Answer:
[tex]\frac{3}{25\pi}[/tex] meters per minute
Step-by-step explanation:
Given:
[tex]V=\pi r^2h[/tex]
[tex]r=5[/tex]
[tex]\frac{dV}{dt}=3[/tex]
[tex]\frac{dh}{dt}=?[/tex]
Substitute r=5:
[tex]V=\pi r^2h[/tex]
[tex]V=\pi(5)^2h[/tex]
[tex]V=25\pi h[/tex]
Differentiate both sides:
[tex]\frac{d}{dt}V=\frac{d}{dt}(25\pi h)[/tex]
[tex]\frac{dV}{dt}=25\pi\frac{dh}{dt}[/tex]
Solve for dh/dt:
[tex]3=25\pi\frac{dh}{dt}[/tex]
[tex]\frac{3}{25\pi}=\frac{dh}{dt}[/tex]
Therefore, the height of the water is increasing at a rate of [tex]\frac{3}{25\pi}[/tex] meters per minute.
