Motivating Questions
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How can we split the acceleration vector into parts in the direction of travel and the direction of turning?
Section 10.6 Splitting the Acceleration Vector
Subsection 10.6.1 Introduction
Position is the central measurement of the study of motion, which is known as kinematics. Velocity and acceleration are the first and second derivatives of position, respectively, and are frequently important properties in the study of physics and mechanics. To better understand velocity as a vector-valued function of time, we examined how the magnitude and direction of position change separately. We separated velocity into its critical parts: speed (magnitude of velocity) and direction of travel (unit tangent vector). Speed is a scalar measurement for how fast the position of an object is changing as a function of time, while \(\vT\) is a unit vector that measures the direction in which position is changing.
As a vector quantity, acceleration is related to many important physical laws and descriptions. For example, Newton’s Second Law of Motion relates the forces acting on an object to its acceleration through the formula \(\vec{F}_{net}= m \vec{a}\text{.}\) This is often summarized as the sum of the forces acting on an object is the mass times the acceleration of the object. This description relates a combination of forces on a object to its position and velocity through integration of \(\va\text{.}\) For an example of these calculations, refer back to Example 10.2.7.
To understand acceleration better as a vector quantity, we will examine acceleration as a change in the magnitude and direction of the velocity vector. In this section, we will use our knowledge of calculus and vector measurements to split the acceleration vector into parts related to the changes in the magnitude and direction of the velocity vector. We will also consider how these measurements relate to physical contexts.
Preview Activity 10.6.1.
All your good work as CEO and lead engineer at Steer Clear has started to pay off, literally. When you showed your work on using the location tracking system (LTS) to develop navigation and telemetry tools to an investment group (your grandparents), they were impressed and decided to give you money to further develop your self-driving car. You decided to use this infusion of cash to buy a gyroscopic accelerometer which will measure acceleration as a vector with magnitude and direction. In order to use your new instrument for information on the “driving” part of your self-driving car, you need to understand how your accelerometer readings will relate to the operations needed to drive a car.
(a)
The image below shows the path you drove on a test drive along with points \(P\) and \(Q\) and the direction of travel. At the points \(P\) and \(Q\text{,}\) we have draw three vectors. The acceleration vector \(\va\) provided by your new gyroscopic accelerometer is shown in magenta. The two blue vectors are the unit tangent vector \(\vT\) and unit normal vector \(\vN\) at that point. Label each of the blue vectors as being \(\vT\) or \(\vN\text{.}\)
A two-dimensional curve with two points indicated. At each point, there are three vectors.
(b)
Draw the projection of \(\va\) on \(\vT\) and the projection of \(\va\) on \(\vN\) at point \(P\) and at point \(Q\) in the graph above.
(c)
Is \(\va \cdot \vT\) is positive, negative, or zero at \(P\text{?}\) What about at \(Q\text{?}\) Write a sentence or two to justify your answers.
(d)
Is \(\va \cdot \vN\) is positive, negative, or zero at \(P\text{?}\) What about at \(Q\text{?}\) Write a sentence or two to justify your answers.
The preview activity shows how you can graphically separate the acceleration vector into parts in the direction of travel and direction of turning. In this section, we will look at how to calculate these separate parts algebraically, understand what is being measured by these quantities, and show how these tools will give us important physical measurements as well as an efficient tool for calculating \(\vN\text{,}\) the unit normal vector to a curve.
Subsection 10.6.2 Splitting Acceleration
Recall that by the definition of \(\vT\), \(\vv\) is parallel to \(\vT\) when \(\vv\) exists and is nonzero. Additionally, Key Idea 10.4.12 states that \(\vT \cdot \vN =0 \text{.}\) In other words, all of the velocity is in a direction parallel to the unit tangent vector and the unit normal vector is always orthogonal to the unit tangent vector to a curve, when they exist. In this subsection, we will look at splitting the acceleration vector into parts in the direction of travel and the direction of turning.
