Why Parametric and Vector-Valued Functions?

Posted on January 16, 2024 by Lin McMullin

Parametric and vector-valued functions are a way of writing and graphing more complicated, and therefore more interesting, curves.

Up to now you have been studying functions where the y-coordinates are found evaluating an expression in terms of the x-coordinates. Parametric and vector equations define both the x– and y-coordinates of a curve in terms of a third variable called a parameter, usually represented by t.

The only difference between the parametric and vector representation is the notation. The parametric form gives two equations, one for x and one for y. For example:

$\displaystyle x\left( t \right)=t-1.2\sin (t)$

$\displaystyle y(t)=1-1.2\cos \left( t \right)$

are the parametric equations of a curve called a prolate cycloid. It is the path of a point on the flange of a train wheel as it rolls along a straight track.

The vector form for this same curve is written:

$\displaystyle \left\langle {t-1.2\sin \left( t \right),1-1.2\cos \left( t \right)} \right\rangle$ .

Notice the two coordinates of the vector are the same as the parametric equations.

In BC calculus you will study parametric and vector-valued equations of motion, that is the path of a point moving according to the parametric and vector equation. Since things are moving, they have a velocity – a vector pulling the point in the direction it is moving at any instant. The velocity vector is tangent to the curve and its length is the speed at which the point is moving. Also, they have an acceleration – a vector pulling the velocity vector. These are found, as I hope you suspect, by differentiating. Vector-valued functions may also be integrated.

The illustration shows the path of a point on the flange of a train wheel rolling along a track. Notice that sometimes the point is moving backwards! The Position vector (from the origin to the point) is in black. Its endpoint is the point defined by the parametric or vector equations. The red vector is the velocity vector “pulling” the point to its next position. The green vector is the acceleration vector “pulling” the velocity vector.

While there are many other uses for parametric and vector-valued functions, the BC Calculus Exam only considers motion situations.

Course and Exam Description Unit 9, Sections 9.1 to 9.6. This is a BC only topic.

Visualizing Unit 9

Posted on February 28, 2023 by Lin McMullin

As you probably realize by now, I think graphs, drawing and other visuals are a great aid in teaching and learning mathematics. Desmos is a free graphing app that many teachers and students use to graph and make other illustrations. Demonstrations can be made in advance and shared with students and other teachers.

Recently, I was looking a some material from Unit 9 Parametric Equation, Polar Coordinates, and Vector-Valued Functions, BC topics, from the current AP Calculus CED. I ended up making three new Desmos illustrations for use in this unit. They will also be useful in a precalculus course introducing these topics. Hope you find them helpful.

Polar Graph Demo

Link

You may replace the polar equation with any polar equation you are interested in. There are directions in the demo. Moving the “a” slider will show a ray rotating around the pole. The “a” value is the angle, $\displaystyle \theta$ , in radians between the ray and the polar axis. On the ray is a segment with a point at its end. This segment’s length is $\displaystyle \left| {r\left( \theta \right)} \right|$ . As you rotate the ray you can see the polar graph drawn. When $\displaystyle r(\theta )<0$ the segment extend in the opposite direction from the ray.

This demo may be used to introduce or review how polar equation work. An interesting extension is to enter something for the argument of the function that is not an integer muntiple of $\displaystyle \theta$ and extend the domain past $\displaystyle 2\pi$ , for example $\displaystyle r=2+4\sin (1.2\theta )$

Basic Parametric and Vector Demo

LInk

A parametric equation and the vector equation of the same curve differ only in notation. So, this demo works for both. Following the directions in the demo, you may see the graph being drawn using the “a” slider. You may turn on (1) the position vector and its components, (2) add the velocity vector attached to the moving point and “pulling” it to its new position, and (3) the acceleration vector “pulling” the velocity vector.

You may enter any parametric/vector equation. When you do, you will also have to enter its first and second derivative. Follow the directions in the demonstration.

Cycloids and their vectors

Link

This demo shows the path on a rolling wheel called a cycloid. The “a” slider moves the position of the point on the wheel. The point may be on the rim of the wheel ( $\displaystyle a=r$ , on the interior of the wheel ( $\displaystyle a<r$ ), or outside the wheel ( $\displaystyle a>r$ – think the flange on a train wheel). Use the “u” slider to animate the drawing. The velocity and acceleration vectors are shown; they may be turned off. The velocity vector is tangent to the curve (not to the circle) and seems to “pull” the point along the curve. The acceleration vector “pulls” the velocity vector. The equation in this demo should not be changed.

