Answer:
The volume is decreasing at a rate 20π cubic meter per hour.
Step-by-step explanation:
We are given the following in the question:
The radius is increasing at a rate 1 meter per hour.
[tex]\dfrac{dr}{dt} = 1\text{ meter per hour}[/tex]
The height is decreasing at a rate 4 meter per hour
[tex]\dfrac{dh}{dt} = -4\text{ meter per hour}[/tex]
At an instant time t,
r = 5 meter
h = 8 meters
Volume of cylinder =
[tex]\pi r^2 h[/tex]
where r is the radius and h is the height of the cylinder.
Rate of change of volume is given by:
[tex]\dfrac{dV}{dt} = \dfrac{d(\pi r^2 h)}{dt}\\\\\dfrac{dV}{dt} = 2\pi rh\dfrac{dr}{dt} + \pi r^2\dfrac{dh}{dt}[/tex]
Putting all the values we get,
[tex]\dfrac{dV}{dt} = 2\pi (5)(8)(1) + \pi (5)^2(-4)\\\\\dfrac{dV}{dt} = 80\pi - 100\pi = -20\pi \approx -62.8[/tex]
Thus, the volume is decreasing at a rate 20π cubic meter per hour or 62.8 cubic meter per hour.
