Question

The circumference of the equator of a sphere was measured to be 82 82 cm with a possible error of 0.5 0.5 cm. Use linear approximation to estimate the maximum error in the calculated surface area to 4 decimal places.

Answer 1

The maximum error in the calculated surface area is approximately 8.3083 square centimeters.

Explanation:

The circumference (
`s`), in centimeters, and the surface area (
`A_(s)`), in square centimeters, of a sphere are represented by following formulas:

`A_(s) = 4\pi\cdot r^(2)` (1)

`s = 2\pi\cdot r` (2)

Where
`r` is the radius of the sphere, in centimeters.

By applying (2) in (1), we derive this expression:

`A_(s) = 4\pi\cdot \left((s)/(2\pi) \right)^(2)`

`A_(s) = (s^(2))/(\pi^(2))` (3)

By definition of Total Differential, which is equivalent to definition of Linear Approximation in this case, we determine an expression for the maximum error in the calculated surface area (
`\Delta A_(s)`), in square centimeters:

`\Delta A_(s) = (\partial A_(s))/(\partial s) \cdot \Delta s`

`\Delta A_(s) = (2\cdot s\cdot \Delta s)/(\pi^(2))` (4)

Where:

`s` - Measure circumference, in centimeters.

`\Delta s` - Possible error in circumference, in centimeters.

If we know that
`s = 82\,cm` and
`\Delta s = 0.5\,cm`, then the maximum error is:

`\Delta A_(s) \approx 8.3083\,cm^(2)`

The maximum error in the calculated surface area is approximately 8.3083 square centimeters.