Re: Related rates problem

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

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

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

now do the same with the circle problem

Re: Related rates problem

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.

Re: Related rates problem

Quote:

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.

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

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

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

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

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