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Thread: I just want to ask if my answer on this problem (area of a plane region) is correct?

  1. #1
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    I just want to ask if my answer on this problem (area of a plane region) is correct?

    Hi everyone. I tried solving this problem:

    "Get the area of the region bounded by the upper branch of the parabola x=y2, the tangent line to the parabola at (1,1) and the x-axis."

    Attached is a picture of my solution to the problem. I just want to ask if I did the right thing, and if not, what did I do wrong and how do I fix it?

    Thank you very much for your help!
    Attached Thumbnails Attached Thumbnails I just want to ask if my answer on this problem (area of a plane region) is correct?-document-2_2-1-.jpg  
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  2. #2
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    Re: I just want to ask if my answer on this problem (area of a plane region) is corre

    You need to find the equation of the tangent line, and then find where it cuts the x-axis (say "a").


    Then subtract the area under the upper branch of the parabola $\displaystyle y=\sqrt{x}$ (from 0 to 1) from the area under the tangent line from "a" to 1.
    Thanks from kaonashi
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  3. #3
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    Re: I just want to ask if my answer on this problem (area of a plane region) is corre

    Quote Originally Posted by Debsta View Post
    You need to find the equation of the tangent line, and then find where it cuts the x-axis (say "a").


    Then subtract the area under the upper branch of the parabola $\displaystyle y=\sqrt{x}$ (from 0 to 1) from the area under the tangent line from "a" to 1.
    Thank you so much for your response! I managed to get the tangent line equation and "a". I'm wondering why I have to subtract the areas though? Based on what I can see on my notes, split areas area always added. Thank you in advance.

    EDIT: Now I understand!
    Last edited by kaonashi; Apr 3rd 2019 at 07:00 AM.
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  4. #4
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    Re: I just want to ask if my answer on this problem (area of a plane region) is corre

    kaonashi,

    you don't have to split areas. You can subtract from right to left. Suppose you worked out the tangent line at (1, 1) to be $\displaystyle \ y = \dfrac{1}{2}x + \dfrac{1}{2}$.

    Solve this for x in terms of y: $\displaystyle \ \ $x = 2y - 1

    Keep the form of $\displaystyle \ x = y^2 \ $ corresponding to the parabola. From y = 0 to y = 1, the upper branch of the parabola $\displaystyle \ x = y^2 \ $ is to the right of the tangent line $\displaystyle \ x = 2y - 1$.

    The integrand is $\displaystyle \ [y^2 \ - \ (2y - 1)]dy. \ $ The limits of integration are from y = 0 up to y = 1.



    By factoring the quantity in the brackets, the antiderivative can be simplified to the point that when you substitute your limits of integration,
    you will be doing the minimum of arithmetic.
    Last edited by greg1313; Apr 3rd 2019 at 08:13 AM.
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