Suppose \(\va(t)\) is the acceleration vector for the motion along a curve in \(\mathbb{R}^3\text{.}\) From Section 10.4, we have the TNB frame as a way of describing key directions related to motion along a curve in space. We would like to split the acceleration vector into a part in each of directions corresponding to \(\vT\text{,}\) \(\vN\text{,}\) and \(\vB\text{.}\) Algebraically, we want to find scalars \(c_1\text{,}\) \(c_2\text{,}\) and \(c_3\) such that \(\va= c_1 \vT + c_2 \vN +c_3 \vB\text{.}\) To do so, we will make use of several key facts from Section 10.4:
\begin{align*}
\vT \cdot \vN \amp=0 \amp \vT \cdot \vB \amp =0 \amp \vN \cdot \vB \amp=0 \amp\vecmag{\vT}\amp=\vecmag{\vN}=\vecmag{\vB}=1\text{.}
\end{align*}
If we compute the dot product of each of \(\vT\text{,}\) \(\vN\text{,}\) and \(\vB\) with the expression for \(\va\text{,}\) we find
\begin{align*}
\va \cdot \vT \amp= (c_1 \vT + c_2 \vN +c_3 \vB)\cdot \vT \\
\amp= c_1 (\vT \cdot \vT) + c_2 (\vN \cdot \vT) +c_3 (\vB \cdot \vT) \\
\amp=c_1 \\
\va \cdot \vN \amp= (c_1 \vT + c_2 \vN +c_3 \vB)\cdot \vN \\
\amp= c_1 (\vT \cdot \vN) + c_2 (\vN \cdot \vN) +c_3 (\vB \cdot \vN) \\
\amp=c_2 \\
\va \cdot \vB \amp= (c_1 \vT + c_2 \vN +c_3 \vB)\cdot \vB \\
\amp= c_1 (\vT \cdot \vB) + c_2 (\vN \cdot \vB) +c_3 (\vB \cdot \vB) \\
\amp=c_3
\end{align*}
These computations demonstrate that because the TNB frame directions are all unit vectors that are orthogonal to each other, we can compute the amount of acceleration in each of these directions with a dot product!
Definition 10.6.1.
The tangential component of acceleration is \(a_{\vT} = \va \cdot \vT \) and the normal component of acceleration is \(a_{\vN} = \va \cdot \vN \text{.}\)
Our use of the word “component” in Definition 10.6.1 is consistent with our use in Section 9.3 because \(\vT\) and \(\vN\) are both unit vectors. Furthermore, because the unit tangent and unit normal vectors are also orthogonal to each other, \(a_{\vT}\) and \(a_{\vN}\) give the amount of acceleration in the direction of travel and the direction of turning, respectively. You may notice that we only gave names for acceleration in the tangential and normal directions above. The acceleration vector for motion along a curve can only have components in the tangential and normal directions to the curve. It is not possible for the acceleration vector of parameterized curve to have any component in a direction, such as that of \(\vB\text{,}\) that does not lie in the plane defined by \(\vT\) and \(\vN\text{,}\) which we state formally and justify below.
Key Idea 10.6.2.
If \(\va(t)\) is the acceleration vector for a parameterization of a curve in \(\mathbb{R}^n\text{,}\) then
\begin{equation*}
\va=a_{\vT} \vT + a_{\vN} \vN
\end{equation*}
Proof.
We claim that the scalar \(a_{\vN}\) measures the acceleration due to change in the direction of \(\vv\) and the scalar \(a_{\vT}\) measures the acceleration due to the change in magnitude of \(\vv\text{.}\) This interpretation of \(a_{\vT}\) and \(a_{\vN}\) follows from a product rule argument based on how we split the velocity vector into speed (magnitude) and \(\vT\) (direction). We start by splitting velocity into its direction and magnitude:
\begin{equation*}
\vv(t) = \text{speed}(t) \cdot \vT(t)
\end{equation*}
Taking the derivative of the velocity using a product rule, we get the following formula for acceleration:
\begin{equation*}
\va(t) = \frac{d\vv}{dt} = \frac{d}{dt}(\text{speed}) \cdot \vT + (\text{speed}) \cdot \frac{d}{dt}(\vT)
\end{equation*}
We can rearrange the terms in our definition for \(\vN\) to solve for \(\vT\, '(t)\text{,}\) which gives
\begin{equation*}
\vN=\frac{\vT\, '}{\vecmag{\vT\, '}} \Rightarrow \vT\, '= \vecmag{\vT'} \vN \text{.}
\end{equation*}
Applying this to our acceleration equation above, we get
\begin{equation}
\va(t) = \frac{d\vv}{dt} = \frac{d}{dt}(\text{speed}) \cdot \vT + (\text{speed}) \cdot \vecmag{\frac{d\vT}{dt}} \vN \text{.}\tag{10.6.1}
\end{equation}
This algebraic splitting of acceleration gives a nice way to interpret \(a_{\vT}\text{.}\) Specifically, \(a_{\vT}\) is the rate of change of speed!
