Question

A spherical balloon is expanding at a constant rate of 32(pi) cm^3/s. The balloon is intially fully deflated. After 9 seconds, the volume of the balloon is 288(pi) cm^3. At this instant, find the rate of increase of the radius of the balloon.
Answer given: 2/9 cm/s

Answer 1

We know the rate of expansion of the balloon.
\[\frac{ dV }{ dt}=32\pi \]
We want to find
\[\frac{ dr }{ dt }\]
so we try to equate the two and see what's missing
\[\frac{ dr }{ dt }=\frac{ dV }{ dt }\times? \]
We easily see that the missing factor is
\[\frac{ dr }{ dV }\]
to find this, we can find the the rate of change of volume with respect to radius from the volume of a sphere
\[V=\frac{ 4 }{ 3 }\pi r ^{3}\]
\[\frac{ dV }{ dr }=4 \pi r ^{2}\]
Flipping it we get
\[\frac{ dr }{ dV }=\frac{ 1 }{ 4 \pi r ^{2} }\]
Now plugging it i gives
\[\frac{ dr }{ dt }=32 \pi \times \frac{ 1 }{ 4 \pi r ^{2} }=\frac{ 8 }{ r ^{2} }\]

Answer 2

At that instance, we know the volume is 288 Pi.
So we can find r from the formula for the volume of a sphere.
r is found to be 6,
Thus, dr/dt = 2/9

Answer 3

@Rapidknight thank you so so much!

Answer 4

You're welcome =)