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Thread: How is the Tanget Unit Vector [T(t)] Orthogonal to T'(t)?

  1. #1
    Mar 2010

    How is the Tanget Unit Vector [T(t)] Orthogonal to T'(t)?

    Hey everyone. I'm having a hard time getting a visual of $\displaystyle \vec{T} $ being perpendicular to $\displaystyle \vec{T'} $. The proof makes sense. Here it is:

    Assume the vector function $\displaystyle |\vec{r(t)}| = c $
    Thus, we take the dot product of the same vector:

    $\displaystyle \vec{r(t)} * \vec{r(t)} = |\vec{r(t)}|^2 = c^2 $

    If we were to take the derivative of this equation, we would get:

    $\displaystyle 0 = \frac{d}{dt}[\vec{r(t)} * \vec{r(t)}] = \vec{r'(t)} * \vec{r(t)} + \vec{r(t)} * \vec{r'(t)} = 2\vec{r'(t)} * \vec{r(t)}$

    Thus, $\displaystyle \vec{r(t)} * \vec{r'(t)} = 0 $

    Now this make sense because we are talking about a position vector and its tangent. This is also easy to visualize: a point on a circle. The position vector is like the centripetal force, it points toward the center. So this makes perfect sense.

    Now let's talk about $\displaystyle \vec{T(t)} $ and $\displaystyle \vec{T'(t)} $

    $\displaystyle |\vec{T(t)}| = 1$ so the proof above applies here too. The problem is that I can't visual this. I mean I can but it contradicts the proof. $\displaystyle \vec{T(t)} $ is not a position vector of $\displaystyle \vec{r(t)} $. It is actually the unit vector of $\displaystyle \vec{r'(t)} $; thus, we're basically saying $\displaystyle \vec{r'(t)} $ is perpendicular to $\displaystyle \vec{r''(t)} $, which is not true. (ie. if f'(x) = x^2, f'"(x) = 2. They are not perpendicular.) HOWEVER, that is in 2D. idk what happens in 3D cuz I am having a hard time visualizing it.

    So the question is, can someone please give me an example of where the tangent and the 2nd derivative of a function are perpendicular? Or somehow help me grasp this. Thanks in advance!
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  2. #2
    Member HappyJoe's Avatar
    Sep 2010
    You consider the example f'(x)=x^2, but notice how this function does not satisfy |f'(x)|=1.

    Only when |T(t)| is constant can you be sure that T(t) and T'(t) are perpendicular (in particular, this is true for |T(t)|=1 for all 1).

    For an easy example, take the circle parametrized by r(t) = (cos(t),sin(t),0) (or you could leave out the last coordinate, if you are comfortable with plane curves).

    Certainly |T(t)|=1. Also, T(t) = (-sin(t),cos(t),0), while T'(t) = (-cos(t),-sin(t),0). Notice how T(t) and T'(t) are perpendicular.
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  3. #3
    Mar 2010
    :O touche. This is what I did:

    I wanted to be consistent with the f(x) = x^2 so I parametrized it:

    x = t, y = t^2 2pi <= t <= 2pi

    took the derivative and got

    x' = 1, y' = 2t 2pi < t < 2pi

    got the length of it and graphed out the parametric and I got a circle! the tangent unit vector was:

    x = (1+4t^2)^(-1/2)
    y = 2t*(1+4t^2)^(-1/2)

    Did it without parametriziaton and got this:

    y = 2x * (x^2 + y^2)^(-1/2)

    ^ couldn't graph that :\ got an error on my graphic calculator saying "Memory"

    So the tangent unit vector is always a circle?
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  4. #4
    MHF Contributor

    Apr 2005
    "So the tangent unit vector is always a circle?"

    ouch!! The tangent unit vector is a vector, at a point on the graph, not a graph itself! It is true that, if you draw any vector of length 1, whether it is a tangent vector or not. with its "tail" at the origin then its "head" will lie on the unit circle- that's precisely what "length 1" means. If, for this problem, you carefully graphed all of the unit tangent vectors with tail at the origin, their heads would be on the unit circle but they would not cover the unit circle. They would cover only the right half of the unit circle because they would go from "almost straight down" (x a very large negative number) to "almost straight up" (x a very large positive number). But as I say, lieing on the unit circle simply reflects that they always have length 1.

