8. Properties of Curves
h. Summary & Examples of Curve Computations
2. Examples of Curve Computations
b. Helix
Next we look at helical motion represented by the position vector \(\vec{r}(\theta)=(a\cos\theta,a\sin\theta,b\theta)\) where \(a\) and \(b\) are any constants. Helical motion is similar to circular motion, but the curve rises as the angle increases. Here are all the properties of this curve and the plot for the special case with \(a=4\), \(b=3\):
Position Vector
\(\vec{r}(\theta)=\langle 4\cos\theta,4\sin\theta,3\theta\rangle\)
Plot
Rotate with your mouse.
Velocity Vector
\(\vec{v}(\theta)=\dfrac{d\vec{r}}{d\theta} =\langle -4\sin\theta,4\cos\theta,3\rangle\)
Acceleration Vector
\(\vec{a}(\theta)=\dfrac{d\vec{v}}{d\theta} =\langle -4\cos\theta,-4\sin\theta,0\rangle\)
Jerk Vector
\(\vec{j}(\theta)=\dfrac{d\vec{a}}{d\theta} =\langle 4\sin\theta,-4\cos\theta,0\rangle\)
Speed
\(\dfrac{ds}{d\theta}=|\vec{v}| =\sqrt{16\sin^2\theta+16\cos^2\theta+9}=5\)
Arclength (from \(\theta=\alpha\) to \(\theta=\beta\))
\(\displaystyle L=\int_{\alpha}^{\beta} |\vec{v}|\,d\theta =\int_{\alpha}^{\beta} 5\,d\theta =5(\beta-\alpha)\)
Notice that this is the hypotenuse of a right triangle whose base is \(4(\beta-\alpha)\) and whose altitude is \(3(\beta-\alpha)\), which has been rolled around a cylinder.
Unit Tangent Vector
\(\hat{T}=\dfrac{\vec{v}}{|\vec{v}|} =\dfrac{1}{5}\langle -4\sin\theta,4\cos\theta,3\rangle =\left\langle -\,\dfrac{4}{5}\sin\theta,\dfrac{4}{5}\cos\theta, \dfrac{3}{5}\right\rangle\)
\(\begin{aligned} \vec{v}\times\vec{a} &=\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -4\sin\theta & 4\cos\theta & 3 \\ -4\cos\theta & -4\sin\theta & 0 \end{vmatrix} =\hat{\imath}(12\sin\theta)-\hat{\jmath}(12\cos\theta) +\hat{k}(16\sin^2\theta+16\cos^2\theta) \\ &=\langle 12\sin\theta,-12\cos\theta,16\rangle \end{aligned}\)
\(|\vec{v}\times\vec{a}| =\sqrt{(12\sin\theta)^2+(-12\cos\theta)^2+(16)^2} =\sqrt{144+256}=\sqrt{400}=20\)
Unit Binormal Vector
\(\hat{B}=\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|} =\dfrac{\langle 12\sin\theta,-12\cos\theta,16\rangle}{20} =\left\langle \dfrac{3}{5}\sin\theta,-\,\dfrac{3}{5}\cos\theta, \dfrac{4}{5}\right\rangle\)
Unit Normal Vector
\(\begin{aligned} \hat{N}&=\hat{B}\times\hat{T} =\dfrac{1}{25}\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3\sin\theta & -3\cos\theta & 4 \\ -4\sin\theta & 4\cos\theta & 3 \end{vmatrix} \\ &=\hat{\imath}\left(-9\cos\theta-16\cos\theta\right) -\hat{\jmath}\left(9\sin\theta+16\sin\theta\right)\\ &\quad+\hat{k}\left(12\sin\theta\cos\theta -12\sin\theta\cos\theta\right) \\ &=\langle -\cos\theta,-\sin\theta,0\rangle \end{aligned}\)
Alternatively
\(\begin{aligned} \hat{N}&=\dfrac{\hat T'(\theta)}{|\hat T'(\theta)|} =\dfrac{\left\langle -\,\dfrac{4}{5}\cos\theta,-\,\dfrac{4}{5}\sin\theta, 0\right\rangle} {\sqrt{\left(-\,\dfrac{4}{5}\cos\theta\right)^2 +\left(-\,\dfrac{4}{5}\sin\theta\right)^2}} \\ &=\dfrac{\left\langle -\,\dfrac{4}{5}\cos\theta,-\,\dfrac{4}{5}\sin\theta,0\right\rangle} {\dfrac{4}{5}} =\langle -\cos\theta,-\sin\theta,0\rangle \end{aligned}\)
So the unit normal is horizontal and points radially inward, directly toward the \(z\)-axis.
Curvature
\(\kappa=\dfrac{|\vec{v}\times\vec{a}|}{|\vec{v}|^{3}}= \dfrac{20}{5^{3}}=\dfrac{4}{25}\)
Torsion
\(\begin{aligned} \tau&=\dfrac{\vec{v}\times\vec{a}\cdot\vec{j}}{|\vec{v}\times\vec{a}|^2} =\dfrac{\langle 12\sin\theta,-12\cos\theta,16\rangle \cdot\langle 4\sin\theta,-4\cos\theta,0\rangle}{16^2} \\ &=\dfrac{48}{256}=\dfrac{3}{16} \end{aligned}\)
Tangential Acceleration
\(\begin{aligned} a_{T}&=\vec{a}\cdot\hat{T} =\langle -4\cos\theta,-4\sin\theta,0\rangle \cdot\left\langle -\,\dfrac{4}{5}\sin\theta,\dfrac{4}{5}\cos\theta, \dfrac{3}{5}\right\rangle \\ &=\dfrac{16}{5}\sin\theta\cos\theta-\,\dfrac{16}{5}\sin\theta\cos\theta=0 \\ a_{T}&=\dfrac{d}{d\theta}|\vec{v}|=\dfrac{d}{d\theta}5=0 \end{aligned}\)
Normal Acceleration
\(\begin{aligned} a_{N}&=\vec{a}\cdot\hat{N} =\langle -4\cos\theta,-4\sin\theta,0\rangle \cdot \langle -\cos\theta,-\sin\theta,0\rangle \\ &=4\cos^2\theta+4\sin^2\theta+0=4 \\ a_{N}&=\kappa|\vec{v}|^2=\dfrac{|\vec{v}|^2}{R}=\dfrac{4}{25}(25)=4 \end{aligned}\)