Question

The measurement of the radius of the end of a log is found to be 9 inches, with a possible error of 1/2 inch. Use differentials to approximate the possible propagated error in computing the area of the end of the log.

Answer 1

`A = (254.34 \pm 28.26) in^2`

Explanation:

radius, r = 9 inches

error = 0.5 inch

The area of the end is

A = 3.14 x r x r = 3.14 x 9 x 9 = 254.34 in^2

`A = \pi r^2\\\\(dA)/(dr)=2\pi r\\dA = 2 pi r dr \\\\dA = 2 * 3.14* 9* 0.5 = 28.26`

So, the area is given by

`A = (254.34 \pm 28.26) in^2`

Answer 2

`\pm 9in^2`

Explanation:

We are given that

Radius of end of a log, r= 9 in

Error,
`\Delta r=\pm 1/2`in

We have to find the error in computing the area of the end of the log by using differential.

Area of end of the log, A=
`pi r^2`

`(dA)/(dr)=2\pi r`

`(dA)/(dr)=2\pi (9)=18\pi in^2`

Now,

Approximate error in area

`dA=(dA)/(dr)(\Delta r)`

Using the values

`dA=18\pi (\pm 1/2)`

`\Delta A\approx dA=\pm 9in^2`

Hence, the possible propagated error in computing the area of the end of the log
`=\pm 9in^2`