Ask Me Help Desk - Rate of change of a volume of a balloon

Inflate a spherical balloon by blowing 1000cm³ of air into it every 5 secs:

a) What is the rate of change of the volume of the balloon (in cm³/sec) at the moment when the radius of the balloon is 5cm?

b) What is the rate of change of the radius of the balloon (in cm/sec) at the moment when the radius of the balloon is 5cm?

a. It appears that the volume increase is simply 1000 cm³ / 5 sec = 200 cm3/sec. The radius doesn't seem to be a factor

b. The volume of the balloon is given by

$V\, = \, \frac 4 3 \pi r^3\$

We solve for V given r=5

$V\, = \, \frac 4 3 \pi r^3\,=\,166.66\,\pi\,\approx\,523.60\,cm^3$

1 second later, the volume is increased by 200 cm³,

$723.60\,cm^3\, \approx \, \frac 4 3 \pi r^3\$

$r\,\approx\,5.138\,cm^3$

The rate of change is simply the change in r (Δr) divided by r

Possibly a better way of solving this is using calculus

$\frac {dV}{dT}=200\,\frac {cm^3}{sec}$

$\frac {dr}{dT}\,=\,\frac {dV}{dT}\,\frac {dr}{dV}$

$r=(\frac {3 \cdot V}{4 \cdot \pi})^{\frac 13}$

therefore $\frac {dr}{dV}\,=\, (\frac {3}{4 \cdot \pi})^{\frac 13}\,\frac 13 V^{-\frac 23}$

$\frac {dr}{dT}=200 \cdot \, (\frac {3}{4 \cdot \pi})^{\frac 13}\,\frac 13 V^{-\frac 23}$

$\frac {dr}{dT}\,\approx\,41.36 \cdot V^{-\frac 23}$

Calculate V at the exact time and plug it into the formula.

(I hope I didn't make any mistakes).

Thank you :)

Quote:

Originally Posted by Perito

a. It appears that the volume increase is simply 1000 cm³ / 5 sec = 200 cm3/sec. The radius doesn't seem to be a factor

b. The volume of the balloon is given by

$V\, = \, \frac 4 3 \pi r^3\$

We solve for V given r=5

$V\, = \, \frac 4 3 \pi r^3\,=\,166.66\,\pi\,\approx\,523.60\,cm^3$

1 second later, the volume is increased by 200 cm³,

$723.60\,cm^3\, \approx \, \frac 4 3 \pi r^3\$

$r\,\approx\,5.138\,cm^3$

The rate of change is simply the change in r (Δr) divided by r

Possibly a better way of solving this is using calculus

$\frac {dV}{dT}=200\,\frac {cm^3}{sec}$

$\frac {dr}{dT}\,=\,\frac {dV}{dT}\,\frac {dr}{dV}$

$r=(\frac {3 \cdot V}{4 \cdot \pi})^{\frac 13}$

therefore $\frac {dr}{dV}\,=\, (\frac {3}{4 \cdot \pi})^{\frac 13}\,\frac 13 V^{-\frac 23}$

$\frac {dr}{dT}=200 \cdot \, (\frac {3}{4 \cdot \pi})^{\frac 13}\,\frac 13 V^{-\frac 23}$

$\frac {dr}{dT}\,\approx\,41.36 \cdot V^{-\frac 23}$

Calculate V at the exact time and plug it into the formula.

(I hope I didn't make any mistakes).