calculus prequisities...

Discussion in 'General Math' started by Antivivisectionist unreal email, Sep 7, 2003.

  1. first, I've moved to another school district on my senior year, and not
    taking/didn't take statistic, but I did take precalculus/trigonometry,
    because in this new school, it's precalc and stats as one class. Anyway,
    aside that story...

    There's a first section of the book that's called "prequisities"
    (spelling?), and its based on precalculus, not sure about statistics. The
    problem I'm having is dealing with a circle formula I've never seen before,
    and having to use MATRIX on a TI calculator (what's worse is that the school
    prefer to use TI-83+ rather than 86 that is used preferably at my other
    school.

    Anyway, the formula for the circle:Ax^2 + Ay^2 + Cx + Dy + F = 0 (zero)

    I've done a few excercise, and it was so so, for the weekend, the teacher
    only assign me one program, it was this:

    x^2 + y^2 - (127/7)x - (15/14)y + (211/14) = 0

    and he wants me to find out what/where the circle is and draw it. I need
    help with this one on how to do it.

    ---

    A similiar problem the teacher done before assigning this was to ask a
    student to pick any random x,y coordinate, 3 of them. It was this:

    (3,-5)
    (4,7)
    (1,-1)

    *then this is what I wrote down from the board. I still can't seem to find
    out what he did:

    (4-1)^2 + (7-(-1))^2 = r^2

    x^2 + y^2 + Ax + By + c = 0
    3A - 5B + C = -34
    4A + 7B + C = -65
    A - B + C = -2

    *then the number he input into the matrix of a ti-86 graph calc:

    -3 -5 1 -34
    4 7 1 -65
    1 -1 1 -2

    *those were all under a big bracket. Sorry for those who aren't using
    outlook express.
    that's basically what I had, he didn't draw any graph at.

    other notes I scramble down when he was talking (about calculator). They
    are/aren't in order, I just wrote down what I hear/saw, it was when class is
    almost over, sorry:

    create circle function....

    draw menu, -> circle

    Circle(1,1,1)
    Circle(1,1,1)
    Circle(1,1,1)
     
    Antivivisectionist unreal email, Sep 7, 2003
    #1
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  2. It is? A formula for a circle centered at (a,b) with radius r is
    (x - a)^2 + (y - b)^2 = r^2
    This is more useful formula for a circle.
    What you mean program? You're supposed to write a computer program to
    solve the problem?
    Push it into the form I've given. Complete the squares
    x^2 - 127x/7 = x^2 - 127x/7 + (127/14)^2 - (127/14)^2
    = (x - 127/14)^2 - (127/14)^2

    Do the same with y and convert equation to one of the form I've given.
    Either he or you are mumbling or not paying attention to clearly state the
    problem.
    This is the equation of a circle centered at (1,-1) with radius r
    I guess he's converting the first equation into the form of the second
    equation.
    Likely he solved the three linear equations given above to find values for
    A,B,C.
    That's quite alright, ask he also not to use the TI-86.
    That isn't math class, it's computer class.
    But that's to be expected, it's supposed to be better, easier and cheaper
    to teach graphic calculator useage that to actually teach any math.

    Different schools, sometime different classes, teaches useage of
    different graphics calculator. It's good for the economy to have student
    buy several graphic calculators and more upon entering work place.
     
    William Elliot, Sep 7, 2003
    #2
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  3. ok, "as is" from the book:

    By squaring and simplifying, the equation (x-h)^2 + (y-k)^2 = r^2 can be
    written in the following <b> general form of the equation of a circle </b>

    (*in a light blue background (highlighting?)): Ax^2 + Ay^2 + Cx + Dy + F =
    0, A (is not equal) to zero

    To convert such an equation to the standard form

    (x-h)^2 + (y-k)^ = p

    you can use a process called <b> completing the square</b>. If p > 0, then
    the graph of the equation is a circle. If p = 0-, then the graph is the
    single point (h,k). If p < 0, then the equation has no graph.

    (end of quotation)

    On the next page, it shows me how to complete the square.
    *problem, i was alittle slacking when I type it.
    actually its math, its just when I said "program", instead of problem.
     
