Question

The circumference of the equator of a sphere was measured to be 82 82 cm with a possible error of 0.5 0.5 cm. Use linear approximation to estimate the maximum error in the calculated surface area to 4 decimal places.

Answer 1

Answer:

The maximum error in the calculated surface area is approximately 8.3083 square centimeters.

Step-by-step explanation:

The circumference ($s$), in centimeters, and the surface area ($A_{s}$), in square centimeters, of a sphere are represented by following formulas:

$A_{s} = 4\pi\cdot r^{2}$ (1)

$s = 2\pi\cdot r$ (2)

Where $r$ is the radius of the sphere, in centimeters.

By applying (2) in (1), we derive this expression:

$A_{s} = 4\pi\cdot \left(\frac{s}{2\pi} \right)^{2}$

$A_{s} = \frac{s^{2}}{\pi^{2}}$ (3)

By definition of Total Differential, which is equivalent to definition of Linear Approximation in this case, we determine an expression for the maximum error in the calculated surface area ($\Delta A_{s}$), in square centimeters:

$\Delta A_{s} = \frac{\partial A_{s}}{\partial s} \cdot \Delta s$

$\Delta A_{s} = \frac{2\cdot s\cdot \Delta s}{\pi^{2}}$ (4)

Where:

$s$ - Measure circumference, in centimeters.

$\Delta s$ - Possible error in circumference, in centimeters.

If we know that $s = 82\,cm$ and $\Delta s = 0.5\,cm$, then the maximum error is:

$\Delta A_{s} \approx 8.3083\,cm^{2}$

The maximum error in the calculated surface area is approximately 8.3083 square centimeters.