The last two demonstrations give a good idea of how the velocity and acceleration vectors affect the movement of the point.

Hope you find these helpful.

Parametric and Vector Equations (Type 8)

Posted on April 5, 2022 by Lin McMullin

AP Questions Type 8: Parametric and Vector Equations (BC Only)

The parametric/vector equation questions only concern motion in a plane. Other topics, such as dot product and cross product, are not tested.

In the plane, the position of a moving object as a function of time, t, can be specified by a pair of parametric equations $\displaystyle x=x\left( t \right)\text{ and }y=y\left( t \right)$ or the equivalent vector $\displaystyle \left\langle {x\left( t \right),y\left( t \right)} \right\rangle$ . 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 $\displaystyle \left\langle {{x}'\left( t \right),{y}'\left( t \right)} \right\rangle$ . 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. Speed = $\displaystyle \sqrt{{{{{\left( {{x}'\left( t \right)} \right)}}^{2}}+{{{\left( {{y}'\left( t \right)} \right)}}^{2}}}}$ . (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 speed $\displaystyle =\left| {v\left( t \right)} \right|=\sqrt{{{{{\left( {{x}'\left( t \right)} \right)}}^{2}}}}$ .)

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

What students should know how to do:

Vectors may be written using parentheses, ( ), or pointed brackets, $\displaystyle \left\langle {} \right\rangle$ , or even $\displaystyle \vec{i},\vec{j}$ 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: Speed = $\displaystyle \sqrt{{{{{\left( {{x}'\left( t \right)} \right)}}^{2}}+{{{\left( {{y}'\left( t \right)} \right)}}^{2}}}}$ .
Use the definite integral for arc length to find the distance traveled $\displaystyle \int_{a}^{b}{{\sqrt{{{{{\left( {{x}'\left( t \right)} \right)}}^{2}}+{{{\left( {{y}'\left( t \right)} \right)}}^{2}}}}}}$ . 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.

When this topic appears on the free-response section of the exam there is no polar equation free-response question and vice versa. When not on the free-response section there are one or more multiple-choice questions on parametric equations.

Free-response questions:

2012 BC 2
2016 BC 2
2021 BC 2
2022 BC2 – slope of tangent line, speed, position, total distance traveled
2023 BC 2 – acceleration vector, speed, tangent line total distance traveled.

Multiple-choice questions from non-secure exams

2003 BC 4, 7, 17, 84
2008 BC 1, 5, 28
2012 BC 2

Visualizing Unit 9 Desmos Demonstrations for Polar, Vector and Parametric Curves

This question typically covers topics from Unit 9 of the CED.

Revised March 12, 2021, April 5, and May 14, 2022, March 6, 2023, June 6, 2023

Unit 9 – Parametric Equations, Polar Coordinates, and Vector-Valued Functions

Posted on January 11, 2022 by Lin McMullin

Unit 9 includes all the topics listed in the title. These are BC only topics (CED – 2019 p. 163 – 176). These topics account for about 11 – 12% of questions on the BC exam.

Comments on Prerequisites: In BC Calculus the work with parametric, vector, and polar equations is somewhat limited. I always hoped that students had studied these topics in detail in their precalculus classes and had more precalculus knowledge and experience with them than is required for the BC exam. This will help them in calculus, so see that they are included in your precalculus classes.

Topics 9.1 – 9.3 Parametric Equations

Topic 9.1: Defining and Differentiation Parametric Equations. Finding dy/dx in terms of dy/dt and dx/dt

Topic 9.2: Second Derivatives of Parametric Equations. Finding the second derivative. See Implicit Differentiation of Parametric Equations this discusses the second derivative.

Topic 9.3: Finding Arc Lengths of Curves Given by Parametric Equations.

Topics 9.4 – 9.6 Vector-Valued Functions and Motion in the plane

Topic 9.4 : Defining and Differentiating Vector-Valued Functions. Finding the second derivative. See this A Vector’s Derivatives which includes a note on second derivatives.

