The radius of a spherical ball is measured at r = 50 cm. Estimate the maximum error in the volume and surface area if r is accurate to within 0.1 cm. (Round your answers to three decimal places.)
0.1 cm means a 0.2% error
so, v can be off by .6% and area by 0.4%
checking for v,
4/3 π(50-0.1)^3 <= v <= 4/3 π(50+0.1)
520463.46194 <= v <= 526746.65563
v(50) = 523598.77560
so, for error e,
3135.31366 <= e <= 3147.88003
so for the relative error er
.005988 <= er <= .00601
or, 0.6%
To estimate the maximum error in the volume and surface area of a spherical ball, we can use the formula for the maximum error in a function due to a small error in the input.
For the volume of a sphere, V = (4/3)πr^3, the differential (dV) is given by:
dV = (4/3)π(3r^2)dr
To estimate the maximum error in the volume, we need to find the maximum possible value for |dr|. Given that r is accurate to within 0.1 cm, the maximum error in the radius would be 0.1 cm. Therefore, we can substitute |dr| = 0.1 cm into the formula.
dV = (4/3)π(3(50 cm)^2)(0.1 cm) = (4/3)π(7500 cm^2)(0.1 cm)
Evaluating this expression:
dV ≈ (4/3)π(750 cm^3)
The maximum error in the volume is approximately 3141.59 cm^3.
For the surface area of a sphere, A = 4πr^2, the differential (dA) can be obtained as:
dA = 4π(2r)dr
Again, to estimate the maximum error in the surface area, we need to find the maximum possible value for |dr|, which in this case is 0.1 cm. Substituting |dr| = 0.1 cm into the formula:
dA = 4π(2(50 cm))(0.1 cm) = 4π(100 cm^2)(0.1 cm)
Simplifying:
dA ≈ 4π(10 cm^2)
The maximum error in the surface area is approximately 125.664 cm^2.
Therefore, the maximum error in the volume is approximately 3141.59 cm^3 and the maximum error in the surface area is approximately 125.664 cm^2.
To estimate the maximum error in the volume and surface area of the spherical ball, you need to use different formulas.
Let's start with the volume. The formula for the volume of a sphere is:
V = (4/3)πr^3
To find the maximum error in the volume, you need to calculate the derivative of the volume with respect to the radius and then multiply it by the maximum error in the radius. In this case, the maximum error is given as 0.1 cm.
Let's calculate it step by step:
1. Find the derivative of the volume formula with respect to r:
dV/dr = 4πr^2
2. Calculate the maximum error in volume:
ΔV = (dV/dr) * Δr
where ΔV is the maximum error in volume, Δr is the maximum error in the radius (0.1 cm), and (dV/dr) is the derivative of the volume with respect to the radius.
ΔV = (4πr^2) * Δr
3. Substitute the given value for r (50 cm):
ΔV = (4π(50^2)) * 0.1
4. Use a calculator to evaluate the expression:
ΔV ≈ 12566.371 cm^3
Therefore, the maximum error in the volume is approximately 12566.371 cm^3.
Now let's calculate the maximum error in the surface area. The formula for the surface area of a sphere is:
A = 4πr^2
To find the maximum error in the surface area, you follow a similar process:
1. Find the derivative of the surface area formula with respect to r:
dA/dr = 8πr
2. Calculate the maximum error in the surface area:
ΔA = (dA/dr) * Δr
where ΔA is the maximum error in the surface area, Δr is the maximum error in the radius (0.1 cm), and (dA/dr) is the derivative of the surface area with respect to the radius.
ΔA = (8πr) * Δr
3. Substitute the given value for r (50 cm):
ΔA = (8π(50)) * 0.1
4. Use a calculator to evaluate the expression:
ΔA ≈ 125.664 cm^2
Therefore, the maximum error in the surface area is approximately 125.664 cm^2.
To summarize:
- The maximum error in the volume is approximately 12566.371 cm^3.
- The maximum error in the surface area is approximately 125.664 cm^2.