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Find unit tangent vector at indicated point

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the unit tangent vector at the indicated point of the vector function

    r(t) = e(19t)costi + e(19t)sintj + e(19t) k


    T(pi/2) = <___i+___j+___k>


    2. Relevant equations

    r'(t) / |r'(t)|

    3. The attempt at a solution


    Answers:



    19e(19*∏/2)(cos(∏/2)-sin(∏/2)) / ((19e(19*∏/2)sin(∏/2))2+(19e(19*∏/2)cos(∏/2))2+(19e(19*∏/2))2)^.5 i

    19e(19*∏/2)(cos(∏/2)+sin(∏/2)) / ((19e(19*∏/2)sin(∏/2))2+(19e(19*∏/2)cos(∏/2))2+(19e(19*∏/2))2)^.5 j (correct)

    19e(19*∏/2) / ((19e(19*∏/2)sin(∏/2))2+(19e(19*∏/2)cos(∏/2))2+(19e(19*∏/2))2)^.5 k (correct)

    What's wrong with i?

    Thanks!
     
    Last edited: Feb 15, 2012
  2. jcsd
  3. Feb 15, 2012 #2

    Dick

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    The overall sign is wrong on the i component.
     
  4. Feb 15, 2012 #3
    Sorry, I didn't mean to put that there. I'm using a program online though and it's still wrong.

    Annoying thing I actually put in:

    19e^(19*pi/2)(cos(pi/2)-sin(pi/2))/((19e^(19*pi/2)sin(pi/2))^2+(19e^(19*pi/2)cos(pi/2))^2+(19e^(19*pi/2))^2)^.5

    The parenthesis are correct because it shows us a much neater version of what we put in with our computer, and everything is as it should look.
     
  5. Feb 15, 2012 #4

    Dick

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    Ok, there is more than the sign wrong. The numerator should be e^(19*pi/2)*(19*cos(pi/2)-sin(pi/2)). Both terms don't have a factor of 19 in them. Check your product rule. There is a similar problem with the j component, but it didn't get caught because cos(pi/2)=0. BTW, you could simplify these expressions a LOT.
     
    Last edited: Feb 15, 2012
  6. Feb 15, 2012 #5

    Thank you so much for your help. I would simplify, but I can't go farther than this on tests because I don't have enough time (this long annoying answer counts as full credit) and I'd like to do it like I'll do it on the test.
     
  7. Feb 15, 2012 #6

    Dick

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    Sure and you are welcome. You can do it anyway you'll get full credit. But at least setting cos(pi/2)=0 and sin(pi/2)=1 will also save you time writing the answer down on a test.
     
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