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Math Help - Tangent & Normal(Q6)

  1. #1
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    Tangent & Normal(Q6)

    Given that \frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-x)^2} for the curve y=\frac{x}{\sqrt{x}-2}, obtain the equation of the normal to the curve y=\frac{x}{\sqrt{x}-2} at the point on the curve where x=1.

    I have attempted the question by finding the value of y when x=1. Then using the formula (y-y_1)=\frac{dy}{dx}(x-x_1). However, i am unable to simplify it...

    Hence, i require a detailed working to examine the problem, thanks!
    Last edited by Punch; February 17th 2010 at 08:09 AM.
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  2. #2
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    Quote Originally Posted by Punch View Post
    Given that \frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-\color{red}2)^2} for the curve y=\frac{x}{\sqrt{x}-2}, obtain the equation of the normal to the curve y=\frac{x}{\sqrt{x}-2} at the point on the curve where x=1.

    I have attempted the question by finding the value of y when x=1. Then using the formula (y-y_1)=\frac{dy}{dx}(x-x_1). However, i am unable to simplify it...

    Hence, i require a detailed working to examine the problem, thanks!
    i think you have a typo (red) , when x=1 , dy/dx=-3/2 , so gradient of normal=2/3

    the normal cuts the curve at (1,-1)

    equation : y+1=(2/3)(x-1)
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  3. #3
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    Hi, mathaddict, please note that there isn't any typo. I have already tried differentiating the curve and proved it to be <br />
\frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-2)^2}<br />
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  4. #4
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    Quote Originally Posted by Punch View Post
    Hi, mathaddict, please note that there isn't any typo. I have already tried differentiating the curve and proved it to be <br />
\frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-2)^2}<br />

    \frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-x)^2} , this is what you wrote initially , then i tried differentiating myself and hv checked with some calculus engines, i got this one

    \frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-2)^2}
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  5. #5
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    Quote Originally Posted by mathaddict View Post
    \frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-x)^2} , this is what you wrote initially , then i tried differentiating myself and hv checked with some calculus engines, i got this one

    \frac{dy}{dx}=\frac{\sqrt{x}-4}{2(\sqrt{x}-2)^2}
    Hi, sorry but I do not see any differences in the 2 deriviative.
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  6. #6
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    Quote Originally Posted by Punch View Post
    Hi, sorry but I do not see any differences in the 2 deriviative.
    compare the denominators , towards the right , do you see the difference ?
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  7. #7
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    Yes, I see it.
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