Answer:
The volume of the cylinder is decreasing at a rate of 83629.2 m³/h.
Step-by-step explanation:
The volume of a is given by:
[tex] V = \pi r^{2}h [/tex]
Where:
r: is the radius = 55 m
h: is the height = 88 m
We can express the rate of change of the volume of the cylinder as follows:
[tex] \frac{dV}{dt} = \pi[h\frac{2rdr}{dt} + r^{2}\frac{dh}{dt}] [/tex]
If dr/dt = 11 m/h and dh/dt = -44m/h, we have:
[tex] \frac{dV}{dt} = \pi[88 m*2*55 m*11 m/h + (55 m)^{2}*(-44 m/h)] = -83629.2 m^{3}/h [/tex]
Therefore, the volume of the cylinder is decreasing at a rate of 83629.2 m³/h.
I hope it helps you!
