The circumference of a sphere was measured to be 76 cm with a possible error of 0.2cm.
1. Use differentials to estimate the maximum error in the calculated surface area.
Please round the answer to the nearest tenth.
2. What is the relative error in the calculated surface area?
_____cm^2
Please round the answer to the nearest tenth.
_________%
3.Use differentials to estimate the maximum error in the calculated volume.
Please round the answer to the nearest tenth.
______cm^3
4.What is the relative error in the calculated volume? Please round the answer to the nearest tenth.
_____%
How do you do? Please help!
i think it is 20
what?
To solve this problem, we'll first need to use the formula for the surface area and volume of a sphere.
The surface area of a sphere is given by the formula:
A = 4πr²
And the volume of a sphere is given by the formula:
V = (4/3)πr³
where r is the radius of the sphere.
Now, let's address each question step by step:
1. To estimate the maximum error in the calculated surface area using differentials, we'll use the concept of differentials. The differential of the surface area, dA, can be approximated by the derivative of the surface area with respect to the radius, multiplied by the change in the radius, dr.
dA = 2πr dr
Given that the radius of the sphere is not directly provided, we need to find it first. Since the circumference of the sphere is given as 76 cm, we can use the formula for the circumference of a sphere to find the radius.
C = 2πr
76 = 2πr
r = 76 / (2π)
Now, let's find the maximum error in the surface area by substituting r ± dr into the differential formula. The maximum error in the radius is given as 0.2 cm.
Maximum error = |2πr dr| = 2πr dr
Substituting the values:
Maximum error = 2π(76 / (2π))(0.2) = 76 × 0.2 = 15.2 cm²
So, the maximum error in the calculated surface area is 15.2 cm².
2. To determine the relative error in the calculated surface area, we need to divide the maximum error by the actual surface area.
The actual surface area can be calculated using the formula A = 4πr² and substituting the value of r obtained earlier.
A = 4π(76 / (2π))² = 4π(76 / (2π))^2 = 4π(76 / 2)^2 = 4π(38)^2 = 4π(1444) = 5776π
Relative error = (Maximum error / Actual surface area) × 100
Relative error = (15.2 / 5776π) × 100
Rounding the answer to the nearest tenth:
Relative error = 0.0044 × 100 ≈ 0.4%
Therefore, the relative error in the calculated surface area is approximately 0.4%.
3. To estimate the maximum error in the calculated volume using differentials, we'll use a similar approach. The differential of the volume, dV, can be approximated by the derivative of the volume with respect to the radius, multiplied by the change in the radius, dr.
dV = 4πr² dr
Using the same process as before, we'll substitute the value of r obtained earlier and calculate the maximum error in the volume.
Maximum error = |4πr² dr| = 4πr² dr
Maximum error = 4π(76 / (2π))²(0.2) = 76² × 0.2 = 19.04 × 76 = 1446.24 cm³
So, the maximum error in the calculated volume is 1446.24 cm³.
4. To determine the relative error in the calculated volume, we need to divide the maximum error by the actual volume.
The actual volume can be calculated using the formula V = (4/3)πr³ and substituting the value of r obtained earlier.
V = (4/3)π(76 / (2π))³ = (4/3)π(76 / 2)^3 = (4/3)(38)^3π = (4/3)(54872)π = 73162.67π
Relative error = (Maximum error / Actual volume) × 100
Relative error = (1446.24 / 73162.67π) × 100
Rounding the answer to the nearest tenth:
Relative error = 0.0167 × 100 ≈ 1.7%
Therefore, the relative error in the calculated volume is approximately 1.7%.