The circumference of a sphere was measured to be 89.000 cm with a possible error of 0.50000 cm.
A. Use linear approximation to estimate the maximum error in the calculated surface area.
B. Estimate the relative error in the calculated surface area.
The circumference of a sphere was measured to be 89.000 cm with a possible error of 0.50000 cm.
A. Use linear approximation to estimate the maximum error in the calculated surface area.
B. Estimate the relative error in the calculated surface area.
To estimate the maximum error in the calculated surface area of a sphere, we can use linear approximation.
A. The surface area of a sphere can be calculated using the formula:
Surface Area = 4πr^2
where r is the radius of the sphere.
Since we are given the circumference, let's find the radius of the sphere using the formula for the circumference:
Circumference = 2πr
Rearranging the formula, we get:
r = Circumference / (2π)
Substituting the given value, we find:
r = 89.000 cm / (2π) ≈ 14.166 cm
Now, let's calculate the variation in the radius due to the error in the measured circumference.
Since the error is given as 0.50000 cm, the maximum change in the radius can be calculated as:
Δr = ΔCircumference / (2π) = 0.50000 cm / (2π) ≈ 0.079577 cm
Next, let's find the variation in the surface area due to the change in the radius.
Using the linear approximation, we have:
ΔSurface Area ≈ 2 × (Surface Area) × (Δr / r)
Substituting the values, we get:
ΔSurface Area ≈ 2 × (4πr^2) × (Δr / r)
Calculating the surface area using the given radius, we find:
Surface Area ≈ 4π(14.166 cm)^2 ≈ 796.51 cm^2
Now, substituting the values, we can calculate the maximum error in the surface area:
ΔSurface Area ≈ 2 × (796.51 cm^2) × (0.079577 cm / 14.166 cm)
≈ 8.922 cm^2
Therefore, the maximum error in the calculated surface area is approximately 8.922 cm^2.
B. To estimate the relative error in the calculated surface area, we divide the maximum error in the surface area by the actual surface area:
Relative Error = (ΔSurface Area) / (Surface Area)
Substituting the values, we get:
Relative Error ≈ (8.922 cm^2) / (796.51 cm^2)
≈ 0.0112 or 1.12%
Therefore, the estimate of the relative error in the calculated surface area is approximately 1.12%.