Question

The circumference of a sphere was measured to be 70 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:

Therefore the maximum error in the surface area of the sphere is 22.27 cm².

Therefore the relative error is 0.014 (approx).

Step-by-step explanation:

Given that, The circumference of a sphere was 70 cm with the possible error 0.5 cm.

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

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

Differentiating with respect to r

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

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

$\Rightarrow dr= \frac{0.5}{2\pi}$    [ relative error = dC= 0.5]

The surface area of the sphere is S= $4\pi r^2$

∴S= $4\pi r^2$

Differentiating with respect to r

$\frac{dS}{dr}=4\pi \times 2r$

$\Rightarrow dS=8\pi r dr$

dS will be maximum when dr is maximum.

Putting the value of r and dr

$\Rightarrow dS= 8\pi \times \frac{C}{2\pi} \times \frac {0.5}{2\pi}$

$\Rightarrow dS=\frac{C}{\pi}$

$\Rightarrow dS= \frac{70}{\pi}$     [ ∵ C= 70 ]

⇒dS= 22.27 (approx)

Therefore the maximum error in the surface area is 22.27 cm².

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

                      $=\frac{\frac{C}{\pi}}{4\pi r^2}$

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

                     $=\frac{1}{C}$

                     $=\frac{1}{70}$

                    =0.014 (approx)

Therefore the relative error is 0.014 (approx).