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Tangent Vectors on a 2D Graph

  1. Jan 26, 2012 #1
    1. The problem statement, all variables and given/known data

    f (x) = e^(3x) + sin(2x) + 3x +1

    (a) Find a vector V that is tangent to the graph of y = f(x) at the point ( 0, 2).
    (b) Find a vector N that is perpendicular to the graph of y = f(x) at the point ( 0, 2).

    2. The attempt at a solution

    The first step I took is to find the derivative of the function, since the problem is asking for a tangent at a point.

    I got this: f'(x) = 3e^(3x) + 2cos(2x) + 3

    However, I am unsure how to continue. The graph of "y = f(x)" is kind of confusing. I am thinking of maybe somehow getting parametric equations for the tangent line, which would allow me to build a vector. But I am unsure how to do this.
     
  2. jcsd
  3. Jan 26, 2012 #2

    ehild

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    Hi Zdravko,

    You need to use the general equation of a straight line -what is it?


    ehild
     
  4. Jan 26, 2012 #3
    Well, the general equations would be y=mx+b
    However, at this point in the class, we are only using parametrics for lines.

    In other words, the only lines we have built so far (involving vectors) have been in this format:

    x = a + bt
    y = a + bt
    etc...
     
  5. Jan 26, 2012 #4

    ehild

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    It is easier to use the first equation. But for both methods, you know one point of the line: (0,2), and need the
    the slope m or the tangent vector of the line: For that, evaluate f''(x) = 3e^(3x) + 2cos(2x) + 3 at x=0.

    ehild
     
  6. Jan 26, 2012 #5
    I think I got it.

    I found the slope of the tangent line at x=0 to be 8.
    After that, I constructed a vector from point (0,2) to next point on tangent line (1,10).
    That vector is V = <1,8> which satisfies the first part of the problem.

    For the second part, I built the equation of the line: y=8x+2

    Then I saw in my notes that if put in standard form (-8x + y = 2), the "a" and "b" are the vector perpendicular to the line.

    So the normal vector to V would be N = <-8,1> .

    Is this correct?
     
  7. Jan 26, 2012 #6

    ehild

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    It is correct. I just noticed that you needed the vectors instead of equations of lines, tangent or perpendicular to the curve.
    You did it right: a tangent vector to f(x) is (vx,vy)=(1, f'(x)), and a normal vector is ((nx,ny)=(-f'(x), 1).
    Anyway, the vectors ( a,b) and (-b, a) are perpendicular to each other. (what is the scalar product of two vectors if they are perpendicular?)

    ehild
     
  8. Jan 26, 2012 #7
    I forgot about that method to check. The scalar/dot product of the two perpendicular vectors should be 0. Thanks for all the help!
     
  9. Jan 26, 2012 #8

    ehild

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    You are welcome :smile:

    ehild
     
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