Using linearization to calculate the thickness of a layer of paint on a spherical ball

Question

The volume of a sphere with radius $r$ is given by the formula $V(r) = \frac{4 \pi}{3} r^3$.

a) If $a$ is a given fixed value for $r$, write the formula for the linearization of the volume function $V(r)$ at $a$.

b) Use this linearization to calculate the thickness $\Delta r$ (in $cm$) of a layer of paint on the surface of a spherical ball with radius $r=52cm$ if the total volume of paint used is $340cm^3$.

The first part is easy to calculate, but I don't know exactly how to get the second part?

Answer 1

The volume $\Delta V$ of paint is approximatively given by $$\Delta V=V(a+\Delta r)-V(a)\doteq V'(a)\>\Delta r=4\pi a^2\>\Delta r\ .\tag{1}$$ In your problem the unknown is the thickness $\Delta r$ of the paint layer. From $(1)$ we immediately get $$\Delta r\doteq{\Delta V\over 4\pi\>a^2}\ .$$

Answer 2

For second part by logarithmic differentiation of volume/radius

$$\dfrac{\Delta V}{V}=3\dfrac{\Delta r}{r}$$

$$\dfrac{340}{\dfrac{4\pi}{3} 52^3}=3\dfrac{\Delta r}{52}$$

from which we calculate the paint thickness $\Delta r \approx 0.01 $ cm.

The same would be the case if instead $a=r$ would represent the side of a cube.