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Math Help - calculus ab word problem

  1. #1
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    Exclamation calculus ab word problem

    At time t, t is greater than or equal to 0, the volume of a sphere is increasing at a rate proportional to the reciprocal of its radius. At t=0, the radius of the sphere is 1 and at t=15 he radius is 2 (the volume V of a sphere with radius r is V=4/3 pi*r^3
    a) Find the radius of this sphere as a function of t.
    b) At what time t will the volume of the sphere be 27 times its volume at t=0?

    I'm completely lost and don't even know where to start. i've been looking at this for about 15 and really need some help. Any pointers would be great
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  2. #2
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    volume of a sphere is increasing at a rate proportional to the reciprocal of its radius
    \frac{dV}{dt} = \frac{k}{r}

    using the formula for a sphere, V = \frac{4}{3} \pi r^3 , ...

    \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

    so ...

    4\pi r^2 \frac{dr}{dt} = \frac{k}{r}

    separate variables ...

    4\pi r^3 \, dr = k \, dt

    integrate and get r as a function of t
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  3. #3
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    so when i integrate is it
    pi*r^4 = kt+c
    if i solve for c is it pi?

    so r= the fourth root of [(1/pi) kt+1]

    is that right or am i completely off track?
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  4. #4
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    r = fourth root of [(1/pi) kt + 1]

    You are on the right track. Just make sure that in the equation

    r = fourth root of [(1/pi) kt + 1], the kt and 1 are separate, and only the kt is being multiplied to the (1/pi).

    other then that, you are correct.
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