Question

The circumference of a sphere was measured to be 78 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:

The maximum error is  $dS = 25 \ cm^{2}$

Relative error is Relative error is = $\frac{1}{78}$

Step-by-step explanation:

Circumference of a sphere C = 2$\pi r$

Differentiate above equation with respect to r

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

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

Given that $dC = 0.5 cm$

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

$dr = \frac{1}{4\pi}$

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

Differentiate above equation with respect to r

$\frac{dS}{dr} = 8 \pi r$

$dS = 4 C dr$

Where C = circumference of the sphere

given that C = 78 cm  & $dr = \frac{1}{4\pi}$

So the maximum error is given by

$dS = 4 (78)(\frac{1}{4\pi} )$

$dS = \frac{78}{\pi}$

$dS = 25 \ cm^{2}$

Now the relative error is given by

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

⇒ $\frac{\frac{78}{\pi} }{4\pi r^{2} }$

Since $r = \frac{C}{2 \pi}$

Relative error is = $\frac{1}{78}$

Therefore the maximum error is  $dS = 25 \ cm^{2}$

Relative error is Relative error is = $\frac{1}{78}$