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Stumbled upon a math problem

  1. May 21, 2010 #1
    Hi everyone,
    I have stumbled upon this maths problem, i have pondered it for a few days without any real work until today. The problem is, i can't seem to comprehend these questions at all... so any help would very much appreciated :biggrin:

    Here is the math problem:

    The curve C has equation y=k*x^3-x^2+x-5, where k is constant.

    A) Find the derivative of the expression with respect to x

    I got 3*k*x^2-2 x+1=y

    The point A with x-coordinate -1/2 lies on c. The tangent c at a is parallel to the line with equation 2*y-7*x+1=0

    Find:

    B) the value of k

    C) the value of the y-coordinate of A.
     
  2. jcsd
  3. May 21, 2010 #2

    Mark44

    Staff: Mentor

    The derivative is correct, but you have mislabelled it as "y". You should have y' = 3kx^2 - 2x + 1 or dy/dx = 3kx^2 - 2x + 1.
    Point A lies on C. The tangent to C at A is parallel to a certain line.

    The derivative you found can be used to find the slope of the tangent line to C at any point. What's the slope of the tangent line (on C) at the point A? What's the slope of the line whose equation is given?
     
  4. May 22, 2010 #3
    So, as far as i understand it, the slope of the line that is tangent to C at point A (-1/2). The second equation given is parallel to the first derivative of C when x=-1/2
    My first thought was to do this:
    [tex] 3 k x^2-2 x+1=2 y+\frac{9}{2} [/tex]
    which is the same as

    [tex] y'\left(-\frac{1}{2}\right) =2 y+\frac{9}{2}[/tex]
    which is equal to the line that is tangent to C. From this you can solve for k, which i got k=2. But i am struggling on how to find y, how could you use the other equation to find y, when the line is the same but rather only parallel. How could i find y in this case?
    So wouldn't the equation look like this with the different y's:
    [tex] \left(y_1\right)'\left(-\frac{1}{2}\right)\to \frac{3 k}{4}+2=2 y_2+\frac{9}{2} [/tex]
    or can we find y by going back to the original equation like this:
    [tex] y=2 \left(-\frac{1}{2}\right)^3-\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{2}\right)-5=-6 [/tex]
    This is why it is confusing me lol i'll get there eventually ;)
     
    Last edited: May 22, 2010
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