Thread: Derivatives: Differential

In a manufacturing process, ball bearings must be made with radius of 0.5 mm, with a maximum error in the radius of ±0.019 mm. Estimate the maximum error in the volume of the ball bearing.

Solution: The formula for the volume of the sphere is 4/3r^3Pi . If an error Dr is made in measuring the radius of the sphere, the maximum error in the volume is DV= 4(r+delta(r))^2pi-4^2pi
Rather than calculating DV, approximate DV with dV, where dV= 4r^2Pidr .

Replacing r with 0.5 and dr=Dr with ± 0.019 gives dV=± _______

The maximum error in the volume is about ________ mm3

The text in red are the answers i got, and the ones that are blank i dont know how to do. Can anyone tell me if the answers i obtained are correct....?

Originally Posted by lemontea

In a manufacturing process, ball bearings must be made with radius of 0.5 mm, with a maximum error in the radius of ±0.019 mm. Estimate the maximum error in the volume of the ball bearing.

Solution: The formula for the volume of the sphere is 4/3r^3Pi .

ok

If an error Dr is made in measuring the radius of the sphere, the maximum error in the volume is DV= 4(r+delta(r))^2pi-4^2pi

wouldn't it be ?

Rather than calculating DV, approximate DV with dV, where dV= 4r^2Pidr .

ok

Replacing r with 0.5 and dr=Dr with ± 0.019

yup

gives dV=± _______

just plug in the values and calculate this. you have the formula for dV, what do you get when you plug in r = 0.5 and dr = 0.019, and then when you plug in r = 0.5 and dr = -0.019 ?

The maximum error in the volume is about ________ mm3

just choose the larger dV value (larger in terms of absolute value, i would think)