The circumference of a sphere was measured to be 74.000 cm with a possible error of 0.50000 cm. Use linear approximation to estimate the maximum error in the calculated surface area.

Estimate the relative error in the calculated surface area. Must be accurate to the fifth decimal place.

A = 4pi r^2

dA = 8pi r dr

Now plug in the numbers. Recall that % error is dA/A * 100

To estimate the maximum error in the calculated surface area, we can use linear approximation. Linear approximation tells us that for small changes in a function, the change in the function can be approximated by the product of the function's derivative at a given point and the change in its input.

The formula to calculate the surface area of a sphere is given by A = 4πr^2, where r is the radius of the sphere.

Let's find the derivative of the surface area function with respect to the radius. Differentiating the formula A = 4πr^2 with respect to r, we get dA/dr = 8πr.

Now, we have the derivative of the surface area function as 8πr.

Given that the circumference of the sphere is 74.000 cm with a possible error of 0.50000 cm, we can find the radius of the sphere using the formula C = 2πr, where C is the circumference and r is the radius.

The radius of the sphere is r = C / (2π) = 74.000 cm / (2π) ≈ 11.778 cm.

To find the maximum error in the calculated surface area, we need to multiply the maximum error in the radius by the derivative of the surface area function.

Using linear approximation, the maximum error in the calculated surface area is ΔA ≈ (8πr)(Δr), where Δr is the maximum error in the radius.

Δr is given as 0.50000 cm.

Substituting the values into the formula, ΔA ≈ (8π * 11.778 cm)(0.50000 cm) ≈ 23.556π cm^2.

To estimate the relative error in the calculated surface area, we need to compute the ratio of the maximum error in the surface area to the actual surface area.

The actual surface area of the sphere can be found using the formula A = 4πr^2.

Substituting the value of r into the formula, A = 4π(11.778 cm)^2 ≈ 1742.778π cm^2.

Therefore, the relative error in the calculated surface area is approximately ΔA / A ≈ (23.556π cm^2) / (1742.778π cm^2) ≈ 0.01349.

To ensure accuracy to the fifth decimal place, the relative error is approximately 0.01349.