The scalar \(a_{\vN}\) is harder to interpret right now because our algebraic splitting shows that \(a_{\vN} = (\text{speed}) \vecmag{\frac{d\vT}{dt}}\text{.}\) Since \(\frac{d\vT}{dt}\) and \(\vecmag{\frac{d\vT}{dt}}\) are as hard to calculate directly as \(\vN\) is, we do not have an intuitive interpretation for the meaning of \(a_{\vN}\text{.}\) We will return to the meaning of \(a_{\vN}\) later in this section, but for now you should think of \(a_{\vN}\) as the acceleration due to a change in the direction of the velocity vector. In other words, \(a_{\vN}\) is the amount of acceleration due to turning.
Note that nothing in the argument for Key Idea 10.6.2 depended on being in three dimensions, and all of the calculus tools used in the argument work for any \(\mathbb{R}^n\text{.}\)
Definition 10.6.3.
The splitting of the acceleration (vector) refers to finding scalars \(a_{\vT}\) and \(a_{\vN}\) such that \(\va = a_{\vT} \vT + a_{\vN} \vN\text{.}\) This allows us to describe the amount of the acceleration that is in the unit tangent direction and the amount that is in the unit normal direction, relative to the motion on the curve.
Figure 10.6.4 shows the splitting of the acceleration vector graphically. Move the slider to change the location on the curve where the splitting of the acceleration is shown. Examine different points along the curve to see how each of \(\vT\text{,}\) \(\vN\text{,}\) \(a_{\vT}\text{,}\) and \(a_{\vN}\) change at different locations. The vector \(a_{\vN} \vN\) is drawn twice to demonstrate its relationship to \(\vN\) and how \(\va\) is a vector sum of \(a_{\vT} \vT\) and \(a_{\vN} \vN\text{.}\)
Activity 10.6.2.
(a)
The tangential component of acceleration is defined as \(a_{\vT} = \va \cdot \vT \text{,}\) while equation (10.6.1) establishes that \(a_{\vT} = \frac{d\text{(speed)}}{dt}\) is also true. Use these formulas to determine whether \(a_{\vT}\) can be zero. Either explain why \(a_{\vT}\) is always nonzero or describe scenarios in which \(a_{\vT}=0\text{.}\)
(b)
Determine whether \(a_{\vT}\) can be negative. Either explain \(a_{\vT}\) is never negative or describe scenarios in which \(a_{\vT}\lt 0\)
(c)
The normal component of acceleration is defined as \(a_{\vN} = \va \cdot \vN \text{,}\) while equation (10.6.1) establishes that \(a_{\vN} = (\text{speed}) \vecmag{\frac{d\vT}{dt}}\) is also true. Use these formulas to determine whether \(a_{\vN}\) can be negative. Either explain why \(a_{\vN}\) is negative or describe scenarios in which \(a_{\vN}\lt 0\text{.}\)
(d)
Determine whether \(a_{\vN}\) can be zero. Either explain why \(a_{\vN}\) is always nonzero or describe scenarios in which \(a_{\vN}=0\text{.}\)
Exercise 4 and Exercise 5 relate the scalar and vector measurements involved in the splitting of acceleration to different modes of transportation and looks at how different actions when riding a bike or driving a car change these vector calculus quantities. Exercise 6 considers how to visually identify the different measurements involved the splitting of the acceleration vector.
