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Math Help - Hyperbolic function

  1. #1
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    Hyperbolic function

    Given that y = sinh x + k cosh x, show that the least value of y is √(k^2 - 1) and that this occurs at x = ln (k-1)/(k+1) where k is a constant and |k| > 1.

    What is mean by the least value?

    My solution:

    <br />
y = \frac {1}{2} (e^x - e^{-x}) + k \frac {1}{2} (e^x + e^{-x})
    = \frac {1}{2}(e^x - e^{-x} + ke^x + ke^{-x})
    = \frac {1}{2} \left(e^x (1+k) - e^{-x} (1-k)\right)
    = \frac {1}{2} \left( \left(\frac {k-1}{k+1}\right)^\frac{1}{2}  (k+1) + \left(-\frac{k-1}{k+1}\right)^\frac{1}{2} (k-1)\right) [Substitute x = ln (k-1)/(k+1) ]

    By simplifying this I got \left(\frac {k-1}{k+1}\right)^\frac {1}{2}

    Is that a right way? Now what to do?
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  2. #2
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    Quote Originally Posted by geton View Post
    Given that y = sinh x + k cosh x, show that the least value of y is √(k^2 - 1) and that this occurs at x = ln (k-1)/(k+1) where k is a constant and |k| > 1.

    What is mean by the least value?

    My solution:

    <br />
y = \frac {1}{2} (e^x - e^{-x}) + k \frac {1}{2} (e^x + e^{-x})
    = \frac {1}{2}(e^x - e^{-x} + ke^x + ke^{-x})
    = \frac {1}{2} \left(e^x (1+k) - e^{-x} (1-k)\right)
    = \frac {1}{2} \left( \left(\frac {k-1}{k+1}\right)^\frac{1}{2}  (k+1) + \left(-\frac{k-1}{k+1}\right)^\frac{1}{2} (k-1)\right) [Substitute x = ln (k-1)/(k+1) ]

    By simplifying this I got \left(\frac {k-1}{k+1}\right)^\frac {1}{2}

    Is that a right way? Mr F says: No.

    Now what to do?
    1. Show that y has a single stationary point by finding the solution single solution to dy/dx = 0.

    2. Show that this stationary point is a minimum turning point by testing its nature.

    3. You have the x-coordinate of the turning point, now find the y-coordinate.
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  3. #3
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    Yes, I solved that. Thanks.
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