free. Note that, in the case of the above examples, In the first case The relative error = . ; and the percentage error = × % . In the second case The relative error = ; and the percentage error = × % .
So the second approximation is a better approximation than the first one. Note that, in order to calculate the relative error or the percentage error one should know the actual value of what we are approximating. Let us consider some examples. Example .
Let us assume that the shape of a soap bubble is a sphere. Use linear approximation to approximate the increase in the surface area of a soap bubble as its radius increases from cm to . cm. Also, calculate the percentage error.
Recall that surface area of a sphere with radius r is given by S r ( ) = . Note that even though we can calculate the exact change using this formula, we shall try to approximate the change using the linear approximation. So, using ( ), we have Change in the surface area S S S ( . ) ( )( .
) ≈ ′ = p ( )( . ) = p cm Exact calculation of the change in the surface gives S S ( . ) = cm Percentage error = relative error % | | Example . A right circular cylinder has radius r = cm.
and height h = cm. Suppose that the radius of the cylinder is increased from cm to . cm and the height does not change. Estimate the change in the volume of the cylinder.
Also, calculate the relative error and percentage error. Recall that volume of a right circular cylinder is given by V r h = π , where r is the radius and h is the height. So we