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Math Help - Differentiation

  1. #1
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    Differentiation

    For the curve with equation y = arsinh (x + 1), find
    (a) the coordinates of its point of inflexion P
    (b) an equation of the normal to the curve at P.


    For (a), I know that if dy/dx = 0 and d^2 y/dx^2 = 0, but d^3 y/dx^3 ≠ 0, then the point is a point of inflexion. But now how can I start to solve this?

    For (b), after finding P, it may be easy.
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  2. #2
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    Quote Originally Posted by geton View Post
    For the curve with equation y = arsinh (x + 1), find
    (a) the coordinates of its point of inflexion P
    (b) an equation of the normal to the curve at P.


    For (a), I know that if dy/dx = 0 and d^2 y/dx^2 = 0, but d^3 y/dx^3 ≠ 0, then the point is a point of inflexion. But now how can I start to solve this?

    For (b), after finding P, it may be easy.
    You start by getting dy/dx ..... Can you do that?
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    Quote Originally Posted by mr fantastic View Post
    You start by getting dy/dx ..... Can you do that?

    Yes. Its done. Thanks.
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    Quote Originally Posted by geton View Post
    Yes. Its done. Thanks.
    Right then.

    \frac{dy}{dx} = \frac{1}{\sqrt{1 + (x + 1)^2}} = \frac{1}{\sqrt{x^2 + 2x + 2}} = (x^2 + 2x + 2)^{-1/2}


    \Rightarrow \frac{d^2y}{dx^2} = -\frac{1}{2} (x^2 + 2x + 2)^{-3/2} \, (2x + 2) = -\frac{1}{2} \, \frac{2x+2}{(x^2 + 2x + 2)^{3/2}}.


    \frac{d^2y}{dx^2} = 0 \Rightarrow 2x+2 = 0 \Rightarrow x = -1.


    Rather than calculating the third derivative, you can test the nature of this solution by examining whether it corresponds to a turning point of \frac{dy}{dx}. This is easily done using the usual sign test .....
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  5. #5
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    Quote Originally Posted by geton View Post
    For the curve with equation y = arsinh (x + 1), find
    (a) the coordinates of its point of inflexion P
    [snip]
    Alternatively, you could use the 'well known' fact that y = arsinh(x) has a point of inflexion at (0, 0) and then apply a translation of 1 unit to the left .....
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Alternatively, you could use the 'well known' fact that y = arsinh(x) has a point of inflexion at (0, 0) and then apply a translation of 1 unit to the left .....
    Ya its good too.
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