Question

The circumference of a sphere was measured to be 70 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated volume. (Round your answer to the nearest integer.)What is the relative error? (Round your answer to three decimal places.)

Answer 1

Answer:

Therefore the error in the calculated volume is 130 $cm^3$.

The relative error is 0.011.

Step-by-step explanation:

Given that,

The circumference of a sphere was 70 cm.

The circumference of a sphere is C= $2\pi r$

C= $2\pi r$

Differentiating with respect to r

$\frac{dC}{dr}=2\pi$

$\Rightarrow \frac{\triangle C}{\triangle r}=2\pi$

$\Rightarrow\triangle r= \frac{\triangle C}{2\pi}$

Given that,the circumference of the sphere was with possible error 0.5 cm.

$\triangle C=0.5$

$\therefore \triangle r= \frac{0.5}{2\pi}$

The volume of the sphere is $V=\frac43 \pi r^3$

$V=\frac43 \pi r^3$

Differentiating with respect to r

$\frac{dV}{dr}=\frac43 \pi r^2$

$\Rightarrow \frac{\triangle V}{\triangle r}=\frac43 \pi r^2$

$\Rightarrow \triangle V}=\frac43 \pi r^2\times \triangle r}$

Putting $\triangle r=\frac{0.5}{2\pi}$

$\Rightarrow \triangle V}=\frac43 \pi r^2\times \frac{0.5}{2\pi}$

$\Rightarrow \triangle V}=\frac13 \pi r^2$

$\Rightarrow \triangle V}=\frac1{3} \pi (\frac C{2\pi})^2$        $[\because r=\frac C{2\pi}]$

$\Rightarrow \triangle V}=\frac1{1 2} \times \frac{C^2}{\pi}$

$\Rightarrow \triangle V}=\frac1{1 2} \times \frac{70^2}{\pi}$        [ C=70 cm]

$\Rightarrow \triangle V}\approx 130 \ cm^3$

Therefore the error in the calculated volume is 130 $cm^3$.

Relative error $=\frac{\triangle V}{V}$

                     $=\frac{\frac13 \pi (\frac{C}{2\pi})^2 }{\frac43 \pi (\frac C{2\pi})^3}$

                     $=\frac{1}{4 (\frac C{2\pi})}$

                    $=\frac{\pi }{4C}$

                    $\approx 0.011$

The relative error is 0.011.