Question

The radius of a right circular cylinder is increasing at a rate of 9 inches per minute and the height is decreasing at a rate of 16 inches per minute. What is the rate of change of the surface area when the radius is 16 inches and the height is 29 inches

Answer 1

Answer:

$\frac{dA}{dt}=1840.97 in^{2}/min$      

Step-by-step explanation:

The equation of the surface of the right circular cylinder is:

$A=2\pi rh+2\pi r^{2}=2\pi(rh+r^{2})$

Now, the rate change of this area will be:

Using the change rule

$\frac{dA}{dt}=\frac{\partial A}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial A}{\partial h}\frac{\partial h}{\partial t}$

Where:                          

• $\frac{\partial h}{\partial t}$ is the rate change of height (-16 in/min)
• $\frac{\partial r}{\partial t}$ is the rate change of radius (9 in/min)

 

$\frac{dA}{dt}=\frac{\partial A}{\partial r}9+\frac{\partial A}{\partial h}(-16)$

$\frac{dA}{dt}=2\pi(h+2r)9-2\pi r16=2\pi((h+2r)9-16r)$

Now, r = 16 and h = 29

$\frac{dA}{dt}=2\pi((29+2*16)9-16*16)$

$\frac{dA}{dt}=1840.97 in^{2}/min$      

I hope it helps you!