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!