Related rates problem

**Benji** · May 8th 2012, 12:32 PM

Hi

I've got two related rates problem that I didn't manage to solve.

The first one is:
The area of a square is decreasing at 2 ft^2/s. How fast is the side length changing when it it 8 ft?

The second one is:
The area of a circle is decreasing at a rate of 2cm^2/min. How fast is the radius of the circle changing when the area is 100 cm^2?

I guess the problems are of the same type. Thankful for all the help I can get

**skeeter** · May 8th 2012, 02:52 PM

$\frac{d}{dt}\left(A = s^2\right)$

$\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}$

sub in your given values and determine the value of $\frac{ds}{dt}$

now do the same with the circle problem

**Benji** · May 9th 2012, 04:54 AM

Thanks for your help

I managed to solve the first problem but not the second one. There is a difference between them. In the first they are asking for the side length over time when the side is 8ft, while they in the second one are asking for the radius over time when the area is 100 cm^2.

**skeeter** · May 9th 2012, 05:29 AM

Originally Posted by Benji

Thanks for your help

I managed to solve the first problem but not the second one. There is a difference between them. In the first they are asking for the side length over time when the side is 8ft, while they in the second one are asking for the radius over time when the area is 100 cm^2.

come on, Benji ... you have to think a little bit.

$\frac{d}{dt}\left(A=\pi r^2\right)$

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

you were given the rate of change of the area, $\frac{dA}{dt}$

you were also given the area ... can you not calculate the value of the radius when $A = 100 \, cm^2$ ?

sub in those values and determine the value of $\frac{dr}{dt}$

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