Parametric Equations and Vectors

Posted on January 22, 2019 by Lin McMullin

In BC calculus the only application parametric equations and vectors is motion in a plane. Polar equations concern area and the meaning of derivatives. See the review notes for more detail and an outline of the topics. (only 3 items here)

Motion Problems: Same Thing Different Context (11-16-2012)

Implicit Differentiation of Parametric Equations (5-17-2014)

A Vector’s Derivative (1-14-2015)

Review Notes

Type 8: Parametric and Vector Equations (3-30-2018) Review Notes

Type 9: Polar Equation Questions (4-3-2018) Review Notes

Roulettes

This is a series of posts that could be used when teaching polar form and curves defined by vectors (or parametric equations). They might be used as a project. Hopefully, the equations that produce the graphs will help students understand these topics. Don’t let the names put you off. Except for one post, there is no calculus here.

Rolling Circles (6-24-2014)

Epicycloids (6-27-2014)

Epitrochoids (7-1-2014) The most common of these are the cycloids.

Hypocycloids and Hypotrochoids (7-7-2014)

Roulettes and Calculus (7-11-2014)

Roulettes and Art – 1 (7-17-2014)

Roulettes and Art – 2 (7-23-2014)

Limaçons (7-28-2014)

The College Board is pleased to offer a new live online event for new and experienced AP Calculus teachers on March 5th at 7:00 PM Eastern.

I will be the presenter.

The topic will be AP Calculus: How to Review for the Exam: In this two-hour online workshop, we will investigate techniques and hints for helping students to prepare for the AP Calculus exams. Additionally, we’ll discuss the 10 type questions that appear on the AP Calculus exams, and what students need know and to be able to do for each. Finally, we’ll examine resources for exam review.

Registration for this event is $30/members and $35/non-members. You can register for the event by following this link: http://eventreg.collegeboard.org/d/xbqbjz

Parametric and Vector Equations

Posted on February 13, 2018 by Lin McMullin

In the plane, the position of a moving object as a function of time, t, can be specified by a pair of parametric equations $x=x\left( t \right)\text{ and }y=y\left( t \right)$ or the equivalent vector $\left\langle x\left( t \right),y\left( t \right) \right\rangle$ . The path is the curve traced by the parametric equations or the tips of the position vector. .

The velocity of the movement in the x- and y-direction is given by the vector $\left\langle {x}'\left( t \right),{y}'\left( t \right) \right\rangle$ . The vector sum of the components gives the direction of motion. Attached to the tip of the position vector this vector is tangent to the path pointing in the direction of motion.

The length of this vector is the speed of the moving object. $\text{Speed }=\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}$ . (Notice that this is the same as the speed of a particle moving on the number line with one less parameter: On the number line $\text{Speed}=\left| v \right|=\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}}$ .)

The acceleration is given by the vector $\left\langle {{x}'}'\left( t \right),{{y}'}'\left( t \right) \right\rangle$ .

What students should know how to do:

Vectors may be written using parentheses, ( ), or pointed brackets, $\left\langle {} \right\rangle$ , or even $\vec{i},\vec{j}$ form. The pointed brackets seem to be the most popular right now, but all common notations are allowed and will be recognized by readers.
Find the speed at time t: $\text{Speed }=\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}$
Use the definite integral for arc length to find the distance traveled $\displaystyle \int_{a}^{b}{\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}}dt$ . Notice that this is the integral of the speed (rate times time = distance).
The slope of the path is $\displaystyle \frac{dy}{dx}=\frac{{y}'\left( t \right)}{{x}'\left( t \right)}$ . See this post for more on finding the first and second derivatives with respect to x.
Determine when the particle is moving left or right,
Determine when the particle is moving up or down,
Find the extreme position (farthest left, right, up, down, or distance from the origin).
Given the position find the velocity by differentiating; given the velocity find the acceleration by differentiating.
Given the acceleration and the velocity at some point find the velocity by integrating; given the velocity and the position at some point find the position by integrating. These are just initial value differential equation problems (IVP).
Dot product and cross product are not tested on the BC exam, nor are other aspects.

Here are two past post on this topic:

Implicit Differentiation of Parametric Equation

A Vector’s Derivatives

Parametric/Vector Question (Type 8 for BC only)

Posted on March 28, 2017 by Lin McMullin

I have always had the impression that the AP exam assumed that parametric equations and vectors were first studied and developed in a pre-calculus course. In fact, many schools do just that. It would be nice if students knew all about these topics when they started BC calculus. Because of time considerations, this very rich topic is not fully developed in BC calculus.

