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Thread: Calculus:Derivatives:Differential (I am stuck on this question help PLEASE)

  1. #1
    nch
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    Calculus:Derivatives:Differential (I am stuck on this question help PLEASE)

    In a manufacturing process, ball bearings must be made with radius of 0.4 mm, with a maximum error in the radius of 0.013 mm. Estimate the maximum error in the volume of the ball bearing.
    Solution: The formula for the volume of the sphere is________. If an error ∆r is made in measuring the radius of the sphere, the maximum error in the volume is ∆V=__________.
    Rather than calculating ∆V, approximate ∆V with dV, where dV=__________.
    Replacing r with___ and dr=∆r with ____ gives dV= ____.
    The maximum error in the volume is about____mm^3.
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  2. #2
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    Hello, nch!

    Some of this is just arithmetic . . .


    In a manufacturing process, ball bearings must be made with $\displaystyle r \,=\,0.4\text{ mm,}$
    with a maximum error in the radius of: .$\displaystyle \Delta r = dr = \pm0.013\text{ mm.}$

    Estimate the maximum error in the volume of the ball bearing.

    Solution: The formula for the volume of the sphere is: .$\displaystyle \boxed{V \:=\:\tfrac{4}{3}\pi r^3}$

    If an error $\displaystyle \Delta r$ is made in measuring the radius of the sphere,
    the maximum error in the volume is: .$\displaystyle \Delta V \:=\:\boxed{ {\color{white}XXX} }$
    The volume of the correct sphere is: .$\displaystyle V_o \:=\:\tfrac{4}{3}\pi (0.4)^3 \:=\:\frac{0.256}{3}\pi\text{ mm}^3$

    With an error of $\displaystyle \Delta r = 0.013$, the radius is: .$\displaystyle r \:=\:0.4 + 0.013 \:=\:0.413\text{ mm.}$
    And the volume is: .$\displaystyle V_1 \;=\;\tfrac{4}{3}\pi(0.413)^3 \;\approx\;\frac{0.28178}{3}\pi\text{ mm}^3$

    The error in volume is: .$\displaystyle \Delta V \;=\;\frac{0.28178}{3}\pi - \frac{0.256}{3}\pi$

    . . Hence: .$\displaystyle \Delta V \;=\;0.00859\pi\text{ mm}^3$




    Rather than calculating $\displaystyle \Delta V$, approximate $\displaystyle \Delta V$ with $\displaystyle dV$, where: .$\displaystyle dV \:=\:\boxed{4\pi r^2dr} $

    Replacing $\displaystyle r$ with $\displaystyle 0.4$,and $\displaystyle dr =\Delta R$ with $\displaystyle \pm0.013$ gives: .$\displaystyle dV \:=\:\boxed{{\color{white}XX}}$
    $\displaystyle dV \;=\;4\pi(0.4)^2(0.13) \;=\;0.00832\pi\text{ mm}^3$

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  3. #3
    nch
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    thankz a lot
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