The circumference of a sphere was measured to be 80 cm with a possible error of 0.5 cm.
1) Use differentials to estimate the maximum error in the calculated volume. (Round your answer to the nearest integer.)
2) What is the relative error? (Round your answer to three decimal places.)
v = 4/3 pi r^3
dv = 4 pi r^2 dr
c = 2pi r
dc = 2pi dr
so, dr = dc/(2pi)
meaning that
dv = 4 pi r^2 dc/(2pi)
= 2 r^2 dr
so, using the given numbers,
dv = 2*(80/2pi)^2 * 0.5 = 1600/pi^2
To estimate the maximum error in the calculated volume, we can use differentials. The formula for the volume of a sphere is given by V = (4/3)πr^3, where r is the radius of the sphere.
1) Estimating the maximum error in the calculated volume:
Let's differentiate the volume formula with respect to the radius:
dV = (4/3)π(3r^2)dr
The maximum error in the circumference is given as 0.5 cm. The radius can be found by dividing the circumference by 2π: r = C / (2π) = 80 cm / (2π) ≈ 12.73 cm.
Now, we can substitute the values into the differential equation:
dV ≈ (4/3)π(3(12.73)^2)dr
≈ 4π(12.73)^2dr
≈ 4π(12.73)(0.5)
≈ 255.46π
Rounding to the nearest integer, the maximum error in the calculated volume is 255π.
2) To calculate the relative error, we can divide the maximum error in the calculated volume by the actual volume of the sphere.
The actual volume of the sphere is V = (4/3)π(12.73)^3.
Relative error = (Maximum error in volume) / (Actual volume)
Relative error = (255π) / [(4/3)π(12.73)^3]
= 0.014 (rounded to three decimal places)
Therefore, the relative error is approximately 0.014.