Skip to main content

equation solving - Finding unit tangent, normal, and binormal vectors for a given r(t)


For my Calc III class, I need to find $T(t), N(t)$, and $B(t)$ for $t=1, 2$, and $-1$, given $r(t)=\{t,t^2,t^3\}$.


I've got Mathematica, but I've never used it before and I'm not sure how to coerce it into solving this. (My professor told us to use a computer, as it starts getting pretty nasty around $T'(t)$. By hand, $$T(t)=\frac{1}{ \sqrt{(1+4t^2+9t^4})}\{1,2t,3t^2\}$$


I've tried defining r as stated above and using D,Norm, and CrossProduct.


However, I get a bunch of Abs in my outputs (am I missing an assumption?). Additionally, I can't seem to google up how to ask Mathematica to sub in a specific value for $t$, once I get the equations worked out properly.




Answer



Mathematica wouldn't be much helpful if one applied only formulae calculated by hand.
Here we demonstrate how to calculate the desired geometric objects with the system having a definition of the curve r[t] :


r[t_] := {t, t^2, t^3}

now we call uT the unit tangent vector to r[t]. Since we'd like it only for real parameters we add an assumption to Simplify that t is a real number. Similarly we can do it for the normal vector vN[t] and the binormal vB[t] :


uT[t_] = Simplify[ r'[t] / Norm[ r'[t] ], t ∈ Reals];
vN[t_] = Simplify[ uT'[t]/ Norm[ uT'[t]], t ∈ Reals];
vB[t_] = Simplify[ Cross[r'[t], r''[t]] / Norm[ Cross[r'[t], r''[t]] ], t ∈ Reals];


let's write down the formulae :


{uT[t], vN[t], vB[t]} // Column // TraditionalForm

enter image description here


Edit


Definitions provided above are clearly more useful than only to write them down. They are powerful enough to animate a moving reper along a curve r[t]. Indeed the vectors uT[t], vN[t] and vB[t] are orthogonal and normalized, e.g.


Simplify[ Norm /@ {uT[t], vN[t], vB[t]}, t ∈ Reals]


 {1, 1, 1}


To demonstrate a moving reper we can use ParametricPlot3D and Arrow enclosed in Animate :


Animate[
Show[ ParametricPlot3D[ {r[t]}, {t, -1.3, 1.3}, PlotStyle -> {Blue, Thick}],
Graphics3D[{ {Thick, Darker @ Red, Arrow[{r[s], r[s] + uT[s]}]},
{Thick, Darker @ Green, Arrow[{r[s], r[s] + vB[s]}]},
{Thick, Darker @ Cyan, Arrow[{r[s], r[s] + vN[s]}]}}],
PlotRange -> {{-2, 2}, {-2, 2}, {-2, 2}}, ViewPoint -> {4, 6, 0},
ImageSize -> 600],
{s, -1, 1}]


enter image description here


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...