You saw in Activity 10.4.4 that calculating \(\vN\) for even a curve with a very “nice” parameterization algebraically using the definition requires extensive algebraic manipulation. Before going into an exampmle of the computations involved in splitting the acceleration vector, we will identify a relationship of the vectors involved in splitting the acceleration vector that yields another way of approaching \(\vN\text{.}\) To do this, we consider the right triangle of vectors in Figure 10.6.5. If we apply the Pythagorean Theorem to this right triangle, we can then use that the unit tangent vector and unit normal vector both have length one to conclude that \(\vecmag{ \va }^2 =(a_{\vT})^2+(a_{\vN})^2\text{.}\)
A plot of the right triangle of vectors made by \(\va, a_{\vT} \vT, a_{\vN} \vN \)
Now we can solve for \(a_{\vN}\) and find
\begin{equation}
a_{\vN}= \sqrt{\vecmag{\va}^2 - (a_{\vT})^2}\text{.}\tag{10.6.2}
\end{equation}
Notice that since \(a_{\vN}\) is always nonnegative, we know that we can take the positive square root when solving here. We will soon use this formula to help find \(a_{\vN}\text{.}\) Since \(\va = a_{\vT}\vT + a_{\vN}\vN\text{,}\) we can also solve for \(\vN\) and have
\begin{equation*}
\vN = \frac{\va-a_{\vT}\vT}{a_{\vN}},
\end{equation*}
which we will use to help find \(\vN\text{.}\)
Example 10.6.6.
In this example we find the splitting of acceleration for a curve parameterized by \(\vr(t)=\langle t, t^2,t^3\rangle\) with \(0\leq t\leq 2\text{.}\) While this curve’s parameterization is simple algebraically, we will find that the formulas involved in the splitting of acceleration become more complex fairly quickly. To get started, we will find the splitting of the acceleration vector for a single parameter value. In the second step, we will look at finding general formulas for \(a_{\vT}\) and \(a_{\vN}\text{.}\)
(a)
In many circumstances, we need to compute the splitting of acceleration for only one parameter value. For this example, we will look at how clean the computations can be if we want to look only at \(t=1\text{.}\) The first steps for calculating \(\vv\) and \(\va\) are to use the definitions and substitute \(t=1\)
\begin{align*}
\vr(t) \amp=\langle t, t^2,t^3\rangle \amp \vr(1) \amp=\langle 1, 1, 1 \rangle \\
\vv(t) \amp=\langle 1, 2t,3t^2\rangle \amp \vv(1) \amp=\langle 1, 2, 4 \rangle\\
\va(t) \amp=\langle 0, 2,6t\rangle \amp \va(1) \amp=\langle 0, 2, 6 \rangle\text{.}
\end{align*}
The next quantity to calculate is \(\vT\text{.}\) Note that
\begin{equation*}
\text{speed}(1)=\vecmag{\vv(1)}= \sqrt{1+4+9}=\sqrt{14}
\end{equation*}
here. Thus,
\begin{equation*}
\vT(1) = \frac{\vv}{\vecmag{\vv}} = \frac{\langle 1, 2, 3 \rangle}{\sqrt{14}} \text{.}
\end{equation*}
This allows us to calculate \(a_{\vT}(1)\) from the definition
\begin{align*}
a_{\vT}(1) = \va(1) \cdot \vT(1) \amp= \langle 0, 2, 6 \rangle \cdot \frac{\langle 1, 2 , 3 \rangle}{\sqrt{14}} \\
\amp= \frac{1}{\sqrt{14}} \left(1(0)+2(2)+3(6)\right)= \frac{22}{\sqrt{14}}\text{.}
\end{align*}
We now use \(a_{\vN}(1) = \sqrt{\vecmag{ \va(1) }^2 -(a_{\vT}(1))^2}\) to efficiently calculate the normal component of the acceleration and have
