Answer:
[tex]\frac{dV}{dt}=-450\pi ft^3/sec[/tex]
Step-by-step explanation:
Let's assume
radius of cylinder as 'r'
height of cylinder as 'h'
so, we have
[tex]\frac{dr}{dt} =3ft/sec[/tex]
[tex]\frac{dh}{dt} =-6ft/sec[/tex]
r=15ft
h=10ft
now, we can use volume of cylinder formula
[tex]V=\pi r^2h[/tex]
we can find derivative with respect to t
[tex]\frac{dV}{dt}=2\pi r h\times\frac{dr}{dt} +\pi r^2\times \frac{dh}{dt}[/tex]
now, we can plug values
[tex]\frac{dV}{dt}=2\pi (15)(10)\times3 +\pi (15)^2\times -6[/tex]
now, we can simplify it
[tex]\frac{dV}{dt}=-450\pi ft^3/sec[/tex]
