Answer:
[tex]\frac{dr}{dt} = 0.14 in/s[/tex]
Explanation:
As the volume of the cylinder is constant here so we can say that its rate of change in volume must be zero
so here we can say
[tex]\frac{dV}{dt} = 0[/tex]
now we have
[tex]V = \pi r^2 h[/tex]
now find its rate of change in volume with respect to time
[tex]\frac{dV}{dt} = 2\pi rh\frac{dr}{dt} + \pi r^2\frac{dh}{dt}[/tex]
now we know that
[tex]\frac{dV}{dt} = 0 = \pi r(2h \frac{dr}{dt} + r\frac{dh}{dt})[/tex]
given that
h = 5 inch
r = 2 inch
[tex]\frac{dh}{dt} = - 0.7 in/s[/tex]
now we have
[tex]0 = 2(5) \frac{dr}{dt} + 2(-0.7)[/tex]
[tex]\frac{dr}{dt} = 0.14 in/s[/tex]