That said, the parametric/vector equation questions only concern motion in a plane. I will try to address the minimum that students need to know to be successful on the BC exam. Certainly, if you can do more and include a unit in a pre-calculus course do so.

Another concern is that most calculus textbooks jump right to vectors in 3-space while the exam only tests motion in a plane and 2-dimensional vectors. (Actually, the equations and ideas are the same with an extra variable for the z-direction)

The acceleration is given by the vector $\left\langle {{x}'}'\left( t \right),{{y}'}'\left( t \right) \right\rangle$ .

What students should know how to do:

Vectors may be written using parentheses, ( ), or pointed brackets, $\left\langle {} \right\rangle$ , or even $\vec{i},\vec{j}$ form. The pointed brackets seem to be the most popular right now, but all common notations are allowed and will be recognized by readers.
Find the speed at time t: $\text{Speed }=\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}$
Use the definite integral for arc length to find the distance traveled $\displaystyle \int_{a}^{b}{\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}}dt$ . Notice that this is the integral of the speed (rate times time = distance).
The slope of the path is $\displaystyle \frac{dy}{dx}=\frac{{y}'\left( t \right)}{{x}'\left( t \right)}$ . See this post for more on finding the first and second derivatives with respect to x.
Determine when the particle is moving left or right,
Determine when the particle is moving up or down,
Find the extreme position (farthest left, right, up, down, or distance from the origin).
Given the position find the velocity by differentiating; given the velocity find the acceleration by differentiating.
Given the acceleration and the velocity at some point find the velocity by integrating; given the velocity and the position at some point find the position by integrating. These are just initial value differential equation problems (IVP).
Dot product and cross product are not tested on the BC exam, nor are other aspects.

Shorter questions on these ideas appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

Friday March 31: For BC Polar Equations (Type 9)

Tuesday April 4: For BC Sequences and Series.(Type 10)

A Vector’s Derivatives

Posted on January 14, 2015 by Lin McMullin

A question on the AP Calculus Community bulletin board this past Sunday inspired me to write this brief outline of what the derivatives of parametric equations mean and where they come from.

The Position Equation or Position Vector

A parametric equation gives the coordinates of points (x, y) in the plane as functions of a third variable, usually t for time. The resulting graph can be thought of as the locus of a point moving in the plane as a function of time. This is no different than giving the same two functions as a position vector, and both approaches are used. (A position vector has its “tail” at the origin and its “tip” tracing the path as it moves.)

For example, the position of a point on the flange of a railroad wheel rolling on a horizontal track (called a prolate cycloid) is given by the parametric equations

$x\left( t \right)=t-1.5\sin \left( t \right)$

$y\left( t \right)=1-1.5\cos \left( t \right)$ .

Or by the position vector with the same components $\left\langle t-1.5\sin \left( t \right),1-1.5\cos \left( t \right) \right\rangle$ .

Derivatives and the Velocity Vector

The instantaneous rate of change in the y-direction is given by dy/dt, and dx/dt gives the instantaneous rate of change in the x-direction. These are the two components of the velocity vector $\displaystyle \vec{v}\left( t \right)=\left\langle \frac{dx}{dt},\frac{dy}{dt} \right\rangle$ .

In the example, $\displaystyle \vec{v}\left( t \right) =\left\langle 1-1.5\cos \left( t \right),1.5\sin \left( t \right) \right\rangle$ . This is a vector pointing in the direction of motion and whose length, $\displaystyle \sqrt{{{\left( \frac{dx}{dt} \right)}^{2}}+{{\left( \frac{dy}{dt} \right)}^{2}}}$ , is the speed of the moving object.

In the video below the black vector is the position vector and the red vector is the velocity vector. I’ve attached the velocity vector to the tip of the position vector. Notice how the velocitiy’s length as well as its direction changes. The velocity vector pulls the object in the direction it points and there is always tangent to the path. This can be seen when the video pauses at the end and in the two figures at the end of this post.

The slope of the tangent vector is the usual derivative dy/dx. It is found by differentiating dy/dt implicitly with respect to x. Therefore, $\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}$ .

