Motivating Questions
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How can we measure the direction of travel on a parametrized curve?
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How can we measure the direction of turning on a parameterized curve?
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How can we measure the axis of rotation along a parameterized curve?
Section 10.4 The TNB Frame
Subsection 10.4.1 Introduction
In Section 10.1, we saw how a vector-valued function of one variable will graphically correspond to a curve in space. While the application of derivatives, integrals, and limits apply to these functions componentwise, we will need to apply our calculus and vector tools together to measure important properties of these curves. Conceptually, you should think about the parameterization of the curve in space as a description on how to travel through those points in space. Remember that parameterizations are not unique: there is more than one way to walk the same path!
For example, consider the curve in space that we are exploring to be a race track as shown in Figure 10.4.1. Not every driver will travel along this racetrack in the same way, but every driver has to go through these same points in space, since we assume that everyone will stay on the track and not take shortcuts. In the vocabulary of this chapter, any parameterization of this curve has to contain the same points, but can possibly go through these \((x,y,z)\)-coordinates at different parameter values. More generally, some properties we measure will depend on the particular way that a driver goes along the course and some properties will be the same for all drivers when measured at the same location on the track.
Preview Activity 10.4.1.
As CEO and Head of Engineering at Steer Clear, you decide that you are almost ready to start testing your self-driving car out on the road. Your navigation and telemetry software will use the location tracking system (LTS) to determine all of the important information about how the car is moving and how adjustments need to be made. Before you start programming your software, you decide to drive on a quiet, country road and collect data from the LTS to use as test data for your programming. In other words, you will drive on a section of road you already have mapped out in order to check that your software is calculating the correct information. The map in Figure 10.4.2 below shows the path you plan to take on the country road (with the direction given by the arrows on the plot). Notice that at points \(B\) and \(F\text{,}\) the road crosses itself to go in a different direction.
The path of a car ride with seven points labeled
(a)
At each of the labeled points on the curve, draw a vector in the direction of travel. Write a sentence about how you are determining this vector at the various points.
(b)
At each point, decide whether the car is turning left or right. Write a sentence about how you are determining this turning based on the plot given.
Preview Activity 10.4.1 highlights the intuitive nature of measuring the direction of travel and the direction of turning given a plot of a curve in space. We will now look at how to use location information in the form of a parameterization of a curve to precisely measure the direction of travel and the direction of turning at a particular point on the curve.
Subsection 10.4.2 Direction of travel
Based on our work in Subsection 10.2.2, if \(\vr(t)\) is a parameterization of a curve in space, then the velocity vector, \(\vv(t)=\vr\, '(t)\text{,}\) is in the direction of travel. Since the speed, \(\vecmag{\vv(t)}\text{,}\) is not always 1 (the velocity is not always a unit vector), we define the direction of travel using a unit vector as follows.
Definition 10.4.3.
The unit tangent vector to a curve, denoted \(\vT(t)\text{,}\) is defined as
\begin{equation*}
\vT(t)=\frac{\vv(t)}{\vecmag{\vv(t)}}
\end{equation*}
and measures the direction of travel along a parameterized curve.
By definition, the unit tangent vector exists as long as the velocity exists and is not the zero vector. If the velocity vector is \(\vec{0}\text{,}\) then the object is not moving, thus the direction of travel would not make sense at that instant. If there is a jump or discontinuity in the derivative of \(\vr\) so that \(\vv\) does not exist, the direction of travel at that instant does not make sense because there is not a consistent way to define how the object is moving.
The definition of \(\vT\) allows us to separate the velocity of a moving object into its magnitude (speed) and direction (unit tangent). Using a unit vector allows us to express the direction of travel as a vector that is separate of the length of any related measurements. Remember that \(\vT\) changes along the curve and is calculated as a function of the parameter.
Example 10.4.4.
For our first example, we will look computing the unit tangent vector of a helix. In particular, we will look at the helix traced out by \(\vr(t)=\langle \cos(t),\sin(t),\frac{t}{\pi}\rangle \text{,}\) which is shown in Figure 10.4.5.