Topic 9.5: Integrating Vector-Valued Functions

Topic 9.6: Solving Motion Problems Using Parametric and Vector-Valued Functions. Position, Velocity, acceleration, speed, total distance traveled, and displacement extended to motion in the plane.

Topics 9.7 – 9.9 Polar Equation and Area in Polar Form.

Topic 9.7: Defining Polar Coordinate and Differentiation in Polar Form. The derivatives and their meaning.

Topic 9.8: Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

Topic 9.9: Finding the Area of the Region Bounded by Two Polar Curves. Students should know how to find the intersections of polar curves to use for the limits of integration.

Timing

The suggested time for Unit 9 is about 10 – 11 BC classes of 40 – 50-minutes, this includes time for testing etc.

Previous posts on these topics:

Parametric and Vector Equations

Implicit Differentiation of Parametric Equations

A Vector’s Derivatives

Adapting 2012 BC 2 (A parametric equation question)

Polar Curves

Polar Equations for AP Calculus

Extreme Polar Conditions

Visualizing Unit 9 Desmos Demonstrations for Polar, Vector and Parametric Curves

Implicit Differentiation of Parametric Equations

Posted on October 31, 2021 by Lin McMullin

I’ve never liked memorizing formulas. I would rather know where they came from or be able to tie it to something I already know. One of my least favorite formulas to remember and explain was the formula for the second derivative of a curve given in parametric form. No longer.

If $\displaystyle y=y(t)$ and, $\displaystyle x=x(t)$ then the traditional formulas give

$\displaystyle \frac{{dy}}{{dx}}=\frac{{dy/dt}}{{dx/dt}}$ , and

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

It is that last part, where you divide by $\displaystyle {\frac{{dx}}{{dt}}}$ , that bothers me. Where did the $\displaystyle {\frac{{dx}}{{dt}}}$ come from?

Then it occurred to me that dividing by $\displaystyle {\frac{{dx}}{{dt}}}$ is the same as multiplying by $\displaystyle {\frac{{dt}}{{dx}}}$

It’s just implicit differentiation!

Since $\displaystyle \frac{{dy}}{{dx}}$ is a function of t you must begin by differentiating the first derivative with respect to t. Then treating this as a typical Chain Rule situation and multiplying by $\displaystyle {\frac{{dt}}{{dx}}}$ gives the second derivative. (There is a technical requirement here that given $\displaystyle x=x(t)$ , then its inverse $\displaystyle t={{x}^{{-1}}}\left( x \right)$ exists.)

In fact, if you look at a proof of the formula for the first derivative, that’s what happens there as well:

$\displaystyle \frac{d}{{dx}}y(t)=\frac{{dy}}{{dt}}\cdot \frac{{dt}}{{dx}}=\frac{{dy/dt}}{{dx/dt}}$

The reason you do it this way is that since x is given as a function of t, it may be difficult to solve for t so you can find dt/dx in terms of x. But you don’t have to; just divide by dx/dt which you already know.

Here is an example for both derivatives.

Suppose that $\displaystyle x={{t}^{3}}-3$ and $\displaystyle y=\ln \left( t \right)$

Then $\displaystyle \frac{{dy}}{{dt}}=\frac{1}{t}$ and $\displaystyle \frac{{dx}}{{dt}}=3{{t}^{2}}$ and $\displaystyle \frac{{dt}}{{dx}}=\frac{1}{{3{{t}^{2}}}}$

Then $\displaystyle \frac{{dy}}{{dx}}=\frac{1}{t}\cdot \frac{{dt}}{{dx}}=\frac{1}{t}\cdot \frac{1}{{3{{t}^{2}}}}=\frac{1}{3}{{t}^{{-3}}}$

And $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\left( {\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)} \right)\cdot \frac{{dt}}{{dx}}=\left( {-{{t}^{{-4}}}} \right)\cdot \left( {\frac{1}{{3{{t}^{2}}}}} \right)=-\frac{1}{{3{{t}^{6}}}}$

Yes, it’s the same thing as using the traditional formula, but now I’ll never have to worry about forgetting the formula or being unsure how to explain why you do it this way.

Revised: Correction to last equation 5/18/2014. Revised: 2/8/2016. Originally posted May 5, 2014.

Adapting 2021 BC 2

Posted on August 3, 2021 by Lin McMullin

Seven of nine. This week we continue our look at the 2021 free-response questions with an eye to ways to adapt and expand the questions. Hopefully, you will find ways to use this and other free-response questions to help your students learn more and be better prepared for the exams.

