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Hyperbolic functions and its tangent

  1. Jun 27, 2006 #1
    Just wanna check my answer.

    Find the equation of the tangent to the curve [tex]y^3 + x^2 \cosh y + \sinh^3 x = 8[/tex] at the point (0, 2)

    I firstly found the derivative and the gradient of the curve at point (0, 2)

    [tex]3y^2 \cdot \frac{dy}{dx} + x \cosh y + x^2 \sinh y \cdot \frac{dy}{dx} + 3 \sinh^3 x \cosh x = 0[/tex]
    [tex]\Rightarrow \frac{dy}{dx} (3y^2 + x^2 \sinh y) = - x \cosh y - 3 \sinh^3 x \cosh x[/tex]
    [tex]\Rightarrow \frac{dy}{dx} = \frac{- x \cosh y - 3 \sinh^3 x \cosh x}{3y^2 + x^2 \sinh y}[/tex]

    Substitute [tex]x = 0\ \mbox{and}\ y = 2[/tex] into the equation and I get a gradient of 0 at point (0, 2).

    y - 2 = 0 (x - 0) \\
    \Rightarrow y = 2

    Therefore, the equation of the tangent at point (0, 2) is [tex]y = 2[/tex]
    Last edited: Jun 27, 2006
  2. jcsd
  3. Jun 30, 2006 #2


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    Homework Helper

    You missed a 2 coeff on the xcoshy term in the first step, I fixed it in the quote: you answer is correct however.

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