There is no need to solve for t in terms of x since dt/dx is the reciprocal of dx/dt, instead of multiplying by dt/dx we can divide by dx/dt: $\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ .

In the example, $\displaystyle \frac{dy}{dx}=\left( 1.5\sin \left( t \right) \right)\frac{dt}{dx}=\left( 1.5\sin \left( t \right) \right)\left( \frac{1}{1-1.5\cos \left( t \right)} \right)=\frac{1.5\sin \left( t \right)}{1-1.5\cos \left( t \right)}$

Second Derivatives and the Acceleration Vector

The components of the acceleration vector are just the derivatives of the components of the velocity vector

In the example, $\displaystyle \vec{a}\left( t \right)=\left\langle 1.5\sin \left( t \right),1.5\cos \left( t \right) \right\rangle$

The usual second derivative $\displaystyle \frac{{{d}^{2}}y}{d{{x}^{2}}}$ is found by differentiating dy/dx, which is a function of t, implicitly with respect to x:

$\displaystyle \frac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \frac{d}{dt}\left( \frac{dy}{dx} \right) \right)\left( \frac{dt}{dx} \right)=\frac{\frac{d}{dt}\left( dy/dx \right)}{dx/dt}$

In the example,

$\displaystyle \frac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \frac{d}{dt}\left( \frac{1.5\sin \left( t \right)}{1-1.5\cos \left( t \right)} \right) \right)\left( \frac{1}{1-1.5\cos \left( t \right)} \right)$

$\displaystyle =\frac{\left( 1-1.5\cos \left( t \right) \right)\left( 1.5\cos \left( t \right) \right)-\left( 1.5\sin \left( t \right) \right)\left( 1.5\sin \left( t \right) \right)}{{{\left( 1-1.5\cos \left( t \right) \right)}^{2}}}\cdot \frac{1}{\left( 1-1.5\cos \left( t \right) \right)}$

$\displaystyle =\frac{1.5\cos \left( t \right)-{{1.5}^{2}}}{{{\left( 1-1.5\cos \left( t \right) \right)}^{3}}}$

The acceleration vector is the instantaneous rate of change of the velocity vector. You may think of it as pulling the velocity vector in the same way as the velocity vector pulls the moving point (the tip of the position vector). The video below shows the same situation as the first with the acceleration vectors in green attached to the tip of the velocity vector.

Here are two still figures so you can see the relationships. On the left is the starting position t = 0 with the y-axis removed for clarity. At this point the red velocity vector is $\left\langle -0.5,0 \right\rangle$ indicating that the object will start by moving directly left. The green acceleration vector is $\left\langle 0,1.5 \right\rangle$ pulling the velocity and therefore the object directly up. The second figure shows the vectors later in the first revolution. Note that the velocity vector is in the direction of motion and tangent to the path shown in blue.

Roulettes and Calculus

Posted on July 11, 2014 by Lin McMullin

Roulettes – 5: Calculus Considerations.

In the first post of this series Roulette Generators (RG) are explained. Here are the files for Winplot or Geometer’s Sketchpad. Use them to quickly see the graphs of these curves by adjusting one or two parameters.

While writing this series of posts I was intrigued by the cusps that appear in some of the curves. In Cartesian coordinates you think of a cusp as a place where the curves is continuous, but the derivative is undefined, and the tangent line is vertical. Cusps on the curves we have been considering are different.

The equations of the curves formed by a point attached to a circle rolling around a fixed circle in the form we have been using are:

$x\left( t \right)=(1+R)\cos \left( t \right)-S\cos \left( \tfrac{1}{R}t+t \right)$

$y\left( t \right)=\left( 1+R \right)\sin \left( t \right)-S\sin \left( \tfrac{1}{R}t+t \right)$

For example, let’s consider the case with $R=S=\tfrac{1}{3}$

Epicycloid with R = S = 1/3

The equations become

$x\left( t \right)=\frac{4}{3}\cos \left( t \right)-\frac{1}{3}\cos \left( 4t \right)$

$y\left( t \right)=\frac{4}{3}\sin \left( t \right)-\frac{1}{3}\sin \left( 4t \right)$

The derivative is

$\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{\cos \left( t \right)-\cos (4t)}{-\sin \left( t \right)+\sin (4t)}$

