Question

The possible error involved in measuring each dimension of a rectangular box is ±0.02 inches. The dimensions of the box are 8 inches by 5 inches by 12 inches. Approximate the propagated error and the relative error in the calculated volume of the box.

Answer 1

Explanation:

Given

Possible dimension error is
`\pm 0.02 in.`

length of box
`L=8 in.`

Width
`W=5 in.`

height
`h=12 in.`

Volume V is given by

`V=LWh`

`dV=LWdh+LhdW+WhdL`

and it is given

`dL=dW=dh=\pm 0.02\ in.`

since error always add therefore

`dV=(8* 5+5* 12+8* 12)\cdot 0.02`

`dV=3.92 in.^3`

Propagated error is
`3.92 in.^3`

relative error
`=(dV)/(V)`

`=(3.92)/(480)`

`=0.00816`

`=0.816 \%`