1 Attachment(s)

Water draining out of spherical bowl

Hi,

Its been 3 years since I've posted here. And I find myself trying to review the basics of calc before moving on to bigger things.

I found this problem in my old math book:

Attachment 29083

Sorry I couldn't rotate it I'm in a rush I tried to look online for some online image utility... oh well.

So, water is flowing out of a half hemisphere blown (the bottom half, lol) at a constant rate of 6 m^3/min. Radius of sphere is R (13 meters), radius of water level is r. height of water is y Volume is V. formula of Volume at height y is:

$\displaystyle V = (pi/3)y^2(3R-y)$

Find an equation for r and find dr/dy at y=8 meters

I assume, that $\displaystyle r=R*sin[(pi/2)(y/R)]$, baseless assumption I admit, so that $\displaystyle dr/dy=cos[(pi/2)(y/r)]$ using chain rule: $\displaystyle f(x)=Rsin[g(x)]$ $\displaystyle g(x)=y/R$

But that doesn't include the volume in any way. If correct it gives me a relationship between r and y (radius of water level vs depth)

I'm also curious if I can derive an actual formula for r with what is given.

This isn't homework I'm an adult returning to higher education and I'm doing a crash course math review on my own before some harder math classes the next semesters coming up. My algebra needs work.

Re: Water draining out of spherical bowl

I don't see how you get $\displaystyle r=R \sin(\frac {\pi} 2 \frac y R)$ - how did you get that? From Pythagoras its apparent that $\displaystyle R^2 = r^2+ (R-y)^2$, so $\displaystyle r = \sqrt {(R^2 - (R-y)^2}$.

Question C asks for the rate of change of r, or dr/dt, not dr/dy. You can get that using the chain rule:

$\displaystyle \frac {dr}{dt} = \frac {dr}{dy} \ \frac {dy} {dV} \ \frac {dV}{dt}$

Re: Water draining out of spherical bowl

Wow that looks obvious. Thank you.