The cusps are evenly spaced one-third of the way around the circle and appear at $t=0,\tfrac{2\pi }{3},\tfrac{4\pi }{3}$ . At the cusps dy/dx is an indeterminate form of the type 0/0. (Note that at $t=\tfrac{2\pi }{3}$ , $\cos \left( 4t \right)=\cos \left( \tfrac{8\pi }{3} \right)=\cos \left( \tfrac{2\pi }{3} \right)$ and likewise for the sine.) Since derivatives are limits, we can apply L’Hôpital’s Rule and find that at $t=\tfrac{2\pi }{3}$

$\displaystyle \frac{dy}{dx}=\underset{t\to \tfrac{2\pi }{3}}{\mathop{\lim }}\,\frac{-\sin \left( t \right)+4\sin \left( 4t \right)}{-\cos \left( t \right)+4\cos \left( 4t \right)}=\underset{t\to \tfrac{2\pi }{3}}{\mathop{\lim }}\,\frac{3\sin \left( t \right)}{3\cos \left( t \right)}=\tan \left( \tfrac{2\pi }{3} \right)=-\sqrt{3}$

This is, I hope, exactly what we should expect. As the curve enters and leaves the cusp it is tangent to the line from the cusp to the origin. (The same thing happens at the other two cusps.) At the cusps the moving circle has completed a full revolution and thus, the line from its center to the center of the fixed circle goes through the cusp and has a slope of tan(t).

The cusps will appear where the same t makes dy/dt = 0 and dx/dt =0 simultaneously.

The parametric derivative is defined at the cusp and is the slope of the line from the cusp to the origin. Now I may get an argument on that, but that’s the way it seems to me.

A look at the graph of the derivative in parametric form may help us to see what is going on. In the next figure $R=S=\tfrac{1}{3}$ with the graph of the curve is in blue. The velocity vector is shown twice (arrows). The first is attached to the moving point and shows the direction and its length shows the speed of the movement. The second shows the velocity vector as a position vector (tail at the origin). The orange graph is the path of the velocity vector’s tip – the parametric graph of the velocity. Note that these vectors are the same (i.e. they have the same direction and magnitude)

The video shows the curve moving through the cusp at. Notice that as the graph passes through the cusp the velocity vector changes from pointing down to the right, to the zero vector, to pointing up to the left. The change is continuous and smooth.

Velocity near a cusp.

Here is the whole curve being drawn with its velocity and the velocity vectors.

Epicycloid with velocity vectors

(If you are using the Winplot file you graph the velocity this way. Open the inventory with CTRL+I, scroll down to the bottom and select, one at a time, the last three lines marked “hidden”, and then click on “Graph.”). The Geometer’s Sketchpad version has a button to show the derivative’s graph and the velocity vectors.

The general equation of the derivative (velocity vector) is

$\displaystyle {x}'\left( t \right)=-\left( 1+R \right)\sin \left( t \right)+S\left( \tfrac{1}{R}+1 \right)\sin \left( \tfrac{1}{R}t+t \right)$

$\displaystyle {y}'\left( t \right)=\left( 1+R \right)\cos \left( t \right)-S\left( \tfrac{1}{R}+1 \right)\cos \left( \tfrac{1}{R}t+t \right)$

Notice that the derivative has to same form as a roulette.

Finally, I have to mention how much seeing the graphs in motion have helped me understand, not just the derivatives, but all of the curves in this series and the ones to come. To experiment, to ask “what if … ?” questions, and just to play is what technology should be used for in the classroom. See what your students can find using the RGs.

Exploration and Challenge:

Consider the epitrochoid $x\left( t \right)=\frac{2}{3}\cos \left( t \right)+\frac{1}{3}\cos \left( 2t \right),y\left( t \right)=\frac{2}{3}\sin \left( t \right)-\frac{1}{3}\sin \left( 2t \right)$ .

Find its derivative as a parametric equation and graph it with a graphing program or calculator. (Straight forward)
Are the graph of the derivative and the graph of the rose curve given in polar form by $r\left( t \right)=\frac{4}{3}\sin \left( 3t \right)$ the same? Justify your answer. (Warning: The graphs certainly look the same. I have not been able to do show they are the same (which certainly doesn’t prove anything), so they may not be the same.) Please post your answer using the “Leave a Reply” box at the end of this post.

Next: Roulette Art.