Based on the parameterization \(r(t)\) for this helix, the velocity vector is \(\vv(t)=\langle -\sin(t),\cos(t),\frac{1}{\pi} \rangle \text{.}\) Thus, the speed is
\begin{equation*}
\text{speed}(t) = \vecmag{\vv(t)} = \sqrt{\left(-\sin(t)\right)^2+\left(\cos(t)\right)^2+\left(\frac{1}{\pi}\right)^2}= \sqrt{1+\frac{1}{\pi^2}}\text{.}
\end{equation*}
One of the useful and interesting ideas of this example is that the speed is constant. That is, no matter what parameter value we consider, the speed is always \(\sqrt{1+\frac{1}{\pi^2}} \text{.}\) Using Definition 10.4.3, we find
\begin{equation*}
\vT(t)= \frac{\vv(t)}{\vecmag{{\vv(t)}}}= \frac{1}{\sqrt{1+\frac{1}{\pi^2}}}\langle -\sin(t),\cos(t),\frac{1}{\pi} \rangle \text{.}
\end{equation*}
To compute some specific examples, we have
\begin{align*}
\vT\left(\frac{\pi}{2}\right)\amp= \frac{1}{\sqrt{1+\frac{1}{\pi^2}}}\left\langle -1,0,\frac{1}{\pi} \right\rangle\\
\vT\left(\frac{5\pi}{4}\right) \amp= \frac{1}{\sqrt{1+\frac{1}{\pi^2}}}\left\langle \frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2},\frac{1}{\pi} \right\rangle\text{.}
\end{align*}
In Figure 10.4.6, \(\vT(\frac{\pi}{2})\) is plotted in red and \(\vT(\frac{5\pi}{4})\) is plotted in blue. Both of these vectors are unit length and tangent to the curve in the direction of travel for that instant.
Note that we can also find \(\vT(\frac{\pi}{2})\) by computing \(\vv(\frac{\pi}{2}) = \langle -1,0,\frac{1}{\pi} \rangle \) and shrinking this vector to be a unit vector. Because our definitions and theorems work for all values of \(t\text{,}\) we can apply them at a specific value of \(t\text{.}\) This has the added advantage that computations tend to be manageable when done by hand. The downside to this calculation is that you only have information for the one parameter value. This will only useful for finding information at a fixed value of \(t\) because you are not concerned with how the measurement is done for other parameter values.
The next activity gives you an opportunity to practice with the velocity and unit tangent vectors and speed.
Activity 10.4.2.
(a)
Find \(\vv\text{,}\) speed, and \(\vT\) as functions of \(t\) for the line parameterized as \(\vr(t)=\langle 3t-1,2-2t,5+t\rangle\text{.}\) Write a few sentences about why your results make sense and why in for this particular curve \(\vT\) does not vary based on the parameter value.
(b)
Find \(\vv\text{,}\) speed, and \(\vT\) for the curve with parameterization \(\vr(t)=\langle t,t^2,t^3\rangle\text{.}\)
(c)
Consider a curve for which we do not know the parameterization. However, we do know that at a point \(P\text{,}\) the tangent line to the curve through \(P\) can be parameterized as \(\langle 2-7t, 3t+1,-4t-1\rangle\text{.}\) What are you able to say about \(\vT\) for this curve at the point \(P\text{?}\)
Subsection 10.4.3 The Direction of Turning
Recalling our analogy in which we think of a parameterization as describing how someone drives along a racetrack, we say that \(\vT\) measures the direction that the car is heading at any given instant. If we use the curve plotted in Figure 10.4.7 as an example of the racetrack, then \(\vT(2)\approx \langle -0.8813, -0.4691, 0.0568 \rangle\text{.}\) The plot in Figure 10.4.7 shows how the direction of travel corresponds to the curve at the point \(\vr(2)\text{.}\) What direction is the car turning at \(t=2\text{?}\) You can see by the plot that the car will be be turning left, but that is a direction relative to the car. We wish to measure the direction that the car is turning at this instant in terms of the locations given by the parameterization in an \(xyz\)-coordinate system?
We first consider \(\vT\, '=\frac{d\vT}{dt}\text{,}\) which measures how quickly \(\vT\) is changing in terms of the parameter. The vector \(\vT\, '=\frac{d\vT}{dt}\) will measure the rate of change of the direction of travel. Since vectors have magnitude and direction, the derivative of a vector-valued function measures both the change in the magnitude and the change in the direction. Since \(\vT\) will always have length 1 (when \(\vT\) exists), all of the change in \(\vT\) corresponds to a change in direction. Thus, the direction of \(\vT'=\frac{d\vT}{dt}\) will be exactly what we are looking for: \(\vT\, '\) will point in the direction of turning. For now, we don’t care about how fast the car is turning; we want to find the direction of the turning. This leads us to define the unit normal vector, denoted \(\vN\text{,}\) as the unit vector in the direction of \(\vT\, '\text{.}\)
Definition 10.4.8.