\begin{align*}
a_{\vN}(1) \amp= \sqrt{\vecmag{ \va(1) }^2 -(a_{\vT}(1))^2} = \sqrt{\frac{76}{14}} = \sqrt{\frac{38}{7}} \text{.}
\end{align*}
We can now compute how much acceleration there is due to turning and use that to explicitly compute \(\vN(1)\text{.}\) We have
\begin{align*}
a_{\vN}(1) \vN(1) \amp= \va(1) - a_{\vT}(1) \vT(1) \\
\amp= \langle 0, 2,6 \rangle - \frac{22}{\sqrt{14}} \frac{\langle 1, 2, 3\rangle}{\sqrt{14}}\\
\amp= \left\langle -\frac{11}{7},-\frac{16}{7},\frac{9}{7} \right\rangle \text{.}
\end{align*}
Thus,
\begin{equation*}
\vN = \frac{\va - a_{\vT} \vT}{a_{\vN}} = \sqrt{\frac{7}{38}} \left\langle -\frac{11}{7},-\frac{16}{7},\frac{9}{7} \right\rangle\text{.}
\end{equation*}
(b)
We now consider splitting the acceleration vector for the same curve, but we do so for an arbitrary parameter value \(t\text{.}\) That is, we will calculate \(\vv\text{,}\) \(\va\text{,}\) \(\vT\text{,}\) \(a_{\vT}\text{,}\) \(a_{\vN}\text{,}\) and \(\vN\) as functions of \(t\text{.}\) Along the way, we will omit a number of straightforward steps in simplifying. Since we already calculated \(\vr(t)\text{,}\) \(\vv(t)\text{,}\) and \(\va(t)\) in the previous part, we can move on to calculate \(\vT\text{.}\) Since \(\text{speed}(t)=\vecmag{\vv(t)}\text{,}\) we have \(\text{speed}(t)= \sqrt{1+4t^2+9t^4}\) here. Thus,
\begin{equation*}
\vT(t) = \frac{\vv}{\vecmag{\vv}} = \frac{\langle 1, 2t,3t^2\rangle}{\sqrt{1+4t^2+9t^4}}\text{.}
\end{equation*}
The definition \(a_{\vT} = \va \cdot \vT\) allows us to calculate \(a_{\vT}\text{:}\)
\begin{align*}
a_{\vT} = \va \cdot \vT \amp= \langle 0, 2,6t\rangle \cdot \frac{\langle 1, 2t,3t^2\rangle}{\sqrt{1+4t^2+9t^4}} \\
\amp= \frac{4t+18t^3}{\sqrt{1+4t^2+9t^4}}\text{.}
\end{align*}
For this parameterization,
\begin{align*}
a_{\vN} \amp= \sqrt{\vecmag{ \va }^2 -(a_{\vT})^2} \\
\amp= \sqrt{(4+36t^2)- \left(\frac{4t+18t^3}{\sqrt{1+4t^2+9t^4}}\right)^2}\\
\amp= \sqrt{\frac{4(9t^4+9t^2+1)}{1+4t^2+9t^4}}\text{.}
\end{align*}
We can now also calculate \(\vN\) as a function of \(t\) without having to go through the complex algebra required to compute \(\frac{d\vT}{dt}\text{.}\) We have
\begin{align*}
a_{\vN} \vN \amp= \va - a_{\vT} \vT \\
\amp= \langle 0, 2,6t \rangle - \frac{4t+18t^3}{\sqrt{1+4t^2+9t^4}} \frac{\langle 1, 2t,3t^2\rangle}{\sqrt{1+4t^2+9t^4}}\text{,}
\end{align*}
so we can write \(\vN\) as
\begin{align*}
\vN \amp= \frac{\va - a_{\vT} \vT}{a_{\vN}}\\
\amp= \sqrt{\frac{1+4t^2+9t^4}{4(9t^4+9t^2+1)}}\left(\langle 0, 2,6t \rangle - \frac{4t+18t^3}{\sqrt{1+4t^2+9t^4}} \frac{\langle 1, 2t,3t^2\rangle}{\sqrt{1+4t^2+9t^4}}\right)\text{.}
\end{align*}
While this process is still algebraically tedious, these steps are less difficult than finding \(\vN\) directly (as in Activity 10.4.4). We also now have insight into how the acceleration is split in the direction of travel and direction of turning. Graphically, we can represent this splitting with Figure 10.6.7.
The method used in the previous example can be applied broadly and shows how to calculate all parts of Definition 10.6.3 in terms of \(\vv\) and \(\va\) (for both an interval or a single value of \(t\)).
