# Related rates problem?

If V is the volume of a cube with edge length x and the cube expands as ime passes, find $\frac{dV}{dt}$ in terms of $\frac{dx}{dt}$.

Help would be greatly appreciated :] I don't even have a clue where to start...

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Try the chain rule. – copper.hat Mar 7 '13 at 18:04

Hint: write the equation of the volume $V$ as a function of the side $x$. Take the derivative of both sides with respect to $t$

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To put this in more concrete terms, do exactly as others have suggested and start with a relation between the Volume of the cube and the length of its side. Basic geometry rules tell us that

V = x^3

Now, take the derivative of both sides with respect to time (t). You find that:

$$dV/dt = d/dt[x^3]$$

Note that when we attempt to differentiat the right side, we are differentiating with respect to t, but there is no t on the right side of the equation. This is an indicator that we have to use the Chain Rule. Specifically,

$$d/dt = dV/dx * dx/dt$$

Don't let the notation confuse you, because we actually know what all these terms are.

$$dV/dx$$ is simply the rate of change of the volume with respect to the rate of change of the length of its side. In other words,

$$dV/dx = d/dx[x^3] = 3x^2$$

And, as for the remaining

$$dx/dt,$$

this value is the rate of change of the length of a side with respect to time. In this particular example, we are not given what dx/dt is, but for the purposes of this question, it is unnecessary to know the exact value of this rate of change.

Now, bringing the problem back full circle, we started with the equation for Volume:

$$V = x^3$$

We differentiated both sides to obtain:

$$dV/dt = d/dt[x^3] = 3x^2 dx/dt$$

And thus, we have provided a relation between dV/dt and dx/dt

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