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Thread: Find the gradient of the normal

  1. #1
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    Find the gradient of the normal

    This question, in a list of past examination questions, has me stuck. It says:


    "For the curve $\displaystyle y= \sqrt{\frac{1-x}{1+x } } $,

    Obtain an expression for $\displaystyle \frac{dy}{dx }$, by first differentiating $\displaystyle \frac{1-x}{1+x }$.

    Thus, show that the gradient of the normal to the curve at the point $\displaystyle (x,y)$ is given by $\displaystyle (1+x)\sqrt{(1-x^2)} $."


    Differentiating $\displaystyle \frac{1-x}{1+x }$ as told, I get $\displaystyle \frac{2x}{(1+x)^2}$.

    Using this and the chain rule to differentiate y, I get $\displaystyle \frac{1}{2}\sqrt{(\frac{1-x}{1+x})}\frac{2x}{(1+x)^2}$.

    Some algebraic gymnastics makes this simplify to $\displaystyle \frac{x}{(1+x)\sqrt{1-x^2}}$, for the gradient of the tangent to the curve...

    But I cannot eliminate the x in the numerator to arrive at the given expression for the gradient of the normal to the curve, even after checking all working carefully. It is produced by the differential to $\displaystyle \frac{1-x}{1+x }$ and that seems correct. Several friends have tried and failed as well. Is there something very obvious that we're missing?
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  2. #2
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    Just realized that differentiating $\displaystyle (1-x)$ gives -1, not 1 as I thought. This eliminates the x term in the numerator. The littlest things get you sometimes!
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