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Thread: Very confused...(derivatives)

  1. #1
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    Very confused...(derivatives)

    I am working on some work that was assigned to me in my calculus class and I don't understand how I am supposed to start this problem. Can anyone guide me in the right direction or get me started off?

    Here's a link to the problem: ImageShack® - Online Photo and Video Hosting

    Not asking for anyone to do it for me, I just need some help getting started .

    Thanks
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  2. #2
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    let a point on the parabola's upper branch be $\displaystyle (x,y)$ ... note that symmetry will take care of the normal line on the lower branch

    slope between $\displaystyle (x,y)$ and $\displaystyle (a,0)$ ...

    $\displaystyle m = \dfrac{y}{x-a}$

    slope of the upper branch at any point $\displaystyle (x,y)$ ... $\displaystyle y' = \dfrac{1}{2y}$

    slope normal to any point $\displaystyle (x,y)$ ... $\displaystyle -\dfrac{1}{y'} = -2y$

    note that the normal slope equals the slope between $\displaystyle (x,y)$ and $\displaystyle (a,0)$ and that $\displaystyle x > 0$

    set up an equation, solve for $\displaystyle x$ in terms of $\displaystyle a$ and then sub into the inequality $\displaystyle x > 0$
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  3. #3
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    would this answer suffice?

    I worked on this for a while last night and came up with this for an answer...have I answered it correctly?


    3a. Let c > 0 be a point along the positive real axis. Then the tangent to the parabola at (c,sqrt(c)) has slope
    1/(2sqrt(c)) => the normal line has slope -2sqrt(c). point slope form: the normal line has equation
    y - sqrt(c) = -2sqrt(c)(x - c)

    this intercepts the x-axis where y = 0 => -sqrt(c) = -2sqrt(c)(a-c) => a = 1/2 + c.

    therefore a must be at least 1/2.

    3b. the normal lines are perpendicular when they have pi/4 angles to the real axis.
    this occurs when -2sqrt(c) = -1 => c = 1/4 => a = 3/4.
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  4. #4
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    That looks good to me.
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