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

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embizze | High School Teacher | (Level 1) Educator Emeritus

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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.

`V=s^3==>dV=3s^2ds`

`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` :

`(dV)/V=(3s^2ds)/s^3=(3ds)/s=3/s(+-.02)`

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

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

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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`

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