The unit normal vector to a curve is denoted by \(\vec{N}(t)\text{.}\) This vector is computed as
\begin{equation*}
\vec{N}(t)=\frac{\vT\, '(t)}{\vecmag{{\vT\, '(t)}}}\text{.}
\end{equation*}
The unit normal vector measures the direction of turning along a parameterized curve.
Based on the definition of the unit normal vector, there are a few different characteristics a curve can have that would cause \(\vN\) to not exist. The next activity leads you to consider three different cases:
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\(\vN\) will not exist if \(\vT\) does not exist
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\(\vN\) will not exist if \(\vT\, '\) does not exist
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\(\vN\) will not exist if \(\vT\, '=\vec{0}\)
Activity 10.4.3.
(a)
Draw an example of a curve such that \(\vT\) does not exist at a point on your curve. Identify the point where \(\vT\) fails to exist. Explain both why the direction of travel does not exist at that point based on your plot and why the direction of turning will not exist at the same point.
(b)
Draw an example of a curve such that \(\vT\, '\) does not exist at a point on your curve but \(\vT\) does exist at that point. Identify the point at which \(\vT\, '\) fails to exist. Explain both why \(\vT\, '\) does not exist at that point based on your plot and why \(\vN\) does not exist at the same point.
(c)
Draw an example of a curve such that \(\vT\, '=\vec{0}\) at a point on your curve. Explain why \(\vT\, '=\vec{0}\) based on your plot and explain why \(\vN\) does not exist at the same point.
Example 10.4.9.
In this example, we will calculate the unit normal vector for a helical path given by \(\vr(t)=\langle \cos(t),\sin(t),\frac{t}{\pi}\rangle \text{.}\) This is the same path as was used in Example 10.4.4 and was plotted in Figure 10.4.5.
From our work in Example 10.4.4, we calculated
\begin{equation*}
\vT(t)= \frac{\vv(t)}{\vecmag{{\vv(t)}}}= \frac{1}{\sqrt{1+\frac{1}{\pi^2}}}\left\langle -\sin(t),\cos(t),\frac{1}{\pi} \right\rangle
\end{equation*}
Using the definition of \(\vN\text{,}\) we need to compute the derivative of \(\vT\) with respect to \(t\text{:}\)
\begin{equation*}
\vT\, '(t)=\frac{d\vT}{dt}= \frac{1}{\sqrt{1+\frac{1}{\pi^2}}}\langle -\cos(t),-\sin(t),0 \rangle
\end{equation*}
Note that in this example, the derivative of \(\vT\) is easily calculated for all time values but this will rarely be the case. Further, making a unit vector in the direction of \(\vT(t)'=\frac{d\vT}{dt}\) is routine and gives us the following calculation:
\begin{align*}
\vec{N}(t) \amp= \frac{\vT\, '(t)}{\vecmag{{\vT'(t)}}} \\
\amp= \frac{ \frac{1}{\sqrt{1+(1/\pi^)}}\langle -\cos(t),-\sin(t),0 \rangle }{\frac{1}{\sqrt{1+1/(\pi^2)}}} \\
\amp= \langle -\cos(t),-\sin(t),0 \rangle
\end{align*}
We can calculate \(\vN(\frac{\pi}{2})=\langle 0,-1,0 \rangle\) and \(\vN(\frac{5\pi}{4})=\langle \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0 \rangle\text{.}\) If we add these vectors to the plot of the helix with the unit tangent vectors, as in Figure 10.4.6, we can see that the direction of turning (as measured by \(\vN\)) is pointing toward the center of the helix. It should make sense the the direction you are turning to stay on the helix is toward the inside of the circular trace. Note that the unit tangent vector has a vertical component which corresponds to going “up” the helix but there is no vertical component to \(\vN\text{.}\) This is because there is vertical motion, but there is no turning in the vertical direction. That is, the \(z\)-component of \(\vT\) is unchanging, which means that the vertical component of \(\vT\, '\) must be be zero.
Example 10.4.9 is the exceptional case where \(\vN\) is easily calculated directly from the definition and the calculations are fairly clean. In the next activity, we will look at a curve with a polynomial parameterization that will demonstrate how easily the calculation of \(\vN\) can go off the rails.
