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Math Help - Derivatives: Differential

  1. #1
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    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....?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lemontea View Post
    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 \frac 43 \pi (r + \Delta r)^3 - \frac 43 \pi r^3 ?

    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)

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