8. Properties of Curves
h. Summary & Examples of Curve Computations
2. Examples of Curve Computations
c. Twisted Cubic
We now look at the twisted cubic given by the position vector \(\vec{r}(t)=(at,bt^2,ct^{3})\) where \(a\), \(b\), and \(c\) are any constants. Here are all the properties of this curve and the plot for the special case with \(a=1\), \(b=1\), and \(c=\dfrac{2}{3}\):
Position Vector
\(\vec{r}(t)=\left\langle t,t^2,\dfrac{2}{3}t^{3}\right\rangle\)
Plot
Rotate with your mouse.
Velocity Vector
\(\vec{v}(t)=\dfrac{d\vec{r}}{dt}=\langle 1,2t,2t^2\rangle\)
Acceleration Vector
\(\vec{a}(t)=\dfrac{d\vec{v}}{dt}=\langle 0,2,4t\rangle\)
Jerk Vector
\(\vec{j}(t)=\dfrac{d\vec{a}}{dt}=\langle 0,0,4\rangle\)
Speed
\(\begin{aligned} \dfrac{ds}{dt}&=|\vec{v}|=\sqrt{1^2+(2t)^2+(2t^2)^2} =\sqrt{1+4t^2+4t^{4}} \\ &=\sqrt{(1+2t^2)^2} =1+2t^2 \end{aligned}\)
Arclength (from \(t=a\) to \(t=b\))
\(\displaystyle \begin{aligned} L&=\int_{a}^{b} |\vec{v}|\,dt =\int_{a}^{b} (1+2t^2)\,dt =\left[t+\dfrac{2}{3}t^{3}\right]_{a}^{b} \\ &=b-a+\dfrac{2}{3}(b^{3}-a^{3}) \end{aligned}\)
Unit Tangent Vector
\(\hat{T}=\dfrac{\vec{v}}{|\vec{v}|} =\dfrac{\langle 1,2t,2t^2\rangle}{1+2t^2} =\left\langle \dfrac{1}{1+2t^2},\dfrac{2t}{1+2t^2}, \dfrac{2t^2}{1+2t^2}\right\rangle\)
\(\vec{v}\times\vec{a} =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2t & 2t^2 \\ 0 & 2 & 4t \end{vmatrix} =\hat{\imath}(8t^2-4t^2)-\hat{\jmath}(4t)+\hat{k}(2) =\langle 4t^2,-4t,2\rangle\)
\(\begin{aligned} |\vec{v}\times\vec{a}| &=\sqrt{(4t^2)^2+(-4t)^2+(2)^2} =\sqrt{16t^{4}+16t^2+4} \\ &=\sqrt{(4t^2+2)^2} =4t^2+2 =2(1+2t^2) \end{aligned}\)
Unit Binormal Vector
\(\hat{B}=\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|} =\dfrac{\langle 4t^2,-4t,2\rangle}{2(1+2t^2)} =\left\langle \dfrac{2t^2}{1+2t^2},\dfrac{-2t}{1+2t^2}, \dfrac{1}{1+2t^2}\right\rangle\)
Unit Normal Vector
\(\begin{aligned} \hat{N}&=\hat{B}\times\hat{T}=\dfrac{1}{(1+2t^2)^2} \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2t^2 & -2t & 1 \\ 1 & 2t & 2t^2 \end{vmatrix} \\ &=\dfrac{1}{(1+2t^2)^2} \left(\hat{\imath}(-4t^{3}-2t) -\hat{\jmath}(4t^{4}-1) +\hat{k}(4t^{3}+2t)\right) \\ &=\dfrac{1}{1+2t^2}\langle -2t,1-2t^2,2t\rangle =\left\langle \dfrac{-2t}{1+2t^2},\dfrac{1-2t^2}{1+2t^2}, \dfrac{2t}{1+2t^2}\right\rangle \end{aligned}\)
Alternatively
\(\begin{aligned} \hat{N}&=\dfrac{T'(t)}{|T'(t)|} =\dfrac{\left\langle \dfrac{-4t}{(1+2t^2)^2},-2\dfrac{2t^2-1}{(1+2t^2)^2}, \dfrac{4t}{(1+2t^2)^2}\right\rangle} {\sqrt{\left(\dfrac{-4t}{(1+2t^2)^2}\right)^2 +\left(-2\dfrac{2t^2-1}{(1+2t^2)^2}\right)^2 +\left(\dfrac{4t}{(1+2t^2)^2}\right)^2}} \\ &=\dfrac{\dfrac{1}{(1+2t^2)^2}\left\langle -4t,2(1+2t^2), 4t\right\rangle}{\dfrac{2}{1-2t^2}} =\left\langle \dfrac{-2t}{1+2t^2},\dfrac{1-2t^2}{1+2t^2}, \dfrac{2t}{1+2t^2}\right\rangle \end{aligned}\)
Curvature
\(\kappa=\dfrac{|\vec{v}\times\vec{a}|}{|\vec{v}|^{3}}= \dfrac{2(1+2t^2)}{(1+2t^2)^{3}} =\dfrac{2}{(1+2t^2)^2}\)
Torsion
\(\tau=\dfrac{\vec{v} \times\vec{a}\cdot\vec{j}}{|\vec{v}\times\vec{a}|^2} =\dfrac{\langle 4t^2,-4t,2\rangle\cdot\langle 0,0,4\rangle}{(2(1+2t^2))^2} =\dfrac{8}{4(1+2t^2)^2} =\dfrac{2}{(1+2t^2)^2}\)
Tangential Acceleration
\(\begin{aligned} a_{T}&=\vec{a}\cdot\hat{T} =\langle 0,2,4t\rangle \cdot\left\langle \dfrac{1}{1+2t^2},\dfrac{2t}{1+2t^2}, \dfrac{2t^2}{1+2t^2}\right\rangle =\dfrac{4t+8t^{3}}{1+2t^2} \\ &=\dfrac{4t(1+2t^2)}{1+2t^2}=4t \\ a_{T}&=\dfrac{d}{dt}|\vec{v}|=\dfrac{d}{dt}(1+2t^2)=4t \end{aligned}\)
Normal Acceleration
\(\begin{aligned} a_{N}&=\vec{a}\cdot\hat{N} =\langle 0,2,4t\rangle \cdot\left\langle \dfrac{-2t}{1+2t^2},\dfrac{1-2t^2}{1+2t^2}, \dfrac{2t}{1+2t^2}\right\rangle =\dfrac{2-4t^2}{1+2t^2}+\dfrac{8t^2}{1+2t^2} \\ &=\dfrac{2+4t^2}{1+2t^2}=2\dfrac{1+2t^2}{1+2t^2}=2 \\ a_{N}&=\kappa|\vec{v}|^2=\dfrac{2}{(1+2t^2)^2}(1+2t^2)^2=2 \end{aligned}\)