Results 1 to 12 of 12

Math Help - Finding the equation of a curve with the givin points?

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    11

    Finding the equation of a curve with the givin points?

    This has been bugging me for quite some time now, I have many uses for this, but that's off-topic.

    How do I find out the equation of a curve using the givin points?

    I have the points for the curve.

    (1, 25)
    (2, 40)
    (3, 60)
    (4, 85)
    (5, 115)
    (6, 150)
    (7, 190)
    (8, 235)
    (9, 285)
    (10, 340)

    I should also mention that I just got into geometry (I'm on my freshman high school year) and my math is a little rough over the break. This is actually something I've been trying to figure out for me, not for the school or any homework. Well, technically homework because of work at home, but not for the school.

    I should also mention that I've tried the methods at which I found while searching on how.
    My answer was way off. Let's just leave it at that.

    Thank you,
    -Tyler W.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Depending on what kind of curve you want fitted will determine the line of best fit. I found a pretty good polynomial model for your data. It is:

    y = 2.5x^2 + 7.5x + 15
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    11
    More or less I'm asking for how did you come up with that answer, not the answer itself.

    -Tyler W.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Fair question, what kind of curve are you looking to fit? My example is polynomial of order 2. You could do this for a high order or even fit a stright line or exponential model.

    For a polynomial model you propose a solution of type

    y = ax^2 + bx + c

    Then you can pick three of the points in your series and substitute them into this model to find a,b & c.

    for example you could pick

    (1, 25), (5, 115) and (10, 340)

    giving

    25 = a(1)^2 + b(1) + c

    115 = a(5)^2 + b(5) + c

    and

    340 = a(10)^2 + b(10) + c

    then solve this system.

    Using different points will produce different models.


    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2009
    Posts
    11
    I've returned after eating out tonight and trying out everything at which you said. The part I'm stuck at right now is the solving the system of polynomial equations.

    I attempted it and tried a method to find 3 variables from 3 equations to see if that's how you solve a system of polynomial equations.

    I might of done the math wrong, I tried my equation from the system that you showed me and it was horribly off.

    If it isn't too much trouble, may you please explain to me the method for solving a system of polynomial equations?

    Here's what I gotten

    f(x)=3.35x^2+33.75x-20

    -Tyler W.

    EDIT: I think I may have found the issue here. I did the square root of 11.25 to simplify a^2=11.25 to a=3.25.
    When I try it out for the first number it's fine. I will report back when I've tested it.

    EDIT2: Nope, resulted in failure after 1.
    f(x)=11.25x^2+33.75x-20
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Tyman2007 View Post
    I attempted it and tried a method to find 3 variables from 3 equations
    That is great. You should try some simple substitution.

    Be mindful that the solution I provided in the first post is only a 'possible' solution to fit the data. The three points I picked in my second post may not yield that exact model.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2009
    Posts
    11
    Thank you kindly, You really relieved a lot of my confusion and brought me one step closer to a higher level of math.

    Again, Thank you,
    -Tyler W.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Jan 2009
    Posts
    591
    Quote Originally Posted by Tyman2007 View Post
    More or less I'm asking for how did you come up with that answer, not the answer itself.

    -Tyler W.
    pickslides is a very good method.

    The methods I use:

    Plot the data and determine the best linear fit
    (least squares)
    Then a log plot of the data.
    Then a log.log plot.
    (& several other types of plot)

    With 10 data points you can get a perfect fit with a polynomial of degree 9.

    You can use the method of differences to determine the approximate degree of polynomial you require.

    (1, 25)
    (2, 40)
    (3, 60)
    (4, 85)
    (5, 115)
    (6, 150)
    (7, 190)
    (8, 235)
    (9, 285)
    (10, 340)

    first difference
    25,40: 15
    40,60: 20
    60,85: 25
    ...
    285,340:55

    2nd difference
    15-20: 5
    20-25: 5
    25-30: 5

    3rd difference
    5-5:0
    etc.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Sep 2009
    Posts
    11
    So, by the looks of the method of differences, I require a polynomial of degree 0? Like I said, just starting geometry, and I really loathe polynomials. looks like finding the equation of a curve is my arch-enemy.

