Question

The circumference of a sphere was measured to be 74 cm with a possible error of 0.5 cm. (a) Use differentials to estimate the maximum error in the calculated surface area. (Round your answer to the nearest integer.) cm2 What is the relative error?

Answer 1

Answer Step-by-step explanation:

Circumference of sphere=74 cm

Error=0.5 cm

a.We know that circumference of sphere=$C=2\pi r$

Using the formula

$74=2\pi r$

$r=\frac{74}{2\pi}=\frac{37}{\pi}$

$C=2\pi r$

Differentiate w.r.t r

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

$dC=2\pi dr$

$dr=\frac{dC}{2\pi}=\frac{0.5}{2\pi}$

Surface area of sphere=$S=4\pi r^2$

Maximum area in the surface area=$dS=8\pi rdr$

$dS=8\pi\times \frac{37}{\pi}\times \frac{0.5}{2\pi}=23.5\approx 24 cm^2$

Hence,the maximum area in the surface area=24 square cm

Relative error=$\frac{\Delta S}{S}\approx \frac{dS}{S}$

$\frac{\Delta S}{S}=\frac{8\pi rdr}{4\pi r^2}=2\frac{dr}{r}=2\times \frac{\frac{0.5}{2\pi}}{\frac{37}{\pi}}=2\times \frac{0.5}{2\pi}\times \frac{\pi}{37}=0.014$

Hence, the relative error=0.014