Rolling Circles

Posted on June 24, 2014 by Lin McMullin

A few weeks ago I covered some trigonometry classes for another teacher. They were studying polar and parametric graphs and the common curves limaçon, rose curves, cardioids, etc. I got to thinking about these curves. the next few posts will discuss what I learned. To help me see what was happening I made a Winplot animation. My “Roulette Generator” (RG) is a rather simple setup and turned out to be very well suited to study a variety of curves: cardioids, epicycloids, epitrochoids, hypochoids, and hypotrochoids to name a few. They are all examples of roulettes – curves generated by a point on a curve as it moves around anther curve. I considered only cases where both curves are circles. Calculus will make it appearance after the first few posts in this series. I hope you will find information here that will let you make a good project or investigation for calculus or precalculus students. The RG can be set to graph any number of situations as I will discuss. So here goes.

Roulettes – 1: Equations and the Roulette Generator.

In this post I will discuss the derivation of the parametric equations used to make the animations and some notes on how to use the RG. Later posts will discuss some of the various curves that result. I began with a simple cardioid.

Cardioid R = S = 1

A cardioid is defined as the locus of a point on a circle as it rolls without slipping around another circle with the same radius.

The setup described here will allow us to change the equations using sliders, for this and a number of related curves. The two circles with the same radii are shown in the figure below. The circle with center at C rolls counterclockwise around the circle with center at the origin, O. The point D traces the cardioid. The blue curve from F to D is the beginning of the cardioid. The equations of the circles are: The circle centered at the origin with radius 1:

$x\left( t \right)=\cos \left( t \right)\text{ and }y\left( t \right)=\sin \left( t \right)$

The moving circle with center at C and radius R:

$x\left( t \right)=(1+R)\cos (t)+R\cos (t)$ and

$y\left( t \right)=(1+R)\sin \left( t \right)+R\sin \left( t \right)$ .

The equation of the locus: In the figure above, the locus of the point marked D, as the moving circle rolls counterclockwise around the fixed circle, will be the path of the curves. A small portion of the curve is shown in blue running from F to D. In our investigations we will eventually want to place the moving point inside, on, or outside the moving circle. To do this we will use S as the distance from the center of the moving circle to the point we are watching. For the moment and for the simple cardioid, we will assume they are the same: $R=S=1$ .

The ratio of the radius of the fixed circle to the radius of the moving circle is 1/R, and will be of interest. The ratio can be adjusted by changing R. We can, therefore, keep the fixed circle’s radius constant and equal to one.

We will use vectors to write the locus. Before writing the individual vectors consider first that the arc length on both circles is always the same for all the curves and combinations of their radii:

$\text{arc }EF=t=\text{arc }DE=R\left( \measuredangle DCE \right)$ .

Therefore, $\measuredangle DCE=\tfrac{1}{R}t$ . Then $\measuredangle BCD=\tfrac{\pi }{2}-\left( \tfrac{1}{R}t+t \right)$ Then the locus of D has the vector equation:

$\overrightarrow{OD}=\overrightarrow{OA}+\overrightarrow{AC}+\overrightarrow{CB}+\overrightarrow{BD}$

$\overrightarrow{OD}=\left\langle (1+R)\cos (t),0 \right\rangle +\left\langle 0,\left( 1+R \right)\sin \left( t \right) \right\rangle +$

$\left\langle -S\sin \left( \tfrac{\pi }{2}-(\tfrac{1}{R}t+t) \right),0 \right\rangle +\left\langle 0,-S\cos \left( \tfrac{\pi }{2}-(\tfrac{1}{R}t+t) \right) \right\rangle$

Notice that $\left( \tfrac{1}{R}t+t \right)$ is the complement of $\left( \tfrac{\pi }{2}-\left( \tfrac{1}{R}t+t \right) \right)$ , so that $\sin \left( \tfrac{\pi }{2}-(\tfrac{1}{R}t+t) \right)=\cos \left( \tfrac{1}{R}t+t \right)$ and $\cos \left( \tfrac{\pi }{2}-(\tfrac{1}{R}t+t) \right)=\sin \left( \tfrac{1}{R}t+t \right)$ . The parametric equations of the path are

$x\left( t \right)=(1+R)\cos \left( t \right)-S\cos \left( \tfrac{1}{R}t+t \right)$

$y\left( t \right)=\left( 1+R \right)\sin \left( t \right)-S\sin \left( \tfrac{1}{R}t+t \right)$

You may enter these equations on any graphing calculator by entering specific values of R and S. You will have to re-type them for each different curve. With the RG you can make the changes easily with sliders.

