The circumference of a sphere was measured to be 86 cm with a possible error of 0.2cm.

Use differentials to estimate the maximum error in the calculated surface area. Please round the answer to the nearest tenth.

What is the relative error in the calculated surface area? Please round the answer to the nearest tenth.

_________%

Use differentials to estimate the maximum error in the calculated volume. Please round the answer to the nearest tenth.

What is the relative error in the calculated volume? Please round the answer to the nearest tenth.

%

For small deviations, the relative area error is twice the relative length (radius or circumference) area, and the relative volume error is three times the relative length error.
In your case the relative length error is 0.2/86 = 0.233%
The relative area error (rounded to 0.1%) is therefore 0.4% and the relative volume error is 0.7%

what is 66.6667 rounded to the nearest tenth

what 1627366.92223517 rounded to the nearest billionths

To find the maximum error in the calculated surface area, we can use the formula for the surface area of a sphere:

Surface Area = 4πr^2

Where r is the radius of the sphere. We need to find the maximum error in the surface area, so we will use differentials to estimate this error.

Let's assume the radius of the sphere is r cm. The circumference of the sphere is given as 86 cm, which means the radius is calculated as:

Circumference = 2πr

86 = 2πr

Solving for r:

r = 86 / (2π)

Now, let's find the differential of the surface area with respect to the radius:

d(Surface Area) = d(4πr^2)

Using the power rule, we can differentiate this equation:

d(Surface Area) = 8πr dr

To find the maximum error in the surface area, we need to find the maximum error in the radius. The possible error in the circumference is given as 0.2 cm. Therefore, the possible error in the radius is:

Possible Error in Radius = 0.2 / (2π) = 0.0318 cm (rounded to 4 decimal places)

Now, we can substitute the values into the differential equation:

d(Surface Area) = 8π(86 / (2π)) (0.0318)

Simplifying:

d(Surface Area) = 8 * 86 * 0.0318

d(Surface Area) ≈ 21.3156 cm^2

Therefore, the maximum error in the calculated surface area is approximately 21.3 cm^2 (rounded to the nearest tenth).

To find the relative error in the calculated surface area, we can divide the maximum error in the surface area by the actual surface area and multiply by 100 to get the percentage:

Relative Error in Surface Area = (21.3 / (4π(86 / (2π))^2)) * 100

Simplifying:

Relative Error in Surface Area = 21.3 / (4 * 86^2) * 100

Relative Error in Surface Area ≈ 0.0617% (rounded to the nearest tenth)

Therefore, the relative error in the calculated surface area is approximately 0.1%.

To estimate the maximum error in the calculated surface area and volume of a sphere, we can use differentials.

First, let's find the maximum error in the calculated surface area. The surface area of a sphere is given by the formula A = 4πr^2, where r is the radius. Since the circumference was measured with a possible error of 0.2 cm, we can estimate the maximum error in the radius.

Since the circumference of a sphere is given by the formula C = 2πr, we can rearrange it to solve for r. Dividing both sides by 2π gives us r = C / (2π), where C is the measured circumference.

Substituting the given values, we have r = 86 cm / (2π) ≈ 13.6709 cm.

To estimate the maximum error in the radius, we can use the relative error formula. The relative error is given by the error divided by the measured value, multiplied by 100%. In this case, the relative error in the circumference is 0.2 cm / 86 cm ≈ 0.0023, or 0.23%. Since the radius is half the circumference, the relative error in the radius is also 0.23%.

Now, we can use differentials to estimate the maximum error in the surface area. The differential of the surface area with respect to the radius is given by dA = 8πr dr. We can approximate the differential ΔA using the maximum error in the radius Δr:

ΔA ≈ 8πr Δr.

Substituting the values, we have ΔA ≈ 8π(13.6709 cm)(0.0023) ≈ 0.9868 cm^2.

Rounding to the nearest tenth, the estimated maximum error in the calculated surface area is 1.0 cm^2.

Next, let's find the relative error in the calculated surface area. The measured surface area is 4π(13.6709 cm)^2 ≈ 2351.7780 cm^2. The relative error is the maximum error divided by the measured value, multiplied by 100%:

Relative error = (0.9868 cm^2 / 2351.7780 cm^2) x 100% ≈ 0.042%.

Rounding to the nearest tenth, the relative error in the calculated surface area is 0.0%.

Now, let's move on to the estimated maximum error in the calculated volume. The volume of a sphere is given by the formula V = (4/3)πr^3. Using differentials, the differential of the volume with respect to the radius is given by dV = 4πr^2 dr.

Similar to the surface area, we can estimate the maximum error in the volume using the maximum error in the radius. The estimated maximum error in the volume is ΔV ≈ 4π(13.6709 cm)^2(0.0023) ≈ 21.8307 cm^3.

Rounding to the nearest tenth, the estimated maximum error in the calculated volume is 21.8 cm^3.

Finally, let's find the relative error in the calculated volume. The measured volume is (4/3)π(13.6709 cm)^3 ≈ 10608.6548 cm^3. The relative error is the maximum error divided by the measured value, multiplied by 100%:

Relative error = (21.8307 cm^3 / 10608.6548 cm^3) x 100% ≈ 0.206%.

Rounding to the nearest tenth, the relative error in the calculated volume is 0.2%.