What rate is the diameter increasing when the diameter is 12m?

Question

a spherical balloon is being inflated so that its volume is increasing at the rate of 5 m3/min. at what rate is the diameter increasing when the diameter is 12m?

(this question is about calculus)

(topic: Related Rates)

(pls. help me answering this question)

(pls. show your solutions)

(thanks)

Answer 1

volume of circle=(4/3)pi r^3..when the diameter is 12m, volume of circle=(4/3)pi 6^3=905m^3...divide by the rate, 905/5=181min...so the rate is 905/181=5m^3/min..which is what it is in the begining...a good way to check...done...

Answer 2

Ok, here's how you do it.

First you get the volume of a sphere. You know that the volume will be:

V=4pir^3/3

but since you want to know how to get the volume in terms of the diameter, then you know that r = d/2 then

V =4*pi*d^3/(3*8) = pi*d^3/6

now all you have to do is to derive V and d like this

dV = 3*pi*d^2dd/6

and multiply it by dt/dt and you get

dV/dt = 3*pi*d^2dd/(6dt)

let's say dV/dt = Vv and dd/dt = Vd. These are the speeds you want to know so

Vv = 3*pi*d^2*Vd/6

now you solve for Vd

Vd = 2*Vv/(pi*d^2)

and substitute

Vd = 2*(5m^3/min)/(3.1416*(12m)^2) = 22.1 mm/min

hope this helps.