The roulette generators: I used Winplot a free graphing program I’ve used for years. Here are the links so you can download Winplot or Winplot for Macs. the Winplot file containing the generator is here: Winplot Roulette Generator. [Sorry, this is no longer available here and WordPress will not accept Winplot files for download. Please contact me at lnmcmullin@aol.com and I’ll send you a copy.] The Winplot equations are discussed here: Notes on the Roulette Generator.

A Geometer’s Sketchpad version may be downloaded Geometer’s Sketchpad Roulette Generator. You will need Geometer’s Sketchpad to run this on a computer or the “Sketch Explorer” app for iPads and other tablets. A big ‘Thanks” to Audrey Weeks who was kind enough to make this for us. Audrey is the author of the Algebra in Motion and Calculus in Motion software. Audrey is my “go to” person when I have math questions.

Other graphers with sliders such as Geogebra and TI-Nspire will probably work as well. For those who wish to adapt this to some other graphing program there are some syntax consideration to making the one circle roll around the other and showing the path as the circle rolls. If you make your own generator on one of these other platforms, please send it along and I’ll post it and give you credit.

Experiment with the R and S sliders. In the next several posts well will do this and learn about various other curves.

Next: Epicycloids

References:

Cardioids: http://en.wikipedia.org/wiki/Cardioid

Roulettes: http://en.wikipedia.org/wiki/Roulette_(curve)

Algebra in Motion

Calculus in Motion

Parametric and Vector Equations

Posted on March 15, 2013 by Lin McMullin

AP Type Questions 8

Particle moving on a plane for BC – the parametric/vector question.

I have always had the impression that the AP exam assumed that parametric equations and vectors were first studied and developed in a pre-calculus course. In fact many schools do just that. It would be nice if students knew all about these topics when they started BC calculus. Because of time considerations, this very rich topic probably cannot be fully developed in BC calculus. I will try to address here the minimum that students need to know to be successful on the BC exam. Certainly if you can do more and include a unit in a pre-calculus course do so.

Another concern is that most textbooks jump right to vectors in 3-space while the exam only test motion in a plane and 2-dimensional vectors.

In the plane, the position of a moving object as a function of time, t, can be specified by a pair of parametric equations $x=x\left( t \right)\text{ and }y=y\left( t \right)$ or the equivalent vector $\left\langle x\left( t \right),y\left( t \right) \right\rangle$ . The path is the curve traced by the parametric equations.

The velocity of the movement in the x- and y-direction is given by the vector $\left\langle {x}'\left( t \right),{y}'\left( t \right) \right\rangle$ . The vector sum of the components gives the direction of motion. Attached to the tip of the position vector this vector is tangent to the path pointing in the direction of motion. The length of this vector is the speed of the moving object. $\text{Speed }=\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}$

The acceleration is given by the vector $\left\langle {{x}'}'\left( t \right),{{y}'}'\left( t \right) \right\rangle$ .

What students should know how to do

Vectors may be written using parentheses, ( ), or pointed brackets, $\left\langle {} \right\rangle$ , or even $\vec{i},\vec{j}$ form. The pointed brackets seem to be the most popular right now, but any notation is allowed.
Find the speed at time t: $\text{Speed }=\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}$
Use the definite integral for arc length to find the distance traveled $\displaystyle \int_{a}^{b}{\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}}dt$ . Notice that this is the integral of the speed (rate times time = distance).
The slope of the path is $\displaystyle \frac{dy}{dx}=\frac{{y}'\left( t \right)}{{x}'\left( t \right)}$ .
Determine when the particle is moving left or right,
Determine when the particle is moving up or down,
Find the extreme position (farthest left, right, up or down).
Given the position find the velocity by differentiating; given the velocity find the acceleration by differentiating.
Given the acceleration and the velocity at some point find the velocity by integrating; given the velocity and the position at some point find the position by integrating. These are really just initial value differential equation problems (IVP).
Dot product and cross product are not tested on the BC exam.

Shorter questions on these ideas appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

Teaching Calculus

The pleasure lies not in discovering the truth, but in searching for it.

Tag Archives: vector

Parametric Equations and Vectors

Parametric and Vector Equations

Parametric/Vector Question (Type 8 for BC only)

A Vector’s Derivatives

Roulettes and Calculus

Rolling Circles

Parametric and Vector Equations

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