    For $\displaystyle y= x^2$, we can use x itself as parameter: $\displaystyle x= t$, $\displaystyle y= t^2$. The "position vector" would be $\displaystyle t\vec{i}+ t^2\vec{j}$.

    The derivative of that is $\displaystyle \vec{i}+ 2t\vec{j}$ which has length $\displaystyle \sqrt{1+ t^2}$.

    That is, the unit tangent vector is [tex]\vec{T}= \frac{1}{\sqrt{1+ t^2}\vec{i}+ \frac{2t}{1+ t^2}\vec{j}[tex], just as you say. Factoring out that root, $\displaystyle \vec{T}= (1+ 4t^2)^{1/2}\left(\vec{i}+ 2\vec{j}\right)$. If we write $\displaystyle (1+ 4t^2)^{-1/2}= (1+ 4t^2)^{-3/2)(1+ 4t^2)$, we can factor out $\displaystyle (1+ 4t^2)^{-3/2}$ in the second term as well leaving $\displaystyle \vec{T}'= (1+ 4t^2)^{-3/2}\left(-4t\vec{i}- 8t^2\vec{j}+ (2+ 8t^2)\vec{j}\right)= (1+ 4t^2)^{-3/2}\left(-4t\vec{i}+ 2\vec{j}\right)$

    The dot product of $\displaystyle \vec{T}$ and $\displaystyle \vec{T}'$ is
    $\displaystyle \left((1+ 4t^2)^{-1/2}(\vec{i}+ 2t\vec{j})\right)\cdot\left((1+ 4t^2)^{-3/2}(-4t\vec{i}+ 2\vec{j})= (1+ 4t^2)^{-2}\left((1)(-4t)+ (2t)(2)\right)= 0$.

    In particular, if t= 0, x= y= 0 so we are at the origin. The unit tangent vector there is $\displaystyle (1+ 4(0^2))^{-1/2}\left(\vec{i}+ 0\vec{j}\right)= \vec{i}$, pointing along the x-axis. It's derivative, $\displaystyle (1+ 4(0^2))^{-3/2}\left(-4(0)\vec{i}+ 2\vec{j}\right)= 2\vec{j}$ points along the y-axis, perpendicular to the tangent vector.

    In fact, $\displaystyle \vec{j}$ is the "unit normal vector" to the curve, and the length of that, 2, is the "curvature" at (0, 0).

    Similarly, at t= 1, x= y= 1 so we are at the point (1, 1) on the parabola. The unit tangent vector is $\displaystyle (1+ 4(1^2))^{-1/2}\left(\vec{i}+ 2(1)\vec{j}\right)= \frac{1}{\sqrt{5}}\left(\vec{i}+ 2\vec{j}\right)$ and its derivative is [tex](1+ 4(1^2))^{-3/2}\left(-4(1)\vec{i}+ 2\vec{j}\right)= \frac{1}{\sqrt{125}}\left(-4\vec{i}+ 2\vec{j}\right). I recommend you plot those, not as a curve but actually draw the vectors with tail at (1, 1). But it is easy to see that their dot product is $\displaystyle \frac{1}{25}((1)(-4)+ 2(2))= 0$. The two vectors are, indeed, perpendicular.
    The length of that derivative, by the way, is $\displaystyle \sqrt{\frac{1}{125}(15+ 4)}= \sqrt{\frac{20}{125}}= \sqrt{\frac{4}{25}}= \frac{2}{5}$. The curvature of the graph of $\displaystyle y= x^2$ is $\displaystyle \frac{2}{5}$ at (1, 1).

    As a simple, example, try this with the circle of radius 5 with center at the origin: parametric equations x= cos(t), y= sin(t). Find the unit tangent vector and its derivative. You should see that the unit tangent vector is always, of course, tangent to the circle and that its derivative always point toward the origin, the center of the circle.

    A very simple example is the
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