The radius of a right circular cylinder is increasing at a rate of 9.5 inches per minute and the height is decreasing at a rate of 11.5 inches per minute. What is the rate of change of the surface area when the radius is 16.5 inches and the height is 31 inches?

The rate of change of the surface area is 2,629 inches per minute.

Step-by-step explanation:

Given that,

The height is decreasing at a rate of 11.5 inches per minute and the radius of a right circular cylinder is increasing at a rate of 9.5 inches per minute.

and

r = radius of the right circular cylinder

h= height of the right circular cylinder

The surface area of the right circular cylinder= 2π(r²+rh)

A=2π(r²+rh)

Differentiating with respect to t

Now plug the value of   and .

The rate of change of the surface area when the height is 31 inches and radius is 16.5 inches.

r = 16.5 inches, h = 31 inches

                      =2,629 inches per minute

The rate of change of the surface area is 2,629 inches per minute.

faster, im not sure why but i know it is faster

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)