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Angle between two tangent lines

  1. Jan 20, 2012 #1
    1. The problem statement, all variables and given/known data

    http://www.xtremepapers.com/Edexcel/Advanced%20Level/Mathematics/Subject%20Sorted/C3/Elmwood%20Papers/Elmwood%20B.pdf [Broken]

    Question 8(b)

    2. Relevant equations

    3. The attempt at a solution

    Ok so I found both values of dy/dx for BOTH EQUATIONS

    y = x2 - 4x → 2x - 4
    y = |4x - x2| → |4 - 2x| could someone please claify this, I have never differentiated a modulus before

    Thus the two gradients of the lines are 4 and -4 BUT HOW DO I GO ON TO FIND THE angle between the tangents??
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 20, 2012 #2

    HallsofIvy

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    [itex]|4x- x^2|= |x(4- x)|[/itex] and is equal to [itex]4x- x^2[/itex] for x between 0 and 4 but equal to [itex]x^2- 4x[/itex] for x< 0 or x> 4. That is, its derivative if 4- 2x for x between 0 and 4 and equal to 2x- 4 for x< 0 or x> 4. At x= 4, there is a "cusp" so technically, there is no derivative. Of course, you are interested in the curve between 0 and 4 so you really want [itex]\lim_{x\to 4^-} 4- 2x= -4[/itex] as you say.

    To find the angle between the lines remember that the derivative is the tangent of the angle the curve makes with the horizontal. And that
    [tex]tan(\theta- \phi)= \frac{tan(\theta)- tan(\phi)}{1+ tan(\theta)tan(\phi)}[/tex]
     
  4. Jan 21, 2012 #3
    I never knew this
    is this A-level maths or beyond?
    could you please explain why this is true?
     
  5. Jan 21, 2012 #4

    SammyS

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    It's usually taught in Calculus when you first learn about derivatives representing the slope of the tangent line.

    It's often taught in trigonometry that the slope of a line is equal to the tangent of the angle the line makes with the x-axis.
     
  6. Jan 21, 2012 #5
    thanks for this :)
    I will note this rule!!
     
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