Connected rates of change

sojeee

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Oct 24, 2017
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The surface area of a closed cylinder is given by A = 2*pi*r^2 + 2*pi*r*h, where h is the height and r is the radius of the base. At the time when the surface area is increasing at the rate of 20*pi cm^2 s^-1, the radius is 4 cm and the height is 1cm and is decreasing at the rate of 2cm s^-1. Find the rate of change of radius at this time.

My working so far:
I know that we need to work out dr/dt
So far I have dA/dt =20pi
Not sure where to go from here.
 

MarkFL

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I would begin with:

\(\displaystyle A=2\pi r^2+2\pi rh\)

Note: all linear measures are in cm and all measures of time are in seconds.

Now, let's implicitly differentiate with respect to time \(t\)...what do you get?
 

MarkFL

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To follow up:

Implicitly differentiating with respect to \(t\), we obtain:

\(\displaystyle \d{A}{t}=2\pi\left(2r\d{r}{t}+\d{r}{t}h+r\d{h}{t}\right)\)

Hence:

\(\displaystyle \d{r}{t}=\frac{\d{A}{t}-2\pi r\d{h}{t}}{2\pi(2r+h)}\)

Plug in the given data:

\(\displaystyle \d{r}{t}=\frac{20\pi\frac{\text{cm}^2}{\text{s}}-2\pi (4\text{ cm})\left(-2\frac{\text{cm}}{\text{s}}\right)}{2\pi(2(4)+1)\text{ cm}}=\frac{18}{9}\,\frac{\text{cm}}{\text{s}}=2\,\frac{\text{cm}}{\text{s}}\)
 
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