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Thread: Derivative Help

  1. #1
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    Derivative Help

    Hello All,

    Here is my equation:$\displaystyle y=\pm \sqrt{4-x^2}$

    I'm supposed to find the value of x where the slope = 1

    Here is what I have so far...

    $\displaystyle y=(4-x^2)^{1/2}$

    $\displaystyle R=(4-x^2)$

    $\displaystyle y=R^{1/2}$

    $\displaystyle \frac{dy}{dR}=\frac{1}{2\sqrt{4-x^2}}$

    $\displaystyle \frac{dR}{dx}=-2x$

    $\displaystyle \frac{dy}{dx}=\frac{-2x}{2\sqrt{4-x^2}}$

    $\displaystyle \frac{dy}{dx}=\frac{-x}{\sqrt{4-x^2}}$

    $\displaystyle 1=\frac{-x}{\sqrt{4-x^2}}$

    Should I rationalize the denominator next?

    $\displaystyle 1=\frac{-x}{\sqrt{4-x^2}}$

    $\displaystyle 1=\frac{-x}{\sqrt{4-x^2}}$ X $\displaystyle \frac{\sqrt{4-x^2}}{\sqrt{4-x^2}}$

    $\displaystyle 1=\frac{-x\sqrt{4-x^2}}{4-x^2}$

    This is where I am stuck. Can anyone give me a tip on where to go next?
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  2. #2
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    No need to rationalize the denominator.

    sqrt(4 - x^2) = -x
    Square both sides. Simplify and solve for x.
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  3. #3
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    $\displaystyle 1=\frac{-x}{\sqrt{4-x^2}}$

    $\displaystyle \sqrt{4-x^2}=-x$

    $\displaystyle (\sqrt{4-x^2})^2=(-x)^2$

    $\displaystyle 4-x^2=x^2$

    $\displaystyle 4=2x^2$

    $\displaystyle 2=x^2$

    $\displaystyle \pm\sqrt{2}=x$

    Is this correct?
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  4. #4
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    Quote Originally Posted by dbakeg00 View Post
    $\displaystyle 1=\frac{-x}{\sqrt{4-x^2}}$

    $\displaystyle \sqrt{4-x^2}=-x$

    $\displaystyle (\sqrt{4-x^2})^2=(-x)^2$

    $\displaystyle 4-x^2=x^2$

    $\displaystyle 4=2x^2$

    $\displaystyle 2=x^2$

    $\displaystyle \pm\sqrt{2}=x$

    Is this correct?
    Well, check it in the original equation! If $\displaystyle x= \sqrt{2}$ then $\displaystyle x^2= 2$ so $\displaystyle \sqrt{4- x^2}= \sqrt{4- 2}= \sqrt{2}$
    $\displaystyle \frac{x}{\sqrt{4- x^2}}= \frac{\sqrt{2}}{\sqrt{2}}= 1$

    On the other hand, if $\displaystyle x= -\sqrt{2}$, $\displaystyle \sqrt{4- x^2}$ is still $\displaystyle \sqrt{2}$ but now we have $\displaystyle \frac{x}{\sqrt{4- x^2}}= \frac{-\sqrt{2}}{\sqrt{2}}= -1$ which is not 1- but your original equation allowed "$\displaystyle \pm$".

    Actually, your original equation had $\displaystyle y= \pm\sqrt{4- x^2}$ which is equivalent to $\displaystyle x^2+ y^2= 4$, a circle with center at (0,0) and radius 2. The tangent line will have slope 1 where that circle intersects the line with slope -1, y= -x. If y= -x, $\displaystyle y^2= x^2$ and $\displaystyle x^2+ y^2= x^2+ x^2= 2x^2= 4$ so $\displaystyle x^2= 2$ and $\displaystyle x= \pm\sqrt{2}$ but notice that the tangent line has slope 1 at $\displaystyle (\sqrt{2}, -\sqrt{2})$ and $\displaystyle (-\sqrt{2}, \sqrt{2}$. The points $\displaystyle (\sqrt{2}, \sqrt{2})$ and $\displaystyle (-\sqrt{2}, \sqrt{2})$ also are on that circle, have $\displaystyle x= \pm\sqrt{2}$ but the tangent lines do not have slope 1 there.

    You know, you really did not need to open a new thread just to continue
    http://www.mathhelpforum.com/math-he...erivative.html

    By the way, what is a "geographic" derivative?
    Last edited by HallsofIvy; Apr 2nd 2010 at 06:23 PM.
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  5. #5
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    By the way, what is a "geographic" derivative?
    I meant geometric derivative...thats what the book I'm self studying from calls it, not my title. Sorry about the typo.

    Thank you for the help. I haven't studied circles much, so equations like that are semi-foreign to me.
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