Related Rates: how fast is water height incr. inside tank?

[Blkmage8]

First of all, I'm not actually taking a calculus class right now but I'm sort of teaching myself. That said, I'm terribly confused on related rates. I was able to a few problems but I'm having a lot of trouble still.

*A cylindrical tank with a radius of 6 meters is filling with fluid at a rate of 108pi m^3/sec. How fast is the height increasing?

I know I need to find dh/dt and I need an equation that relates volume to height. I used:

V = (pi)(r^2)(h), so then

dV/dt = pi [(r^2 dh/dt) + (h * 2r dr/dt)]

I'm a little unsure of if I did that right but if I did, then regardless of substitution for the radius, I still can't solve for dh/dt without knowing the h or dr/dt. I can't figure out a way to find the height or eliminate it from the problem. I couldn't figure out a way to solve for dr/dt using a different equation(s) either. Any help would be appreciated.

Also, if any has any general tips for relating the functions that'd be great

 

[galactus]

Re: Related Rates: Word Problems

The radius of a cylinder remains constant, therefore, dr/dt=0

 

[Blkmage8]

Oh I see now. I just got into the habit of monotonously just solving for the rates I wasn't really paying attention to what dr/dt really represented. So I got:

dV/dt = pi [(r^2 dh/dt} + (h * 2r dr/dt)]
108pi = pi [(36 dh/dt)]
3 m/sec ?

If you're willing to help with one more: (I'm really starting to get the hang of it now I think)

The sides of an equilateral triangle are increasing at the rate of 27 in/sec. How fast is the triangle's area increasing when the sides of the triangle are each 18 inches long?

I know I need to find dA/dt but I'm entirely sure what 27 in/sec is representing. I mean, I couldn't find an equation relating sides to area.

A = 1/2(bh)
dA/dt = 1/2 [(b dh/dt) + (h db/dt)]

Since it's equilateral then the base, b, is given by b=s=18 inches. But then that makes things more complicated
I'm just really bad at setting up these equations and area was all I could go by. Since it's asking how area is changing then I'm assuming that dh/dt and db/dt are probably changing as wlel so they can't be 0.

 

[galactus]

The area of an equilateral triangle is given by \(\displaystyle A=\frac{\sqrt{3}}{4}s^{2}\). Where s is the length of a side

So, you have \(\displaystyle \frac{ds}{dt}=27\)

\(\displaystyle \frac{dA}{dt}=\frac{\sqrt{3}}{2}s\frac{ds}{st}\)

Finish it?.