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Math Help - Curve equation

  1. #1
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    Curve equation

    Hi, I would like to know if its possible to find a equation for a curve if 3 points were given. Say

    1. 0, 0
    2. 5, 2
    3. 10, 0

    If so, how would i go about solving this.

    Thanks all
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  2. #2
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    Quote Originally Posted by mrwelcam View Post
    Hi, I would like to know if its possible to find a equation for a curve if 3 points were given. Say

    1. 0, 0
    2. 5, 2
    3. 10, 0

    If so, how would i go about solving this.

    Thanks all
    Sure - try a quadratic y = ax^2+bx+c. Sub. in your points and solve a system of equations for a, b and c.
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  3. #3
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    Could you possibly show me an example using the points above.

    Thanks for your help.
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  4. #4
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    Or would it make more sense to have the vertex at x=0.

    say

    -5, 0
    0, 2
    5, 0
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  5. #5
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    Hello, mrwelcam!

    I would like to know if its possible to find a equation for a curve
    if 3 points were given, say: .A(0,0), B(5,2), C(10,0)

    If so, how would i go about solving this?
    Code:
          |
          |       B
          |       o
          |
      . . o - - - - - - - o - -
          A               C

    With only three points, there is an infinite number of equations.

    The first idea would be a down-opening parabola.

    But there are an infinite number of cubics, quartics, etc. that contain those points
    . . as well as zillions of circles, ellipses, sine/cosine functions . . .

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  6. #6
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    Any chance of an example
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  7. #7
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    Quote Originally Posted by mrwelcam View Post
    Any chance of an example
    The quadratic is probably the easiest.

     <br />
\text{At}\; (0,0)\;\; 0 = 0 a + 0 b + c<br />
     <br />
\text{At}\; (5,2)\;\; 2 = 25a + 5b + c<br />
     <br />
\text{At}\; (10,0) \;\; 0 = 100 a + 10 b + c<br />

    Solving gives a = - \frac{2}{25},\;\; b = \frac{4}{5},\;\; c= 0

    and so we have

     <br />
y =- \frac{2}{25} x^2 + \frac{4}{5} x.<br />

    or

     <br />
y = \frac{2}{25} x (10-x)<br />

    Other possibilities

    y = \frac{2}{5^{p+q}} x^p(10-x)^q where p, q \in \mathbb{R^+}.

    A trig function

     <br />
y = 2 \sin \frac{\pi x}{10}<br />

    An ellipse

     <br />
\frac{(x-5)^2}{25} + \frac{y^2}{4} = 1<br />

    Here's a circle

     <br />
(x-5)^2 + \left(y + \frac{21}{4} \right)^2 = \left(\frac{29}{4}\right)^2<br />
.

    Like Soroban said - the list is exhaustive.
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  8. #8
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    Thanks everyone. If i wanted to expand that to be given 5 points would a quadratic equation still be the best (simplest) option.
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  9. #9
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    You have to go up in order. So for 5 points try

     <br />
y = ax^4 + bx^3 + cx^2 + dx + e<br />

    However, you might get lucky and find a polynomial less the degree 4 that works.
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  10. #10
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    whatever the degree of the polynomial is I believe you need that many + 1 points to get it exactly.

    so 3 points would give you any x^2 degree equation. So the answer is yes for parabolas but no for any curve.
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  11. #11
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    Hi. I'm found this on the web:

    A quadratic equation has the form:

    Y = A * X^2 + B * X + C
    Since we have three points, we have three sets of values for X and Y. That gives us three equations and three unknowns: A, B, and C.
    Plugging in the points (X1, Y1), (X2, Y2), and (X3, Y3) gives:
    Y1 = AX1^2 + BX1 + C Y2 = AX2^2 + BX2 + C Y3 = AX3^2 + BX3 + C
    Subtracting equation 1 from the other two gives:
    Y2 - Y1 = A(X2^2 - X1^2) + B(X2-X1) Y3 - Y1 = A(X3^2 - X1^2) + B(X3-X1)
    Multiplying the first equation by (X1-X3) and the second by (X2-X1) and adding them gives:
    (Y2-Y1)(X1-X3) = (X1-X3)A(X2^2 - X1^2) + B(X2-X1)(X1-X3) (Y3-Y1)(X2-X1) = (X2-X1)A(X3^2 - X1^2) + B(X3-X1)(X2-X1)
    Or
    (Y2-Y1)(X1-X3) + (Y3-Y1)(X2-X1) = A[(X1-X3)(X2^2-X1^2) + (X2-X1)(X3^2-X1^2)]
    Solving for A gives:
    A = [(Y2-Y1)(X1-X3) + (Y3-Y1)(X2-X1)]/[(X1-X3)(X2^2-X1^2) + (X2-X1)(X3^2-X1^2)]
    Plugging into the previous equation gives:
    B = [(Y2 - Y1) - A(X2^2 - X1^2)] / (X2-X1)
    Plugging into the original equation gives:
    C = Y1 - AX1^2 - BX1

    This formula seems to be working well for me. How would i go about expanding this formula to allow for 5 points.

    Thanks for your time.
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  12. #12
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    Quote Originally Posted by mrwelcam View Post
    Hi. I'm found this on the web:

    A quadratic equation has the form:

    Y = A * X^2 + B * X + C
    Since we have three points, we have three sets of values for X and Y. That gives us three equations and three unknowns: A, B, and C.
    Plugging in the points (X1, Y1), (X2, Y2), and (X3, Y3) gives:
    Y1 = AX1^2 + BX1 + C Y2 = AX2^2 + BX2 + C Y3 = AX3^2 + BX3 + C
    Subtracting equation 1 from the other two gives:
    Y2 - Y1 = A(X2^2 - X1^2) + B(X2-X1) Y3 - Y1 = A(X3^2 - X1^2) + B(X3-X1)
    Multiplying the first equation by (X1-X3) and the second by (X2-X1) and adding them gives:
    (Y2-Y1)(X1-X3) = (X1-X3)A(X2^2 - X1^2) + B(X2-X1)(X1-X3) (Y3-Y1)(X2-X1) = (X2-X1)A(X3^2 - X1^2) + B(X3-X1)(X2-X1)
    Or
    (Y2-Y1)(X1-X3) + (Y3-Y1)(X2-X1) = A[(X1-X3)(X2^2-X1^2) + (X2-X1)(X3^2-X1^2)]
    Solving for A gives:
    A = [(Y2-Y1)(X1-X3) + (Y3-Y1)(X2-X1)]/[(X1-X3)(X2^2-X1^2) + (X2-X1)(X3^2-X1^2)]
    Plugging into the previous equation gives:
    B = [(Y2 - Y1) - A(X2^2 - X1^2)] / (X2-X1)
    Plugging into the original equation gives:
    C = Y1 - AX1^2 - BX1

    This formula seems to be working well for me. How would i go about expanding this formula to allow for 5 points.

    Thanks for your time.
    The same applies but obviously, more complicated. You could also try Cramer's Rule

    Cramer's rule - Wikipedia, the free encyclopedia
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