The radius of a right circular cylinder is increasing

Ответ:

cxttiemsp021

26.01.2022

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.

$\frac{dr}{dt}=9.5\ inches/min$

and

$\frac{dh}{dt}=-11.5 \ inches/min$

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

$\frac{dA}{dt}=2\pi(2r\frac{dr}{dt}+r\frac{dh}{dt}+h\frac{dr}{dt})$

Now plug the value of $\frac{dr}{dt}$ and $\frac{dh}{dt}$ .

$\frac{dA}{dt}=2\pi(2r\times 9.5+r\times (-11.5)+h\times 9.5)$

$=2\pi(19r-11.5r+9.5h)$

$=2\pi(7.5r+9.5h)$

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

$\frac{dA}{dt}|_{r=16.5, h=31 }=2\pi (7.5 \times 16.5+9.5\times 31)$

=2,629 inches per minute

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

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Ответ:

jazz9876

14.11.2020

Mathematics

-58185.70868

Step-by-step explanation:

Ello, so basically yh

the way u work this out is by: go on google nd ask this same maths question

and then the anwer will come.

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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?

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