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Simple Unit Vector Problem Part 2

  1. Jul 7, 2014 #1
    1. The problem statement, all variables and given/known data

    This is almost exactly the same problem as my earlier post, however different equation and
    point.

    Find the unit vector parallel to the graph f(x) = tan(x) and the point (pi/4 , 1) , and the
    unit vector perpendicular to the graph and point (pi/4 , 1 ).

    2. Relevant equations

    Alright so following some advice from earlier, f'(x) = sec(x)tan(x)
    at the point (pi/4 , 1), the slope of the function is f'(pi/4) = (1/(sqrt(2))/2)*(1) = (2)/(sqrt(2))


    3. The attempt at a solution

    So therefore, the parallel vector should have a base of < sqrt(2) , 2 >
    and since the magnitude is sqrt (2 + 4) = sqrt (6)
    then the unit vector parallel would then be <sqrt(2/6) , 2/sqrt(6) >

    AND
    the perp. vector is based off of the negative reciprocal of the parallel slope, equals -sqrt(2)/2
    perp vector = < 2, -sqrt(2) >

    unit perp. vector = <2/sqrt(6), -sqrt(1/3) >

    Attached is a photo.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jul 7, 2014 #2
    Since when is the derivative of tan(x) equal to sec(x)tan(x)?

    Chet
     
  4. Jul 7, 2014 #3
    Since my derivatives and integrals are starting to blur together ^.^

    Clearly then, f'(x) = sec^2(x)
    During which, when x = pi/2, f'(x) = 2,

    Vector par =< 1i, 2j >
    Vhat = <1/sqrt(5) , 2/sqrt(5) >

    Vector perp. = <2i, -1j >
    Vp hat = <2/sqrt(5) , -1/sqrt(5) >

    Bamn, web assign accepts, thank you!!
    :D
     
  5. Jul 7, 2014 #4

    LCKurtz

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    Don't mix notations like that. Either write < 1,2> or write i + 2j.
     
  6. Jul 7, 2014 #5
    Hmmm. I did not know this. Alright so if you have a vector ai + bj, it can be stated as <a, b> OR ai + bj .

    May have been making up my own thing there. Please disregard previous method of writing them up.
     
  7. Jul 8, 2014 #6

    Fredrik

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    Right, because by definition, i=<1,0> and j=<0,1>. So <a,b> = <a,0>+<0,b> = a<1,0>+b<0,1> = ai+bj.
     
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