. At what rate does the diameter decrease when the diameter is 10 cm? And yes, this is all the info I was given.
    I have little time more before I have to go for work. I just want to practice on this. I'm groping right now, rusty.

    To find: rate of decrease of diameter
    So dD/dt, where D means diameter

    Sphere:
    V = (4/3)(pi)(r^3)

    in terms of D,
    V = (4/3)(pi)[(D/2)^3]
    V = (1/6)(pi)(D^3)
    Differentiate both sides with respect to time t,
    dV/dt = (pi/6)[3 D^2 dD/dt]
    dV/dt = (pi/2)(D^2) dD/dt ----------**

    So when D = 10cm, and since dV/dt = -1 cu.cm,
    -1 = (pi/2)(10^2) dD/dt
    -1 = 50pi dD/dt
    dD/dt = -1/(50pi)

    Therefore, when the diameter is 10cm, the diameter is decreasing at the rate of 1/(50pi) cm/sec. ---------------answer.


    No time to review. If there are mistakes, I'll rectify after work---about 11 hrs from now.
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