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Find the equation of the curve

  1. Aug 1, 2007 #1
    1. The problem statement, all variables and given/known data
    The tangent to the curve [tex]y=f(x)[/tex] at the point (0,1) has equation [tex]y=x+1[/tex]. The gradient function of the curve is given by [tex]\frac{{dy}}{{dx}} = ax + b [/tex]. Find the equation of the curve given that it also passes through the point (1,3)

    2. The attempt at a solution
    Well we have the curve y=f(x) which I am assuming to be a quadratic or something of the like, and at the point (0,1) it has the tangent equation [tex]y=x+1[/tex].
    The gradient function dy/dx, at the point (0,1):
    [tex]\frac{{dy}}{{dx}} = ax + b [/tex]

    [tex]\therefore[/tex] [tex]1 = a(0) + b[/tex]
    hence b=1

    Can we now solve for 'a' by saying that:
    [tex]ax+1 = x+1[/tex]
    from that a=1 and [tex]y=x+1[/tex] as before.
    But the integral of [tex]y=x+1[/tex] is [tex]y=1/2 x^2 +x + C[/tex]
    @ (1,3)
    [tex]y=1/2 x^2 +x + 1/2[/tex]

    The answer is:
    [tex]y=1/2 x^2 +x + 1[/tex]
    Where is my mistake?!?!?!?

  2. jcsd
  3. Aug 1, 2007 #2


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    Homework Helper

    the ans is clearly wrong for y = 0.5 x^2 + x +1
    at (1,3) you get: 3 = 0.5 + 1 + 1 which is NOT equal.
    in fact your calc was wrong and it should be 3/2 and not 1/2
  4. Aug 1, 2007 #3
    ok sorry, just wrote that in wrong lol
    [tex] y = 0.5 x^2 + x + 3/2 [/tex]
    the answer is clearly wrong then I take it?
  5. Aug 1, 2007 #4


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    Homework Helper

    now that you have the answer, you can now plug in back in to all the conditions and see if they all agree. If so, you ans must be allowed. the only question is whether this is the unique soln. but I guess you are not interest in that sort of thing and i don't believe there could be other answers either.
  6. Aug 1, 2007 #5
    Yep putting it all back in, the answer stands ...

    thanks mjsd for your help :D
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