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Word problem involving function expression

  1. Mar 1, 2012 #1
    1. The problem statement, all variables and given/known data

    A 300-foot-long cable, originally of diameter 5 inches, is submerged in seawater. Because of corrosion, the
    surface area of the cable diminishes at the rate of 1250 in 2 /year. Express the diameter d of the cable as a
    function of time t (in years).

    2. Relevant equations

    C=π*d

    Surface area=C*length (in inches)

    3. The attempt at a solution

    I understand (I hope correctly) that the form of the equation has to be

    [itex]d=initial value - the rate of deminution[/itex] , hence

    [itex]d=5 - the rate of diminution[/itex]

    How do I construct the formula for the rate of diminution?

    [itex]\frac{πd*l}{1250*t}[/itex] ?
     
  2. jcsd
  3. Mar 1, 2012 #2

    HallsofIvy

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    The surface area of the cable is [itex]\pi d h[/itex] and are given that h=300 and does not change so [itex]A= 300\pi d[/itex] If we let [itex]\Delta d[/itex] be the change in diameter in one year, then the change in surface are is [itex]\Delta A= 300\pi (d+ \Delta d)- 300\pi d= 300\pi\delta d= -625[/itex].
    So [itex]\Delta d= \frac{6}{12\pi}[/itex].

    Now, what will the diameter be after t years?
     
    Last edited: Mar 1, 2012
  4. Mar 1, 2012 #3
    Before trying to answer your question, could you explain the formula: [itex]\Delta A= 300\pi (d+ \Delta d)- 300\pi d= 300\pi\delta d= -625[/itex]

    Namely: why do you add [itex]\Delta d[/itex]? And what is this business with " - 300\pi d= 300\pi\delta d= -625 " ... how did you construct this? I guess that you use a (pseudo-) difference quotient and taking a limit so the delta becomes the small delta? But I have to remind you that it is for a reason that I wrote this problem in pre-calculus section.
     
    Last edited: Mar 1, 2012
  5. Mar 2, 2012 #4
    I still need a helping hand here ... can someone help?
     
  6. Mar 2, 2012 #5
    Ok, so thus far I have discerned.

    (1) We assume that only d changes because of the corrosion, therefore

    (2) ΔA is due only to Δd.

    (3) We know ΔA's numeric value, therefore it is useful to express it in terms of what is known e. g. [itex]ΔA=πΔdL[/itex] (L is lenght). This is true since only d contributes to the changing A by (1).

    (4) Since [itex]A=πdL[/itex] we get

    (5) [itex]A-ΔA=πdL-πΔdL[/itex] (The reason why we have to do it is unclear to me, I can not give sufficient grounds for it, the only one is that it is the only possible way to achieve the result of getting d). But this is the same as:

    (6) [itex]-ΔA=πdL+πΔdL-A[/itex] This is

    (7) [itex]-ΔA=πL(d+Δd)-A[/itex]

    (8) [itex]-ΔA=πL(d+Δd)-πLd[/itex]

    (9) [itex]-ΔA=πL((d+Δd)-d)[/itex]

    (10) Now I convert 300 ft into 3600 inches and get

    (11) [itex]-\frac{1250}{3600π}=Δd[/itex]

    Why is this negative? Is the procedure correct?
     
  7. Mar 2, 2012 #6
    Sorry there was a mistake in the problem statement. The part: "cable diminishes at the rate of 1250 in 2 /year" should be: "cable diminishes at the rate of 1250 in^2/year" (in^2 - added). That is one of the reasons of misunderstanding.
     
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