Activity 10.4.4.
In general, calculating \(\vN\) as a function of the parameter \(t\) ends up being very difficult because there are multiple compositions of functions involved which means there is a nesting of chain rules involved in the definition. In this activity, we will examine how to go through the direct calculation of \(\vN\) for a curve parameterized by \(\vr(t) =\langle t,t^2,t^3\rangle \text{.}\)
(a)
Calculate velocity \(\vv(t)\) and \(\text{speed}(t)\) for the parameterization \(\vr(t) =\langle t,t^2,t^3\rangle \text{.}\)
(b)
(c)
Calculate the first component of \(\vT\, '=\frac{d\vT}{dt}\text{.}\)
(d)
Calculate the second component of \(\vT\, '\text{.}\)
(e)
Calculate the third component of \(\vT\, '\text{.}\)
(f)
Put together your work for the three components of \(\vT\, '\) and compute \(\vecmag{\frac{d\vT}{dt}}\text{.}\)
(g)
In the previous part, you likely decided that trying to simplify your formulas for \(\frac{d\vT}{dt}\) and \(\vecmag{\frac{d\vT}{dt}}\) was more complicated than you had space for on your paper. Instead of doing more intricate algebraic computations, describe how you would calculate \(\vN\) if you had nice formulas for \(\frac{d\vT}{dt}\) and \(\vecmag{\frac{d\vT}{dt}}\text{.}\)
As talked about in Subsection 10.4.2, the unit tangent vector \(\vT\) measures the direction of travel along a parameterized curve. The unit normal vector \(\vN\) measures the direction that an object is turning in order to stay on the curve. In Figure 10.4.10, we observed that the unit tangent and unit normal vectors as at each point along the curve were orthogonal to each other.
Question 10.4.11.
We first answer this question with a conceptual, geometric argument. By Definition 10.4.8, the unit normal vector is parallel to \(\vT\, '=\frac{d\vT}{dt}\text{.}\) The derivative of the unit tangent vector will measure the change in \(\vT\text{,}\) and since \(\vT\) is a vector, this includes change in both the magnitude and the direction of \(\vT\text{.}\) Since the length of \(\vT\) does not change, all of \(\vT\, '\) represents change in the direction of \(\vT\text{.}\) This is an intuitive argument for why \(\vT\) and \(\vN\) are orthogonal. We next state this formally for future reference and present a formal, algebraic argument.
Key Idea 10.4.12.
The unit tangent vector \(\vT\) and the unit normal vector \(\vN\) are orthogonal so that
\begin{equation*}
\vT\cdot \vN = 0\text{.}
\end{equation*}
Proof.
Here we give a rigorous, algebraic argument. However, this argument does not provide insight into the measurements or relationship between \(\vT\) and \(\vN\) that the paragraph before this Key Idea presents. For this algebraic argument, we consider at the derivative of \(\vecmag{\vT}^2\text{.}\) Because the length of \(\vT\) is constant, we must have that \(\frac{d}{dt}\left(\vecmag{\vT}^2 \right)=0\text{.}\) To simplify \(\frac{d}{dt}\left(\vecmag{\vT}^2 \right)\text{,}\) we use properties of the dot product related to the magnitude, the derivative product rule for the dot product, and finall the commutative property of the dot product. Hence, we see
\begin{align*}
\frac{d}{dt}\left(\vecmag{ \vT }^2 \right) \amp=\frac{d}{dt}\left(\vT \cdot \vT \right) \\
\amp= \vT' \cdot \vT + \vT \cdot \vT' \\
\amp= 2 \vT \cdot \vT'
\end{align*}
If \(0=\frac{d}{dt}\left(\vecmag{\vT}^2\right) = 2 \vT \cdot \vT'\text{,}\) then \(\vT\) and \(\vT'\) are orthogonal. Since \(\vN\) is parallel to \(\vT'\text{,}\) we must also have that \(\vT\) and \(\vN\) are orthogonal.
Subsection 10.4.4 The Binormal Vector
Since the title of this section is The TNB Frame and we have defined \(\vT\) and \(\vN\text{,}\) you likely are expecting the definition of a vector called \(\vB\text{,}\) which we do in this subsection. Since we saw in Key Idea 10.4.12 that the unit tangent and unit normal vectors are orthogonal, this final vector will be defined in a way that completes a right-handed coordinate system with \(\vT\) and \(\vN\text{.}\)
Definition 10.4.13.
