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Homework Help: Tricky problem please HELP

  1. Nov 27, 2006 #1
    tricky problem please HELP :( -SOLVED by "courtrigrad"-

    [SOLVED, thanks to "courtrigrad"!]

    I've tried to solve this problem for over 3 or 4 hours but I can't figure out how, please help me.

    Ok so it's a curve ( parable? ) y = (ax - b) / ( 1 - x^2 )

    You are supposed to solve what 'a' and 'b' should be if the curve goes through the point (-2,1) AND that the tangents* direction of that same point (-2,1) is 45 degrees ( so the tangents 'k' value should be 1, wich is 45 degrees ).

    *I'm not sure but I might have the wrong name in English for 'tangent', but it is a line with the same 'k' value as the single point in the curve... if you know what I mean?


    Ok, so we all know that to get the 'k' value of a point in a curve you must begin with taking the derivate of the curve, wich in this case would be:

    MAOL s.43 : D f/g = ( gDf - fDg ) / ( g^2 )

    #1. y' = ( ( 1-x^2 )*a - ( ax-b )*( 2x ) ) / ( 1 - x^4 )

    and then you should put the derivate in a function of the x value, something like: y'(x) = ... , wich would in this case be: y'(-2) = ... .

    so we take that y'(-2) = ... and equal it to 1, because that's the 'k' value we had to have for that point.

    and thus we should get an equation with both 2 unknown variables, 'a' and 'b' , in wich we should somehow figure out what they should be to get this to work... wich is where I fail.

    The answer for this problem is that 'a' and 'b' both = 1.

    I've tried to do the countings myself etc but I can't come up with it...

    Could someone please help me!?
    Last edited: Nov 27, 2006
  2. jcsd
  3. Nov 27, 2006 #2
    You know that [tex] 1 = \frac{-2a-b}{-3} \Rightarrow -2a-b = -3 [/tex]

    Also [tex] y'(x) = \frac{(1-x^{2})(a)-(ax-b)(-2x)}{(1-x^{2})^{2}} [/tex]

    Thus [tex] y'(-2) = 1 = \frac{-3a+8a+4b}{9} \Rightarrow 5a + 4b = 9 [/tex]

    So the two equations are: [tex] -2a - b = -3 [/tex] and [tex] 5a+4b = 9 [/tex]

    Solve these two equations for [tex] a,b [/tex] and you should get the answer.
  4. Nov 27, 2006 #3

    Thanks alot!

    Forgot completaly that you could use the first equation to compare it to the result of y'(-2) = ... . Not to mention that I made a few mistakes in the calculating also such as (1-x^2) would be (1-x^4) wich is totally wrong...

    Thanks again, helped me tremendously!! <3 <3
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