2021 BC 2

This is a Parametric and Vector Equation (Type 8) question and contains topic from Unit 9 of the current Course and Exam Description. The vector equation of the velocity of a particle moving in the xy-plane is given along with the position of the particle at t = 0. No units were given.

The stem for 2021 BC 2 is next. (Note the $\displaystyle \left\langle \text{ } \right\rangle$ notation for vectors. Any of the usual notations may be used by students, but be sure to show them the others in case the one their book usage is different than the exam’s.)

Part (a): Students were asked to find the speed and acceleration of the particle at t = 1.2. This is a calculator active questions and the students were expected, but not actually required, to use their calculator. With their calculator in parametric mode, students should begin by entering the velocity as xt1(t) and yt1(t).

Discussion and ideas for adapting this question:

There is little I can suggest here other than changing the time.
At the given time and other times, you can ask in what direction is the particle moving and which way the acceleration is pulling the velocity.
Ask student to do this without using their calculator. The answer need not be simplified or expressed as a decimal.

Part (b): Asked the students to find the total distance traveled by the particle over a given the time interval. This must be done on a calculator. Be sure your students know how to enter the expression using the already entered values for xt1(t) and yt1(t). The calculator entry should look like this.

$\displaystyle \int_{0}^{{1.2}}{{\sqrt{{{{{\left( {\text{xt}1(t)} \right)}}^{2}}+{{{\left( {\text{yt1}(t)} \right)}}^{2}}}}}}dt$

Discussion and ideas for adapting this question:

Use different intervals.
Discuss the similarities with the number line distance formula. In linear motion, the distance is simply the integral of the absolute value of the velocity. Since $\displaystyle \int_{a}^{b}{{\left| {v\left( t \right)} \right|}}dt=\int_{a}^{b}{{\sqrt{{{{{\left( {v(t)} \right)}}^{2}}}}}}dt$ , this is the same formula reduced to one dimension.

Part (c): The situation is reduced to a one-dimensional problem: students were asked to find the coordinates of the point at which the particle is farthest left and explain why there is no point farthest to the right.

Discussion and ideas for adapting this question:

Discuss how to do this and how students should present their answer and explanation.
Show that this is the same as an extreme value problem and done the same way (i.e., find where the derivative is zero, and show that this is a minimum (farthest left), etc.).
Discuss how you know there is no maximum and interpret this in the context of the equation.

For further exploration. Try graphing the path of the particle. Discuss how to do that with your class. See what they suggest. Here a few approaches.

The first thought may be to integrate the velocity vector as an initial value problem. Unfortunately, this cannot be done. Neither the x-component nor the y-component can be integrated in terms of Elementary Functions. Even WolframAlpha.com is no help.
Having entered the velocity vector as xt1(t) and yt1(t), as suggested above, enter something like this depending on your calculator’s syntax and then graph in a suitable window. Compare the graph with the previous analysis in part (c)?

$\displaystyle \text{x2t}(t)=-2+\int_{0}^{t}{{\text{x1t}(t)dt}}$

$\displaystyle \text{y2t}(t)=5+\int_{0}^{t}{{\text{y2t}(t)dt}}$

You may also try expressing the components of velocity as a Taylor series centered at some positive number, a, not at zero. Integrate that to get an approximation to graph. Be sure to adjust things so the initial point is on the graph. WolframAlpha will help here. The one problem here is that the y-component is not defined for negative numbers. Therefore, zero cannt be then center and the largest the interval of convergence can be is [0, 2a] (Why?) and may not even by that large. This is an interesting approach mathematically but will not help with most of the graph.

Personal opinion: I do not think much of this question because all the first two parts require is entering the formula in your calculator and computing the answer, and the third part is really an AB level question. Just my opinion.

Next week 2021 BC 5

I would be happy to hear your ideas for other ways to use this question. Please use the reply box below to share your ideas.

Parametric Equations, Polar Coordinates, and Vector-Valued Functions – Unit 9

Posted on January 12, 2021 by Lin McMullin

Unit 9 includes all the topics listed in the title. These are BC only topics (CED – 2019 p. 163 – 176). These topics account for about 11 – 12% of questions on the BC exam.