If the surface area of a sphere increases by 44%, by what approximate percent does the radius increase?
20%
The surface area of a sphere is \(A = 4\pi r^2\), so the area depends on the square of the radius, i.e. \(A \propto r^2\).
Because \(4\pi\) is constant, the ratio of new area to old area equals the ratio of the squared radii: \(\dfrac{A_2}{A_1} = \left(\dfrac{r_2}{r_1}\right)^2\).
An increase of 44% means the new area is \(1 + 0.44 = 1.44\) times the original.
Therefore \(\left(\dfrac{r_2}{r_1}\right)^2 = 1.44\).
Taking the positive square root, \(\dfrac{r_2}{r_1} = \sqrt{1.44} = 1.2\).
So the radius becomes 1.2 times its original value, a rise of \(1.2 - 1 = 0.2\), i.e. 20%.
The key concept: when a quantity scales as the square of another, a factor change in the whole is the square root of that factor in the base variable.
The radius increases by approximately 20%.
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