1. ## Quick Derivative problem

I totally blanked on this. I need to find the derivative of the volume of a hemispherical balloon. From the picture i can tell that the volume, $v$= $\pi r^2h$ $+$ $\frac{4}{3}\pi (r^3)$. For some reason I totally blanked on how to get the derivative of a sphere and cylnder. This is what i got, but I am not sure it is correct. Could someone double check please? Thanks

$v'$= $(\pi)(2r)(\frac{dr}{dt})h+$ $4\pi(r^2)(2\frac{dr}{dt})$

I think i messed up somewhere but not sure where, just as a clarification, $\frac{dr}{dt}$ is the rate of change of the radius in relation to time.

Thanks for helping!

2. Originally Posted by OnMyWayToBeAMathProffesor
I totally blanked on this. I need to find the derivative of the volume of a hemispherical balloon. From the picture i can tell that the volume, $v$= $\pi r^2h$ $+$ $\frac{4}{3}\pi (r^3)$. For some reason I totally blanked on how to get the derivative of a sphere and cylnder. This is what i got, but I am not sure it is correct. Could someone double check please? Thanks

$v'$= $(\pi)(2r)(\frac{dr}{dt})h+$ $4\pi(r^2)(2\frac{dr}{dt})$

I think i messed up somewhere but not sure where, just as a clarification, $\frac{dr}{dt}$ is the rate of change of the radius in relation to time.

Thanks for helping!
is h constant here? please post the original problem. you may have to write h in terms of r depending on the problem

3. I'm sorry, I should have done that in the first place.

The balloon shown above is in the shape of a cylinder with hemispherical ends of the same radius as that of the cylinder. The balloon i being inflated at the rate of $261\pi$ cubic centimeters per minutes. At the instant the radius of the cylinder is 3 centimeters, the volume of the balloon is $144\pi$ cubic centimeters and the radius of the cylinder is increasing at the rate of 2 centimeters per minute.

(a) at this instant, what is the height of the cylinder?

(b) at this instant, how fast is the height of the cylinder increasing?

so from this problem i got that $\frac{dv}{dt}=261\pi$, $r=3$, $v=144\pi$, $\frac{dr}{dt}=2$

so yes, i think h is a constant in part a and then once h is found, you can find $\frac{dh}{dt}$ for part b, correct?

4. So was my original formula correct then? If h is a constant, then $\frac{dv}{dt}=(\pi)(2r)(\frac{dr}{dt})h+4\pi(r^2)( 2\frac{dr}{dt})$ holds true?

5. i see my mistake, i get it now. nvm.

6. Originally Posted by OnMyWayToBeAMathProffesor
i see my mistake, i get it now. nvm.
okie dokie