Question

Helium is pumped into a spherical balloon at a rate of 3 cubic feet per second. How fast is the radius increasing after 2 minutes? Note: The volume of a sphere is given by V=43πr3V=43πr3. Rate of change of radius (in feet per second) =

Answer 1

Answer:

The rate of change of radius is 0.0530 ft/sec.

Step-by-step explanation:

Given : Helium is pumped into a spherical balloon at a rate of 3 cubic feet per second.

To find : How fast is the radius increasing after 2 minutes?

Solution :

The volume of the sphere is $V=\frac{4}{3}\pi r^3$

Differentiating w.r.t. r,

$\frac{dV}{dr}=4\pi r^2$

We have given, Helium is pumped into a spherical balloon at a rate of 3 cubic feet per second.

i.e. $\frac{dV}{dt}=3$

Integrate w.r.t. t,

$V=3t+C$

Assuming V=0 when  t=0 then C=0

So, $\frac{4}{3}\pi r^3=3t$

$r^3=\frac{t}{4\pi}$

$r=\sqrt[3]{\frac{t}{4\pi}}$

Now, we apply chain rule

i.e.  $\frac{dV}{dt}=\frac{dV}{dr}\cdot \frac{dr}{dt}$

$3=4\pi(\frac{t}{4\pi})^{\frac{2}{3}}\cdot \frac{dr}{dt}$

The radius increasing after 2 minutes i.e. at $t=2\times 60=120$

$3=4\pi(\frac{120}{4\pi})^{\frac{2}{3}})\cdot \frac{dr}{dt}$

$3=56.56\cdot \frac{dr}{dt}$

$\frac{dr}{dt}=\frac{3}{56.56}$

$\frac{dr}{dt}=0.0530$

Therefore, the rate of change of radius is 0.0530 ft/sec.