# Visualizing Unit 9

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

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

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

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 ($\displaystyle a), or outside the wheel ($\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.

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# Motion Problems: Same Thing, Different Context

Calculus is about things that are changing. Certainly, things that move are changing, changing their position, velocity, and acceleration. Most calculus textbooks deal with things being dropped or thrown up into the air. This is called uniformly accelerated motion since the acceleration is due to gravity and is constant. While this is a good place to start, the problems are by their nature somewhat limited. Students often know all about uniformly accelerated motion from their physics class.

The Advanced Placement exams take motion problems to a new level. AB students often encounter particles moving along the x-axis or the y-axis (i.e. on a number line) according to a function that gives the particle’s position, velocity, or acceleration.  BC students often encounter particles moving around the plane with their coordinates given by parametric equations or their velocity given by a vector. Other times the information is given as a graph or even in a table of the position or velocity. The “particle” may become a car, or a rocket or even chief readers riding bicycles.

While these situations may not be all that “real”, they provide excellent ways to ask both differentiation and integration questions. but be aware that they are not covered that much in some textbooks; supplementing the text may be necessary.

The main derivative ideas are that velocity is the first derivative of the position function, acceleration is the second derivative of the position function and the first derivative of the velocity. Speed is the absolute value of velocity. (There will be more about speed in the next post.) The same techniques used to find the features of a graph can be applied to motion problems to determine things about the moving particle.

So, the ideas are not new, but the vocabulary is. The table below gives the terms used with graph analysis and the corresponding terms used in motion problem.

Vocabulary: Working with motion equations (position, velocity, acceleration) you really do all the same things as with regular functions and their derivatives. Help students see that while the vocabulary is different, the concepts are the same.

Function                                Linear Motion
Value of a function at x               position at time t
First derivative                            velocity
Second derivative                       acceleration
Increasing                                   moving to the right or up
Decreasing                                 moving to the left or down
Absolute Maximum                    farthest right
Absolute Minimum                     farthest left
yʹ = 0                                        “at rest”
yʹ changes sign                          object changes direction
Increasing & cc up                     speed is increasing
Increasing & cc down                speed is decreasing
Decreasing & cc up                   speed is decreasing
Decreasing & cc down              speed is increasing
Speed                                       absolute value of velocity

Here is a short quiz on this idea.

Revised and updated from a post originally published on November 16, 2012

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.

# Type 2 Questions: Linear Motion

We continue the discussion of the various type questions on the AP Calculus Exams with linear motion questions.

“A particle (or car, or bicycle) moves on a number line ….”

These questions may give the position equation, the velocity equation (most often), or the acceleration equation of something that is moving, along with an initial condition. The questions ask for information about motion of the particle: its direction, when it changes direction, its maximum position in one direction (farthest left or right), its speed, etc.

The particle may be a “particle,” a person, car, a rocket, etc.  Particles don’t really move in this way, so the equation or graph should be considered to be a model. The question is a versatile way to test a variety of calculus concepts since the position, velocity, or acceleration may be given as an equation, a graph, or a table; be sure to use examples of all three forms during the review.

Many of the concepts related to motion problems are the same as those related to function and graph analysis (Type 3). Stress the similarities and show students how the same concepts go by different names. For example, finding when a particle is “farthest right” is the same as finding the when a function reaches its “absolute maximum value.” See my post for November 16, 2012 for a list of these corresponding terms. There is usually one free-response question and three or more multiple-choice questions on this topic.

The position, s(t), is a function of time. The relationships are:

• The velocity is the derivative of the position, ${s}'\left( t \right)=v\left( t \right)$. Velocity is has direction (indicated by its sign) and magnitude. Technically, velocity is a vector; the term “vector” will not appear on the AB exam.
• Speed is the absolute value of velocity; it is a number, not a vector. See my post for November 19, 2012.
• Acceleration is the derivative of velocity and the second derivative of position, $\displaystyle a\left( t \right)={v}'\left( t \right)={{s}'}'\left( t \right)$. It, too, has direction and magnitude and is a vector.
• Velocity is the antiderivative of the acceleration
• Position is the antiderivative of velocity.

What students should be able to do:

• Understand and use the relationships above.
• Distinguish between position at some time and the total distance traveled during the time period.
• The total distance traveled is the definite integral of the speed (absolute value of velocity) $\displaystyle \int_{a}^{b}{\left| v\left( t \right) \right|}\,dt$.
• The net distance traveled, displacement, is the definite integral of the velocity (rate of change): $\displaystyle \int_{a}^{b}{v\left( t \right)}\,dt$. Note that “displacement” has not been used preciously on AP exam, but (as per the new Course and Exam Description) may be used now. Be sure your students know this term.
• The final position is the initial position plus the displacement (definite integral of the rate of change from xa to x = t): $\displaystyle s\left( t \right)=s\left( a \right)+\int_{a}^{t}{v\left( x \right)}\,dx$ Notice that this is an accumulation function equation (Type 1).
• Initial value differential equation problems: given the velocity or acceleration with initial condition(s) find the position or velocity. These are easily handled with the accumulation equation in the bullet above.
• Find the speed at a given time. The speed is the absolute value of the velocity.
• Find average speed, velocity, or acceleration
• Determine if the speed is increasing or decreasing.
• If at some time, the velocity and acceleration have the same sign then the speed is increasing.If they have different signs the speed is decreasing.
• If the velocity graph is moving away from (towards) the t-axis the speed is increasing (decreasing). See my post for November 19, 2012.
• There is also a worksheet on speed here
• THe analytic approach to speed: A Note on Speed
• Use a difference quotient to approximate derivative.
• Riemann sum approximations.
• Units of measure.
• Interpret meaning of a derivative or a definite integral in context of the problem

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

This may be an AB or BC question. The BC topic of motion in a plane, (Type 8: parametric equations and vectors) will be discussed in a later post.

