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Calculus Word Question

  1. Jan 15, 2007 #1
    Okay, the following is just me trying to confirm if I did the problem correctly. It seems like I made a mistake somewhere, but the answer does sort of fit. So I suppose I'm just look for a check to see if I did this correctly. The equations I got from the world problem are top right, I then move into the problem on the left side and finish on the right. :D

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

    [​IMG]

    3. The attempt at a solution

    [​IMG]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 15, 2007 #2

    HallsofIvy

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    At one point you have
    which is correct. The next line is
    which is wrong: to multiply a fraction by 3, you multiply the numerator, not both numerator and denominator!
    Again wrong: you divide by the 3. But since that 3 is wrong to begin with, you've just made it worse. What you should have is P= 2x+ 3600 x-1.

    Next you have
    that "5400" is your 10800 divided by 2. Apparently you have divided through by 2. But you have to divide both sides of the equation by 2. This would be P/2, not P. Anyway, it's best to leave it as P= 2x+ 3600x-1 and differentiate that.
    Next, you have
    No, the "-1" exponent applies only to x, not the "5400". There is, in any case, no point in moving the 1/x up to x-1 only to put it back in the denominator. Go ahead and differentiate P= 2x+ 3600x-1.

    Unfortunately, having got the incorrect fraction, you add the fractions to get
    which is incorrect. I only mention it because you then have
    No, no, no!! The derivative of [itex]\frac{f}{g}[/itex] is NOT [itex]\frac{f'}{g'}[/itex]!
    As if that weren't bad enough, from
    you go to
    and solve that to get x= 1/2. No! 5400*0= 0 not 5400!

    Better is to take P= 2x+ 3600x-1 and differentiate it directly.
    P'= 2- 3600x-2= 0. Then x2= 1800. Solve that for x.
     
  4. Jan 15, 2007 #3
    Thanks for the reply HallsofIvy, I just wanted to check in on something. All of your corrections make sense, and clearly my brain isn't functioning today because those are all mostly things I should have caught myself doing.

    But here's what happens when I try to go forward with P' = 2 - 3600x^-2

    [​IMG]

    So from there sub 0 for the P'. At which poin I can move the -3600x^-2 over to the left which gives me 3600x^-2 = 2

    To isolate x do I not divide by 3600 both sides and then take the -2root of the right side? Perhaps, I'm just looking at everything wrong, I'm not sure what it is that's throwing me off.

    I know If I go with your equation above:
    x^2= 1800
    then x = 42.4 but I'm hoping to find out what I'm doing wrong that I can't even get to that point.
     
  5. Jan 15, 2007 #4

    cristo

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    You are at that point! When you work out your value for x it will be 42.4! But, to do it the way that HoI did (and to keep things as fractions, which is always more a more exact way of calculating things) you have the line

    Note that [tex]\frac{2}{3600}=\frac{1}{1800}[/tex]. Now, take the reciprocal of both sides, and you will obtain x2=1800.

    A quick comment about your notation in the penultimate line: The right hand side would generally be written as [itex](5.56 \times 10^{-4})^{-1/2}[/itex]
     
  6. Jan 15, 2007 #5

    HallsofIvy

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    Just seconding what cristo said: you have 3600x-2- 2= 0 so
    3600x-2= 2 or x-2= 2/3600= 1/1800. That means that x2= 1800= (2)9(100). [itex]x= \pmsqrt{2}(3)(10)= \pm30\sqrt{2}[/itex]. That's approximately 42.4. (Since x is a distance, it must be positive.)
     
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