Parametric and Vector Equations (Type 8)

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

Multiple-choice questions from non-secure exams

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

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


Revised March 12, 2021, April 5, and May 14, 2022

Parametric and Vector Equations (Type 8)

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 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.

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

Multiple-choice questions from non-secure exams

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

This question typically covers topics from Unit 9 of the 2019 CED .


 

 

 

 

 

 

Revised March 12, 2021

2019 CED Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions

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 Equations

Vector Valued Functions

Polar Form

Type 8: Parametric and Vector Questions

The parametric/vector equation questions only concern motion in a plane.

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.

When this topic appears on the free-response section of the exam there is no polar equation 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

Multiple-choice questions from non-secure exams

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

Schedule of review postings:

 

 

 

 


 

Parametric Equations and Vectors

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


 

 

 

 

 


 

Implicit Differentiation

Often a relation (an expression in x and y), that has a graph but is not a function, needs to be analyzed. But the relation is not or cannot be solved for y. What to do? The answer is to use the technique of implicit differentiation. Assume there is a way to solve for y and differentiate using the Chain Rule. Whenever you get to the y,“differentiate” it by writing dy/dx. Then solve for dy/dx

Here are several previous posts on this topic and how to go about using it.

Implicit Differentiation

Implicit Differentiation and Inverses

Implicit differentiation of parametric equations   These are BC topics

A Vector’s Derivative  These are BC topics

_____________________________________________________

Parametric and Vector Equations

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