1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    Hi Zdravko,

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

  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
  5. Jan 26, 2012 #4


    User Avatar
    Homework Helper

    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.

  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


    User Avatar
    Homework Helper

    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?)

  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


    User Avatar
    Homework Helper

    You are welcome :smile:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook