3. Properties of Curves

Exercises

Vector Functions, Position, Velocity and Plots

Consider the position vector \(\vec{r}(t)=\langle t,t^2+1\rangle\).
1. Find the position at \(t=3\)
  
  Just substitute \(t=3\) into \(\vec{r}(t)=\langle t,t^2+1\rangle\).
  
  \(\vec{r}(3)=\langle 3,10\rangle\)
2. Find the velocity at \(t=3\)
  
  The velocity is \(\vec{v}(t)=\dfrac{d\vec{r}}{dt}\). Substitute in \(t=3\).
  
  \(\vec{v}(3)=\langle 1,6\rangle\)
  
  The velocity is \(\vec{v}(t)=\dfrac{d\vec r}{dt}=\langle 1,2t\rangle\). At \(t=3\) this is \(\vec{v}(3)=\langle 1,6\rangle\).
  
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3. Plot the parametric curve.
  
  Express \(y\) as a function of \(x\) by setting \(t=x\).
  
  The coordinate form of the parametric curve is \[ x=t \qquad y=t^2=1 \] Eliminating the parameter we have \(y=x^2+1\), whose graph is:
  
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4. Add the plot of the position vector at \(t=3\)
5. Add the plot of the velocity vector at \(t=3\)
Consider the position vector \(\vec{r}(t) =\langle 2t,3t^2,8t-4\rangle\)
1. Find the position at \(t=3\)
  
  Just substitute \(t=3\) into \(\vec{r}(t)=\langle 2t,3t^2,8t-4\rangle\).
  
  \(\vec{r}(3)=\langle 6,27,20\rangle\)
2. Find the velocity at \(t=3\)
  
  The velocity is \(\vec{v}(t)=\dfrac{d\vec{r}}{dt}\). Substitute in \(t=3\).
  
  \(\vec{v}(3)=\langle 2,18,8\rangle\)
  
  The velocity is \(\vec{v}(t)=\dfrac{d\vec r}{dt}=\langle 2,6t,8\rangle\). At \(t=3\) this is \(\vec{v}(3)=\langle 2,18,8\rangle\).
  
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3. Plot the parametric curve in 3 dimensions.
4. Plot the position vector at \(t=3\)
5. Add the plot of the velocity vector at \(t=3\)

Limits and Derivatives of Vector Functions

Find the derivative of the vector function \(\vec{f}(t)=\langle 3t-1,2t^{5}+3t+1,6-t\rangle\).

Find the derivative of each component with respect to \(t\)

\(\vec f''(t)=\langle 3,10t^4+3,-1\rangle\)

\[\begin{aligned} \vec f'(t) &=\dfrac{d\vec{f}}{dt} =\left\langle \dfrac{df_1}{dt},\dfrac{df_2}{dt},\dfrac{df_3}{dt}\right\rangle \\ &=\dfrac{d}{dt}\langle 3t-1,2t^{5}+3t+1,6-t\rangle =\langle 3,\,10t^4+3,\,-1\rangle \end{aligned}\]

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Find the derivative of the vector function \(\vec{f}(t)=\langle -9t,3t^2+5t+4,2t^3+8t^2-6t\rangle\).

\(\vec f'(t)=\langle -9,6t+5,6t^2+16t-6\rangle\)

\[ \vec f'(t)= \dfrac{d\vec{f}}{dt} =\left\langle \dfrac{df_1}{dt},\dfrac{df_2}{dt},\dfrac{df_3}{dt}\right\rangle \] \[ \dfrac{d}{dt}\langle -9t,3t^2+5t+4,2t^3+8t^2-6t\rangle =\langle -9,\,6t+5,\,6t^2+16t-6\rangle \]

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Find the derivative of the vector function \(\vec{f} (t)=\langle 2t,t^2+t+1,7-t\rangle\) at the point when \(t=2\)

\(\vec f'(2)=\langle 2,5,-1\rangle\)

\[ \vec f'(t) =\dfrac{d\vec{f}}{dt} =\left\langle \dfrac{df_1}{dt},\dfrac{df_2}{dt},\dfrac{df_3}{dt}\right\rangle \] \[ \dfrac{d}{dt}\langle 2t,t^2+t+1,7-t\rangle =\langle 2,2t+1,-1\rangle \] \[ \vec f'(2) =\langle 2,5,-1\rangle \]

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Find \(\vec f'(-3)\) if \(\vec{f}(t)=\langle 6,2t^2+t,-4t\rangle\)

\(\vec f'(-3)=\langle 0,-11,-4\rangle\)

\[ \vec f't) =\dfrac{d\vec{f}}{dt} =\left\langle \dfrac{df_1}{dt},\dfrac{df_2}{dt},\dfrac{df_3}{dt}\right\rangle \] \[ \dfrac{d}{dt}\langle 6,2t^2+t,-4t\rangle =\langle 0,4t+1,-4\rangle \] \[ \vec f'(-3) =\langle 0,-11,-4\rangle \]

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Velocity, Acceleration, and Jerk

Consider the parabola \(y=8-\,\dfrac{1}{2}(x-4)^2\)
1. Find the position vector. (Take \(x=t\).)
  
  \( \vec{r}(t)=\left\langle t,8-\,\dfrac{1}{2}(t-4)^2\right\rangle\)
2. Find the velocity vector
  
  \( \vec{v}(t)=\langle 1,4-t\rangle \)
  
  To find the velocity vector, we simply take the derivative of the position vector as \[ \vec{v}(t)=\dfrac{d\vec{r}}{dt} =\left\langle \dfrac{dx}{dt},\dfrac{dy}{dt}\right\rangle =\langle 1,4-t\rangle \]
  
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3. Find the acceleration vector
  
  \(\vec{a}(t)=\langle 0,-1\rangle\)
  
  We have already found the velocity vector \(\vec{v}(t)=\langle 1,4-t\rangle\) for the parabola. To find the acceleration vector, we simply take the derivative as \[ \vec{a}(t)=\dfrac{d\vec{v}}{dt}=\langle 0,-1\rangle \]
  
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4. Find the jerk vector
  
  \(\vec{j}(t)=\langle 0,0\rangle\)
  
  We have already found the acceleration vector \(\vec{a}(t)=\langle 0,-1\rangle\) for the parabola. To find the jerk vector, we simply take the derivative as \[ \vec{j}(t)=\dfrac{d\vec{a}}{dt}=\langle 0,0\rangle \]
  
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5. Plot the original parabola from \(t=0\) to \(t=8\). At three different points along the curve, plot the curve's corresponding velocity and acceleration vectors.
  
  The plot shows the parabola along with its corresponding velocity and acceleration vectors at \(7\) points.
  
  Note that the acceleration is a constant vector on the curve which points straight down (like gravity). This means the velocity is always bending down: an object that followed along this curve would be decelerating on the way up and accelerating on the way down.
  
  \(\vec{r}(t)\) is Blue,
  \(\vec{v}(t)\) is Green,
  \(\vec{a}(t)\) is Orange.
Consider the position vector \(\vec{r}(t)= \langle t^3+2t^2+t,t-3,t^4+6\rangle\)
1. Find the velocity vector
  
  \(\vec{v}(t)=\dfrac{d\vec{r}}{dt} =\left\langle \dfrac{dr_{1}}{dt},\dfrac{dr_{2}}{dt}, \dfrac{dr_{3}}{dt}\right\rangle\)
  
  \(\vec{v}(t)=\langle 3t^2+4t+1,1,4t^3\rangle\)
  
  \[\begin{aligned} \vec{v}(t)=\dfrac{d\vec{r}}{dt} =\left\langle \dfrac{dr_{1}}{dt},\dfrac{dr_{2}}{dt}, \dfrac{dr_{3}}{dt} \right\rangle \\ \end{aligned}\] \[\begin{aligned} \dfrac{d}{dt}\langle t^3+2t^2+t,t-3,t^4+6\rangle =\langle 3t^2+4t+1,1,4t^3\rangle \end{aligned}\]
  
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2. Find the acceleration vector
  
  \(\vec{a}(t)=\dfrac{d\vec{v}}{dt} =\left\langle \dfrac{dv_{1}}{dt},\dfrac{dv_{2}}{dt}, \dfrac{dv_{3}}{dt} \right\rangle\)
  
