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Thread: Cylinder inscribed in sphere, diff word qn

  1. #1
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    Cylinder inscribed in sphere, diff word qn

    A circular cylinder is inscribed in a fixed sphere with radius 18cm such that all the points on the circ of both cylindrical ends are always lying on the spherical surface. Given that the cylinder has a ht of h cm and cylindrical base radius of r cm, show that $\displaystyle h^2+4r^2=1296$ (done)

    Initially the ht of the cylinder is 24cm and is decreasing at constant rate of 0.25cm/s
    (ii) show that rate at which radius is changing at this instant is $\displaystyle \frac {\sqrt5} {20} $ cm/s (done)
    (iii) Find the time taken for the volume of the cylinder to reach its maximum value. (you need not verify that the volume is maximum) Ans:12.9s


    ok part (iii) is the one i need help in. highlight beside it for ans.

    i know that max value for volume means $\displaystyle \frac {dv} {dr}=0$ but then i'll just get r=0 when i sub in h=24.
    then if i use the h and r eqn, i just get r=0 again.
    in truth i'm clueless how to continue... is max volume just taking the given values of h and r, then using that divide by $\displaystyle \frac {dv} {dt} $? if so then i couldn't find that either :/ maybe i'm just thinking too much x.x
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  2. #2
    Super Member malaygoel's Avatar
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    Either you are thinking too much or you are commiting a mistake in solving equation.
    Volume,V=$\displaystyle \pi r^2h$
    $\displaystyle =\pi r^2\sqrt{1296-4r^2}$...now you will not get r=0 on differentiating.

    Infavt, if you use $\displaystyle \frac{dv}{dh}$ for determining maximum volume, steps will be easier.
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  3. #3
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    ty
    yep it was some careless mistake (d'oh!)
    haha ok got r now though it's a strange decimal, how do i find the time taken in the first place? ><
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  4. #4
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    You are told that the initial height of the cylinder was 24 cm and you know, from (a), that [tex]h^2+ 4r^2= 1296[tex] so the initial radius is given by $\displaystyle 24^2+ 4r^2= 1296$. [tex]4r^2= 1296- 576= 720[tex]. $\displaystyle r^2= 180$, and $\displaystyle r= \sqrt{180}= \sqrt{9*4*5}= 6\sqrt{5}$. Subtract that from your value of r that gives a maximum volume. Since you know that r "is decreasing at constant rate of 0.25cm/s" dividing that difference by .25 gives the time in seconds.

    (I would be inclined to leave the value of r that gives maximum volume in terms of radicals rather than as "a strange decimal".)
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  5. #5
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    FINALLY got the ans! ty v much
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