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

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...

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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.

Approved by eNotes Editorial Team