Related Rates (Derivatives)

Question

How to show that if the volume of a balloon is decreasing at a rate proportional to its surface area, the radius of the balloon is shrinking at a constant rate.

Answer 1

Volume and surface area of a sphere are given by:

V = 4/3 πr³

^and

S = 4πr².

Taking the derivative of volume with respect to time (t):

dV/dt = 4πr² dr/dt

We are also given that dV/dt = k(4πr²) ; where k < 0 (rate of change of the volume is proportional to surface area and decreasing).

By substitution:

k(4πr²)= 4πr² dr/dt,

Solving for dr/dt results in:

dr/dt = k, k <0 , therefore the radius is decreasing at a constant rate (k is a negative constant).

Answer 2

If balloon is a sphere then dV/dt=d/dt(4/3Pir³)=4pir²dr/dt=KS where K is a constant and S = surface area. However the surface area of a sphere =4pir². Doing the algebra we see that 4pir²dr/dt=K(4pir²) simplifies to dr/dt=K

Answer 3

I assume that the balloon is a sphere.

The radius of the sphere is r(t) -- a function of time t.

The formulas for volume and area of a sphere are

V = 4πr³/3

S = 4πr²

The rate of volume change is is the derivative

dV/dt = 4πr² dr/dt

Let k be the constant of proportionality between dV/dt and S.

4πr² dr/dt = k4πr²

From that we derive

dr/dt = k

That is, the change of radius is the constant k.