Free-response examples:

• Equation stem 2017 AB 5,
• Graph stem: 2009 AB1/BC1,
• Table stem 2015 AB 3/BC3

Multiple-choice examples from non-secure exams:

• 2012 AB 6, 16, 28, 79, 83, 89
• 2012 BC 2, 89

Updated January 31, 2019, March 12, 2021

# Speed

Speed is the absolute value of velocity: speed = $\left| v\left( t \right) \right|$ .

This is the definition of speed, but hardly enough to be sure students know about speed and its relationship to velocity and acceleration.

Velocity is a vector quantity; that is, it has both a direction and a magnitude. The magnitude of velocity vector is the speed. Speed is a non-negative number and has no direction associated with it. Velocity has a magnitude and a direction. Speed has the same value and units as velocity; speed is a number.

The question that seems to trouble students the most is to determine whether the speed is increasing or decreasing. The short answer is

Speed is increasing when the velocity and acceleration have the same sign.

Speed is decreasing when the velocity and acceleration have different signs.

You should demonstrate this in some real context, such as driving a car (see below). Also, you can explain it graphically.

The figure below shows the graph of the velocity $v\left( t \right)$ (blue graph) of a particle moving on the interval $0\le t\le f$. The red graph is $\left| v\left( t \right) \right|$, the speed. The sections where $v\left( t \right)<0$ are reflected over the x-axis. (The graphs overlap on [b, d].) It is now quite east to see that the speed is increasing on the intervals [0,a], [b, c] and [d,e].

Another way of approaching the concept is this: the speed is the non-directed length of the vertical segment from the velocity’s graph to the t-axis. Picture the segment shown moving across the graph. When it is getting longer (either above or below the t-axis) the speed increases.

Thinking of the speed as the non-directed distance from the velocity to the axis makes answering the two questions below easy:

1. What are the values of t at which the speed obtains its (local) maximum values? Answer: x = a, c, and e.
2. When do the minimum speeds occur?  What are they? Answer: the speed is zero at b and d

Students often benefit from a verbal explanation of all this. Picture a car moving along a road going forwards (in the positive direction) its velocity is positive.

• If you step on the gas, acceleration pulls you in the direction you are moving and your speed increases. (v > 0, a > 0, speed increases)
• Going too fast is not good, so you put on your brakes, you now accelerate in the opposite direction (decelerate?), but you are still moving forward, but slower. (v > 0, a < 0, speed decreases)
• Finally, you stop. Then you shift into reverse and start moving backwards (negative velocity) and you push on the gas to accelerate in the negative direction, so your speed increases. (v < 0, a < 0, speed increases)
• Then you put on the breaks (accelerate in the positive direction) and your speed decreases again. (v < 0, a > 0, speed decreases)

Here is an activity that will help your students discover this relationship. Give Part 1 to half the class and Part 2 to the other half. Part 3 (on the back of Part 1 and Part 2) is the same for both groups.  – Added 12-19-17

Also see: A Note on Speed for the purely analytic approach.

Update: “A Note on Speed” added 4-21-2018

# Motion Problems: Same Thing, Different Context

Calculus is about things that are changing. Certainly, things that move are changing, changing their position, velocity and acceleration. Most calculus textbooks deal with things being dropped or thrown up into the air. This is called uniformly accelerated motion since the acceleration is due to gravity and is constant. While this is a good place to start, the problems are by their nature, somewhat limited. Students often know all about uniformly accelerated motion from their physics class.

The Advanced Placement exams take motion problems to a new level. AB students often encounter particles moving along the x-axis or the y-axis (i.e. on a number line) according to some function that gives the particle’s position, velocity or acceleration.  BC students often encounter particles moving around the plane with their coordinates given by parametric equations or its velocity given by a vector. Other times the information is given as a graph or even in a table of the position or velocity. The “particle” may become a car, or a rocket or even chief readers riding bicycles.

While these situations may not be all that “real”, they provide excellent ways to ask both differentiation and integration questions. but be aware that they are not covered that much in some textbooks; supplementing the text may be necessary.

The main derivative ideas are that velocity is the first derivative of the position function, acceleration is the second derivative of the position function and the first derivative of the velocity. Speed is the absolute value of velocity. (There will be more about speed in the next post.) The same techniques used to find the features of a graph can be applied to motion problems to determine things about the moving particle.

So the ideas are not new, but the vocabulary is. The table below gives the terms used with graph analysis and the corresponding terms used in motion problem.

Vocabulary: Working with motion equations (position, velocity, acceleration) you really do all the same things as with regular functions and their derivatives. Help students see that while the vocabulary is different, the concepts are the same.

Function                                Linear Motion
Value of a function at x               position at time t
First derivative                            velocity
Second derivative                       acceleration
Increasing                                   moving to the right or up
Decreasing                                 moving to the left or down
Absolute Maximum                    farthest right
Absolute Minimum                     farthest left
yʹ = 0                                         “at rest”
yʹ changes sign                          object changes direction
Increasing & cc up                     speed is increasing
Increasing & cc down                speed is decreasing
Decreasing & cc up                   speed is decreasing
Decreasing & cc down              speed is increasing
Speed                                       absolute value of velocity