8. Properties of Curves
h. Summary & Examples of Curve Computations
2. Examples of Curve Computations
d. Cubic Spiral Helix
Finally we look at the cubic spiral helix represented by the position vector \(\vec{r}(t)=\langle at\cos t,at\sin t,bt^{3}\rangle\) where \(a\) and \(b\) are any constants. Here are all the properties of this curve and the plot for the special case with \(a=2\), \(b=\dfrac{1}{3}\):
Position Vector
\(\vec{r}(t)=(2t\cos t,2t\sin t,\dfrac{1}{3}t^{3})\)
Plot
Rotate with your mouse.
Velocity Vector
\(\vec{v}(t)=\dfrac{d\vec{r}}{dt} =\langle 2\cos t-2t\sin t,2\sin t+2t\cos t,t^2\rangle\)
Acceleration Vector
\(\vec{a}(t)=\dfrac{d\vec{v}}{dt} =\langle -4\sin t-2t\cos t,4\cos t-2t\sin t,2t\rangle\)
Jerk Vector
\(\vec{j}(t)=\dfrac{d\vec{a}}{dt} =\langle -6\cos t+2t\sin t,-6\sin t-2t\cos t,2\rangle\)
Speed
\(\begin{aligned} \dfrac{ds}{dt} &=|\vec{v}| =\sqrt{(2\cos t-2t\sin t)^2 +(2\sin t+2t\cos t)^2+(t^2)^2} \\ &=\sqrt{4+4t^2+t^{4}}=2+t^2 \end{aligned}\)
Arclength (from \(t=0\) to \(t=3\))
\(\displaystyle L=\int_0^{3} |\vec{v}(t)|\,dt =\int_0^{3} (2+t^2)\,dt =\left[2t+\dfrac{1}{3}t^{3}\right]_0^{3}=15\)
Unit Tangent Vector
\(\hat{T}=\dfrac{\vec{v}}{|\vec{v}|} =\left\langle \dfrac{2\cos t-2t\sin t}{2+t^2}, \dfrac{2\sin t+2t\cos t}{2+t^2}, \dfrac{t^2}{2+t^2}\right\rangle\)
\(\begin{aligned} \vec{v}\times\vec{a} &=\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2\cos t-2t\sin t & 2\sin t+2t\cos t & t^2 \\ -4\sin t-2t\cos t & 4\cos t-2t\sin t & 2t \end{vmatrix} \\ &=\hat{\imath}(2t(2+t^2)\sin t) -\hat{\jmath}(2t(2+t^2)\cos t) +\hat{k}(8+4t^2) \\ &=2(2+t^2)\langle t\sin t,-t\cos t,2\rangle \end{aligned}\)
\(\begin{aligned} |\vec{v}\times\vec{a}| &=2(2+t^2)\sqrt{(t\sin t)^2+(-t\cos t)^2+2^2} \\ &=2(2+t^2)\sqrt{4+t^2} \end{aligned}\)
Unit Binormal Vector
\(\begin{aligned} \hat{B}&=\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|} =\dfrac{1}{2(2+t^2)\sqrt{4+t^2}}\langle 2t(2+t^2)\sin t, -2t(2+t^2)\cos t,4(2+t^2)\rangle \\ &=\left\langle\dfrac{t\sin t}{\sqrt{4+t^2}}, \dfrac{-t\cos t}{\sqrt{4+t^2}},\dfrac{2}{\sqrt{4+t^2}}\right\rangle \end{aligned}\)
Unit Normal Vector
\(\begin{aligned} \hat{N}&=\hat{B}\times\hat{T} =\dfrac{1}{(2+t^2)\sqrt{4+t^2}} \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ t\sin t & -t\cos t & 2 \\ 2\cos t-2t\sin t & 2\sin t+2t\cos t & t^2 \end{vmatrix} \\ &=\dfrac{1}{(2+t^2)\sqrt{4+t^2}} \left( \begin{array}{llll} \hat{\imath}(-t^{3}\cos t-2(2\sin t+2t\cos t)) \\ -\hat{\jmath}(t^{3}\sin t-2(2\cos t-2t\sin t)) \\ +\hat{k}(t\sin t(2\sin t+2t\cos t)+t\cos t(2\cos t-2t\sin t)) \end{array} \right) \\ &=\dfrac{1}{(2+t^2)\sqrt{4+t^2}} \langle -4\sin t-(4t+t^{3})\cos t,4\cos t-(4t+t^{3})\sin t,2t\rangle \end{aligned}\)
Curvature
\(\kappa=\dfrac{|\vec{v}\times\vec{a}|}{|\vec{v}|^{3}}= \dfrac{2(2+t^2)\sqrt{4+t^2}}{(2+t^2)^{3}}= \dfrac{2\sqrt{4+t^2}}{(2+t^2)^2}\)
\(\begin{aligned} \vec{v}\times\vec{a}\cdot\vec{j} &=2(2+t^2)\langle t\sin t,-t\cos t,2\rangle \cdot\langle -6\cos t+2t\sin t,-6\sin t-2t\cos t,2\rangle \\ &=2(2+t^2)(2t^2+4) =4(2+t^2)^2 \end{aligned}\)
Torsion
\(\tau=\dfrac{\vec{v}\times\vec{a}\cdot\vec{j}}{|\vec{v}\times\vec{a}|^2} =\dfrac{4(2+t^2)^2}{4(2+t^2)^2(4+t^2)}=\dfrac{1}{4+t^2} \)
Tangential Acceleration
\(\begin{aligned} a_{T}&=\vec{a}\cdot\hat{T} \\ &=\langle -4\sin t-2t\cos t,4\cos t-2t\sin t,2t\rangle \\ &\quad\cdot\dfrac{1}{2+t^2}\langle 2\cos t-2t\sin t,2\sin t+2t\cos t,t^2\rangle \\ &=\dfrac{1}{2+t^2} \left( \begin{array}{l} -8\sin t\cos t+8t\sin^2t-4t\cos^2t+4t^2\sin t\cos t \\ +8\sin t\cos t+8t\cos^2t-4t\sin^2t-4t^2\sin t\cos t+2t^{3} \end{array} \right) \\ &=\dfrac{4t+2t^{3}}{2+t^2}=2t \\ a_{T}&=\dfrac{d|\vec{v}|}{dt}=\dfrac{d}{dt}(2+t^2)=2t \end{aligned}\)
Normal Acceleration
\(\begin{aligned} a_{N}\,&=\vec{a}\cdot\hat{N} \\ &=\langle -4\sin t-2t\cos t,4\cos t-2t\sin t,2t\rangle \\ &\quad\cdot\dfrac{1}{(2+t^2)\sqrt{4+t^2}} \langle -4\sin t-(4t+t^{3})\cos t,4\cos t -(4t+t^{3})\sin t,2t\rangle \\ &=\dfrac{2(2+t^2)(4+t^2)}{(2+t^2)\sqrt{4+t^2}} =2\sqrt{4+t^2} \\ a_{N}&=\kappa|\vec{v}|^2=\dfrac{2\sqrt{4+t^2}}{(2+t^2)^2}(2+t^2)^2 =2\sqrt{4+t^2} \end{aligned}\)