Answer:
[tex]\frac{1}{9\pi} cm/s \ or \ 0.0354cm/s[/tex]
Step-by-step explanation:
question tests our knowledge of differentiation:
[tex]\frac{dV}{dt} =100\frac{cm^3}{s}\\d=2r=30\\=>r=15[/tex]
Find [tex]\frac{dr}{dt}[/tex]
Given the formula for a sphere's volume is
[tex]V=\frac{4}{3}\pi r^3[/tex]
Take derivatives on both sides:
[tex]\frac{d}{dt}(V)=\frac{d}{dt}(\frac{4}{3}\pr r^3)}\\\frac{dV}{dt}=\frac{4}{3}\pi (3r^2)\frac{dr}{dt}=4\pi r^2 \frac{dr}{dt}[/tex]
Hence;
[tex]=\frac{dr}{dt}=\frac{1}{4\pi r^2}.\frac{dV}{dt}\\=\frac{dr}{dt}=\frac{1}{4\pi(15^2)}.100\\=\frac{1}{9\pi}[/tex]
Hence, the balloon radius increases at the rate of
[tex]\frac{1}{9\pi} cm/s \ or \ 0.0354cm/s[/tex]
