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Math Help - differentiation 4 problem

  1. #1
    Senior Member furor celtica's Avatar
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    differentiation 4 problem

    find d^2/dx^2 as a function of x if siny + cosy = x
    i tried to solve this problem this way
    differentiating once makes cos y (dy/dx) - sin y (dy/dx) = 1
    dy/dx (cos y - siny) = 1
    differentiating this makes d^2y/dx^2 (cosy - siny) -dy/dx (sin y + cos y) = 0
    d^2y/dx^2 (cos y -siny) - x(dy/dx) = 0
    from there i isolate d^2y/dx^2, but apparently its wrong
    can anyone help?
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello furor celtica
    Quote Originally Posted by furor celtica View Post
    find d^2/dx^2 as a function of x if siny + cosy = x
    i tried to solve this problem this way
    differentiating once makes cos y (dy/dx) - sin y (dy/dx) = 1
    dy/dx (cos y - siny) = 1
    differentiating this makes d^2y/dx^2 (cosy - siny) -dy/dx (sin y + cos y) = 0
    d^2y/dx^2 (cos y -siny) - x(dy/dx) = 0
    from there i isolate d^2y/dx^2, but apparently its wrong
    can anyone help?
    Thanks for showing us your attempt. Your second differentiation is wrong:

    \frac{dy}{dx}(\cos y - \sin y )=1

    \Rightarrow \frac{d^2y}{dx^2}(\cos y - \sin y) - \frac{dy}{dx}(\sin y + \cos y)\frac{dy}{dx}=0

    Can you continue?

    (By the way, it would be much easier to read your work in LaTeX.)

    Grandad

    PS Had you thought of expressing \sin y + \cos y as \sqrt2\sin(y+\pi/4)? That way, you can get an explicit equation for y in terms of \arcsin x, which will probably make the final result easier to express in terms of x.
    Last edited by Grandad; November 23rd 2009 at 01:32 AM. Reason: Add PS
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  3. #3
    Senior Member furor celtica's Avatar
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    can you explain that part i got wrong? i dont get it. and isnt it superfluous to express siny + cosy as a compound angle sine when it is given 'siny + cosy = x'?
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  4. #4
    MHF Contributor
    Grandad's Avatar
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    Quote Originally Posted by furor celtica View Post
    can you explain that part i got wrong? i dont get it. and isnt it superfluous to express siny + cosy as a compound angle sine when it is given 'siny + cosy = x'?
    When you differentiated the second term of the product \frac{dy}{dx}(\cos y - \sin y), you wrote -(\sin y + \cos y), whereas it should be (as I wrote) -(\sin y +\cos y)\frac{dy}{dx}, because you're differentiating with respect to x, not y.

    \sin y + \cos y = x

    \Rightarrow \sqrt2\sin(y+\pi/4)=x

    \Rightarrow \sin(y+\pi/4)=\frac{x}{\sqrt2}

    \Rightarrow y+\pi/4 = \arcsin\left(\frac{x}{\sqrt2}\right)

    \Rightarrow y= \arcsin\left(\frac{x}{\sqrt2}\right)-\pi/4

    Differentiate this twice, and you're done.

    Grandad
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