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Tangent line

  1. Oct 25, 2006 #1
    Hi, im having problems with the following problem. The main issue is actually starting the problem.

    Find the two points on the curve y = x^4 - 2x^2 - x that have a common tangent line.
     
  2. jcsd
  3. Oct 25, 2006 #2
    Say I'd give you a point p0=(x0,y0), where x0 and y0 are certain, fixed real values. How would you calculate the tangent line?
     
  4. Oct 25, 2006 #3
    You first need to find the derivative of y = x^4 - 2x^2 - x
     
  5. Oct 26, 2006 #4
    I took the derivative of the equation and got:

    y`= 4x^3 - 4x - 1 and put that into the equation of a tangent to the curve:

    y=(4x^3 - 4x - 1)x + c ; where c is the intercept.

    Then replaced y with the original equation:

    x^4 - 2x^2 - x = (4x^3 -4x - 1)x + c

    Then solved for the intercept:

    c= -3x^4 + 2x^2

    Need 2 points which have a common tangent, (a,b) and (c,d) then:

    4a^3 -4a -1 = 4c^3 -4c -1 and

    -3a^4 + 2a^2 = -3c^4 - 2c^2

    Now here's where I am stuck. I don't know how to solve these equations which would then give me the points (a,b) and (c,d) that i need for a common tangent line.
     
  6. Oct 26, 2006 #5

    HallsofIvy

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    Try writing
    [tex]4a^3 -4a -1 = 4c^3 -4c -1[/tex]
    and
    [tex]-3a^4 + 2a^2 = -3c^4 - 2c^2[/tex]
    as
    [tex]4(a^3- c^3)- 4(a-c)= 0[/tex]
    and
    [tex]-3(a^4- c^4)+ 2(a^2- c^2)= 0[/tex]
    and factor:
    [tex]a^4- c^4= (a- c)(a^3+ a^2b+ ab^2+ b^3)[/tex]
    [tex]a^3- c^3= (a- c)(a^2+ ab+ b^2)[/tex]
    [tex]a^3- b^2= (a- c)(a+ b)[/tex]
    Obviously a= c so a- c= 0 satisfies both of your original equations. What else will?
     
    Last edited: Oct 26, 2006
  7. Oct 26, 2006 #6
    Take the derivative of the f(x).

    y' = 4x^3 - 4x - 1

    That is your tangant line function, now we need to find only two points that have the same tangent line. Find the tangent line at a point. In this case, we'll find it for 1.

    y' = 4(1)^3 - 4(1) - 1
    y' = -1

    Now, let's plug that back into the equation and try to find the other values that have the same slope.

    -1 = 4x^3 - 4x - 1
    0 = 4x^3 - 4x
    0 = 4x(x^2 - 1)
    0 = 4x(x-1)(x+1)

    Which means that it will have a slope of -1 at (-1,-2); (1,-2); and (0,0).

    Let's check that out.

    f'(-1) = 4x^3 - 4x - 1 = -1
    f'(0) = 4x^3 - 4x - 1 = -1
    f'(1) = 4x^3 - 4x - 1 = -1 (From previous steps)

    Confirmed.

    Those points up above are points that have the same tangent line.
     
  8. Oct 27, 2006 #7

    HallsofIvy

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    No, that gives points that have parallel tangent lines, not necessarily the same tangent line.
     
  9. Oct 28, 2006 #8
    Oh, I see. They want the points where they have the same linear tangent line function. Well, in that case:

    y + 2 = -1(x -1) => y = -x -1
    y + 2 = -1(x + 1) => y = -x -3
    y + 0 = -x

    In which case, those points do not have equal tangent line functions. Yeah, I would use a different method to find things of equal tangent line functions.
     
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