The Student Room Group

Rates of Change -C3

hey, please help me to understand/ answer if you can :smile:

" A baloon is filled with air at a constant rate of 80cm^3.
Assuming the baloon is spherical as it is filled, find the rate at which the radius is increasing at the instant when the radius is 6cm"



My (probably useless) workings so far:


dv/dt = 80 .

v= 4/3 pi r^3

dv/dr=4pi r^2

merciverymuchos x
Reply 1
I think you need to find dr/dt ?
Reply 2
You can use drdv=1(dvdr)\frac{dr}{dv}=\dfrac{1}{\left(\frac{dv}{dr}\right)} and drdt=drdv×dvdt\frac{dr}{dt}=\frac{dr}{dv} \times \frac{dv}{dt} by the chain rule.
Reply 3
Original post by ttoby
You can use drdv=1(dvdr)\frac{dr}{dv}=\dfrac{1}{\left(\frac{dv}{dr}\right)} and drdt=drdv×dvdt\frac{dr}{dt}=\frac{dr}{dv} \times \frac{dv}{dt} by the chain rule.


thanks. dr/dt = 20 pi r^2

so do i then plug in 6 in to got dr/dt = 720pi as the answer?
Reply 4
Original post by jim666666

Original post by jim666666
thanks. dr/dt = 20 pi r^2

so do i then plug in 6 in to got dr/dt = 720pi as the answer?


Not quite, you should get dr/dt = 20/ (pi r^2). Then you plug in r=6.

Quick Reply

Latest