Question

the surface area of a cylinder is increasing by 2 pi square inches per hour and the height is decreasing by 0.1 inches per hour when the radius is 16 inches and the height is 7 inches. how fast is the radius of the cylinder changing?

Answer 1

The radius of the cylinder is changing at a rate of approximately 0.0344 inches per hour.

Step-by-step explanation:

To find the rate of change of the radius of the cylinder, we can use the formula for the surface area of a cylinder: SA = 2πr(r +h), where SA is the surface area, r is the radius, and h is the height.

Taking the derivative of this formula with respect to time gives us dSA/dt = 2π(2r(dr/dt) + dh/dt).

Given that dSA/dt = 2π square inches per hour, dh/dt = -0.1 inches per hour, r = 16 inches, and h = 7 inches, we can substitute these values into the derivative formula to solve for dr/dt.

dSA/dt = 2π(2(16)(dr/dt) + (-0.1))

= 2π(32(dr/dt) - 0.1) = 2π square inches per hour.

Setting this equation equal to 2π and solving for dr/dt gives us:

32(dr/dt) - 0.1 = 1

32(dr/dt) = 1.1

dr/dt = 1.1/32

dr/dt ≈ 0.0344 inches per hour.

Answer 2

The decreasing part area :0.1×2×Pi×r
＝0.2 Pi r
The increasing part is 2 pi square.
the top and bottom area now: 2pi ×(16＋X)
X means the increasing inches.
Then use the top and bottom part miners the decreasing equals to the the increasing part. Then the answer you can calculate by yourself....