Efficient Calculation of \(a_{\vT}, a_{\vN}, \vT, \vN\) in terms of \(\vv\) and \(\va\).
\begin{align*}
\vT \amp= \frac{\vv}{\vecmag{\vv}} \\
a_{\vT} \amp= \frac{\va\cdot\vv}{\vecmag{\vv}} \Rightarrow a_{\vT} \vT = \frac{\va\cdot \vv}{\vv \cdot \vv} \vv\\
a_{\vN} \amp= \vecmag{ \va - a_{\vT} \vT } = \sqrt{\vecmag{ \va } - (a_{\vT})^2}= \sqrt{(\va \cdot \va)- \frac{(\va\cdot\vv)^2}{\vecmag{\vv}}} \\
a_{\vN} \vN \amp= \va - a_{\vT} \vT \\
\vN \amp= \frac{\va - a_{\vT} \vT}{a_{\vN}}
\end{align*}
Activity 10.6.3.
Compute the splitting of the acceleration for the curve given by \(\vr(t)= \langle t, \frac{t^2}{2},\frac{t^3}{6} \rangle\) at \(t=1\text{.}\) You should calculate \(a_{\vT}\text{,}\) \(a_{\vN}\text{,}\) \(\vT\text{,}\) and \(\vN\text{.}\) Additionally, you should verify \(\va = a_{\vT} \vT +a_{\vN} \vN \) and that \(\vT\) is orthogonal to \(\vN\text{.}\)
Exercise 4, Exercise 5, and Exercise 6 offer conceptual and physical meaning to the different measurements involved in the splitting of acceleration. We will take a moment now to relate \(a_{\vN}\text{,}\) the acceleration due to a change in the direction of travel, and \(\kappa\text{,}\) the curvature, which measures rate of change for the direction of travel on a curve.
The previous statement may seem a bit like we said the same thing twice, but \(a_{\vN}\) and \(\kappa\) are not the same. Theorem 10.5.7 shows that
\begin{equation*}
\kappa = \frac{\vecmag{ \frac{d\vT}{dt} }}{ \vecmag{ \vr\, '(t) }}\text{.}
\end{equation*}
We can solve for \(\vecmag{ \frac{d\vT}{dt} }\) in terms of \(\kappa\) to get \(\vecmag{ \frac{d\vT}{dt} }= \kappa \cdot \text{speed}\text{.}\) We saw earlier in this section that
\begin{equation*}
a_{\vN}= \text{speed}\cdot \vecmag{\frac{d\vT}{dt}}
\end{equation*}
Thus, we can write \(a_{\vN}\) in terms of speed and curvature as
\begin{equation}
a_{\vN}=\kappa\cdot (\text{speed})^2 \text{.}\tag{10.6.3}
\end{equation}
Equation (10.6.3) helps demonstrate the difference between \(a_{\vN}\) and \(\kappa\text{.}\) Specifically, \(\kappa\) measures how fast the path is curving and \(a_{\vN}\text{,}\) the normal component of acceleration, measures the acceleration felt due to turning. Equation (10.6.3) shows that if two drivers are driving along the same road but the first driver is going twice as fast as the second driver is going, then the first driver feels four times as much acceleration due to turning as the second driver does, even though both drivers are following the exact same road. Curvature is how fast the road is turning and \(a_{\vN}\) is how fast it feels like the driver is turning based on speed, which is determined by the parameterization.
Activity 10.6.4. The Driver or the Road?
In this activity, you will determine if the measurements related to the splitting of acceleration are properties of the driver or properties of the road. This is a continuation of Activity 10.3.3. As a reminder, a measurement is a property of the driver if the value of that measurement can be different for different drivers when measured at the same location on the racetrack. A measurement is a property of the road when different drivers must have the same value(s) when measured at the same location on the racetrack.
(a)
Pedal Usage: We measure the pedal usage as \(+\text{,}\) \(-\text{,}\) or \(0\) in the following way:
-
\(+\) if the gas pedal is being used
-
\(-\) if the brake pedal is being used
-
\(0\) if no pedal is being used
Since our drivers are safety minded, they do not use more than one pedal at a time. Note that this measurement is a sign rather than a numerical value. Which vector calculus quantity’s sign corresponds to the pedal usage? Write a sentence to explain your answer.
Hint.
The vector calculus quantities related to the motion along a curve we have covered are time, position, velocity, acceleration, speed, arc length, unit tangent vector, unit normal vector, binormal vector, tangential acceleration component, normal acceleration component, curvature, and osculating circle.
(b)
Describe how the value of \(a_{\vec{N}}\) would be felt by the driver in our analogy.
(c)
In terms of the race car analogy, explain why \(a_{\vec{N}}\) can’t be negative.
(d)
Write a few sentences about whether \(a_{\vT}\) is a property of the driver or a property of the road.
(e)
Write a few sentences about whether \(a_{\vN}\) is a property of the driver or a property of the road.
Summary
-
The acceleration vector for a parameterization of a curve, \(C\text{,}\) can be split into parts in the direction of travel and direction of turning\begin{equation*} \va=a_{\vT}\vT+a_{\vN}\vN \end{equation*}where \(a_{\vT}=\va \cdot \vT\) and \(a_{\vN}=\va \cdot \vN\text{.}\)
-
The scalar \(a_{\vT}\) measures the rate of change of the speed. The scalar \(a_{\vN}\) measure the acceleration due to change in direction and can be calculated by\begin{equation*} a_{\vN}=\kappa (speed)^2 \end{equation*}where \(\kappa\) is the curvature.
Exercises 10.6.3 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.
In this exercise, we will compute and compare elements of the splitting of acceleration for three different people (Nick, Jane, and Tony) as they drive around the same racetrack.
(a)
Suppose that at time \(t=b\text{,}\) Jane is at a location which we will denote \(P_1\) in her race car. The velocity and acceleration vectors for Jane are \(\langle{2,0,-2}\rangle\) and \(\langle{3,2,-3}\rangle\text{.}\) Calculate the following:
-
\(\displaystyle \vec{T}(b)\)
-
\(\displaystyle a_{\vec{T}}(b)\)
-
\(\displaystyle a_{\vec{N}}(b)\)
-
\(\displaystyle \vec{N}(b)\)
(b)
Suppose that at time \(t=c\text{,}\) Nick is at a location \(P_1\) in his race car. Without any calculation at all, answer the following:
-
\(\displaystyle \vec{T}(c)\)
-
\(\displaystyle \vec{N}(c)\)
(c)
The velocity and acceleration vectors for Nick (at time \(c\)) are \(\langle{1,0,-1}\rangle\) and \(\langle{-1,2,1}\rangle\text{.}\) Calculate the following:
-
\(\displaystyle a_{\vec{T}}(c)\)
-
\(\displaystyle a_{\vec{N}}(c)\)
(d)
Is Jane or Nick turning the steering wheel farther? Explain how you know this based on your calculations above.
(e)
Is Jane using the gas pedal or the brake pedal when she is at \(P_1\text{?}\) Explain how you know this based on your calculations above.
(f)
Is Nick using the gas pedal or the brake pedal when he is at \(P_1\text{?}\) Explain how you know this based on your calculations above.
(g)
Nick and Jane talk to their friend Tony who claims that his velocity and acceleration vectors at position \(P_1\) are \(\langle{4,0,6}\rangle\) and \(\langle{8,9,10}\rangle\text{.}\) Explain how Jane and Nick know he is lying.
2.
(a)
Compute the splitting of the acceleration for the curve given by \(\vr(t)= \langle t, \frac{t^2}{2},\frac{t^3}{6} \rangle\) at \(t=-1\text{.}\) You should calculate \(a_{\vT}\text{,}\) \(a_{\vN}\text{,}\) \(\vT\text{,}\) and \(\vN\text{,}\) then verify \(\va = a_{\vT} \vT +a_{\vN} \vN \) and that \(\vT\) is orthogonal to \(\vN\text{.}\)
(b)
Compute the splitting of the acceleration for the curve given by \(\vr(t)= \langle t, e^t,t e^t \rangle\) at \(t=0\text{.}\) You should calculate \(a_{\vT}\text{,}\) \(a_{\vN}\text{,}\) \(\vT\text{,}\) and \(\vN\text{,}\) then verify \(\va = a_{\vT} \vT +a_{\vN} \vN \) and that \(\vT\) is orthogonal to \(\vN\text{.}\)
3.
