Answer to Question #189997 in Calculus for Devlin

Question

How fast is the surface area of a spherical balloon increasing when the radius is 10 cm and the volume is increasing at 15 cm3 /sec?

Answer 1

Solution:

$\dfrac{dV}{dt}=15 cm^3/sec,\ r=10 cm, \dfrac{dA}{dt}=?$

$V=\dfrac43\pi r^3$

On differentiating w.r.t $r$,

$\dfrac{dV}{dt}=4\pi r^2 \dfrac{dr}{dt} \\ \Rightarrow 15=4\pi (10)^2 \dfrac{dr}{dt} \\ \Rightarrow \dfrac{dr}{dt}=\dfrac{3}{80\pi} cm/sec$

Now, $A=4\pi r^2$

On differentiating w.r.t $r$,

$\dfrac{dA}{dt}=8\pi r \dfrac{dr}{dt}=8\pi(10)(\dfrac{3}{80\pi})=3 cm^2/sec$