122k views
1 vote
The radius of a right circular cylinder is increasing at the rate of 8 in./sec, while the height is decreasing at the rate of 10 in./sec. At what rate is the volume of the cylinder changing when the radius is 11 in. and the height is 10 in.?

a) 660π in³/sec
b) 580π in³/sec
c) 700π in³/sec
d) 500π in³/sec

1 Answer

3 votes

Final answer:

The rate at which the volume of a right circular cylinder is changing, with a radius increasing at 8 in./sec and height decreasing at 10 in./sec when the radius is 11 in. and height is 10 in., is calculated to be 550π in.³/sec using the differentiation of the volume formula.

Step-by-step explanation:

The student is asking about the rate of change in volume of a right circular cylinder when the radius and height are changing at given rates. To find this rate of change, we use the formula for the volume of a cylinder, which is V = πr²h, and apply the related rates differentiation technique with respect to time (t).

Given:
The radius (r) is increasing at 8 in./sec (dr/dt = +8),
The height (h) is decreasing at 10 in./sec (dh/dt = -10),
The radius at a particular instance (r) = 11 in,
The height at that instance (h) = 10 in.

Using these values in the differentiation of the volume formula with respect to time, we can calculate the rate of change of the volume (dV/dt). Upon calculation, we find:

dV/dt = π(2rh dr/dt + r² dh/dt)
dV/dt = π(2 * 11 * 10 * 8 + 11² * -10)
dV/dt = π(1760 - 1210)
dV/dt = π * 550
dV/dt = 550π in.³/sec

Therefore, the answer is 550π in.³/sec, which does not match any of the provided options (a-d).

User Smocking
by
6.9k points