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Homework Help: Graph Analysis

  1. Aug 1, 2008 #1

    Air

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    1. The problem statement, all variables and given/known data:
    A curve for for [itex]z>-1[/itex] is given by:

    [itex]x=\ln(1+z), \ y=e^{z^2}[/itex].

    Find [itex]\frac{\mathrm{d}y}{\mathrm{d}x}[/itex] and [itex]\frac{\mathrm{d}^2y}{\mathrm{d}x^2}[/itex] in terms of [itex]z[/itex] and show that the curve has only one turning point and that this must be a minimum.


    2. The attempt at a solution:
    [itex]\frac{\mathrm{d}z}{\mathrm{d}x} = \frac{1}{1+z}[/itex]

    [itex]\frac{\mathrm{d}z}{\mathrm{d}y} = 2ze^{z^2}[/itex]

    [itex]\therefore \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{(1+z)(2ze^{z^2})}[/itex]


    3. The problem I have encountered:
    Can you check that it is corerct so far and provide tips on how to differentiate further to find [itex]\frac{\mathrm{d}^2y}{\mathrm{d}x^2}[/itex].
     
  2. jcsd
  3. Aug 1, 2008 #2
    On the second line of your attempt, you have dz/dy = 2ze^(z^2) but the right hand side is the result of differentiating dy by dz. That should fix your third line.
     
    Last edited: Aug 1, 2008
  4. Aug 1, 2008 #3

    Air

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    [itex]\frac{\mathrm{d}x}{\mathrm{d}z} = \frac{1}{1+z}[/itex]

    [itex]\frac{\mathrm{d}y}{\mathrm{d}z} = 2ze^{z^2}[/itex]

    [itex]\therefore \frac{\mathrm{d}y}{\mathrm{d}x} = (1+z)(2ze^{z^2})[/itex]
     
    Last edited: Aug 1, 2008
  5. Aug 1, 2008 #4

    Air

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    Is this correct:

    [itex]\frac{\mathrm{d}^2y}{\mathrm{d}z^2} = 2e^{z^2} + 4z^2e^{z^2}[/itex]

    [itex]\frac{\mathrm{d}^2x}{\mathrm{d}z^2} = -\frac{1}{(1+z)^2}[/itex]

    [itex]\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4z^4e^{z^2} + 8z^3e^{z^2} + 6z^2e^{z^2} + 4ze^{z^2} + 2e^{z^2}[/itex]
     
  6. Aug 1, 2008 #5
    The dy/dx is correct. You have a slight typo on the second line (should be dy/dz), no biggie.

    Anyways I've never encountered such a second derivative but I would assume you would then take d/dx [(1+z)(2ze^(z^2))] and implicitly differentiate. This could get ugly, so first take the constant 2 out and probably would be a good idea to distribute the ze^(z^2) factor and applying the product rule twice.

    I think this way works, though now you do need dz/dx, which is just e^x. And then you could express everything in terms of z in the end. However I wouldn't doubt that there is a better way of doing this.
     
  7. Aug 1, 2008 #6

    Air

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    I've done it by using [itex]\frac{\mathrm{d}^2y}{\mathrm{d}z^2}[/itex] and [itex]\frac{\mathrm{d}^2x}{\mathrm{d}z^2}[/itex] then finding [itex]\frac{\mathrm{d}^2y}{\mathrm{d}x^2} [/itex]. Are my answer's posted above correct?
     
  8. Aug 1, 2008 #7
    I don't think that manipulation works for derivatives higher than the first, i.e. dy/dx = (dy/dz)/(dy/dz) but you have to differentiate the resulting expression with respect to x after that. Though I'm working out the final answer at the moment.
     
  9. Aug 1, 2008 #8
    OK, after taking d/dx [(1+z)(2ze^(z^2))] with the fact that dz/dx = e^x = e^[ln(z+1)] = z+1 and checking it, I got

    [tex]\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=[/tex]

    2[e^(z^2)](z+1)(2z^3 + 2z^2 + 2z +1)
     
  10. Aug 1, 2008 #9

    HallsofIvy

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    [tex]\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)= \frac{d}{dx}(1+z)(2ze^{z^2})[/tex]
    [tex]= \frac{dz}{dx}(2ze^{z^2})+ z\frac{d(2ze^z)}{dx}[/tex]
     
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