Thread: 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

and the radius of the inner circle is increasing at a rate of . Find the rate of change of the area enclosed by the two circles.











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

I get

the correct answer is

Can someone tell me where I am going wrong? Thanks.

Hey,

First write down the equation for the enclosed area:



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

dA/dt = *(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)

Originally Posted by arash

Hey,

First write down the equation for the enclosed area:



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

dA/dt = *(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

where did the 2pi come from?

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)

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,

thanks for your help.