Respuesta :
Answer:
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.
[tex]\frac{dr}{dt}=9.5\ inches/min[/tex]
and
[tex]\frac{dh}{dt}=-11.5 \ inches/min[/tex]
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
[tex]\frac{dA}{dt}=2\pi(2r\frac{dr}{dt}+r\frac{dh}{dt}+h\frac{dr}{dt})[/tex]
Now plug the value of [tex]\frac{dr}{dt}[/tex] and [tex]\frac{dh}{dt}[/tex].
[tex]\frac{dA}{dt}=2\pi(2r\times 9.5+r\times (-11.5)+h\times 9.5)[/tex]
[tex]=2\pi(19r-11.5r+9.5h)[/tex]
[tex]=2\pi(7.5r+9.5h)[/tex]
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
[tex]\frac{dA}{dt}|_{r=16.5, h=31 }=2\pi (7.5 \times 16.5+9.5\times 31)[/tex]
=2,629 inches per minute
The rate of change of the surface area is 2,629 inches per minute.
