the radius of a cylinder is increasion at a rate of 1 meter per hour, and the height of the cylinder is decreasing at a rate of 4 meters per hour. at a certain instant, the base radius is 5 meters and the height is 8 meters. what is the rate of change of the volume of the cylinder at the instant? (Note the formula for volume of a cylinder is V r h)

Answer:The volume is decreasing at a rate 20 cubic meters per hour.Step-by-step explanation:We are given the following in the question:Rate of change of radius = [tex]\displaystyle\frac{dr}{dt} = 1 \text{ meter per hour}[/tex]Rate of change of height = [tex]\displaystyle\frac{dr}{dt} = -4 \text{ meter per hour}[/tex]At an instant,radius, r = 5 metersHeight, h = 8 metersVolume of cylinder , V= [tex]\pi r^2 h[/tex]where r is the radius of cylinder and h is the height of cylinder.Rate of change of volume of cylinder =[tex]\displaystyle\frac{dV}{dt} = \frac{d(\pi r^2 h)}{dt} = \pi\Big(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\Big)[/tex]Putting the value, we get,[tex]\displaystyle\frac{dV}{dt} = \pi\Big(2(5)(1)(8) + (5)^2(-4)\Big) = -20\text{ cubic meters per hour}[/tex]Thus, the volume is decreasing at a rate 20 cubic meters per hour.