# The circumference of a sphere was measured to be 78 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated surface area. (Round your answer to the nearest integer.) cm^{2} What is the relative error?

**Solution:**

It is given that

Circumference of a sphere = 78 cm

Error = 0.5 cm

(a) The formula to find the circumference of sphere is

C = 2πr

Substituting the values

78 = 2πr

r = 78/2π = 39/π

C = 2πr

By differentiating with respect to r

dC/dr = 2π

dC = 2πdr

dr = dC/2π = 0.5/2π

We know that

Surface area of sphere S = 4πr^{2}

The maximum surface area is dS = 8πrdr

dS = 8π × 39/π × 0.5/2π

dS = 24.82 = 25 cm^{2}

Maximum area in the surface area = 25 cm^{2}

Here

Relative error = ΔS/S

ΔS/S = 8πrdr/4πr^{2}

So we get

ΔS/S = 2 dr/r

Substituting the values

ΔS/S = 2 × [0.5/2π]/[39/π]

ΔS/S = 2 × 0.5/2π × π/39

ΔS/S = 0.0128

Therefore, the relative error is 0.0128.

## The circumference of a sphere was measured to be 78 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated surface area. (Round your answer to the nearest integer.) cm^{2} What is the relative error?

**Summary:**

The circumference of a sphere was measured to be 78 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated surface area. (Round your answer to the nearest integer.) cm^{2} . The relative error is 0.0128.