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Thread: Another Related Rates

  1. March 12th 2008, 05:55 PM #1
    smschaefer
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    Another Related Rates

    A 100ft. long cable of diameter 4 inches is submerged in seawater. Due to corrosion, the surface area of the cable decreases at a rate of 750 in^2 / year. Ignoring the corrosion at the ends of the cable, find the rate at which the diameter is decreasing.
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  2. March 12th 2008, 08:38 PM #2
    o_O
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    Surface area (S) of a cylinder: S = 2\pi r^{2} + 2\pi r h

    Since you are dealing with the diameter, it would be nice to express the radius in terms of its diameter (D):

    S = 2\pi \left(\frac{1}{2}D\right)^{2} + 2 \pi \left(\frac{1}{2}D\right)h
    S = \frac{\pi}{2}D^{2} + \pi Dh

    Differentiate with respect to time:
    \frac{dS}{dt} = \frac{\pi}{2} \underbrace{\left(2D \frac{dD}{dt}\right)}_{\mbox{Chain Rule}} + \pi\underbrace{\left(h \frac{dD}{dt} + D\frac{dh}{dt}\right)}_{\mbox{Product Rule}}

    And solve for \frac{dD}{dt}
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  3. March 13th 2008, 05:03 AM #3
    colby2152
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    Quote Originally Posted by smschaefer View Post
    A 100ft. long cable of diameter 4 inches is submerged in seawater. Due to corrosion, the surface area of the cable decreases at a rate of 750 in^2 / year. Ignoring the corrosion at the ends of the cable, find the rate at which the diameter is decreasing.
    Always nice to see a fellow Schaeffer (or any other spelling) here!
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  4. March 13th 2008, 06:32 AM #4
    smschaefer
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    Quote Originally Posted by o_O View Post
    Surface area (S) of a cylinder: S = 2\pi r^{2} + 2\pi r h

    Since you are dealing with the diameter, it would be nice to express the radius in terms of its diameter (D):

    S = 2\pi \left(\frac{1}{2}D\right)^{2} + 2 \pi \left(\frac{1}{2}D\right)h
    S = \frac{\pi}{2}D^{2} + \pi Dh

    Differentiate with respect to time:
    \frac{dS}{dt} = \frac{\pi}{2} \underbrace{\left(2D \frac{dD}{dt}\right)}_{\mbox{Chain Rule}} + \pi\underbrace{\left(h \frac{dD}{dt} + D\frac{dh}{dt}\right)}_{\mbox{Product Rule}}

    And solve for \frac{dD}{dt}
    but would you use the lateral surface area of a cylinder (2 pi r h) or the regular surface area of a cylinder, because corrosion doesnt happen at the ends
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