    So what pickslides was saying, I just keep trying different points until I get a good equation that works out?

    -Tyler W.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Tyman2007 View Post
    So, by the looks of the method of differences, I require a polynomial of degree 0?
    A polynomial with degree 0 is as follows

    y = ax^0 = a as x^0 = 1

    This will not model your data. I would suggest the lowest order polynomial would be of order 1 which is linear and has the general form of

    y = ax+b where a is the slope and b is your y-intercept.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Jan 2009
    Posts
    591
    Quote Originally Posted by Tyman2007 View Post
    So, by the looks of the method of differences, I require a polynomial of degree 0? Like I said, just starting geometry, and I really loathe polynomials. looks like finding the equation of a curve is my arch-enemy.

    So what pickslides was saying, I just keep trying different points until I get a good equation that works out?

    -Tyler W.
    by the looks of the method of differences, I require a polynomial of degree 0?
    NO.
    You have first differences & 2nd differences & all 3rd differences are zero, so you will need an equation NO higher than 2nd degree.

    So what pickslides was saying, I just keep trying different points until I get a good equation that works out?
    ONLY if you want to.

    Try this
    from the y values:
    difference from one Y value to the next Y value
    first differences
    y1 - y0 = d10 = ? (it is not given, so we don't know)
    y2 - y1 = d21 = 40-25 = 15
    y3 - y2 = d32 = 60-40 = 20
    y4 - y3 = d43 = 85-60 = 25
    y5 - y4 = d54 = 115-85 = 30
    y6 - y5 = d65 = 150-115 = 35
    You only need to do this until you see what's occurring.

    2nd DIFFERENCES (difference between 1st diffs)
    d32-d21 = dif3221 = 20-15 = 5
    d43-d32 = dif4332 = 25-20 = 5
    d54-d43 = dif5443 = 30-25 = 5
    d65-d54 = dif6554 = 35-30 = 5

    3rd Differences will all be zero
    That means you will have an equation of degree 2
    THIS FORM:
    y = ax^2 + bx + c


    from above
    the difference between y1 & y0
    will be the difference between
    y2 and y1 (which is 15) MINUS (since we're going backwards) the 2nd difference (which is 5)
    which makes
    y1 - y0 = 10
    y1 = 25
    y1 - 10 = y0 = 15
    We know that y0 = 15; when x=0 , y=15; We have the constant!
    y = ax^2 + bx + 15
    --
    solely for ease of compuation
    reduce all of the y values by 15
    v0 = 0
    v1 = 10
    v2 = 25
    v3 = 45
    v4 = 70
    etc

    work out some values for the x's
    at x=1: v1 =  10 = a \cdot 1^2 + b \cdot 1
     10 = a + b

    at x=2: v2 =  25 = a \cdot 2^2 + b \cdot 2
     25 = a \cdot 4 + b \cdot 2

    we have two equations with 2 unknowns
    10 = a + b ;thus 10-a = b
    25 = 4a + 2b
    substituting
    25 = 4a + 2(10-a)
    25 = 4a + 20 - 2a
    25 = 2a + 20
    5 = 2a
    a =2.5

    since 10 = a + b
    10 = 2.5 + b
    7.5 = b

    We have all the coefficients
    and can write the equation:

    Y = 2.5X^2 + 7.5X + 15

    which is exactly the equation that pickslides supplied in his first post.
    The procedure he gave in his other post is an excellent method of deriving the polynomial.

    .
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Sep 2009
    Posts
    11
    Wow... Just... Wow...
    I'm in very much debt to you.

    You deserve my thanks, both of you.

    -Tyler W.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: August 24th 2011, 11:35 AM
  2. Replies: 7
    Last Post: June 4th 2010, 08:28 AM
  3. Finding singular points of a curve in affine space
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 29th 2010, 03:35 AM
  4. finding points on the curve
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 19th 2010, 10:21 AM
  5. Replies: 8
    Last Post: October 8th 2007, 04:29 PM

Search Tags


/mathhelpforum @mathhelpforum