Results 1 to 4 of 4

Math Help - tricky problem please HELP

  1. #1
    Newbie
    Joined
    Nov 2006
    Posts
    3

    Unhappy tricky problem please HELP [SOLVED]

    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?

    THE BELOW IS WHAT I'VE TRIED TO DO:

    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!?

    EDIT:

    And could someone clarify why I got an Infraction for spamming Advertisements in this post?? As far as I know I havn't adversided anything!
    ????
    Last edited by me_maths; November 27th 2006 at 02:12 PM. Reason: to let others know the topic is solved
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by me_maths View Post
    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?

    THE BELOW IS WHAT I'VE TRIED TO DO:

    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!?
    Your method is good, but in your derivative you've got a sign problem and you squared the denominator incorrectly.
    y = \frac{ax - b}{1 - x^2}

    y' = \frac{(a)(1 - x^2) - (ax - b)(-2x)}{(1 - x^2)^2}

    Note the "-" on the (-2x) factor. Also (1 - x^2)^2 = 1 - 2x^2 + x^4, NOT 1 - x^4.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2006
    Posts
    3
    Quote Originally Posted by topsquark View Post
    Your method is good, but in your derivative you've got a sign problem and you squared the denominator incorrectly.
    y = \frac{ax - b}{1 - x^2}

    y' = \frac{(a)(1 - x^2) - (ax - b)(-2x)}{(1 - x^2)^2}

    Note the "-" on the (-2x) factor. Also (1 - x^2)^2 = 1 - 2x^2 + x^4, NOT 1 - x^4.

    -Dan
    ahh, thanks for the help topsquark

    but it still leaves 2 unknown variables in the equation.

    I cleaned up the equations and I got it to be:

    y'(-2) = \frac{a(1-4)-(-2a-b)(4)}{1-8+16}
    <=>
    y'(-2) = \frac{5a+4b}{9}

    so I put that equal to 1, wich was the needed angle ( k value ) of the point on the curve:

    y'(-2) = \frac{5a+4b}{9} = 1

    and here is where i'm stuck then...

    I get a = \frac{9-4b}{5} etc, wich doesn't really help at all now does it?

    so, where should I continue from here?

    Once again, the solution to the problem is that 'a' and 'b' should both equal 1, but I have problems in getting there myself , help is much appreciated!
    Last edited by me_maths; November 27th 2006 at 02:00 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2006
    Posts
    3
    Ahh, now I've got it!

    I take the first equation:

    y = \frac{ax - b}{1 - x^2}

    put in the point it should be on (-2,1)

    -2a - b = -3

    wich makes it:

    -2a - b = -3 or 2a + b = 3, wichever you want.

    As I said, I take the first equation and compare it to the result I got in the last post:

    1 = \frac{5a + 4b}{9}

    <=>

    5a + 4b = 9

    Then I take both of them and compare them:

    #1. 2a + b = 3
    #2. 5a + 4b = 9

    I'll use the addition method so I take #1. times (-4) and I get:

    #1. -8a - 4b = -12
    #2. 5a + 4b = 9

    then I kill the b's and I'm left with:

    #1. -8a = -12
    #2. 5a = 9

    wich then is added to become:

    -3a = -3
    <=>
    -a = -1
    <=>
    a = 1

    and then I just put that a = 1 into for instance here:

    #1. 2a + b = 3
    <=>
    #1. 2 + b = 3
    <=>
    #1. b = 3-2
    <=>
    #1.  b = 1

    Voila, thanks for all the help guys
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tricky (ln) problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 23rd 2011, 01:03 PM
  2. Tricky Problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 6th 2009, 05:59 PM
  3. Another tricky problem
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: October 8th 2009, 06:08 PM
  4. Tricky problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 12th 2009, 03:15 PM
  5. Replies: 6
    Last Post: March 30th 2008, 10:39 AM

Search Tags


/mathhelpforum @mathhelpforum