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Calc Help!

  1. Sep 6, 2008 #1
    1. The problem statement, all variables and given/known data
    Find a unit vector (a) parallel to and (b) normal to the graph of f(x) at the given point.

    F(x)= -x^2 + 5
    Point= (1,4)

    2. Relevant equations

    no equations really.

    3. The attempt at a solution

    Im not exactly sure what find a unit vector to the parallel and to the normal even means. I sketched the graph and plotted the points but im not sure where to go from there. I know a unit vector is of length one. If i draw a unit vector off of a parabola, in the normal (assuming the normal is straight out perpendicular of the parabola). I would get a triangle (45,45,90) and sqrt(2)/2 sides and a 1 hypot. Im not really sure where to go from there...
     
  2. jcsd
  3. Sep 6, 2008 #2

    Dick

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    You know how to find the slope of a tangent using differentiation, right? Then if i and j are the x and y unit vectors, a tangent vector is a vector a*i+b*j that has the same slope. So find one. Now make it unit. Then rotate it 90 degrees to find the normal.
     
  4. Sep 7, 2008 #3

    To find the slope of the tangent i would just take the derivative and plug in the value of x where the point is correct? I would then just use that slope to find the x and y components of the tangent? I could be totally wrong here. sorry, i am just having trouble visualizing this problem
     
  5. Sep 7, 2008 #4

    Dick

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    Yes. Try it out. Don't just worry about it.
     
  6. Sep 7, 2008 #5

    Ok, so the first step is to take the deriv. of -x^2+5, which is -2x. At the point (1,4) the slope is -2. So that means the norm is sqrt(5). So plotting the point from that point would be... P= (1,4) and Q= (2,2). When i put it into vector component form i get PQ= <1,-2>. To turn it into a unit vector i divide the vector by the norm, which is sqrt(5). So <1,-2>/sqrt(5). Which gives <1/sqrt(5), -2/sqrt(5)> for the final vector at the origin.
     
  7. Sep 7, 2008 #6

    Dick

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    Perfect. What about the normal?
     
  8. Sep 7, 2008 #7

    It would be the same except the V2 would turn from negative to positive because it is rotated above the x axis... So the normal would = <1/sqrt(5), 2/sqrt(5)> correct?
     
  9. Sep 7, 2008 #8

    Dick

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    Nope. To rotate 90 degrees you interchange x and y and then make one of them negative. Draw a picture.
     
  10. Sep 7, 2008 #9

    Alright, i went back and found where i went wrong. So it would just switch the two values and change the sign of one... So <2/sqrt(5), 1/sqrt(5). Im not 100% sure where you got the negative from because both the x and y components of the vector are in quadrant one.
     
  11. Sep 7, 2008 #10

    HallsofIvy

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    The point of "switch the two values and change the sign of one" is because two vectors are perpendicular if and only if their dot product is 0. If we start with <a, b>, "switch the two values and change the sign of one" we get either <-b, a> or <b, -a>. It doesn't matter which because <a, b>.<-b, a>= -ab+ ab= 0 and <a, b>.<b, -a>= ab- ab= 0.
     
  12. Sep 7, 2008 #11

    OHHH!!!! Wow, i totally forgot about that, thanks a bunch for all of the help ivy and dick!
     
  13. Sep 7, 2008 #12

    HallsofIvy

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    Finding perpendiculars in 3 dimensions is a lot more complicated! At a point on a line there is a whole plane of perpendicular vectors.
     
  14. Sep 7, 2008 #13
    Ahh, sounds fun! Maybe one day you'll be guiding me through a problem involving 3-d perpendiculars!! haha!
     
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