# If r is positive and increasing, for what value of r is the rate of increase of r^3 twelve times that of r?

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Let's assume that r is a function of t, so `r^3` is also a function of t.

The rate of increase of r is given by, `(dr)/(dt)` wrt t.

The rate of increase of `r^3` is `(d(r^3))/(dt)` .

But we know that rate of increase of `r^3` is 12 times that of r.

`(d(r^3))/(dt) = 12(dr)/(dt)`

But we know, `(d(r^3))/(dt) = 3r^2 (dr)/(dt)`

Therefore,

`3r^2 (dr)/(dt) = 12 (dr)/(dt)`

This gives,

`3r^2=12`

`r^2 = 4`

This gives `r = +-sqrt(4) = +-2`

**But we have been told that r is positive, therefore the answer is,**

**r =2.**

I'm assuming you know differentiation here, since you've tagged this with calculus. That makes things easier :)

First, what's the *rate *of* change* of a function with respect to something? It's *defined* as the derivative with respect to that something. So:

- the rate of change of r is just the derivative of r with respect to r.

(d/dr) r = 1*r^0 = 1 (power in front, take one away from the power).

- the rate of change of r^3 is a little more interesting:

(d/dr) r^3 = 3*r^2

So the rate of change of r depends on r squared. That gets fast!

We're told that r is increasing: this means that the derivative is *positive* (what would a negative derivative mean?).

We're implicitly given the rate of change of r^3 in terms of the rate of change of r: (d/dr)r^3 = 12*(d/dr) r.

But we know what those derivatives are:

3r^2 = 12.

Solve this, and you have your answer :)