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Math Help - Determining equation from given graph

  1. #1
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    Determining equation from given graph


    An arched window with base width 2b and height h is set into a wall. The arch
    is to be either an arc of a parabola or a half-cycle of a cosine curve.

    a. If the arch is an arc of a parabola, write an equation for the parabola
    relative to the coordinate system shown in the figure. (x-intercepts are
    (−b,0) and (b,0). y-intercept is (0,h).)
    b. If the arch is a half-cycle of a cosine curve, write an equation for the
    cosine curve relative to the coordinate system shown in the figure.
    c. Of these two window designs, which has the greater area? Justify your
    answer.

    a.) I got y=-x^2+b^2?
    b.) I know that it might be something like cos(pi*bx/2) possibly?
    c.) I'm clueless how to do this one. Do i take the integral of each equation?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Standard Form Of a Quadratic Equation

    You know the standard equation of a quadratic equation?

    a(x-h)^2+k where (h,k) is the vertex of the parabola.

    Well this is obviously a Quadratic because the graph is a downward parabola which is symetrical with respect to the y-axis.

    Can you find a way to write this equation in standard form without anymore of my help?

    Hint: h is the last term of your eqution!
    Last edited by VonNemo19; May 1st 2009 at 07:03 PM. Reason: 1
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  3. #3
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    Oh I haven't done that for so long. It is a(x)^2+h?
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  4. #4
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    remember a is not defined in the given graph.

    Spoiler:
    try
    .

    lol. sorry this dosent work. how do you add a spoiler?

    Last edited by adhyeta; May 1st 2009 at 11:55 PM. Reason: spoiled.
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  5. #5
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    Oh ok that makes more sense. I knew b^2 had to be part of it.
    Last edited by ment2byours; May 1st 2009 at 08:28 PM. Reason: 1
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by ment2byours View Post

    An arched window with base width 2b and height h is set into a wall. The arch
    is to be either an arc of a parabola or a half-cycle of a cosine curve.

    a. If the arch is an arc of a parabola, write an equation for the parabola
    relative to the coordinate system shown in the figure. (x-intercepts are
    (−b,0) and (b,0). y-intercept is (0,h).)
    b. If the arch is a half-cycle of a cosine curve, write an equation for the
    cosine curve relative to the coordinate system shown in the figure.
    c. Of these two window designs, which has the greater area? Justify your
    answer.

    a.) I got y=-x^2+b^2?
    b.) I know that it might be something like cos(pi*bx/2) possibly?
    c.) I'm clueless how to do this one. Do i take the integral of each equation?
    a.) Well the two zeroes are b and -b, and the parabola is concave down, so the equation is f(x)=-k(x-b)(x+b). Since you know that f(0)=-k(-b)(b) = kb^2 = h, we can conclude that k=\frac{h}{b^2} So the equation you're looking for is f(x)=-\frac{h}{b^2}(x-b)(x+b).


    b.) If it's a cosine curve, and it takes 2b to complete a half-cycle, then it takes 4b to complete a full cycle. Thus we have that the period is \frac{2\pi}{4b}=\frac{\pi}{2b}. So the equation is g(x)=k\cos\left(\frac{\pi}{2b}x\right). Since g(0)=k=h, we have that g(x)=h\cos\left(\frac{\pi}{2b}x\right)


    c.) To see which has the greatest area, simply calculate the area for each one.

    -\frac{h}{b^2}\int_{-b}^b x^2-b^2\,dx = -\frac{h}{b^2}\left[\frac{x^3}{3}-b^2 x\right]_{-b}^b = \frac{4}{3}hb

    h\int_{-b}^b \cos\left(\frac{\pi}{2b}x\right)\,dx = h\left[\frac{2b}{\pi}\sin\left(\frac{\pi}{2b}x\right)\rig  ht]_{-b}^b = \frac{4}{\pi}hb

    Comparing \frac{4}{3}hb and \frac{4}{\pi}hb, we see that \frac{4}{3}hb > \frac{4}{\pi}hb, so the area of the parabola is bigger.


    adhyeta: You add a spoiler by wrapping spoiler tags around it "(spoiler)" and "(/spoiler)" (using square brackets of course, but I couldn't write them properly here or otherwise the system would actually put a spoiler in this explanation!)
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  7. #7
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    Quote Originally Posted by redsoxfan325 View Post

    adhyeta: You add a spoiler by wrapping spoiler tags around it "(spoiler)" and "(/spoiler)" (using square brackets of course, but I couldn't write them properly here or otherwise the system would actually put a spoiler in this explanation!)
    i tried that. dosent work. is the code case sensitive?
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  8. #8
    Super Member redsoxfan325's Avatar
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    I don't think so. In your post above, you have the close tag as [ /spoiler]. Delete the extra space to make it [/spoiler].
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  9. #9
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    Quote Originally Posted by redsoxfan325 View Post
    I don't think so. In your post above, you have the close tag as [ /spoiler]. Delete the extra space to make it [/spoiler].
    -wow-



    thnx!
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