# Rate of change problem.

• May 19th 2010, 10:24 AM
Tweety
Rate of change problem.
At a given instant, the radii of two concentric circles aree 8cm and 12cm. The radius of the outer circle is increasing at a rate of $1cms^{-1}$

and the radius of the inner circle is increasing at a rate of $2cms^{-1}$. Find the rate of change of the area enclosed by the two circles.

$r = 12, \frac{dr}{dt} = 1$

$r = 8 \frac{dr}{dt} = 2$

$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$

$A = \pi r^{2}$

$\frac{dA}{dr} = 2\pi r$

is this all correct? So I would just add the two da/dt for each circle?

I get $56\pi$

the correct answer is $8\pi$

Can someone tell me where I am going wrong? Thanks.
• May 19th 2010, 12:01 PM
arash
Hey,

First write down the equation for the enclosed area:

$A=pi*(r_1^2-r_2^2)$

Then take derivative with respect to t, you'll get:

dA/dt = $2*pi$*(r1*dr1/dt-r2*dr2/dt)

Then sub in the values to get dA/dt:

dA/dt = 2*pi*(12*1 - 8*2) = -8*pi

So the rate of change is 8*pi (negative because the enclosed area is decreasing)
• May 19th 2010, 12:14 PM
Tweety
Quote:

Originally Posted by arash
Hey,

First write down the equation for the enclosed area:

$A=pi*(r_1^2-r_2^2)$

Then take derivative with respect to t, you'll get:

dA/dt = $2*pi$*(r1*dr1/dt-r2*dr2/dt)

Then sub in the values to get dA/dt:

dA/dt = 2*pi*(12*1 - 8*2) = -8*pi

So the rate of change is 8*pi (negative because the enclosed area is decreasing)

so thats $2\pi ( r^{1} \frac{dr^{1}}{dt} - r^{2} \frac{dr^{2}}{dt} )$

where did the 2pi come from?
• May 19th 2010, 12:20 PM
arash
2*Pi comes from differentiating pi*(r1^2-r2^2).

A=pi*(r1^2-r2^2) = pi*r1^2 - pi*r2^2

dA/dt = 2*pi*r1*dr1/dt - 2*pi*r2*dr2/dt = 2*pi*(r1*dr1/dr - r2*dr2/dt)
• May 19th 2010, 12:31 PM
Tweety
Quote:

Originally Posted by arash
2*Pi comes from differentiating pi*(r1^2-r2^2).

A=pi*(r1^2-r2^2) = pi*r1^2 - pi*r2^2

dA/dt = 2*pi*r1*dr1/dt - 2*pi*r2*dr2/dt = 2*pi*(r1*dr1/dr - r2*dr2/dt)

Yes sorry I realised after I had posted,