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 that of r.
`(d(r^3))/(dt) = 12(dr)/(dt)`
But we know, `(d(r^3))/(dt) = 3r^2 (dr)/(dt)`
`3r^2 (dr)/(dt) = 12 (dr)/(dt)`
`r^2 = 4`
This gives `r = +-sqrt(4) = +-2`
But we have been told that r is positive, therefore the answer is,
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 :)