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 [tex]V=\frac{4}{3}\pi r^3[/tex]
Differentiating w.r.t. r,
[tex]\frac{dV}{dr}=4\pi r^2[/tex]
We have given, Helium is pumped into a spherical balloon at a rate of 3 cubic feet per second.
i.e. [tex]\frac{dV}{dt}=3[/tex]
Integrate w.r.t. t,
[tex]V=3t+C[/tex]
Assuming V=0 when t=0 then C=0
So, [tex]\frac{4}{3}\pi r^3=3t[/tex]
[tex]r^3=\frac{t}{4\pi}[/tex]
[tex]r=\sqrt[3]{\frac{t}{4\pi}}[/tex]
Now, we apply chain rule
i.e. [tex]\frac{dV}{dt}=\frac{dV}{dr}\cdot \frac{dr}{dt}[/tex]
[tex]3=4\pi(\frac{t}{4\pi})^{\frac{2}{3}}\cdot \frac{dr}{dt}[/tex]
The radius increasing after 2 minutes i.e. at [tex]t=2\times 60=120[/tex]
[tex]3=4\pi(\frac{120}{4\pi})^{\frac{2}{3}})\cdot \frac{dr}{dt}[/tex]
[tex]3=56.56\cdot \frac{dr}{dt}[/tex]
[tex]\frac{dr}{dt}=\frac{3}{56.56}[/tex]
[tex]\frac{dr}{dt}=0.0530[/tex]
Therefore, the rate of change of radius is 0.0530 ft/sec.
