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Chapter 3: Problem 15

A spherical snowball melts at a rate proportional to its surface area. Show that the rate of change of the radius is constant. (Hint: Surface area \(=4 \pi r^{2}\) )

Short Answer

Expert verified
Answer: Yes, the rate of change of the radius of a melting spherical snowball is constant, as demonstrated by the differential equation \(\frac{dr}{dt} = -\frac{k}{4\pi}\) where k is a constant and 4蟺 is also constant.

Step by step solution

01

Express the melting rate in terms of surface area

Let's denote the melting rate as \(\frac{dV}{dt}\) (the rate of change of Volume with respect to time t) and the constant of proportionality as k. According to the problem, the melting rate is proportional to the surface area (S), so we can write: \(\frac{dV}{dt} = -kS\) where S = 4蟺\(r^2\) (surface area) and the negative sign represents the decrease in volume.
02

Find the expression for volume of the snowball

The volume (V) of a sphere is given by the formula: \(V = \frac{4}{3}\pi r^{3}\)
03

Differentiate the volume function with respect to time

To find the rate of change of the volume, we'll differentiate the volume function with respect to time using the chain rule. \(\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} \) Differentiating the volume function with respect to the radius gives: \(\frac{dV}{dr} = 4\pi r^{2}\)
04

Substitute the expressions found in Steps 1 and 3

Now, we'll substitute the expression for \(\frac{dV}{dr}\) from Step 3 and the surface area into our equation from Step 1: \(-k(4\pi r^2) = (4\pi r^{2})\cdot \frac{dr}{dt} \)
05

Solve for the rate of change of radius

Now, let's isolate \(\frac{dr}{dt}\) to find the rate of change of the radius: \(\frac{dr}{dt}=-\frac{k}{4\pi}\cdot(4\pi r^{2})\div(4\pi r^{2})\) \(\frac{dr}{dt}=-\frac{k}{4\pi}\) Now we have found the rate of change of the radius with respect to time, and since k and 4蟺 are constants, we have demonstrated that the rate of change of the radius is constant.

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