Related Rates - Water poured in a bowl

[danago]

The volume of a cap, of depth h cm, cut from a sphere of radius a cm (a>h) is given by $$V = {\textstyle{1 \over 3}}\pi h^2 (3a - h)$$
. A bowl is in the shape of a cap of depth a/2 cm cut from a spherical shell of radius a cm. Water is being poured into the bowl at a constant rate of $$\frac{{\pi a^3 }}{{24}}$$
. Show that, when the depth of the water in the bowl is x cm, $$24\frac{{dx}}{{dt}} = \frac{{a^3 }}{{x(2a - x)}}$$
. Hence show that, if the bowl is initially empty, it will take 5 seconds to fill up.

I managed to do the first part of it:
$$\begin{array}{c}<br /> V = {\textstyle{1 \over 3}}\pi x^2 (3a - x) \\ <br /> = \pi ax^2 - {\textstyle{1 \over 3}}\pi x^3 \\ <br /> \frac{{dV}}{{dt}} = 2\pi ax\frac{{dx}}{{dt}} - \pi x^2 \frac{{dx}}{{dt}} \\ <br /> = \frac{{dx}}{{dt}}(2\pi ax - \pi x^2 ) \\ <br /> \frac{{dx}}{{dt}} = \frac{{\frac{{dV}}{{dt}}}}{{(2\pi ax - \pi x^2 )}} \\ <br /> = \frac{{\pi a^3 }}{{24(2\pi ax - \pi x^2 )}} \\ <br /> = \frac{{a^3 }}{{24(2ax - x^2 )}} \\ <br /> 24\frac{{dx}}{{dt}} = \frac{{a^3 }}{{x(2a - x)}} \\ <br /> \end{array}$$

But with the second part, it says "Hence, show that if the bowl...". What I am not getting is how to use the part above to show that it takes 5 seconds?

I managed to do it like this: (i think it is correct, anyway)


The maximum volume the bowl can hold is given by
$$\begin{array}{c}<br /> <br /> V = {\textstyle{1 \over 3}}\pi h^2 (3a - h) \\ <br /> = {\textstyle{1 \over 3}}\pi ({\textstyle{a \over 2}})^2 (3a - {\textstyle{a \over 2}}) \\ <br /> = \frac{{\pi a^2 (3a - 0.5a)}}{{12}} \\ <br /> = \frac{{5\pi a^3 }}{{24}} \\$$

The bowl is full at time T
$$\begin{array}{l}<br /> \int_0^T {\frac{{\pi a^3 }}{{24}}} dt = \frac{{5\pi a^3 }}{{24}} \\ <br /> \left[ {\frac{{\pi a^3 }}{{24}}t} \right]_0^T = \frac{{5\pi a^3 }}{{24}} \\ <br /> \frac{{\pi a^3 }}{{24}}T = \frac{{5\pi a^3 }}{{24}} \\ <br /> T = 5 \\ <br /> \end{array}$$


But i don't know how to do it using the fact that $$24\frac{{dx}}{{dt}} = \frac{{a^3 }}{{x(2a - x)}}$$.

Any help greatly appreciated.

Thanks,
Dan.

 

[chanvincent]

$$24\frac{{dx}}{{dt}} = \frac{{a^3 }}{{x(2a - x)}}$$

after some simple manipulation, this will become:

$$\frac{24x(2a-x)}{a^3} dx = dt$$

then integrate over both size...

$$\int_0^{a/2}\frac{24x(2a-x)}{a^3} dx = \int_0^Tdt$$

solve for T and you will get the desired answer...

 

[danago]

Ah ok, thanks for that