  \(\vec{a}(t)=\langle 6t+4,0,12t^2\rangle \)
  
  \[\begin{aligned} \vec{a}(t)=\dfrac{d\vec v}{dt} =\left\langle \dfrac{dv_{1}}{dt},\dfrac{dv_{2}}{dt}, \dfrac{dv_{3}}{dt} \right\rangle \\ \end{aligned}\] \[\begin{aligned} \dfrac{d}{dt}\langle 3t^2+4t+1,1,4t^3\rangle =\langle 6t+4,0,12t^2\rangle \end{aligned}\]
  
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3. Find the jerk vector
  
  \(\vec{j}(t)=\dfrac{d\vec{a}}{dt}= \left\langle \dfrac{da_{1}}{dt}, \dfrac{da_{2}}{dt},\dfrac{da_{3}}{dt}\right\rangle\)
  
  \(\vec{j}(t)=\langle 6,0,24t\rangle\)
  
  \[\begin{aligned} \vec{j}(t)=\dfrac{d\vec{a}}{dt} =\left\langle \dfrac{da_{1}}{dt},\dfrac{da_{2}}{dt}, \dfrac{da_{3}}{dt}\right\rangle \\ \end{aligned}\] \[\begin{aligned} \dfrac{d}{dt}\langle 6t+4,0,12t^2\rangle =\langle 6,0,24t\rangle \end{aligned}\]
  
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Consider the position vector \(\vec{r}(t) =\left\langle e^{t/2},t^{5/2}+1,\sin t\right\rangle\)
1. Find the velocity vector
  
  \(\vec{v}(t) =\left\langle \dfrac{1}{2}e^{t/2}, \dfrac{5}{2}t^{3/2},\cos t\right\rangle\)
  
  \[\begin{aligned} \vec{v}(t)=\dfrac{d\vec{r}}{dt} =\left\langle \dfrac{dr_{1}}{dt},\dfrac{dr_{2}}{dt}, \dfrac{dr_{3}}{dt}\right\rangle \\ \end{aligned}\] \[\begin{aligned} \dfrac{d}{dt}\left\langle e^{t/2} ,t^{5/2}+1,\sin t \right\rangle =\left\langle \dfrac{1}{2}e^{t/2}, \dfrac{5}{2}t^{3/2},\cos t\right\rangle \end{aligned}\]
  
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2. Find the acceleration vector
  
  \(\vec{a}(t) =\left\langle \dfrac{1}{4}e^{t/2}, \dfrac{15}{4}t^{1/2},-\sin t\right\rangle\)
  
  \[\begin{aligned} \vec{a}(t)=\dfrac{d\vec v}{dt} =\left\langle \dfrac{dv_{1}}{dt},\dfrac{dv_{2}}{dt}, \dfrac{dv_{3}}{dt}\right\rangle \\ \end{aligned}\] \[\begin{aligned} \dfrac{d}{dt}\left\langle \dfrac{1}{2}e^{t/2}, \dfrac{5}{2}t^{3/2},\cos t\right\rangle =\left\langle \dfrac{1}{4}e^{t/2}, \dfrac{15}{4}t^{1/2},-\sin t\right\rangle \end{aligned}\]
  
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3. Find the jerk vector
  
  \(\vec{j}(t)=\left\langle \dfrac{1}{8}e^{t/2}, \dfrac{15}{8}t^{-1/2},-\cos t\right\rangle\)
  
  \[\begin{aligned} \vec{j}(t)=\dfrac{d\vec{a}}{dt} =\left\langle \dfrac{da_{1}}{dt},\dfrac{da_{2}}{dt}, \dfrac{da_{3}}{dt}\right\rangle \\ \end{aligned}\] \[\begin{aligned} \dfrac{d}{dt}\left\langle \dfrac{1}{4}e^{t/2}, \dfrac{15}{4}t^{1/2},-\sin t\right\rangle =\left\langle \dfrac{1}{8}e^{t/2}, \dfrac{15}{8}t^{-1/2},-\cos t\right\rangle \end{aligned}\]
  
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Find the tangent vector to the curve \(\vec{r}(t)=(4t+5,2t^2-2,7)\) at \(t=3\)

Which vector (position, velocity, or acceleration) lies tangent to the curve?

\(\vec{v}(3)=\langle 4,12,0\rangle\)

The tangent vector to the curve \(\vec r(t)\) is \(\vec v(t)\). \[\begin{aligned} \vec{v}(t)=\dfrac{d\vec{r}}{dt} =\left\langle \dfrac{dr_{1}}{dt},\dfrac{dr_{2}}{dt}, \dfrac{dr_{3}}{dt}\right\rangle \\ \end{aligned}\] \[\begin{aligned} \dfrac{d}{dt}\langle 4t+5,2t^2-2,7\rangle =\langle 4,4t,0\rangle \end{aligned}\] \[\begin{aligned} \vec{v}(3)=\langle 4,12,0\rangle \end{aligned}\]

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Arc Length, Arc Length Parameter, and Speed

Find the length of the curve \(\vec{r}(t)=(3t+32,4t+1,-12t-1)\) from \(t=-2\) to \(t=4\).

\(\displaystyle L=\int_{a}^{b} |\vec{v}|\,dt\)

\(L=78\)

\[ \vec v(t) =\langle 3,4,-12\rangle \qquad |\vec v|=\sqrt{3^2+4^2+12^2}=13 \] \[\begin{aligned} L&=\int_{a}^{b} |\vec{v}|\,dt =\int_{-2}^4 13\,dt \\ &=13(4-(-2))=78 \\ \end{aligned}\]

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Find the length of the curve \(\vec{r}(t)=(\sin t,-\cos t,3t+2)\) from \( t=0\) to \(t=\dfrac{\pi}{2}\)

\(L=\dfrac{\pi\sqrt{10}}{2}\approx 4.97\)

\[ \vec v(t) =\langle \cos t,\sin t,3\rangle \qquad |\vec v|=\sqrt{\cos^2 t+\sin^2 t+3^2}=\sqrt{10} \] \[\begin{aligned} L&=\int_{a}^{b} |\vec{v}|\,dt =\int_0^{\pi/2} \sqrt{10}\,dt \\ &=\sqrt{10}\left(\dfrac{\pi}{2}-0\right) =\dfrac{\pi\sqrt{10}}{2}\approx 4.97 \end{aligned}\]

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Reparametrize \(\vec{r}(t)=(3t+7,4\sin t-5,4\cos t-1)\) with respect to arc length, \(s\), starting from \(t=0\).

To reparametrize \(\vec{r}(t)\) with respect to arc length \(s\) starting from \(t=0\), solve for \(s\) in terms of \(t\) using \(\displaystyle s(t)=\int_0^t |\vec{v}|\,dt\). Then solve for \(t(s)\), and plug \(t(s)\) into \(\vec{r}(t)\).

\(\vec r(s) =\left\langle 3\left(\dfrac{s}{5}\right)+7, 4\sin\left(\dfrac{s}{5}\right)-5, 4\cos\left(\dfrac{s}{5}\right)-1\right\rangle\)

\[\begin{aligned} \vec v(t)&=\langle 3,4\cos t,-4\sin t\rangle \\ |\vec v(t)|&=\sqrt{3^2+4^2\cos^2 t+4^2\sin^2 t}=\sqrt{9+16}=5 \end{aligned}\] \[ s(t) =\int_0^t |\vec{v}|\,dt =\int_0^{t} 5\,dt=5t \] So \(t=\dfrac{s}{5}\) and: \[ \vec r(s) =\left\langle 3\left(\dfrac{s}{5}\right)+7, 4\sin\left(\dfrac{s}{5}\right)-5, 4\cos\left(\dfrac{s}{5}\right)-1\right\rangle \]

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Find the speed of the position vector \(\vec{r} (t)=\langle 3t^2+2,2t^3-5t,9t-2\rangle\) at \(t=1\)

\(\text{speed}(t)=|\vec v(t)|\)

\(\text{speed}(1)=\sqrt{118}\approx 10.9\)

\[\begin{aligned} \vec{v}(t)&=\langle 6t,6t^2-5,9\rangle \\ \vec{v}(1)&=\langle 6,1,9\rangle \\ \end{aligned}\] \[\begin{aligned} \text{speed}(1)=|\vec{v}(1)|=\sqrt{6^2+1^2+9^2}=\sqrt{118}\approx 10.9 \end{aligned}\]

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For the parametric curve \(\vec r(t)=\left\langle2t^{1/2},\sqrt{6}t,2t^{3/2}\right\rangle\) find the arc length between \(t=1\) and \(t=4\).

In finding the speed, the quantity inside the square root is a perfect square.

\(L=16\)

Since the position is: \[r=\left\langle2t^{1/2},\sqrt{6}t,2t^{3/2}\right\rangle\] the velocity is: \[v=\left\langle t^{-1/2},\sqrt{6},3t^{1/2}\right\rangle\] Therefore, the speed is: \[\begin{aligned} |v|&=\sqrt{t^{-1}+6+9t} =\sqrt{(t^{-1/2}+3t^{1/2})^2} \\ &=t^{-1/2}+3t^{1/2} \end{aligned}\] To find the arc length, we take the integral of the speed between \(t=1\) and \(t=4\). \[ \begin{aligned} L&=\int_1^4\left|\vec v\right| dt =\int_1^4\left(t^{-1/2}+3t^{1/2}\right) dt \\ &=\left[2t^{1/2}+2t^{3/2}\right]_1^4 =(4+16)-(2+2)=16 \end{aligned} \]

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Tangent, Normal, and Binormal Vectors

For the parametric curve \(\vec r(t)=\left\langle2t^{1/2},\sqrt{6}t,2t^{3/2}\right\rangle\) find \(\hat T\), \(\hat N\) and \(\hat B\).

The tangent vector is \(\hat T=\dfrac{\vec v}{|\vec v|}\).

The binormal vector is \(\hat B=\dfrac{\vec v\times\vec a}{|\vec v\times\vec a|}\).

Finally, the normal vector is \(\hat N=\vec B\times\vec T\).

\(\vec T =\dfrac{v}{|v|} =\left\langle\dfrac{1}{3t+1},\dfrac{\sqrt{6}t^{1/2}}{3t+1}, \dfrac{3t}{3t+1}\right\rangle\)
\(\vec N =\left\langle-\,\dfrac{\sqrt{6}t^{1/2}}{3t+1},\dfrac{1-3t}{3t+1}, \dfrac{\sqrt{6}t^{1/2}}{3t+1}\right\rangle\)
\(\vec B =\left\langle\dfrac{3t}{3t+1},-\,\dfrac{\sqrt{6}t^{1/2}}{3t+1}, \dfrac{1}{3t+1}\right\rangle\)

From a previous problem, the velocity is: \[v=\left\langle t^{-1/2},\sqrt{6},3t^{1/2}\right\rangle\] and its length is \[\begin{aligned} |v|&=t^{-1/2}+3t^{1/2}=\dfrac{3t+1}{t^{1/2}} \end{aligned}\] Therefore, the unit tangent vector is: \[\begin{aligned} \vec T&=\dfrac{v}{|v|} =\dfrac{t^{1/2}}{3t+1}\left\langle t^{-1/2},\sqrt{6},3t^{1/2}\right\rangle \\ &=\left\langle\dfrac{1}{3t+1},\dfrac{\sqrt{6}t^{1/2}}{3t+1}, \dfrac{3t}{3t+1}\right\rangle \end{aligned}\] To find the binormal, we need \(\vec v\times\vec a\) and its length. The acceleration is: \[ a=\left\langle -\dfrac{1}{2}t^{-3/2},0,\dfrac{3}{2}t^{-1/2}\right\rangle \] So: \[\begin{aligned} \vec v\times\vec a &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ t^{-1/2} & \sqrt{6} & 3t^{1/2} \\ -\dfrac{1}{2}t^{-3/2} & 0 & \dfrac{3}{2}t^{-1/2} \end{vmatrix} \\ &=\hat\imath\left(\dfrac{3}{2}\sqrt{6}t^{-1/2}\right) -\hat\jmath\left(\dfrac{3}{2}t^{-1}--\dfrac{3}{2}t^{-1}\right) +\hat k\left(\dfrac{1}{2}\sqrt{6}t^{-3/2}\right) \\ &=\left\langle\dfrac{3}{2}\sqrt{6}t^{-1/2},-3t^{-1}, \dfrac{1}{2}\sqrt{6}t^{-3/2}\right\rangle \\ \end{aligned}\] \[\begin{aligned} |\vec v\times\vec a| &=\sqrt{\left(\dfrac{3}{2}\sqrt{6}t^{-1/2}\right)^2+(-3t^{-1})^2 +\left(\dfrac{1}{2}\sqrt{6}t^{-3/2}\right)^2} \\ &=\sqrt{\dfrac{27}{2}t^{-1}+9t^{-2}+\dfrac{3}{2}t^{-3}} \\ &=\dfrac{\sqrt{6}}{2t^{3/2}}\sqrt{9t^2+6t+1} \\ &=\dfrac{\sqrt{6}(3t+1)}{2t^{3/2}} \end{aligned}\] Therefore, the unit binormal vector is: \[ \begin{aligned} \vec B &=\dfrac{\vec v\times\vec a}{|\vec v\times\vec a|} \\ &=\dfrac{2t^{3/2}}{\sqrt{6}(3t+1)}\left\langle\dfrac{3}{2}\sqrt{6}t^{-1/2},-3t^{-1},\dfrac{1}{2}\sqrt{6}t^{-3/2}\right\rangle \\ &=\left\langle\dfrac{3t}{3t+1},-\,\dfrac{\sqrt{6}t^{1/2}}{3t+1},\dfrac{1}{3t+1}\right\rangle \end{aligned} \] Finally, the unit normal vector is: \[\begin{aligned} \vec N&=\hat B\times\hat T=\dfrac{1}{(3t+1)^2} \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 3t & -\sqrt{6}t^{1/2} & 1 \\ 1 & \sqrt{6}t^{1/2} & 3t \end{vmatrix} \\ &=\dfrac{1}{(3t+1)^2}\left[ \hat\imath\left(-3\sqrt{6}t^{3/2}-\sqrt{6}t^{1/2}\right) -\hat\jmath\left(9t^2-1\right) +\hat k\left(3\sqrt{6}t^{3/2}+\sqrt{6}t^{1/2}\right) \right] \\ &=\dfrac{1}{(3t+1)^2}\left\langle -\sqrt{6}t^{1/2}(3t+1),(1-3t)(1+3t), \sqrt{6}t^{1/2}(3t+1)\right\rangle \\ &=\dfrac{1}{(3t+1)}\left\langle -\sqrt{6}t^{1/2},1-3t,\sqrt{6}t^{1/2}\right\rangle \\ &=\left\langle-\,\dfrac{\sqrt{6}t^{1/2}}{3t+1},\dfrac{1-3t}{3t+1}, \dfrac{\sqrt{6}t^{1/2}}{3t+1}\right\rangle \end{aligned}\]

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To see \(\vec T\), \(\vec N\), and \(\vec B\) are orthogonal, we check their dot products are zero: \[ \begin{aligned} \vec T\cdot\vec N &=\left\langle\dfrac{1}{3t+1},\dfrac{\sqrt{6}t^{1/2}}{3t+1}, \dfrac{3t}{3t+1}\right\rangle\cdot\left\langle-\,\dfrac{\sqrt{6}t^{1/2}}{3t+1}, \dfrac{1-3t}{3t+1},\dfrac{\sqrt{6}t^{1/2}}{3t+1}\right\rangle \\ &=\dfrac{-\sqrt{6}t^{1/2}+\sqrt{6}t^{1/2}(1-3t)+3t\sqrt{6}t^{1/2}}{(3t+1)^2} \\ &=\dfrac{-\sqrt{6}t^{1/2}+\sqrt{6}t^{1/2}-3t\sqrt{6}t^{1/2}+3t\sqrt{6}t^{1/2}}{(3t+1)^2} =0 \\ \vec N\cdot\vec B &=\left\langle-\,\dfrac{\sqrt{6}t^{1/2}}{3t+1},\dfrac{1-3t}{3t+1}, \dfrac{\sqrt{6}t^{1/2}}{3t+1}\right\rangle\cdot\left\langle\dfrac{3t}{3t+1}, -\,\dfrac{\sqrt{6}t^{1/2}}{3t+1},\dfrac{1}{3t+1}\right\rangle \\ &=\dfrac{-\sqrt{6}t^{1/2}3t-(1-3t)\sqrt{6}t^{1/2}+\sqrt{6}t^{1/2}}{(3t+1)^2} \\ &=\dfrac{-3\sqrt{6}t^{3/2}-\sqrt{6}t^{1/2}+3\sqrt{6}t^{3/2}+\sqrt{6}t^{1/2}}{(3t+1)^2} =0 \\ \vec B\cdot\vec T &=\left\langle\dfrac{3t}{3t+1},-\,\dfrac{\sqrt{6}t^{1/2}}{3t+1}, \dfrac{1}{3t+1}\right\rangle\cdot\left\langle\dfrac{1}{3t+1}, \dfrac{\sqrt{6}t^{1/2}}{3t+1},\dfrac{3t}{3t+1}\right\rangle \\ &=\dfrac{3t-6t+3t}{(3t+1)^2}=0 \end{aligned} \]
Find the unit tangent vector of the position vector \(\vec{r}(t)=\left\langle t+2,\sqrt{\dfrac{5}{2}}(t^2-4),\dfrac{5}{3}t^3+3\right\rangle\) at:
1. a general value of \(t\)
  
  The unit tangent vector is \(\hat{T}=\dfrac{\vec{v}}{|\vec{v}|}\).
  
  \(\hat{T}=\left\langle \dfrac{1}{1+5t^2},\dfrac{\sqrt{10}t}{1+5t^2}, \dfrac{5t^2}{1+5t^2} \right\rangle\)
  
  Since the position is: \[r=\left\langle t+2,\sqrt{\dfrac{5}{2}}(t^2-4),\dfrac{5}{3}t^3+3\right\rangle\] the velocity is: \[v=\left\langle 1,\sqrt{10}t,5t^2 \right\rangle\] and the speed is: \[\begin{aligned} |v|&=\sqrt{1^2+\left(\sqrt{10}t\right)^2+(5t^2)^2} \\ &=\sqrt{1+10t^2+25t^4} \\ &=\sqrt{(1+5t^2)^2} \\ &=1+5t^2 \end{aligned}\] Therefore, the unit tangent vector is: \[\begin{aligned} \hat T&=\dfrac{v}{|v|} \\ &=\dfrac{1}{1+5t^2} \cdot\left\langle 1,\sqrt{10}t,5t^2 \right\rangle \\ &=\left\langle \dfrac{1}{1+5t^2},\dfrac{\sqrt{10}t}{1+5t^2},\dfrac{5t^2}{1+5t^2} \right\rangle \end{aligned}\]
  
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2. \(t=1\)
  
  \(\hat{T} =\left\langle\dfrac{1}{6},\dfrac{\sqrt{10}}{6},\dfrac{5}{6}\right\rangle\)
  
  From part (a), we found: \[ \hat T =\left\langle \dfrac{1}{1+5t^2},\dfrac{\sqrt{10}t}{1+5t^2}, \dfrac{5t^2}{1+5t^2} \right\rangle \] Plugging \(t=1\), we get: \[ \hat T=\left\langle \dfrac{1}{1+5\cdot1^2},\dfrac{\sqrt{10}\cdot1}{1+5\cdot1^2}, \dfrac{5\cdot1^2}{1+5\cdot1^2} \right\rangle =\left\langle\dfrac{1}{6},\dfrac{\sqrt{10}}{6},\dfrac{5}{6}\right\rangle \]
  
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Find the binormal vector of the curve \(\vec{r} (t)=\langle 3t+2,3t^2+5,2t^3+4\rangle\) at:
1. a general value of \(t\)
  
  The unit binormal vector is \(\hat{B} =\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|}\)
  
  \(\hat{B}=\dfrac{\langle 2t^2,-2t,1\rangle}{2t^2+1}\)
  
  We know that the binormal is: \[ \hat{B}=\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|} \] So we need to know \(\vec{v}\times\vec{a}\) and its length. Since the position is \[ \vec{r}=\langle 3t+2,3t^2+5,2t^3+4\rangle \] The velocity is: \[ \vec{v}=\langle 3,6t,6t^2\rangle \] And the acceleration is: \[ \vec{a}=\langle 0,6,12t\rangle \] So, \(\vec{v}\times\vec{a}\) is: \[\begin{aligned} \vec{v}\times\vec{a}&= \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 3 & 6t & 6t^2 \\ 0 & 6 & 12t \end{vmatrix} =\hat\imath(72t^2-36t^2)-\hat\jmath(36t-0)+\hat k(18-0) \\ &=\langle 36t^2,-36t,18\rangle \end{aligned}\] and its length is: \[\begin{aligned} |\vec{v}\times\vec{a}| &=\sqrt{(36t^2)^2+(36t)^2+18^2} =\sqrt{(36t^2)^2+2\cdot36t^2\cdot18+18^2} \\ &=36t^2+18 \\ \end{aligned}\] So, the unit binormal vector is: \[\begin{aligned} \hat{B}&=\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|} =\dfrac{\langle 36t^2,-36t,18\rangle}{36t^2+18} =\dfrac{\langle 2t^2,-2t,1\rangle}{2t^2+1} \\ \end{aligned}\]
  
  lh,ad
2. \(t=2\)
  
  \(\hat{B}=\left\langle \dfrac{8}{9},-\,\dfrac{4}{9},\dfrac{1}{9}\right\rangle\)
  
  Plugging \(t=2\) into \(\hat{B}=\dfrac{\langle 2t^2,-2t,1\rangle}{2t^2+1}\) we get \[\hat{B}=\dfrac{\langle 8,-4,1\rangle}{9} =\left\langle \dfrac{8}{9},-\,\dfrac{4}{9},\dfrac{1}{9}\right\rangle\]
  
  lh,ad
Find the normal vector of the curve \(\vec{r}(t)=\left\langle 2t+4,2t^2+11,\dfrac{4}{3}t^3\right\rangle\) at:
1. a general value of \(t\)
  
  The unit normal vector is \(\hat{N}=\hat{B}\times\hat{T}\).
  
  \(\hat{N} =\dfrac{\langle -2t,1-2t^2,2t\rangle}{2t^2+1}\)
  
  We know that the unit normal vector is: \[ \hat{N}=\hat{B}\times\hat{T} \] So we need \(\hat{T}\) and \(\hat{B}\), which means we need \(\vec{v}\), \(\vec{v}\times\vec{a}\) and their lengths. Since the position is: \[ \vec{r}=\left\langle 2t+4,2t^2+11,\dfrac{4}{3}t^3\right\rangle \] The velocity is: \[ \vec{v}=\langle 2,4t,4t^2\rangle \] And the acceleration is: \[ \vec{a}=\langle 0,4,8t\rangle \] So the speed is: \[\begin{aligned} |\vec{v}|&=\sqrt{4+16t^2+16t^4} \\ &=\sqrt{(2+4t^2)^2}=2+4t^2 \end{aligned}\] Therefore, the unit tangent vector is: \[ \hat{T}=\dfrac{\vec{v}}{|\vec{v}|}=\dfrac{\langle 2,4t,4t^2\rangle}{2+4t^2} =\dfrac{\langle 1,2t,2t^2\rangle}{1+2t^2} \] To find \(\hat{B}\), we need \(\vec{v}\times\vec{a}\) and its length. So \(\vec{v}\times\vec{a}\) is: \[\begin{aligned} \vec{v}\times\vec{a}&= \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 2 & 4t & 4t^2 \\ 0 & 4 & 8t \end{vmatrix} \\ &=\hat\imath(32t^2-16t^2)-\hat\jmath(16t)+\hat k(8) \\ &=\langle 16t^2,-16t,8\rangle \end{aligned}\] And its length is: \[\begin{aligned} |\vec{v}\times\vec{a}|&=\sqrt{(16t^2)^2+(16t)^2+8^2} \\ &=\sqrt{(16t^2+8)^2}=16t^2+8 \end{aligned}\] Therefore, the unit binormal vector is: \[\begin{aligned} \hat{B}&=\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|} \\ &=\dfrac{\langle 16t^2,-16t,8\rangle}{16t^2+8} =\dfrac{\langle 2t^2,-2t,1\rangle}{2t^2+1} \end{aligned}\] Finally, the unit normal is: \[\begin{aligned} \hat{N}&=\hat{B}\times\hat{T} =\dfrac{1}{(2t^2+1)^2} \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 2t^2 & -2t & 1 \\ 1 & 2t & 2t^2 \end{vmatrix} \\ &=\dfrac{1}{(2t^2+1)^2}\left[\rule{0pt}{10pt}\hat\imath(-4t^3-2t)-\hat\jmath(4t^4-1)\right. \\ &\qquad\qquad\qquad\left.\rule{0pt}{10pt}+\hat k(4t^3+2t)\right] \\ &=\dfrac{1}{(2t^2+1)^2}\left[\rule{0pt}{10pt}\hat\imath(-2t)(2t^2+1)-\hat\jmath(2t^2-1)(2t^2+1)\right. \\ &\qquad\qquad\qquad\left.\rule{0pt}{10pt}+\hat k(2t)(2t^2+1)\right] \\ &=\dfrac{\langle -2t,1-2t^2,2t\rangle}{2t^2+1} \\ \end{aligned}\]
  
  lh,ad
  
  We check that \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) are mutually perpendicular unit vectors. \[\begin{aligned} \hat{T}\cdot\hat{T} &=\dfrac{1}{(2t^2+1)^2}\langle 1,2t,2t^2\rangle\cdot\langle 1,2t,2t^2\rangle \\ &=\dfrac{1}{(2t^2+1)^2}\left(1+4t^2+4t^4\right) =1 \\ \hat{N}\cdot\hat{N} &=\dfrac{1}{(2t^2+1)^2}\langle -2t,1-2t^2,2t\rangle\cdot\langle -2t,1-2t^2,2t\rangle \\ &=\dfrac{1}{(2t^2+1)^2}\left(4t^2+(1-2t^2)^2+4t^2\right) \\ &=\dfrac{1}{(2t^2+1)^2}\left(8t^2+1-4t^2+4t^4\right) =1 \\ \hat{B}\cdot\hat{B} &=\dfrac{1}{(2t^2+1)^2}\langle 2t^2,-2t,1\rangle\cdot\langle 2t^2,-2t,1\rangle \\ &=\dfrac{1}{(2t^2+1)^2}\left(4t^4+4t^2+1\right) =1 \end{aligned}\] \[\begin{aligned} \hat{T}\cdot\hat{N} &=\dfrac{1}{(2t^2+1)^2}\langle 1,2t,2t^2\rangle\cdot\langle -2t,1-2t^2,2t\rangle \\ &=\dfrac{1}{(2t^2+1)^2}\left(-2t+2t-4t^3+4t^3\right) =0 \\ \hat{T}\cdot\hat{B} &=\dfrac{1}{(2t^2+1)^2}\langle 1,2t,2t^2\rangle\dot\langle 2t^2,-2t,1\rangle \\ &=\dfrac{1}{(2t^2+1)^2}\left(2t^2-4t^2+2t^2\right) =0 \\ \hat{N}\cdot\hat{B} &=\dfrac{1}{(2t^2+1)^2}\langle -2t,1-2t^2,2t\rangle\cdot\langle 2t^2,-2t,1\rangle \\ &=\dfrac{1}{(2t^2+1)^2}\left(-4t^3-2t(1-2t^2)+2t\right) =0 \end{aligned}\]
2. \(t=2\)
  
  \(\hat{N}=\left\langle-\,\dfrac{4}{9},\dfrac{7}{9},\dfrac{4}{9}\right\rangle\)
  
  Plugging \(t=2\) into \(\hat{N}=\dfrac{\langle -2t,1-2t^2,2t\rangle}{2t^2+1}\), we get: \[ \hat{N}=\dfrac{\left\langle -4,-7,4\right\rangle}{9} =\left\langle-\,\dfrac{4}{9},-\,\dfrac{7}{9},\dfrac{4}{9}\right\rangle \]
  
  lh,ad

PY Checked to here.Renumber below.

Curvature and Torsion

Find the curvature \(\kappa\) for the curve \(\vec{r} (t)=(3t^4,t+5,4t^3+3)\)
1. for a general value of \(t\)
  
  \(\kappa=\dfrac{|\vec{v}\times\vec{a}|} {|\vec{v}|^3}=\left|\dfrac{d\hat{T}}{ds}\right|\)
  
  \(\kappa=\dfrac{12t\sqrt{144t^6+9t^2+4}}{(144t^6+144t^4+1)^{3/2}}\)
  
  \[\begin{aligned} \kappa &=\dfrac{|\vec{v}\times\vec{a}|}{ |\vec{v}|^3} \\ \vec{v}&=\langle 12t^3,1,12t^2 \rangle \\ |\vec{v}|&=\sqrt{144t^6+1+144t^4} \\ \vec{a}&=\langle 36t^2,0,24t \rangle \\ \vec{v}\times\vec{a}&= \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 12t^3 & 1 & 12t^2 \\ 36t^2 & 0 & 24t \end{vmatrix} =\hat\imath(24t-0)-\hat\jmath(288t^4-432t^4)+ \hat k(0-36t^2) \\ &=\langle 24t,144t^4,-36t^2 \rangle \\ |\vec{v}\times\vec{a}| &=\sqrt{(24t)^2+(144t^4)^2+(36t^2)^2}\\ &=\sqrt{(12t)^2(2^2+(12t^3)^2+(3t)^2)}=12t\sqrt{144t^6+9t^2+4} \\ \kappa &=\dfrac{12t\sqrt{144t^6+9t^2+4}} {\left(\sqrt{144t^6+144t^4+1}\right)^3} =\dfrac{12t\sqrt{144t^6+9t^2+4}}{(144t^6+144t^4+1)^{3/2}} \end{aligned}\]
  
  lh
2. at \(t=1\)
  
  \(\kappa=\dfrac{12\sqrt{157}}{17^3}\approx 0.0306\)
  
  We plug \(t=1\) into \[\kappa=\dfrac{12t\sqrt{144t^6+9t^2+4}}{(144t^6+144t^4+1)^{3/2}}\] And get \[\kappa=\dfrac{12\sqrt{144+9+4}}{(144+144+1)^{3/2}} =\dfrac{12\sqrt{157}}{17^3}\approx 0.0306\]
  
  lh
Find the torsion \(\tau\) for the curve in the previous problem
1. for a general value of \(t\)
  
  \(\tau =\dfrac{\vec{v}\times\vec{a}\cdot\vec{j}}{|\vec{v}\times\vec{a}|^2}\)
  
  \(\tau=\dfrac{6}{144t^6+9t^2+4}\)
  
  \[\begin{aligned} \vec{v}&=\langle 12t^3,1,12t^2\rangle \\ \vec{a}&=\langle 36t^2,0,24t\rangle \\ \vec{v}\times\vec{a} &=\langle 24t,144t^4,-36t^2\rangle \\ |\vec{v}\times\vec{a}| &=12t\sqrt{144t^6+9t^2+4} \\ \vec{j}&=\langle 72t,0,24\rangle \\ \tau &=\dfrac{\vec{v}\times\vec{a}\cdot\vec{j}}{|\vec{v}\times\vec{a}|^2} =\dfrac{\langle 24t,144t^4,-36t^2\rangle \cdot\langle 72t,0,24 \rangle}{(12t\sqrt{144t^6+9t^2+4})^2} \\ &=\dfrac{1728t^2+0-864t^2}{144t^2(144t^6+9t^2+4)} =\dfrac{864t^2}{144t^2(144t^6+9t^2+4)} =\dfrac{6}{144t^6+9t^2+4} \end{aligned}\]
  
  lh
2. at \(t=1\)
  
  \(\tau=\dfrac{6}{157}\)
  
  We plug in \(t=1\) into \[\tau=\dfrac{6}{144t^6+9t^2+4}\] And get \[\tau=\dfrac{6}{144+9+4} =\dfrac{6}{157}\]
  
  lh

Tangential and Normal Acceleration

Find the tangential and normal acceleration to the curve \(\vec{r}(t)=( \tfrac{1}{3}t^3+5,2t+1,t^2-3)\)
1. for a general value of \(t\)
  
  \(a_{T}=\dfrac{d}{dt}|\vec{v}|\)
  \(a_{N}=a\cdot\hat{N}\)
  
  \(a_{T}=2t\)
  \(a_{N}=2\)
  
  To find the tangential acceleration: \[\begin{aligned} \vec{v}&=\langle t^2,2,2t \rangle \\ |\vec{v}|&=\sqrt{t^4+4t^2+4}=\sqrt{(t^2+2)^2} =(t^2+2) \\ a_{T}&=\dfrac{d|\vec{v}|}{dt}=2t \end{aligned}\] To find the normal acceleration: \[\begin{aligned} \hat T&=\dfrac{\vec v}{|\vec v|}=\dfrac{\langle t^2,2,2t \rangle} {(t^2+2)}=\left\langle \dfrac{t^2}{t^2+2},\dfrac{2}{t^2+2}, \dfrac{2t}{t^2+2} \right\rangle \\ T'&=\left\langle \dfrac{(t^2+2)2t-(t^2)(2t)}{(t^2+2)^2}, \dfrac{4t}{(t^2+2)^2},\dfrac{(t^2+2)(2)-(2t)(2t)}{(t^2+2)^2} \right\rangle \\ &=\left\langle \dfrac{4t}{(t^2+2)^2},\dfrac{-4t}{(t^2+2)^2},\dfrac{-2(t^2-2)}{ (t^2+2)^2} \right\rangle \\ |T'|&=\sqrt{\dfrac{16t^2+16t^2+4(t^2-2)^2}{(t^2+2)^4}} =\dfrac{\sqrt{4t^4+16t^2+16}}{(t^2+2)^2} =\dfrac{2\sqrt{(t^2+2)^2}}{(t^2+2)^2} =\dfrac{2(t^2+2)}{(t^2+2)^2}=\dfrac{2}{t^2+2} \\ \hat{N}&=\dfrac{T'(t)}{|T'(t)|} =\dfrac{t^2+2}{2}\left\langle \dfrac{4t}{(t^2+2)^2}, \dfrac{-4t}{(t^2+2)^2},\dfrac{-2(t^2-2)}{(t^2+2)^2} \right\rangle \\ \hat{N} &=\left\langle \dfrac{2t}{(t^2+2)},\dfrac{-2t}{(t^2+2)}, \dfrac{-(t^2-2)}{(t^2+2)}\right\rangle \\ \vec a &=\langle 2t,0,2\rangle \\ a_{N}&=\vec a\cdot\hat{N}=\langle 2t,0,2 \rangle \cdot\left\langle \dfrac{2t}{(t^2+2)},\dfrac{-2t}{(t^2+2)}, \dfrac{-(t^2-2)}{(t^2+2)}\right\rangle \\ a_{N}&=\dfrac{4t^2}{(t^2+2)}+0+\dfrac{-2(t^2-2)}{(t^2+2)}=\dfrac{ 4t^2-2t^2+4}{(t^2+2)} \\ a_{N}&=\dfrac{2(t^2+2)}{(t^2+2)}=2 \end{aligned}\]
  
  lh
2. at \(t=1\)
  
  \[\begin{aligned} a_{T}(t=1)=2 \\ a_{N}(t=1)=2 \end{aligned}\]
3. Which component (tangential or normal) of the acceleration is independent of the position on this curve?
  
  Think about: which component (tangential or normal) is a constant for any given value of \(t\)?
  
  Normal
Find the tangential and normal acceleration to the curve \(\vec{r} (t)=(2\sin t,3t^2+4,2\cos t)\)
1. for a general value of \(t\)
  
  \[\begin{aligned} a_{T}&=\dfrac{d|\vec{v}|}{dt}=\dfrac{1}{(9t^2+1)^{1/2}} \\ a_{N}&=\dfrac{\sqrt{288t^2\sin^2t-288t^2\sin t\cos t+144t^2\cos ^2t+16+288t\sin t\cos t+144\cos^2t}}{\sqrt{4+36t^2}} \end{aligned}\]
  
  To find the tangential acceleration: \[\begin{aligned} \vec{v}&=\langle 2\cos t,6t,-2\sin t\rangle \\ |\vec{v}|&=\sqrt{4\cos^2t+36t^2+4\sin^2t}=\sqrt{ 4+36t^2} \\ &=2\sqrt{9t^2+1}=2(9t^2+1)^{1/2} \\ a_{T}&=\dfrac{d|\vec{v}|}{dt}=\dfrac{18t}{(9t^2+1)^{1/2}} \end{aligned}\] To find the normal acceleration: \[\begin{aligned} a_{N}&=\kappa|\vec{v}|^2 =\dfrac{|\vec{v}\times\vec{a}|}{|\vec{v}|} \\ \vec a&=\langle -2\sin t,6,-2\cos t \rangle \\ |\vec{v}\times\vec{a}|&= \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 2\cos t & 6t & -2\sin t \\ -2\sin t & 6 & -2\cos t \end{vmatrix} \\ &=|\hat\imath (-12t\cos t+12\sin t) -\hat\jmath (-4\cos^2t-4\sin^2t) +\hat k(12\cos t+12t\sin t)| \\ &=|\langle 12(\sin t-t\cos t),4,12(t\sin t+\cos t)\rangle| \\ |\vec{v}\times\vec{a}| &=\sqrt{144(\sin^2t-2t\sin t\cos t+\cos^2t)+16 +144(t^2\sin^2t+2t\sin t\cos t+\cos^2t)} \\ &=\sqrt{(144\sin^2t+144\cos^2t)+(144t^2\sin^2t+144t^2\cos^2t)+16} \\ &=\sqrt{144t^2+160} \\ a_{N}&=\dfrac{\sqrt{144t^2+160}}{\sqrt{4+36t^2}} \\ &=\dfrac{\sqrt{36t^2+40}}{\sqrt{1+9t^2}} \\ \end{aligned}\]
  
  lh
2. at \(t=0\)
  
  \(a_{T}=\dfrac{d|\vec{v}|}{dt}=0\)
  \(a_{N}=\dfrac{\sqrt{40}}{\sqrt{1}}=\sqrt{40}\)

Summary & Examples of Curve Computations

Consider the position vector \(\vec{r}(t)=\langle 6\cos t,6\sin t,8t\rangle\)
1. Is this a helical or circular curve? How can you tell?
  
  This is a helical curve because the \(z\) component is a linear polynomial. If it was a constant, then it would be a circular curve. The \(x\) and \(y\) components make a circle in the \(xy\)-plane, therefore the helix spirals in the \(z\) direction as shown to the right:
2. Compute the velocity of the curve at \(t=\pi/2\)
  
  \(\vec{v}(t)=\langle -6\sin t,6\cos t,8\rangle\)
  \(\vec{v}(\pi/2)=\langle -6,0,8\rangle\)
Consider the position vector \(\vec{r}(t)=\langle 12\cos t,9,12\sin t\rangle\).
1. Is this a helical or circular curve? How can you tell?
  
  This is a circular curve because the \(x\) and \(z\) components are the parameters for a circle while the y component is constant. The \(y\) component is \(9\), meaning that there is a circle in the \(xz\)-plane at \(y=7\) as shown to the right:
2. Compute the acceleration of the curve at \(t=2\pi\)
  
  \(\vec{v}(t)=\langle -12\sin t,0,12\cos t\rangle\)
  \(\vec{v}(2\pi)=\langle 0,0,12\rangle\)
Consider the position vector \(\vec{r}(t)=(5t,12\sin t,12\cos t)\)
1. Is this a helical or circular curve? How can you tell?
  
  This is a helical curve because the \(x\) component is a linear polynomial. If it was a constant, then it would be a circular curve. The \(y\) and \(z\) components are the parameters for a circle, meaning that in the \(yz\) plane a circle is formed when looking from the \(x\) axis. The helix spirals in the \(x\) direction, as shown to the right:
2. What is the length of the curve from \(t=0\) to \(t=\pi\)?
  
  \(\displaystyle L=\int_a^b |\vec{v}|\,dt\)
  
  \(L=13\pi\)
  
  The arclength is calculate from integrating the speed of the curve over the distance along the parameter \(t\): \[L=\int_{a}^{b} |\vec{v}|\,dt\] First we must find the speed (the magnitude of the velocity vector): \[\begin{aligned} \vec{v}&=\vec{r\,}'(t) =\langle 5,12\cos t,-12\sin t\rangle \\ |\vec{v}| &=\sqrt{5^2+12^2\cos^2t+(-12)^2\sin^2t} \\ &=\sqrt{25+144(\cos^2t+\sin^2t)}=\sqrt{25+144}=13 \\ \end{aligned}\] Finally we can input this into the arclength formula and easily solve for the length of the curve: \[L=\int_0^{\pi} 13\,dt=13\pi\]
  
  lh

Frenet Frames

Verify the Frenet equations for the derivatives of \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) for the circle: \[ \vec{r}(\theta)=(a\cos\theta,a\sin\theta,b) \] where \(a\) and \(b\) are constants.

