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Tangent unit vector of a curve

  1. Jul 30, 2010 #1
    1. The problem statement, all variables and given/known data
    See first figure.


    2. Relevant equations



    3. The attempt at a solution
    See second figure.

    When I set [tex]t = 0[/tex] in [tex]\vec{r(t)}[/tex] I get [tex]0\hat{i} +2\hat{j} + 1\hat{k}[/tex].

    I know this is a vector and not a point but it has the same "coordinates" as the point they are asking us to find the unit tanget at. Is there a reason for this?

    So I took the derivative of [tex]\vec{r(t)}[/tex] to get [tex]\vec{v(t)}[/tex]. I tried to get the magnitude of [tex]|\vec{v(t)}|[/tex], but I end up with a pretty messy expression.

    At this point I wasn't to sure how to proceed so I tried to make sense of things the best I could.

    When I evaluate [tex]\vec{r(0)}[/tex] it brings me to the point (the tip of the vector [tex]\vec{r(0)}[/tex]) at which I want to find the unit tangent vector.

    So if I evaluate [tex]\frac{\vec{v(0)}}{\vec{|v(0)}|}[/tex] I should be able to get the unit tangent vector at the desired point.

    I have a feeling this is wrong because [tex]\vec{r(0)}[/tex] is still a vector and this is not the same as a point.

    Does anyone have any suggestions for me? Or can correct my thought process?
     

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    Last edited: Jul 30, 2010
  2. jcsd
  3. Jul 30, 2010 #2
    You are wrongly interchanging the curve C and the vector r(t). Whereas r(t) is a different vector for different values of t, the curve C is the set of all points mapped by r(t). We are not finding a vector tangent to r(t) but rather to the curve C!! In this case, we are finding the unit tangent vector to a specific point (0, 2, 1) on C.

    Yes, the unit tangent is [tex]\frac{\vec{v(0)}}{\vec{|v(0)}|}[/tex]. However, you made finding v(t) more complicated than necessary and I don't think you got it right in the end. Factor out a -e^(-t) from v(t) and then find its norm (magnitude). This gets rid of quite a bit of algebra. Inside the radical, the 2sintcost terms should cancel out and so there's no need to use the double angle identity. You should be able to finish the rest from here.
     
  4. Jul 30, 2010 #3
    ...if I did the algebra right in my head (that's a noteworthy IF)... |v(0)| = sqrt(2).

    As for part b, do you know the formula for arc length? If you don't, try to come up with it by yourself :D! It's quite obvious that it will be the integral of ds (where ds represents an infinitesimal portion of the curve just as dx is infinitesimal portion of x). Imagine ds as a tiny tangent line to a point on the curve. But we want to get ds in terms of t and dt. By the Pythagorean theorem, (ds)^2 = (dx)^2 + (dy)^2 + (dz)^2. I'll leave it to you from here!
     
    Last edited: Jul 31, 2010
  5. Jul 30, 2010 #4
    I keep getting,

    [tex] \vec{|v(0)|} = i\sqrt{2}[/tex]

    Simplifying the expression,

    [tex]\sqrt{-e^{-2t}(1 - sin(2t) + 1 +sin(2t)}[/tex]

    If I set t = 0 then,

    [tex] \sqrt{-2} = i\sqrt{2}[/tex]

    Can someone double check this for me?

    As far part b, I'll get to that when I finish part a.

    Thanks again!
     
  6. Jul 30, 2010 #5

    Dick

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    |v(0)| is sqrt(x'(t)^2+z'(t)^2). The quantity inside the square root CAN'T be negative. (-1)*(-1)=1. Check your algebra.
     
  7. Jul 30, 2010 #6
    You forgot to square the negative in front of e. [tex] (-e^{-t})^2 = +e^{-2t} [/tex]
     
  8. Jul 30, 2010 #7
    Whoops! :redface:

    Alrighty so I'm finally convinced that,

    [tex]\vec{|v(0)|} = \sqrt{2}[/tex]

    So,

    [tex]\vec{T} = \frac{\vec{v(0)}}{\vec{|v(0)|}} = \frac{\hat{i} - \hat{k}}{\sqrt{2}}[/tex]

    Hopefully this is correct now!
     
  9. Jul 31, 2010 #8
    Now for part b,

    Since,

    [tex] L = \int^{b}_{a} \vec{|r'(0)|}du[/tex]

    then,

    [tex] L = \int^{2\pi}_{0} \sqrt{2}du = 2\pi\sqrt{2}[/tex]
     
  10. Jul 31, 2010 #9
    That's not right. Where'd you get that formula from? Refer back to my other post about finding ds in terms of dx, dy, and dz.
     
  11. Jul 31, 2010 #10
    A simple way to semi-check it is to find the magnitude of T. If [tex] |\vec{T}| = 1, [/tex] then you most likely have it right (unless you just get lucky and it works out to be 1 haha).
     
  12. Jul 31, 2010 #11
    I got it from the Stewart textbook.(see figure)

    The point,
    [tex](0,2,1)[/tex] is when [tex]t=0[/tex]

    and the point,

    [tex](0,2,e^{-2\pi}[/tex] is when [tex]t=2\pi[/tex]

    So they are asking me to find the length from 0 to 2pi in terms of t right?

    So why doesnt,

    [tex] \int^{2\pi}_{0} \vec{|r'(0)|}dt[/tex] work?
     

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    Last edited: Jul 31, 2010
  13. Jul 31, 2010 #12
    That's not what you have. There's a HUGE difference between r'(t) and r'(0) :P!

    But really, before using that formula, you should know how it was derived (refer back to my (ds)^2 = (dx)^2 + (dy)^2 + (dz)^2 post). All these results aren't just magical shortcuts to finding the answer. There's an important distinction between learning and memorizing.
     
  14. Jul 31, 2010 #13
    Another quick way to realize this can't be correct is that it would mean the arc length is independent of t and only dependent on the two limits. From this formula, you would be saying that the arc length from t = 0 to t = 1 is the same as that from t = 1 to t = 2 is same as...etc.
     
  15. Jul 31, 2010 #14
    This would be ds correct? (see figure)
     

    Attached Files:

  16. Jul 31, 2010 #15
    [tex]S = \int^{t}_{0} \sqrt{2}dt = \sqrt{2}t[/tex]

    So,

    [tex]t(S) = \frac{S}{\sqrt{2}}[/tex]

    I don't know where I'm suppose to go from here.
     
  17. Jul 31, 2010 #16
    No...like I said, there's a difference between r'(t) and r'(0)! Whereas r'(0) is a constant vector, r'(t) is a vector dependent on t. r'(t) = v(t) from part a.

    [tex]S = \int^{2\pi}_{0} |r'(t)| dt [/tex]
     
  18. Jul 31, 2010 #17
    In other words, [tex] |r'(0)| = \sqrt{2} [/tex] but [tex] |r'(t)| \neq \sqrt{2} [/tex].
     
  19. Jul 31, 2010 #18
    So,

    [tex]S = \int_{0}^{2\pi} \sqrt{2e^{-2t}}dt[/tex]

    ?
     
    Last edited: Aug 1, 2010
  20. Jul 31, 2010 #19
    yep, though you forgot the dt at the end.
     
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