Let's organize our given information.
[tex]\frac{dh}{dt}=-1[/tex]
[tex]\frac{dV}{dt}=-2341[/tex]
[tex]\frac{dr}{dt}=?, h=1010, V=577[/tex]
[tex]V=\pi r^2h[/tex]
What I'd like to do is solve for [tex]r[/tex] first, it'll help later.
[tex]577=\pi r^2(1010)[/tex]
[tex]r=\sqrt{\frac{577}{1010\pi}}[/tex]
[tex]r=0.426[/tex]
Now, we can differentiate our formula for the volume of a cylinder to find the rate of change of the radius.
[tex]V=\pi r^2h[/tex]
Use the product rule of differentiation.
[tex]\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2\frac{dh}{dt}[/tex]
Plug in the information we are given.
[tex]-2341=2\pi (0.426)(1010)\frac{dr}{dt}+\pi (0.426)^2(-1)[/tex]
Solve for [tex]\frac{dr}{dt}[/tex].
[tex]-2341=860.52\pi\frac{dr}{dt}+(0.426)^2\pi[/tex]
[tex]-2341.57=860.52\pi\frac{dr}{dt}[/tex]
[tex]\frac{dr}{dt}=-0.866[/tex]
The rate of change of the radius is -0.866 centimeters per minute.
Hope this helps!
