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)

February 2016

Posted on January 31, 2016 by Lin McMullin

As I hope you’ve noticed there is a new pull-down on the navigation bar called “Website.” For some years I’ve had a website at linmcmullin.net that lately I’ve been neglecting. I decided to close it in the next few days, and therefore, I move most of the material that is there to this new tab. The main items of interest are probably those under “Calculus”, “Winplot”, and “CAS.” If you used that website you should be able to find what you need here. If you cannot find something, then please write and I’ll try to help.

In my post entitled January 2016 are listing of post for the applications of integration for both AB and BC calculus. This month’s posts are BC topics on sequences, series, and parametric and polar equations.

Posts from past Februarys

Sequences and Series

February 9, 2015 Amortization A practical application of sequences.

February 8, 2013: Introducing Power Series 1

February 11, 2013: Introducing Power Series 2

February 13, 2013: Introducing Power Series 3

February 15, 2013 New Series from Old 1

February 18, 2013: New Series from Old 2

February 20, 2013: New Series from Old 3

February 22, 2013: Error Bounds

May 20, 2015 The Lagrange Highway

Polar, Parametric, and Vector Equations

March 15, 2013 Parametric and Vector Equations

March 18, 2013 Polar Curves

May 17, 2014 Implicit Differentiation of Parametric Equations

A series on ROULETTES some special parametric curves (BC topic – enrichment):

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 Art – 1

Posted on July 17, 2014 by Lin McMullin

For the last few post, we have been exploring roulettes using the roulette generator (RG) for either Winplot or Geometer’s Sketchpad. These files use the equations

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

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

The derivative is given by the equations

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

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

Notice that the derivative is also a form of roulette.

To generate various roulettes by changing the values of R and S as explained in the first post in this series.

When my friend Audrey Weeks finished making the Sketchpad RG for me she sent these three designs that she made using the generator. Suddenly, we were into art!

: R = 0.713, S = 0.436,
Revolutions = 17

: R = – 0.332, S = +/- 0.3,
Revolutions = 17

: R = 0.506, S = 0.728,
Revolution = 17

The designs may make a nice project for students studying parametric curves and help them learn a little more about the curves and their graphs. Here are some hints about how to use the Winplot RG to make these designs. (Most of the hints will also do for the Sketchpad RG except for changing colors, and generating several curves at once.)

Hint 1: Begin by opening the RG file and saving it with a different name so you can makes the changes and still have the original available. In the new file,

Open the Inventory (CRTL+I) and delete everything except the one marked “Roulette” and the one marked “Velocity Graph” by selecting each and clicking the “delete” button. This will remove the circles and other lines from the final drawings.
Then make a duplicate of the two remaining files by selecting them and clicking “dupl.” For the duplicates, click “edit” and change the sign preceding the S. This will let you draw graphs for S and –S at the same time. More on this below.
Click CTRL+G and turn the axes off.

Hint 2: Number of revolutions: We learned in the first post in this series that from the number R expressed as a reduced fraction $\left| R \right|=\tfrac{n}{d}$ , that d is the number of dips, loops, or cusps in the graph and n full revolutions will draw the entire graph (i. e. after n revolutions the same graph will be redrawn).

When using a large value of d the parts of the graph overlap each other and add to the design. So, use a large d to get a “fancier” graph.

If you use n revolutions your graph will have a number of rotational symmetries offset by $2\pi /n$ radians.

But you also get nice designs by using less than n revolutions. This draws part of the graph and can also make a pleasing design. See the captions to the figures in this post to get an idea of how this works.

Hint #3: Color: You can change the color by selecting the equation in the Inventory list and clicking “edit” and then “color.” The background color can be changed by clicking “Misc” in the top bar and then “Background.”

So, let’s try one. First graph: I chose R = –0.321 and S = 0.440. Since 321/1000 does not reduce, there will be 1000 loops in 321 revolutions. But I graphed only about 6.5 revolutions (A = 40.212). Second Graph: includes the derivative of the first made by using the velocity equation in the Inventory.

: R = –0.321 and S = 0.440,
Revolutions = 6.5

: R = –0.321 and S = 0.440,
Revolutions = 6.5

These were just a few hints to get you started. In the next post we’ll look at some much more fancy designs. Meanwhile try some of your own and post them as comments. (Please include the R, S, and revolution values you used.).

Preview of the next post Roulettes and Art – 2:

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.

Teaching Calculus

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

Tag Archives: parametric equations

Parametric Equations and Vectors

Parametric and Vector Equations

Parametric/Vector Question (Type 8 for BC only)

February 2016

A Vector’s Derivatives

Roulettes and Art – 1

Roulettes and Calculus

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