Question

The radius of a right circular cylinder is increasing at the rate of 3 ft/sec, while the height is decreasing at the rate of 6 ft/sec. At what rate is the volume of the cylinder changing when the radius is 15 ft and the height is 10 ft? Remember to use the product rule when you find the expression for dV/dt.
30 ft3/sec
−900 ft3/sec
−450π ft3/sec
−900π ft3/sec

Answer 1

Answer:

$\frac{dV}{dt}=-450\pi ft^3/sec$

Step-by-step explanation:

Let's assume

radius of cylinder as 'r'

height of cylinder as 'h'

so, we have

$\frac{dr}{dt} =3ft/sec$

$\frac{dh}{dt} =-6ft/sec$

r=15ft

h=10ft

now, we can use volume of cylinder formula

$V=\pi r^2h$

we can find derivative with respect to t

$\frac{dV}{dt}=2\pi r h\times\frac{dr}{dt} +\pi r^2\times \frac{dh}{dt}$

now, we can plug values

$\frac{dV}{dt}=2\pi (15)(10)\times3 +\pi (15)^2\times -6$

now, we can simplify it

$\frac{dV}{dt}=-450\pi ft^3/sec$