Answer:
[tex]\frac{3}{16\pi}in/sec[/tex]
Explanation:
We are given that
[tex]\frac{dv}{dt}=3in^3/s[/tex]
r=2 in when t=4 s
We have to find the rate of change of radius.
We know that
Volume of sphere=[tex]V=\frac{4}{3}\pi r^3[/tex]
Differentiate w.r.t t
[tex]\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}[/tex]
Substitute the values
[tex]3=4\pi(2)^2\times \frac{dr}{dt}[/tex]
[tex]\frac{dr}{dt}=\frac{3}{4\pi(2)^2}[/tex]
[tex]\frac{dr}{dt}=\frac{3}{16\pi}in/sec[/tex]
