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...
See This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Already a member? Log in here.
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.