The binormal vector, denoted \(\vec{B}\text{,}\) for a parameterized curve is defined as
\begin{equation*}
\vec{B}=\vT \times \vec{N}\text{.}
\end{equation*}
This definition of \(\vB\) creates a three-dimensional right-handed coordinate system that is relative to the motion along the curve. Notice that while we had to divide by magnitude to ensure that \(\vT\) and \(\vN\) were unit vectors, the binormal vector is a unit vector because \(\vecmag{\vT \times \vN } = \vecmag{\vT} \vecmag{\vN} \sin(\theta)\) where \(\theta\) is the angle between \(\vT\) and \(\vN\text{.}\) Since \(\vT\) and \(\vN\) are unit vectors and orthogonal to each other, we see that \(\vecmag{\vT}\vecmag{\vN} \sin(\theta)= 1\text{.}\) The binormal vector \(\vB\) gives the axis of rotation for the motion along the curve. This measure of rotation follows our right-handed measure of rotation. If you put the fingers of your right hand in the direction of travel (\(\vT\)) and curl your fingers in the direction of turning (\(\vN\)), then your thumb will correspond to the axis of rotation based on how you are moving along the curve.
The TNB frame is also called the Frenet frame and is frequently used in fields such as aviation and space travel because much of the important flight information is expressed in terms of relative measurements to the craft and its motion.
Example 10.4.14.
In this example, we complete the calculation of the TNB frame for the helix given by \(\vr(t)=\langle \cos(t),\sin(t)\frac{t}{\pi}\rangle \text{.}\) From Example 10.4.4 and Example 10.4.9 we have the following:
\begin{align*}
\vT(t) \amp= \frac{\vv(t)}{\vecmag{{\vv(t)}}}= \frac{1}{\sqrt{1+\frac{1}{\pi^2}}}\left\langle -\sin(t),\cos(t),\frac{1}{\pi} \right\rangle \\
\vN \amp= \langle -\cos(t),-\sin(t),0\rangle
\end{align*}
Using the definition for \(\vB\text{,}\) we have
\begin{equation*}
\vB = \vT \times \vN = \frac{1}{\sqrt{1+\frac{1}{\pi^2}}} \left\langle -\frac{1}{\pi}\sin(t), -\frac{1}{\pi}\cos(t), \sin(t)^2+\cos(t)^2 \right\rangle
\end{equation*}
Factoring out \(\frac{1}{\pi}\) from each component and simplifying the coefficient gives
\begin{equation*}
\vB = \frac{1}{\sqrt{\pi^2+1}} \langle -\sin(t), -\cos(t), \pi \rangle \text{.}
\end{equation*}
Figure 10.4.15 shows the helix with the TNB frame depicted at a single point. You can adjust the location of the TNB frame by moving the slider.
Note that as you change the \(t\)-value and move around the helix, the unit tangent points in the direction of travel, the unit normal points in the direction the object is turning (always to the left while going up the helix), and the binormal vector is the axis of rotation for the \(\vT\) and \(\vN\) vectors. If you curl the fingers on your right hand from the direction of \(\vT\) toward the direction of \(\vN\text{,}\) notice that your thumb will point in the direction of \(\vB\text{.}\) This sets up the TNB frame as a right-handed coordinate system that changes relative to the motion of the object.
Subsection 10.4.5 The Driver or the Road: TNB Frame
To conclude this section, we consider whether the measurements of \(\vT\text{,}\) \(\vN\text{,}\) and \(\vB\) are properties of the driver or the road. A parameterized curve has a stated orientation (direction of travel). Remember that a measurement involving a parameterized curve is a property of the driver if two different parameterizations can have different measurements at the same location on the curve. It is important to remember that this does not mean that you look at the same parameter value on the curve, but rather the same location on the curve. A measurement on a parameterized curve is a property of the road if every parameterization of this curve must have the same measurement for a fixed location on the curve. These are the same ideas that were first discussed in Activity 10.3.3. In the analogy of a parameterization being a description for how a particular racecar travels around a track as a function of time, a measurement is a property of the driver if that measurement can have different values for different drivers when measured at the same point on the racetrack. A measurement is a property of the road if that measurement must have the same value for all drivers when measured at the same point on the racetrack.
Activity 10.4.5.
(a) Direction of Travel.
