Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Use the equation of the straight line to predict future deaths.

  1. Jul 15, 2006 #1
    Hello,

    I have a set of data points to plot as graph.

    X axis: Year
    Y axis: Number of deaths due to cancer.

    The graph is of parabolic shape opening to the right.


    Following is my question:

    I have been asked to plot only the first and last points and connect those points with a straight line. Use the equation of the straight line to predict future deaths.


    Can this graph be considered as a function? I am not sure how to answer this. My answer is yes and No.

    Yes because, by the definition of a function, you have a certain output for a certain in put. The definition does not worry about the accuracy of the output.

    No because, two points in a set of data can not accurately predict the future outcomes.

    Experts... what are your thoughts?


    Thank You in advance.

    Gamma
     
    Last edited by a moderator: Jan 7, 2014
  2. jcsd
  3. Jul 15, 2006 #2
    Hmm..seems like you're just plotting a linear function of x

    This might sound a bit daft, but to answer your question another way lets say I decide that the best way to graph a cosine wave between [itex] -\pi [/itex] and [itex] \pi [/itex] is to use the following functions...
    [itex] y = (\frac {2}{\pi})^2 (x+ \pi)^2 -1 [/itex] on the interval [ [itex] -\pi,\frac {-\pi}{2} ] [/itex]

    [itex]y = -(\frac {2}{\pi})^2 x^2 +1 [/itex] on the interval [ [itex] \frac {-\pi}{2},\frac {\pi}{2}[/itex] ]and

    [itex] y = (\frac {2}{\pi})^2 (x- \pi)^2 -1 [/itex] on the interval [[itex] \frac{\pi}{2}, \pi [/itex]]

    Just because the method I employ is totally ridiculous does it make [itex] y = (\frac {2}{\pi})^2 x^2 +1 [/itex], [itex] y = (\frac {2}{\pi})^2 (x+ \pi)^2 -1 [/itex], or[itex] y = (\frac {2}{\pi})^2 (x- \pi)^2 -1 [/itex] any less functions of x? :smile:
     
    Last edited: Jul 15, 2006
  4. Jul 15, 2006 #3
    A function does not have to predict accurately the future outcomes. This a problem of modeling intelligently a situation.

    A funtions is roughly defined as being a relation between to variables such as for each x, there is one and only one y wich is associated to.

    Hence, a linear relation between to variables is a function.

    A circle does is not a function because for each x, there are two values of y that are associated. You must therefore take the upper (0,Pi) OR the lower (Pi, 2Pi) part of the circle to accurately describe it as a function.
     
    Last edited: Jul 15, 2006
  5. Jul 19, 2006 #4
    Make sense. Thank you both for your thoughts. Regards,


    Gamma.
     
  6. Jul 19, 2006 #5
    To be correct, a circle is a function, but is a function of two variables (x and y). A function is simply a mapping from one set (call it A) to another set (call it B) such that every element of A corresponds to only 1 element of B.
     
  7. Jul 19, 2006 #6

    HallsofIvy

    User Avatar
    Science Advisor

    No, a "circle" is not a function! It is a geometric object. What, exactly, is the function of two variables, f(x,y), that you are associating with the circle?
     
  8. Jul 19, 2006 #7
    Well it seems to me that if you want to extrapolate the function then clearly it is not the best approach to use a linear function that includes the first and last point of your original function.

    That depends on the function that is inter/exra-polated. :smile:
    If you were to plot number of cancer deaths per 1000 people per year, the function will become a lot flatter. Then you could go one step further and plot the growth of that number per year. Then your linear extrapolation would become a bit more meaningful.
     
    Last edited: Jul 19, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook