4.2 Linear approximations and differentials  (Page 4/12)

1. The measurement of the side length is accurate to within $\text{±}0.1$ cm. Therefore,
   
   $\mathrm{-0.1}\le dx\le 0.1.$
   
   The volume of a cube is given by $V=x^3,$ which leads to
   
   $dV=3x^{2}dx.$
   
   Using the measured side length of 5 cm, we can estimate that
   
   $\mathrm{-3}{(5)}^2(0.1)\le dV\le 3{(5)}^2(0.1).$
   
   Therefore,
   
   $\mathrm{-7.5}\le dV\le 7.5.$
2. If the side length is actually 4.9 cm, then the volume of the cube is
   
   $V(4.9)={(4.9)}^3=117.649{\text{cm}}^3.$
   
   If the side length is actually 5.1 cm, then the volume of the cube is
   
   $V(5.1)={(5.1)}^3=132.651{\text{cm}}^3.$
   
   Therefore, the actual volume of the cube is between 117.649 and 132.651. Since the side length is measured to be 5 cm, the computed volume is $V(5)=5^3=125.$ Therefore, the error in the computed volume is
   
   $117.649-125\le \text{Δ}V\le 132.651-125.$
   
   That is,
   
   $\mathrm{-7.351}\le \text{Δ}V\le 7.651.$
   
   We see the estimated error $dV$ is relatively close to the actual potential error in the computed volume.

If the measurement of the radius is accurate to within $\text{±}80,$ we have

Since the volume of a sphere is given by $V=\left(\frac{4}{3}\right)\pi r^3,$ we have

Using the measured radius of 4000 mi, we can estimate

To estimate the relative error, consider $\frac{dV}{V}.$ Since we do not know the exact value of the volume $V,$ use the measured radius $r=4000\text{mi}$ to estimate $V.$ We obtain $V\approx \left(\frac{4}{3}\right)\pi {\left(4000\right)}^3.$ Therefore the relative error satisfies

which simplifies to

The relative error is 0.06 and the percentage error is $6\text{\%}.$