Each race car has a 1 meter arrow attached to the hood of the car which points straight ahead. This arrow will measure the three dimensional vector that is the direction of travel. Is the direction of travel a property of the driver or the road? Be sure to explain your answer.
(b) Turn Signal.
The turn signal is measured by checking to see if the steering wheel is turned left or right. We are not measuring how much the steering wheel is turned in either direction, just whether the wheel is turned left or right. If the steering wheel is not turned either way at the instant we measure , then we say there is no turn signal because the car isn’t turning at that instant. Is the turn signal a property of the driver or the road? Be sure to explain your answer.
(c) Direction of Turning.
The direction of turning is measured by using a 1 meter arrow that sticks perpendicularly out of the left side of the car when the steering wheel is turned left and out of the right side of the car when the steering wheel is turned right. We are not measuring how much the wheel is turned in either direction, just whether the wheel is turned left or right. If the steering wheel is not turned either way at the instant we measure the direction of turning, then we say there is no direction of turning because the car isn’t turning at that instant. Is the direction of turning as a vector in \(\mathbb{R}^3\) a property of the driver or the road? Be sure to explain your answer.
(d)
Is \(\vB\) a property of the driver or the road? Justify your ideas.
Summary
-
The unit tangent vector, \(\vT(t)=\frac{\vv}{\vecmag{\vv}}\text{,}\) measures the direction of travel for an object along a parameterized curve.
-
The unit normal vector, \(\vN(t)=\frac{\vT\, '}{\vecmag{\vT\, '}}\text{,}\) measures the direction that an object is turning in order to stay on a parameterized curve.
-
The binormal vector, \(\vB(t)= \vT \times \vN\text{,}\) is the direction of the axis for which \(\vT\) and \(\vN\) are rotating around.
Exercises 10.4.6 Exercises
The WeBWorK problems are written by many different authors. Some authors use parentheses when writing vectors, e.g., \((x(t),y(t),z(t))\) instead of angle brackets \(\langle x(t),y(t),z(t) \rangle\text{.}\) Please keep this in mind when working WeBWorK exercises.
1.
2.
3.
Consider the moving particle whose position at time \(t\) in seconds is given by the vector-valued function \(\vr\) defined by \(\vr(t) = 5t \vi + 4\sin(3t) \vj + 4\cos(3t) \vk\text{.}\) Use this function to answer each of the following questions.
-
Find the unit tangent vector, \(\vT(t)\text{,}\) to the space curve traced by \(\vr(t)\) at time \(t\text{.}\) Write one sentence that explains what \(\vT(t)\) tells us about the particle’s motion.
-
Determine the speed of the particle moving along the space curve with the given parameterization.
-
Find the exact distance traveled by the particle on the time interval \([0,\pi/3]\text{.}\)
-
Find the average velocity of the particle on the time interval \([0, \pi/3]\text{.}\)
-
Determine the parameterization of the given curve with respect to arc length.
4.
Consider the standard helix parameterized by \(\vr(t) = \cos(t) \vi + \sin(t) \vj + t \vk\text{.}\)
-
Recall that the unit tangent vector, \(\vT(t)\text{,}\) is the vector tangent to the curve at time \(t\) that points in the direction of motion and has length 1. Find \(\vT(t)\text{.}\)
-
Explain why the fact that \(| \vT(t) | = 1\) implies that \(\vT\) and \(\vT'\) are orthogonal vectors for every value of \(t\text{.}\) (Hint: note that \(\vT \cdot \vT = |\vT|^2 = 1,\) and compute \(\frac{d}{dt}[\vT \cdot \vT]\text{.}\))
-
For the given function \(\vr\) with unit tangent vector \(\vT(t)\) (from (a)), determine \(\vN(t) = \frac{1}{|\vT'(t)|} \vT'(t)\text{.}\)
-
What geometric properties does \(\vN(t)\) have? That is, how long is this vector, and how is it situated in comparison to \(\vT(t)\text{?}\)
-
Let \(\vB(t) = \vT(t) \times \vN(t)\text{,}\) and compute \(\vB(t)\) in terms of your results in (a) and (c).
-
What geometric properties does \(\vB(t)\) have? That is, how long is this vector, and how is it situated in comparison to \(\vT(t)\) and \(\vN(t)\text{?}\)
-
Sketch a plot of the given helix, and compute and sketch \(\vT(\pi/2)\text{,}\) \(\vN(\pi/2)\text{,}\) and \(\vB(\pi/2)\text{.}\)
