If the sides of a cube are measured with an error of 2% use differentials to estimate the relative error in the volume.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Let `s` represent the side length of the cube. We want to estimate the relative error in measuring the volume if the error in measuring the sides is 2%.

The propagated error can be estimated by `DeltaV~~dV` where `DeltaV` is the actual error in the volume.


`ds=+-.02` so `dV=+-.06s^2`

Thus an estimate for the error in the volume is `+-.06s^2`

To get the relative error we consider `(dV)/V` :


Thus the relative error depends on the measure of the side length.


The relative error is `3/s(+-.02)`


Ex Suppose the side is measured as 2in, `+-` .02in. Then `DeltaV=.242408` while the estimate yields `dV=3(2)^2(.02)=.24`

The relative error would be `.242408/8=.030301` while the estimate yields `.06/2=.03`

Approved by eNotes Editorial Team