We must verify: \[\begin{aligned} \dfrac{d\hat{T}}{ds}&=\qquad \quad \kappa \hat{N} \\ \dfrac{d\hat{N}}{ds}&=-\kappa \hat{T} \quad + \quad \tau \hat{B} \\ \dfrac{d\hat{B}}{ds}&=\qquad -\tau \hat{N} \end{aligned}\] Compute \(\hat{T}\), \(\hat{N}\), \(\hat{B}\), \(\kappa\) and \(\tau\). To compute \(\dfrac{d\hat{T}}{ds}\), \(\dfrac{d\hat{T}}{ds}\) and \(\dfrac{d\hat{T}}{ds}\). Remember: \[ \dfrac{d}{ds}=\dfrac{d\theta}{ds}\dfrac{d}{d\theta} =\dfrac{1}{|\vec v|}\dfrac{d}{d\theta} \]

First, we compute \(\hat T\). The velocity and its length are: \[\begin{aligned} \vec{v} &=\langle -a\sin\theta,a\cos\theta,0\rangle \\ |\vec{v}| &=\sqrt{a^2\sin^2\theta+a^2\cos^2\theta}=a \end{aligned}\] So the unit tangent vector is: \[ \hat{T}=\dfrac{\vec{v}}{|\vec{v}|} =\langle -\sin\theta,\cos\theta,0\rangle \] Second, we compute \(\hat B\). The acceleration, the cross product, \(\vec v\times\vec a\), and its length are: \[\begin{aligned} \vec{a} &=\langle -a\cos\theta,-a\sin\theta,0\rangle \\ \vec{v}\times\vec{a} &=\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -a\sin\theta & a\cos\theta & 0 \\ -a\cos\theta & -a\sin\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(0) -\hat{\jmath}(0) +\hat{k}(a^2\sin^2\theta+a^2\cos^2\theta) \\ &=\langle 0,0,a^2\rangle \\ |\vec{v}\times\vec{a}| &=a^2 \\ \end{aligned}\] So the unit binormal vector is: \[ \hat{B} =\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|} =\langle 0,0,1\rangle \] Third, we compute \(\hat N\). \[\begin{aligned} \hat N&=\hat B\times\hat T =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 0 & 0 & 1 \\ -\sin\theta & \cos\theta & 0 \\ \end{vmatrix} \\ &=\hat{\imath}(-\cos\theta) -\hat{\jmath}(\sin\theta) +\hat{k}(0) \\ &=\langle -\cos\theta,-\sin\theta,0\rangle \\ \end{aligned}\] To determine the curvature and torsion, we compute the jerk and triple product: \[\begin{aligned} \vec j =\langle a\sin\theta,-a\cos\theta,0\rangle \\ \vec v\times\vec a\cdot\vec j =0 \end{aligned}\] So the curvature and torsion are: \[\begin{aligned} \kappa&=\dfrac{|\vec{v}\times\vec{a}|}{|\vec{v}|^3} =\dfrac{a^2}{a^3} =\dfrac{1}{a} \\ \tau&=\dfrac{\vec{v}\times\vec{a}\cdot\vec{j}}{|\vec{v}\times\vec{a}|^2} =0 \end{aligned}\] Finally, we must verify the Frenet equations: \[\begin{aligned} \dfrac{d\hat{T}}{ds}&=\qquad \quad \kappa \hat{N} \\ \dfrac{d\hat{N}}{ds}&=-\kappa \hat{T} \quad + \quad \tau \hat{B} \\ \dfrac{d\hat{B}}{ds}&=\qquad -\tau \hat{N} \end{aligned}\] We compute each side. First for \(\hat{T}\): \[\begin{aligned} \dfrac{d\hat{T}}{ds}&=\dfrac{1}{|\vec{v}|}\dfrac{d\hat{T}}{d\theta} =\dfrac{1}{a}\langle -\cos\theta,-\sin\theta,0\rangle \\ \kappa\hat N &=\dfrac{1}{a}\langle -\cos\theta,-\sin\theta,0\rangle \end{aligned}\] which agree. Second for \(\hat{N}\): \[\begin{aligned} \dfrac{d\hat{N}}{ds}&=\dfrac{1}{|\vec{v}|}\dfrac{d\hat{N}}{d\theta} =\dfrac{1}{a}\langle \sin\theta,-\cos\theta,0\rangle \\ -\kappa\hat T+\tau\hat B &=\dfrac{-1}{a}\langle -\sin\theta,\cos\theta,0\rangle +0\langle 0,0,1\rangle \\ &=\dfrac{1}{a}\langle \sin\theta,-\cos\theta,0\rangle \end{aligned}\] which also agree. And finally for \(\hat{B}\): \[\begin{aligned} \dfrac{d\hat{B}}{ds}&=\dfrac{1}{|\vec{v}|}\dfrac{d\hat{B}}{d\theta} =\dfrac{1}{a}\langle 0,0,0\rangle=\langle 0,0,0\rangle \\ -\tau\hat{N} &=0\langle -\cos\theta,-\sin\theta,0\rangle=\langle 0,0,0\rangle \end{aligned}\] which also agree.

lh
Verify the Frenet equations for the derivatives of \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) for a general helix: \[ \vec{r}(\theta)=(a\cos\theta,a\sin\theta,b\,\theta) \] where \(a\) and \(b\) are constants.

First, we compute \(\hat T\). The velocity and its length are: \[\begin{aligned} \vec{v} &=\langle -a\sin\theta,a\cos\theta,b\rangle \\ |\vec{v}| &=\sqrt{a^2\sin^2\theta+a^2\cos^2\theta+b^2} =\sqrt{a^2+b^2} \end{aligned}\] So the unit tangent vector is: \[ \hat{T}=\dfrac{\vec{v}}{|\vec{v}|} =\dfrac{1}{\sqrt{a^2+b^2}} \langle -a\sin\theta,a\cos\theta,b\rangle \] Second, we compute \(\hat B\). The acceleration, the cross product, \(\vec v\times\vec a\), and its length are: \[\begin{aligned} \vec{a} &=\langle -a\cos\theta,-a\sin\theta,0\rangle \\ \vec{v}\times\vec{a} &=\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -a\sin\theta & a\cos\theta & b \\ -a\cos\theta & -a\sin\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(ab\sin\theta) -\hat{\jmath}(ab\cos\theta) +\hat{k}(a^2\sin^2\theta+a^2\cos^2\theta) \\ &=\langle ab\sin\theta,-ab\cos\theta,a^2\rangle \\ |\vec{v}\times\vec{a}| &=\sqrt{a^2b^2\sin^2\theta+a^2b^2\cos^2\theta+a^4} \\ &=a\sqrt{a^2+b^2} \\ \end{aligned}\] So the unit binormal vector is: \[\begin{aligned} \hat{B} &=\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|} =\dfrac{1}{a\sqrt{a^2+b^2}}\langle ab\sin\theta,-ab\cos\theta,a^2\rangle \\ &=\dfrac{1}{\sqrt{a^2+b^2}}\langle b\sin\theta,-b\cos\theta,a\rangle \\ \end{aligned}\] Third, we compute \(\hat N\). \[\begin{aligned} \hat N&=\hat B\times\hat T =\dfrac{1}{\sqrt{a^2+b^2}^{\,2}}\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ b\sin\theta & -b\cos\theta & a \\ -a\sin\theta & a\cos\theta & b \\ \end{vmatrix} \\ &=\dfrac{1}{a^2+b^2}\left(\begin{matrix} \hat{\imath}(-b^2\cos\theta-a^2\cos\theta) \\ -\hat{\jmath}(b^2\sin\theta+a^2\sin\theta) \\ +\hat{k}(ab\sin\theta\cos\theta-ab\sin\theta\cos\theta) \end{matrix}\right) \\ &=\langle -\cos\theta,-\sin\theta,0\rangle \\ \end{aligned}\] To determine the curvature and torsion, we compute the jerk and triple product: \[\begin{aligned} \vec j &=\langle a\sin\theta,-a\cos\theta,0\rangle \\ \vec v\times\vec a\cdot\vec j &=a^2b\sin^2\theta+a^2b\cos^2\theta=a^2b \end{aligned}\] So the curvature and torsion are: \[\begin{aligned} \kappa&=\dfrac{|\vec{v}\times\vec{a}|}{|\vec{v}|^3} =\dfrac{a\sqrt{a^2+b^2}}{(a^2+b^2)^{3/2}} =\dfrac{a}{a^2+b^2} \\ \tau&=\dfrac{\vec{v}\times\vec{a}\cdot\vec{j}}{|\vec{v}\times\vec{a}|^2} =\dfrac{a^2b}{\left(a\sqrt{a^2+b^2}\right)^2} =\dfrac{b}{a^2+b^2} \end{aligned}\] Finally, we must verify the Frenet equations: \[\begin{aligned} \dfrac{d\hat{T}}{ds}&=\qquad \quad \kappa \hat{N} \\ \dfrac{d\hat{N}}{ds}&=-\kappa \hat{T} \quad + \quad \tau \hat{B} \\ \dfrac{d\hat{B}}{ds}&=\qquad -\tau \hat{N} \end{aligned}\] We compute each side. First for \(\hat{T}\): \[\begin{aligned} \dfrac{d\hat{T}}{ds}&=\dfrac{1}{|\vec{v}|}\dfrac{d\hat{T}}{d\theta} =\dfrac{1}{a^2+b^2}\langle -a\cos\theta,-a\sin\theta,0\rangle \\ \kappa\hat N &=\dfrac{a}{a^2+b^2}\langle -\cos\theta,-\sin\theta,0\rangle \end{aligned}\] which agree. Second for \(\hat{N}\): \[\begin{aligned} \dfrac{d\hat{N}}{ds}&=\dfrac{1}{|\vec{v}|}\dfrac{d\hat{N}}{d\theta} =\dfrac{1}{\sqrt{a^2+b^2}}\langle \sin\theta,-\cos\theta,0\rangle \\ -\kappa\hat T+\tau\hat B &=\dfrac{-a}{(a^2+b^2)^{3/2}}\langle -a\sin\theta,a\cos\theta,b\rangle \\ &\quad+\dfrac{b}{(a^2+b^2)^{3/2}}\langle b\sin\theta,-b\cos\theta,a\rangle \\ &=\dfrac{1}{\sqrt{a^2+b^2}}\langle \sin\theta,-\cos\theta,0\rangle \end{aligned}\] which also agree. And finally for \(\hat{B}\): \[\begin{aligned} \dfrac{d\hat{B}}{ds}&=\dfrac{1}{|\vec{v}|}\dfrac{d\hat{B}}{d\theta} =\dfrac{1}{a^2+b^2}\langle b\cos\theta,b\sin\theta,0\rangle \\ -\tau\hat{N} &=\dfrac{-b}{a^2+b^2}\langle -\cos\theta,-\sin\theta,0\rangle \end{aligned}\] which also agree.

