I am studying maths as a hobby and am doing a chapter on calculus and small changes and errors. I am trying to understand the following problem. I cannot get the answer in the text book, which is 6%. I have not tried to calculate the area change as I obviously am making a fundamental error.
If a 2% error is made in measuring the diameter of a sphere, find approximately the resulting percentage errors in the volume and surface area.
I have said:
Let V = volume and D = diameter
$V = \frac{4 \pi r^3}{3} = \frac{\pi D^3}{6}$
Now if $\delta V, \delta D$ represent small changes in V and D respectively:
$\frac{\delta V}{\delta D} \approx \frac{dV}{dD}$ and $\frac{dV}{dD} = \frac{\pi D^2}{2}$ so
$\delta V = \frac{\delta D.dV}{dD} = \frac{\delta D\pi D^2}{2} = \frac{2}{100}.\frac{\pi D^2}{2} = \frac{\pi D^2}{100}$
Percentage error = $\frac{\delta V}{V} = \frac{\pi D^2}{100} \div \frac{\pi D^3}{6} = \frac{\pi D^2}{100}.\frac{6}{\pi D^3} = \frac{6}{100D}$ but the answer is actually 6%.
Where have I gone wrong.
Best Answer
$\delta V = \frac{\delta D.dV}{dD} = \frac{\delta D\pi D^2}{2} = \frac{2}{100} \times D \times \frac{\pi D^2}{2} = \frac{\pi D^3}{100}$
Please note $\delta D = \frac{2}{100} \times D $ (New $D$ minus old $D$).
$\frac{\delta V}{V} = \frac{\pi D^3}{100} \div \frac{\pi D^3}{6} = \frac{\pi D^3}{100}.\frac{6}{\pi D^3} = \frac{6}{100}$