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Thread: Related Rates Questions

  1. #1
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    Exclamation Related Rates Questions

    A spherical ballon is being inflated at a rate of 3 pi m^3/min. Twelve minutes after inflation first begins, what is the rate of increase of the diameter?

    i solved for dr/dt and then adjuster for diameter, but i dont understand why we were given t = 12 min...

    thx for any help u can giv
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  2. #2
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    Quote Originally Posted by a.a View Post
    A spherical ballon is being inflated at a rate of 3 pi m^3/min. Twelve minutes after inflation first begins, what is the rate of increase of the diameter?

    i solved for dr/dt and then adjuster for diameter, but i dont understand why we were given t = 12 min...

    thx for any help u can giv
    Let diameter be D. Then $\displaystyle \frac{dD}{dt} = \frac{dD}{dV} \times \frac{dV}{dt}$.


    Given: $\displaystyle \frac{dV}{dt} = 3 \pi \, m^3/min$.

    Common knowledge: $\displaystyle V = \frac{4}{3} \, \pi r^3 = \frac{\pi}{6} \, D^3$. Therefore $\displaystyle \frac{dV}{dD} = \frac{\pi}{2} \, D^2$.

    Therefore: $\displaystyle \frac{dD}{dt} = \frac{2}{\pi D^2} \times 3 \pi = \frac{6}{D^2}$.


    You were given t = 12 minutes because that's the question!!! Find the value of $\displaystyle \frac{dD}{dt}$ when t = 12 minutes!

    So you have to substitute the value of D at t = 12. Clearly $\displaystyle V = (12)(3 \pi) = 36 \pi \, m^3 $ and you know that $\displaystyle V = \frac{\pi}{6} \, D^3$. Therefore D = ....... at t = 12 minutes.

    Therefore $\displaystyle \frac{dD}{dt}$ when t = 12 minutes is equal to ......
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  3. #3
    a.a
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    Thx a million.
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  4. #4
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    Question final ans...

    umm.. for the final answer i got 1/6 m/min for dD/dt, is tha right?
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  5. #5
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    Quote Originally Posted by a.a View Post
    umm.. for the final answer i got 1/6 m/min for dD/dt, is tha right?
    Yes.
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