The radius of the base of a cylinder is decreasing at a rate of 9 millimeters per hour and the height of the cylinder is increasing at a rate of 2 millimeters per hour. At a certain instant, the base radius is 8 millimeters and the height is 3 millimeters. What is the rate of change of the surface area of the cylinder at that instant in square millimeters per hour? a) 310pi
b) 155pi
c) -155pi
d) -310pi

d) -310pi

Step-by-step explanation:

Instantaneous Rate of Change

Is the change in the rate of change of a function at a particular instant. It's the same as the derivative value at a specific point.

The surface area of a cylinder of radius r and height h is:

We need to calculate the rate of change of the surface area of the cylinder at a specific moment where:

The radius is r=8 mm

The height is h=3 mm

The radius changes at r'=-9 mm/hr

The height changes at h'=+2 mm/hr

Find the derivative of A with respect to time:

Substituting the values:

Calculating:

d) -310pi

10 down (bang1)

5 up

5down (2)

2.5 up

2.5 down (3)

1.25 up

1.25 down (4)

now add them up : )

hoped i lol