Answer
percentage change in radius is 1%
Work Step by Step
We are given that the % error change in volume is 3%
x is diameter of sphere
radius $r=\frac{x}{2}$
Volume of sphere $V=\frac{4}{3}\pi r^3= (\frac{4}{3})\pi (\frac{x^3}{8})= (\frac{1}{6})\pi {x^3}$
on differentiating the above equation
$dV=\frac{\pi}{6}(3x^2dx))$
so
%change in volume =change in volume/original volume
$\frac{dV}{V}=\frac{\frac{\pi}{6}(3x^2(dx))}{ (\frac{1}{6})\pi x^3} $
$3=\frac{\frac{\pi}{6}(3(100)^2(1))}{ (\frac{1}{6})\pi (100)^3} $
$\frac{dx}{x}=1$
so the percentage change in radius is 1%