lh

PY Checked 25 & 26

Verify the Frenet equations for the derivatives of \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) for a twisted cubic.

Circular motion is represented by the position vector: \[\vec{r}(\theta)=\langle at,bt^2,ct^3\rangle\] Where \(a\) and \(b\) are abitiary constants. First, find \(\hat{T}=\dfrac{\vec{v}}{|\vec{v}|}\) \[\begin{aligned} \vec{v}=\langle a,2bt,3ct^2\rangle \\ |\vec{v}|=\sqrt{a^2+4b^2t^2+9c^2t^4} \\ \hat{T}=\dfrac{\vec{v}}{|\vec{v}|}=\dfrac{1}{\sqrt{ a^2+4b^2t^2+9c^2t^4}}\langle a,2bt,3ct^2\rangle \\ \end{aligned}\] Next, find \(\hat{B}=\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|}\) \[\begin{aligned} \vec{v}\times\vec{a}&= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ a & 2bt & 3ct^2 \\ 0 & 2b & 6ct \end{vmatrix} =\hat{\imath}(12bct^2-6bct^2) -\hat{\jmath}(6cat-0)+\hat{k}(2ab-0) \\ &=\langle 6bct^2,-6cat,2ab\rangle \\ |\vec{v}\times\vec{a}| &=\sqrt{(6bct^2)^2-(6cat)^2+(2ab)^2}=a^2 \hat{B}=\dfrac{1}{a^2}\ft\langle 0,0,a^2\rangle =\langle 0,0,1\rangle \\ \end{aligned}\] Then, find \(\hat{N} =\dfrac{\hat{T}'(\theta)}{|\hat{T}'(\theta)|}\) \[\begin{aligned} \hat{T}'=\langle -\cos(\theta),-\sin(\theta),0\rangle \\ |\hat{T}'(\theta)| =\sqrt{(\cos^2(\theta)+\sin^2(\theta))}=1 \\ \hat{N}=\dfrac{\hat{T}'(\theta)}{|\hat{T}'^{\prime}'(\theta)|} =\langle -\cos(\theta),-\sin(\theta),0\rangle \\ \end{aligned}\] Based on previously found values, we can determine that \(\kappa=\dfrac{|\vec{v}\times\vec{a}|}{|\vec{v}| ^3}\) \(=\dfrac{a^2}{a^3}=\dfrac{1}{a}\).
Additionally, we can find that: \[ \tau=\dfrac{\vec{v}\times\vec{a}\cdot\vec{j}} {|\vec{v}\times\vec{a}|^2} =\dfrac{\langle 0,0,a^2\rangle \cdot\langle a\sin\theta,-a\cos\theta,0\rangle}{a^4}=0 \] Finally, we must verify that: \[\begin{aligned} \dfrac{d\hat{T}}{ds}&=\qquad \quad \kappa \hat{N} \\ \dfrac{d\hat{N}}{ds}&=-\kappa \hat{T} \quad + \quad \tau \hat{B} \\ \dfrac{d\hat{B}}{ds}&=\qquad -\tau \hat{N} \end{aligned}\] We compute each side. First for \(\hat{T}\): \[\begin{aligned} \dfrac{d\hat{T}}{ds}&=\dfrac{1}{|\vec{v}|}\dfrac{d \hat{T}}{dt} =\dfrac{1}{a}\dfrac{d}{dt}\langle -\sin\theta,\cos\theta,0\rangle \\ &=\dfrac{1}{a}\langle -\cos\theta,-\sin\theta,0\rangle \qquad \text{while} \\ \kappa \hat{N}&=\dfrac{1}{a}\langle -\cos\theta,-\sin\theta,0\rangle \qquad \text{which agrees.} \end{aligned}\] Second for \(\hat{N}\): And finally for \(\hat{B}\):

lh

Review Exercises

Find the derivative of the vector function \(\vec{f}(t)=\langle \ln t,t^2+1,\tan t\rangle\) at the point when \(t=\pi\).

\(\vec{f\,'}(\pi)=\langle \dfrac{1}{\pi},2\pi,1\rangle\)
Consider the position vector \(\vec{r}(t)=(6\cos t,6\sin t,8t)\). Find:
1. the velocity vector
  
  \(\vec{v}(t)=\left\langle -6\sin t,6\cos t,8 \right\rangle\)
2. the acceleration vector
  
  \(\vec{a}(t)=\left\langle -6\cos t,-6\sin t,0 \right\rangle\)
3. the jerk vector
  
  \(\vec{j}(t)=\left\langle 6\sin t,-6\cos t,0 \right\rangle\)
4. the arc length from an arbitrary \(a\) to an arbitrary \(b\)
  
  \(L=10(b-a)\)
5. the arclength parameter \(s=s(t)\)
  
  \(s=10\,t\)
6. the unit tangent vector
  
  \(\hat{T}=\left\langle -\,\dfrac{3}{5}\sin t,\dfrac{3}{5}\cos t,\dfrac{4}{5} \right\rangle\)
7. the unit binormal vector
  
  \(\hat{B}=\left\langle \dfrac{4}{5}\sin t,-\,\dfrac{4}{5}\cos t,\dfrac{3}{5} \right\rangle\)
8. the unit normal vector
  
  \(\hat{N}=\left\langle -\cos t,-\sin t,0 \right\rangle\)
9. the curvature of the curve
  
  \(\kappa=\dfrac{3}{50}\)
10. the torsion of the curve
  
  \(\tau=\dfrac{2}{25}\)
11. the component of the acceleration that is tangent to the curve
  
  \(a_{T}=0\)
  
  \[\begin{aligned} a_{T}&=\vec{a}\cdot\hat{T} \\ &=\langle -6\cos t,-6\sin t,0 \rangle \cdot\langle -\,\dfrac{3}{5}\sin t,\dfrac{3}{5}\cos t,\dfrac{4}{5}\rangle \\ &=\dfrac{18}{5}\sin t\cos t-\,\dfrac{18}{5}\sin t\cos t =0 \\ \end{aligned}\] or \[ a_{T}=\dfrac{d}{dt}|\vec{v}|=\dfrac{d}{dt}10=0 \]
  
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12. the component of the acceleration that is normal to the curve
  
  \(a_{N}=6\)
  
  \[\begin{aligned} a_{N}&=\vec{a}\cdot\hat{N} \\ &=\langle -6\cos t,-6\sin t,0\rangle\cdot\langle -\cos t,-\sin t,0\rangle \\ &=6\cos^2t+6\sin^2t =6 \end{aligned}\] or \[ a_{N}=\kappa|\vec{v}|^2=\dfrac{3}{50}\cdot100=6 \]
  
  lh

3. Properties of Curves

Exercises

Review Exercises

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