related rates.. weeeee!!!

[maeveoneill]

Okay this is a question from my homework I am stuck on. It is concerning related rates:

A circular wading pool with sides sloping at a slope of 1/10 upward from its centre point is being filled by a hose at a rate of 1/4 L/s.

a) How fast is the depth at the centre increasing when that depth is 15cm *I was able to solve this and got 1/90pi cm/s and that is the right answer*
b) How fast is the top surface area of the water increasing at the same instant?

*I was able to solve part a) and got 1/90pi cm/s which is the right answer. I need help with b. if you could just tell me how to set up the equation that would be enough. Thank you.

 

[maeveoneill]

this is using derivativees.. related rates.

 

[soroban]

Hello, maeveoneill!

A circular wading pool with sides sloping at a slope of 1/10 upward from its centre point
is being filled by a hose at a rate of 1/4 L/s. . <-- irrelevant! (and incorrect)
When the depth is 15cm, the depth at the centre is increasing at 1/90 pi cm/s.
How fast is the top surface area of the water increasing at the same instant (15 cm)?

  *---------------------------*---------------------------*
       *                      |                      *
            *                 |     R           *
                  * - - - - - + - - - - - *
                        *    h|     *         m = 1/10
                              *

The top surface area is a circle of radius \(\displaystyle R:\;\;A\:=\:\pi R^2\)

. . Its rate of change is: .\(\displaystyle \frac{dA}{dt}\,=\,2\pi R\left(\frac{dR}{dt}\right)\) [1]

From the diagram we have: /\(\displaystyle R\,=\,10h\;\;\Rightarrow\;\;\frac{dR}{dt}\,=\,10\left(\frac{dh}{dt}\right)\)

. . When \(\displaystyle h\,=\,15,\;R\,=\,150\) . . . When \(\displaystyle \frac{dh}{dt}\,=\,\frac{\pi}{90},\;\frac{dR}{dt}\,=\,\frac{10\pi}{9}\)


Substitute into [1]: .\(\displaystyle \L\frac{dA}{dt}\:=\:2\pi(150)\left(\frac{10\pi}{9}\right)\:=\;\frac{1000\pi}{3}\text{ cm}^2/\text{s}\)

 

[maeveoneill]

thanks. but when i looked it over i think dr/dt should be pi/9