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 ma

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12 days ago

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ximum error in the calculated surface area. (Round your answer to the nearest integer.) cm2 What is the relative error

1 answer:

Leona [12.2K] · Answer 1 · 2026-03-29T04:51:00+03:00

Therefore, the maximum possible error in the sphere's surface area is 22.27 cm².

Consequently, the relative error approximates to 0.014.

Step-by-step explanation:

Given that, the measured circumference of a sphere was 70 cm with a possible deviation of 0.5 cm.

The circumference is C $=2 \pi r$

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

Differentiating concerning r

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

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

[ relative error = dC= 0.5]

$\Rightarrow dr= \frac{0.5}{2\pi}$ The surface area of the sphere is S= $4\pi r^2$

∴S= $4\pi r^2$

Differentiating with respect to r

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

$\Rightarrow dS=8\pi r dr$

The change in surface area (dS) will be maximal when dr is maximal.

Putting in the values of r and dr

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

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

[ ∵ C= 70 ] $\Rightarrow dS= \frac{70}{\pi}$ ⇒dS= 22.27 (approx)

Thus, the maximum surface area error 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)

Hence, the relative error is approximately 0.014.