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Math Help - Equation of a line tangent

  1. #1
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    Equation of a line tangent

    What is the value of the constant c so that x-y=c is the equation of the line tangent to the graph f(x)=In(Sin(x)) where 0<x<pi

    I know that line x-y=c has a slope of 1, so therefore f'x = 1
    I do not what to do next
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by 05holtel View Post
    What is the value of the constant c so that x-y=c is the equation of the line tangent to the graph f(x)=In(Sin(x)) where 0<x<pi

    I know that line x-y=c has a slope of 1, so therefore f'x = 1
    I do not what to do next
    Take the derivative:

    \frac{d}{dx}\left[\ln{(\sin{x})}\right]=\cot{x}\Rightarrow\cot{x}=1\Rightarrow\cos{x}=\si  n{x}\Rightarrow{x}=\frac{\pi}{4}
    Last edited by VonNemo19; July 26th 2009 at 06:27 PM. Reason: Made a boo-boo.
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    Take the derivative:

    \frac{d}{dx}\left[\ln{(\sin{x})}\right]=\cot{x}\Rightarrow\cot{x}=1\Rightarrow\cos{x}=\si  n{x}\Rightarrow{x}=\frac{\sqrt{2}}{2}
    correction ...

    x = \frac{\pi}{4}
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by skeeter View Post
    correction ...

    x = \frac{\pi}{4}
    Thank you. Got a little ahead of myself...
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