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Math Help - Maxima/ Minima

  1. #1
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    Maxima/ Minima

    The Question:
    A curve for for z>-1 is given by:

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

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

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    The attempt at a solution:
    \frac{\mathrm{d}z}{\mathrm{d}x} = \frac{1}{1+z}

    \frac{\mathrm{d}z}{\mathrm{d}y} = 2ze^{z^2}

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

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    My problem I have encountered:
    Can you check that it is correct so far and provide tips on how to differentiate further to find \frac{\mathrm{d}^2y}{\mathrm{d}x^2} and how I can find the turning point to prove that this must be a minimum.
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  2. #2
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    Quote Originally Posted by Air View Post
    The Question:
    A curve for for z>-1 is given by:

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

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

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    The attempt at a solution:
    \frac{\mathrm{d}{\color{red}x}}{\mathrm{d}{\color{  red}z}} = \frac{1}{1+z}

    \frac{\mathrm{d}{\color{red}y}}{\mathrm{d}{\color{  red}z}} = 2ze^{z^2}

    \therefore \frac{\mathrm{d}y}{\mathrm{d}x} = {\color{red}\frac{dy}{dz} \cdot \frac{dz}{dx}} = {\color{red}2 z e^{z^2} \, (1 + z)}

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    My problem I have encountered:
    Can you check that it is correct so far and provide tips on how to differentiate further to find \frac{\mathrm{d}^2y}{\mathrm{d}x^2} and how I can find the turning point to prove that this must be a minimum.
    Note the corrections (in red).

    To find the turning point, solve \frac{\mathrm{d}y}{\mathrm{d}x} = 0: ~ z = 0 (since z > -1) ....
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  3. #3
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    Quote Originally Posted by Air and revised by Mr F (see post #2) View Post
    The Question:
    A curve for for z>-1 is given by:

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

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

    __________________
    The attempt at a solution:
    \frac{\mathrm{d}x}{\mathrm{d}z} = \frac{1}{1+z}

    \frac{\mathrm{d}y}{\mathrm{d}z} = 2ze^{z^2}

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

    __________________
    My problem I have encountered:
    Can you check that it is correct so far and provide tips on how to differentiate further to find \frac{\mathrm{d}^2y}{\mathrm{d}x^2} and how I can find the turning point to prove that this must be a minimum.
    \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d} x} \left[ \frac{\mathrm{d}y}{\mathrm{d}x} \right] = \frac{\mathrm{d}}{\mathrm{d} z} \left[ \frac{\mathrm{d}y}{\mathrm{d}x} \right] \cdot \frac{\mathrm{d}z}{\mathrm{d}x} = \, ....

    (where I used the chain rule).

    Once you have \frac{\mathrm{d}^2y}{\mathrm{d}x^2} you can use the second derivative test to test the nature of the stationary point at z = 0.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d} x} \left[ \frac{\mathrm{d}y}{\mathrm{d}x} \right] = \frac{\mathrm{d}}{\mathrm{d} z} \left[ \frac{\mathrm{d}y}{\mathrm{d}x} \right] \cdot \frac{\mathrm{d}z}{\mathrm{d}x} = \, ....

    (where I used the chain rule).

    Once you have \frac{\mathrm{d}^2y}{\mathrm{d}x^2} you can use the second derivative test to test the nature of the stationary point at z = 0.
    \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2e^{z^2}(2z^3+2z^2+2z+1)(1+z) ?
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  5. #5
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    Quote Originally Posted by Air View Post
    \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2e^{z^2}(2z^3+2z^2+2z+1)(1+z) ?
    That's what I get too (and what are the chances we'd both be wrong ).

    By the way ..... I hope you differentiated \frac{\mathrm{d}y}{\mathrm{d}x} by re-writing it as e^{z^2} (2z + 2z^2) \, ....
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    That's what I get too (and what are the chances we'd both be wrong ).

    By the way ..... I hope you differentiated \frac{\mathrm{d}y}{\mathrm{d}x} by re-writing it as e^{z^2} (2z + 2z^2) \, ....
    If you've got the same answer, it's bound to be correct.

    Yes, differentiated very similarly but took out the constant 2.

    So I had u=e^{z^2}, \ v=z+z^2 and then when derivative was found, just multiplied it by 2 in the end.
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