Finding the Rate of Change of Water in a Trough

[footprints]

Imagine a trough filled with water (I can't put up a picture).
The water in the tank at time t seconds is given by V = 12x^2. Given that water is flowing into the trough at the rate of 60 cm^3/s, find the rate at which x is increasing when x = 10.

\frac{dV}{dx} = 24x
\frac{d?}{ds} = 60 cm^3/s
Then what do I do?

 

[dextercioby]

Who's "x" and what does your second equation stand for...?

Daniel.

 

[footprints]

If it shouldn't be x, can you tell me what it should be? I don't know what to do for the second equation. I just know its \frac{d\text{ something}}{dx} = 60 cm^3/s. Again, if I'm wrong please tell me what it should be.

 

[dextercioby]

I don't know about "x",what is its point...But:
\frac{dV}{dt}=24x \frac{dx}{dt}

,uding the chain rule...Now plug in the values and solve for the rate of "x"...

Daniel.

 

[footprints]

Oh now I know why you asked me about the second equation. I always get mixed up and put ds instead of dt .
So I got 1440x. I know that's not the final answer because there's still the x = 10 part. And I'm confused on that part.

 

[dextercioby]

Well,\frac{dV}{dt}=60cm^{3}s^{-1}...x=10,you must find \frac{dx}{dt}...

Daniel.

 

[footprints]

I thought \frac{dx}{dt} = 60 cm^3/s and I've to find \frac{dV}{dt} which is 1440x?

 

[dextercioby]

Nope,it's the other way aroung,the volume is increasing at the rate given in the problem...

Daniel.

 

[HallsofIvy]

The point is that
\frac{dV}{dt}= \frac{dV}{dx}\frac{dx}{dt}
(the chain rule).

You are given \frac{dV}{dt} and you can (and did) calculate \frac{dV}{dx}. Put them into that equation and solve for \frac{dx}{dt}.

 

[footprints]

Oh... Got it! Thank you!