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
Duplicate post. See "CALC"
C = 2 pi r
dr/dC = 1/(2pi)
A = 4 pi r^2
dA/dr = 8 pi r
dA/dC = dA/dr * dr/dC
dA/dC = 8 pi r *1/(2pi)
= 4 r
but r = C/(2 pi)
so
dA/dC = 4 C/(2 pi)
= 2 C/pi
dA = 2 * 74/pi * .5 = 74/pi
To estimate the maximum error in the calculated surface area, we can use linear approximation.
The formula for the surface area of a sphere is given by:
A = 4πr^2
where A is the surface area and r is the radius of the sphere.
First, we need to find the radius of the sphere using the circumference measurement. The formula for the circumference of a sphere is given by:
C = 2πr
where C is the circumference and r is the radius. Rearranging this formula, we have:
r = C / (2π)
Substituting the given circumference of 74.000 cm, we find:
r = 74.000 cm / (2π)
r ≈ 11.775 cm
Now that we have the radius, we can calculate the surface area of the sphere.
A = 4π(11.775 cm)^2
A ≈ 1732.40 cm^2
To estimate the maximum error in the calculated surface area, we need to consider the maximum error in the circumference measurement. In this case, the possible error is given as 0.50000 cm.
Since the error in the circumference measurement propagates to the surface area calculation, we can use linear approximation to estimate the maximum error in the surface area.
The linear approximation states that the relative error in a function of one variable is approximately equal to the relative error in the variable itself, multiplied by the absolute value of the derivative of the function with respect to the variable.
In this case, the function is the surface area A and the variable is the radius r. The derivative of the surface area function with respect to the radius is given by:
dA/dr = 8πr
Applying linear approximation, we can estimate the maximum error in the surface area as:
ΔA ≈ (dA/dr) * Δr
Substituting the values we found earlier, we get:
ΔA ≈ (8π * 11.775 cm) * (0.50000 cm)
ΔA ≈ 293.528 cm^2
Therefore, using linear approximation, we estimate the maximum error in the calculated surface area to be approximately 293.528 cm^2.