    Antivivisectionist unreal email, Sep 7, 2003
    #3
  4. *yikes*, did I say ti-86? its 83+
     
    Antivivisectionist unreal email, Sep 7, 2003
    #4
  5. what I tried (and hope its what the teacher's looking for):

    original problem: x^2 + y^2 - (127/7)x - (15/14)y + (211/14) = 0

    So i complete the square...

    x^2 + y^2 - (127/7)x - (15/14)y = (211/14)

    (x^2 - 127x/7 + (blank))^2 + (y^2 - 15y/14 + (blank))^2 = 211/14
    + (blank) + (blank)

    (x^2 - 127x/7 + 82.3)^2 + (y^2 - 15y/14 + .27)^2 = 211/14 + 82.3
    + .27
    *half of (-127x/7), then squared; same with -15/14

    (x - 9.07)^2 + (y - .5357)^2 = 97.6414
    *(x-h)^2 + (y-k)^2 = r^2
    so...

    (x - 9.07) + (y - .5357) = 9.881

    so in conclusion, its x = 9.07, y = .53, and r = 9.881

    Can anyone confirm this? again, I hope its what the teacher's looking for.

    And I have the drawing of the circle on a x,y graph also.
     
    Antivivisectionist unreal email, Sep 8, 2003
    #5
  6. I actually looked in the book and do it similiar to the way they show. I
    don't understand what the teacher's trying to teach, making me do matrix and
    circle and all those things on those horrifying piece of technology.
     
    Antivivisectionist unreal email, Sep 8, 2003
    #6
  7. From: Antivivisectionist unreal email <>
    Newsgroups: alt.math, alt.math.undergrad
    Subject: Re: calculus prequisities...
    You want a,b such that
    (x - a)^2 = x^2 - 2ax + a^2 = x^2 - 127x/7 + a^2
    (y - b)^2 = y^2 - 2by + b^2 = x^2 - 15/14 + b^2

    Thus a = 127/14, b = 15/28 and
    (x - a)^2 + (x-b)^2 - a^2 - b^2 + 211/14 = 0

    (x^2 - 127x/7 + a)^2 is bogus.
    Your use of algebra much too sloppy.
    Yes, circle is centered at about (9,1/2)
    Use of decimals assures inaccurate answers.
    I get a different answer for r.

    ----
     
    William Elliot, Sep 8, 2003
    #7
  8. Antivivisectionist unreal email

    Dana Guest

    Dana, Sep 12, 2003
    #8
  9. Antivivisectionist unreal email

    Dana Guest

    It's been awhile for me doing circles, but this looks familiar.
    Here's what Mathematica gives...

    Here are your terms...
    a = 1; c = 1; d = -127/7; e = -15/14; f = 211/14;

    x value
    (-d/2)*a
    9.071428571428571

    y value
    (-e/2)*a
    0.5357142857142857

    Radius
    Sqrt[(d^2 + e^2)/(4*a^2) - f/a]
    8.216226478805243

    HTH if it is correct.
     
    Dana, Sep 12, 2003
    #9
  10. I forgot to reply with the answer, but my teacher didn't convert it to
    decimal, he kept it rational, but the same process applies
     
    Antivivisectionist unreal email, Sep 12, 2003
    #10
  11. Antivivisectionist unreal email

    Dana Guest

    Not sure how the TI calculator works, so I don't know if this will help.
    Perhaps you could break the problem down into 4 '3x3' determinants.
    Here, I assume your calculator can do Determinants...

    Here, each group of { } represents a row in your 3x3 determinante.

    a = {{x1, y1, 1}, {x2, y2, 1}, {x3, y3, 1}};

    d = -{{x1^2 + y1^2, y1, 1}, {x2^2 + y2^2, y2, 1}, {x3^2 + y3^2, y3, 1}};

    e = {{x1^2 + y1^2, x1, 1}, {x2^2 + y2^2, x2, 1}, {x3^2 + y3^2, x3, 1}};

    f = -{{x1^2 + y1^2, x1, y1}, {x2^2 + y2^2, x2, y2}, {x3^2 + y3^2, x3, y3}};

    Here are your 3 points...
    {x1, y1} = {3., -5.};
    {x2, y2} = {4., 7.};
    {x3, y3} = {1., -1.};

    Now, use your calculator to find the determinate:

    a = Det[a]
    28.

    d = Det[d]
    -508.

    e = Det[e]
    -30.

    f = Det[f]
    422.

    X Center:
    -(d/(2*a))
    9.07143

    Y Center
    -e/(2*a)
    0.535714

    Radius:
    Sqrt[(d^2 + e^2)/(4*a^2) - f/a]
    8.21623

    It appears that the 3 points represent the same circle as given in the other
    part of your question:
     
    Dana, Sep 13, 2003
    #11
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