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Simple equation.

  1. Jan 20, 2007 #1
    1. The problem statement, all variables and given/known data
    For the graph of equation (y^2)=a(x^2), where a is a positive constant, whic of the following is true?
    1) y increases linearly with x
    2) y increases as the square of x
    3) y increases as y^1/2
    4) If x increases, y can increase or decrease.


    2. Relevant equations
    y is directly proportional to x. as a is a constant, the first option would be right!


    3. The attempt at a solution

    Is my approach right?
     
  2. jcsd
  3. Jan 20, 2007 #2

    arildno

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    Not all, I'm afraid. Rewrite your equation as:
    [tex](y-(\sqrt{a})x)(y+(\sqrt{a})x)=0[/tex]
    Now, two questions:
    1. Why is this rewriting permissible?
    2. What must you conclude from it?
     
    Last edited: Jan 20, 2007
  4. Jan 20, 2007 #3
    Ans 1 because the difference of two squares can be factorised as suc:
    a^2-b^2=(a+b)(a-b)

    Ans 2. I dont know what can be concluded! Please be clear!
     
  5. Jan 20, 2007 #4

    arildno

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    Well, if two numbers multiplied together is zero, what can we say about those numbers?
     
  6. Jan 20, 2007 #5
    That any one of them is equal to zero or both are zero.
     
  7. Jan 20, 2007 #6

    arildno

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    So,
    EITHER:
    [tex]y=\sqrt{a}x[/tex]
    OR
    [tex]y=-\sqrt{a}x[/tex]

    Given that knowledge, which statement is true?
     
  8. Jan 20, 2007 #7
    The fourth option??? :approve:
     
  9. Jan 20, 2007 #8

    arildno

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  10. Jan 20, 2007 #9
    but sir. I agree totally with you. But if you look at it straight away, it seems that y is directly proprtional to x. hence they will have a linear relationship. How will you criticise this point? Thanks for the effort!!!! :approve:
     
  11. Jan 20, 2007 #10

    arildno

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    Three flaws:

    1. As for direct proportionality (i.e, ratio constant), remember that the proportionality constant in such a relation can be positive or negative, so that from mere direct proportionality, you cannot conclude that y will increase with x, it could equally well decrease with x.

    2. If we assume that the y-coordinate is to be regarded as a FUNCTION of the x-coordinate (so that we limit ourselves to only a part of the permissible solution pairs of the original equation), we might have a discontinuous function relationship in that in one x-region, the function values followed the upper line, and in an adjoining x-region, the function values follow the lower line.
    In this case, we see that the ratio x/y is NOT constant, that is direct proportionality do not hold between x and y

    3. There is no reason why we should assume that there exist any function relating x-values to y-values. Indeed, the solution set to the equation do not permit a function construction in general.

    To be sure, if we limit ourselves to just a part of the solution set, then we may, indeed, have some functional relation between the picked (x,y)-values.
    But that is a totally different issue.
     
  12. Jan 21, 2007 #11

    HallsofIvy

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    This is really only echoing arildno but:

    I would "criticise" that point by saying "Yes, it true- but off the point." The answer did not say "linearly related", it said "increases linearly". That was arildno's point.
     
  13. Jan 22, 2007 #12

    If I read the question again, it says that a is "positive constant". Are you stressing on the point that the squares of a negative number is always positive??


    Whats the difference between "inearly related" and "increases linearly"?
     
  14. Jan 22, 2007 #13

    HallsofIvy

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    Yes, and so with y2= ax2 we can have either
    [itex]y= \sqrt{a}x[/itex] or [itex]y= -\sqrt{a}x[/itex]


    In both [itex]y= \sqrt{a}x[/itex] and [itex]y= -\sqrt{a}x[/itex] y and x are "linearly related". In the first, y "increases linearly" with x but in the second y "decreases linearly" with x.
     
  15. Jan 22, 2007 #14
    Thanks a lot for the efforts. o:)
     
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