Area of Rectangle As Function of x

Discussion in 'Algebra' started by nycmathguy, Jul 4, 2021.

  1. nycmathguy

    nycmathguy

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    See attachment.

    I say the area in terms of x and y is
    A = length • width or x•y.

    I know that y = sqrt{36 - x^2}.

    A(x) = x•y

    Replace y with sqrt{36 - x^2}.

    A(x) = x•sqrt{36 - x^2}

    Is this function?

    If not, why not?

    See attachment for domain.

    I say the domain is (-00, -6] U [6, 00).

    Yes?

    20210702_031754.jpg 20210703_190846.jpg
     
    nycmathguy, Jul 4, 2021
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  2. nycmathguy

    MathLover1

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    A(x) = x•sqrt{36 - x^2}=> correct


    => x=6 or x=-6=> points are (6,0) and (-6,0)-> so, both are in domain
    so, the domain is is [-6,6]
    or
    {x element R : -6<=x<=6}
     
    MathLover1, Jul 4, 2021
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  3. nycmathguy

    nycmathguy

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    The domain is on the line y = 0 from -6 to 6.
     
    nycmathguy, Jul 4, 2021
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  4. nycmathguy

    MathLover1

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    The domain is on the line x from -6 to 6 (take a look at graph)
     
    MathLover1, Jul 4, 2021
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  5. nycmathguy

    nycmathguy

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    The x-axis is the line y = 0.

    Yes?
     
    nycmathguy, Jul 4, 2021
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  6. nycmathguy

    MathLover1

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    correction to the formula for area, there was a mistake

    length= x units left fro origin and x units right from origin: so L=x+x=2x

    width=W=y=sqrt(36-x)^2

    we know that area of rectangle is length*width

    A=L*W

    now, we can plug values

    A=2x*sqrt(36-x)^2 see attached to understand why is 2x in formula
     

    Attached Files:

    MathLover1, Jul 4, 2021
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  7. nycmathguy

    nycmathguy

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    My equation is missing number 2. Your graph makes clear why 2x and not x alone is multiplied by
    sqrt{36 - x^2} to form the needed formula.

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    Note 1: I will use Precalculus With Unit-Circle Trigonometry by the late David Cohen and our Ron Larson Precalculus textbook. Both textbooks will be used to deeply explore precalculus as we make our way to calculus 1.

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    I also have decided to join calculusworkshop (yearly plan) to watch precalculus videos by math expert Jenn. I am going to do everything within my power to "master" or better yet, increase my knowledge of this wonderful course.
     
    nycmathguy, Jul 4, 2021
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