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1 month ago

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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

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

1 answer:

Leona [12.6K]1 month ago

5 0

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.

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