Given a curve with a parameterization such that \(\vv(11)= \langle 3,4 \rangle\) and \(\va(11) = \langle -2,1 \rangle \text{.}\)
(a)
Find \(\vT(11)\)
(b)
Find \(a_{\vT}(11)\)
(c)
Find \(a_{\vT}(11) \vT(11)\)
(d)
Find \(a_{\vN}(11)\)
(e)
Find \(a_{\vN}(11) \vN(11)\)
(f)
Find \(\vN(11)\)
4.
In this exercise we want to relate the scalar and vector measurements involved in the splitting of acceleration to different modes of transportation. While everyone has different experiences with piloting different modes of transportation, we will phrase this question in terms of driving a car. Feel free to substitute another mode of transportation like riding a bike, riding a scooter, driving a boat, rowing a boat, flying a plane, or flying a spaceship.
(a)
In the context of driving a car (or substitute your mode of transportation), write a couple sentences about what vector calculus quantity is controlled by the brake pedal. Be specific which vector calculus quantity is involved and how the brake pedal changes this measurement.
(b)
In the context of driving a car (or substitute your mode of transportation), write a couple sentences about what vector calculus quantity is controlled by the accelerator pedal. Be specific which vector calculus quantity is involved and how the accelerator pedal changes this measurement.
(c)
In the context of driving a car (or substitute your mode of transportation), write a couple sentences about what vector calculus quantity is controlled by the direction of the steering wheel (consider this to be either left, right, or straight). Be specific which vector calculus quantity is involved and how the the direction of the steering wheel changes this measurement.
5.
In this exercise we want to relate the scalar and vector measurements involved in the splitting of acceleration to different modes of transportation. While everyone has different experiences with piloting different modes of transportation, we will phrase this question in terms of riding a bike. Feel free to substitute another mode of transportation like driving a car, riding a scooter, driving a boat, rowing a boat, flying a plane, or flying a spaceship.
(a)
In the context of riding a bike, what does the unit tangent vector, \(\vT\text{,}\) measure?
(b)
In the context of riding a bike, what does the unit normal vector, \(\vN\text{,}\) measure?
(c)
In the context of riding a bike, how would you increase \(a_{\vT}\text{?}\)
(d)
In the context of riding a bike, how would you decrease \(a_{\vT}\text{?}\)
(e)
In the context of riding a bike, how would you decrease \(a_{\vN}\text{?}\)
(f)
In the context of riding a bike, how would you increase \(a_{\vN}\text{?}\)
6.
In this exercise, we will look at how to visually identify the different measurements involved the splitting of the acceleration vector. While each of \(\va\text{,}\) \(\vT\text{,}\) \(\vN\text{,}\) \(a_{\vT} \vT\text{,}\) and \(a_{\vN} \vN\) are drawn, the colors have been randomized so you will need to look at how each element is related to the motion along the curve.
(a)
The vectors \(\va\text{,}\) \(\vT\text{,}\) \(\vN\text{,}\) \(a_{\vT} \vT\text{,}\) and \(a_{\vN} \vN\) are drawn in different colors in Figure 10.6.8. Match each color with proper vector.
A 2D curve with splitting of acceleration shown using randomized colors
(b)
The vectors \(\va\text{,}\) \(\vT\text{,}\) \(\vN\text{,}\) \(a_{\vT} \vT\text{,}\) and \(a_{\vN} \vN\) are drawn in different colors in Figure 10.6.9. Match each color with proper vector.
A 2D curve with splitting of acceleration shown using randomized colors
(c)
The vectors \(\va\text{,}\) \(\vT\text{,}\) \(\vN\text{,}\) \(a_{\vT} \vT\text{,}\) and \(a_{\vN} \vN\) are drawn in different colors in Figure 10.6.10. Match each color with proper vector.
A 2D curve with splitting of acceleration shown using randomized colors
