[Math] Linear approximation (surface area question)

calculus

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The surface area of a sphere of radius $r$ is given by $A=4{\pi}r^2$. Use the linear approximation or differentials to compute the approximate percent change in the surface area if the radius of the sphere decreases by $3$%

I attached my solution.

Best Answer

$A = 4\pi r^2$

r decrease by 3%, Hence the new r = 0.97r

The new area $A_1 = 4\pi )(.97r)^2 = 4\pi (0.9409) r^2$

Change in area $ = (0.9409-1) (4\pi r^2)$

Percentage change$ = \frac{-0.0591 (4\pi r^2)}{4\pi r^2} = -5.91$%

This is the right answer.

You can also use what you have mentioned which is $\frac{dA}{A} = \frac{8\pi rdr}{4\pi r^2} = \frac{2dr}{r}$

But you know $\frac{dr}{r} = -0.03$ Hence percentage change $= 2\times (-0.03)=-0.06 = -6$% $ \approx -0.0591 = -5.91$% Thanks

Satish

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