# Related Rates Questions

• Mar 24th 2008, 01:38 AM
a.a
Related Rates Questions
A spherical ballon is being inflated at a rate of 3 pi m^3/min. Twelve minutes after inflation first begins, what is the rate of increase of the diameter?

i solved for dr/dt and then adjuster for diameter, but i dont understand why we were given t = 12 min...

thx for any help u can giv :)
• Mar 24th 2008, 02:57 AM
mr fantastic
Quote:

Originally Posted by a.a
A spherical ballon is being inflated at a rate of 3 pi m^3/min. Twelve minutes after inflation first begins, what is the rate of increase of the diameter?

i solved for dr/dt and then adjuster for diameter, but i dont understand why we were given t = 12 min...

thx for any help u can giv :)

Let diameter be D. Then $\frac{dD}{dt} = \frac{dD}{dV} \times \frac{dV}{dt}$.

Given: $\frac{dV}{dt} = 3 \pi \, m^3/min$.

Common knowledge: $V = \frac{4}{3} \, \pi r^3 = \frac{\pi}{6} \, D^3$. Therefore $\frac{dV}{dD} = \frac{\pi}{2} \, D^2$.

Therefore: $\frac{dD}{dt} = \frac{2}{\pi D^2} \times 3 \pi = \frac{6}{D^2}$.

You were given t = 12 minutes because that's the question!!! Find the value of $\frac{dD}{dt}$ when t = 12 minutes!

So you have to substitute the value of D at t = 12. Clearly $V = (12)(3 \pi) = 36 \pi \, m^3$ and you know that $V = \frac{\pi}{6} \, D^3$. Therefore D = ....... at t = 12 minutes.

Therefore $\frac{dD}{dt}$ when t = 12 minutes is equal to ......
• Mar 24th 2008, 10:44 AM
a.a
Thx a million. :)
• Mar 24th 2008, 03:32 PM
a.a
final ans...
umm.. for the final answer i got 1/6 m/min for dD/dt, is tha right?
• Mar 24th 2008, 07:36 PM
mr fantastic
Quote:

Originally Posted by a.a
umm.. for the final answer i got 1/6 m/min for dD/dt, is tha right?

Yes.