Fiveable
Fiveable

or

Log in

Find what you need to study


Light

Find what you need to study

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

8 min readjanuary 23, 2023

Sumi Vora

Sumi Vora

Kashvi Panjolia

Kashvi Panjolia

Sumi Vora

Sumi Vora

Kashvi Panjolia

Kashvi Panjolia

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(565).png?alt=media&token=7dc52252-995a-4df2-96bc-e31057973a52

There are many different kinds of functions in math because not everything in the world exists on a plane with two variables. So far, everything we have been doing has been on the Cartesian plane: ℝ^2. This is also known as the xy-plane. However, some functions that model the world around us are better graphed using other types of planes, which we will explore in this unit. This unit makes up 11-12% of the AP Calculus BC Exam.

As you are reading through this guide, pay special attention to the formulas mentioned. This unit is very formula-heavy, and ideally you should have all these formulas memorized, but some of them you can derive on the exam.

9.1 Defining and Differentiating Parametric Equations

Parametric functions are a way to express a relationship between variables in the form of an equation that involves time. We will often use parametric functions to express the position of an object moving in space, or to describe the shape of a curve.

A parametric equation is typically written in the form:

x = f(t) y = g(t)

where x and y are the coordinates of a point on the curve, and t represents time. By changing the value of t, we can trace out the entire curve defined by the . In a parametric function, both the x and y variables are dependent variables, and time is the independent variable.

To find the derivative of a parametric function, we need to find the derivative of x(t) and y(t) and set y'(t) over x'(t). When we do this, the dt's cancel out and we are left with the derivative dy/dx.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1072).png?alt=media&token=6aa39361-ae1a-4311-95ca-d1e78289129b

9.2 Second Derivatives of Parametric Equations

As with equations in the , we can take the second derivative of a parametric function. The process for finding the second derivative is a bit different than the process you are used to. We use the chain rule after finding the first derivative to arrive at this equation for the second :

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1156).png?alt=media&token=b1735297-ce41-4edc-a2be-d557c5f5ef18

Notice how inside the parentheses, the formula states we need to find dy/dx, not dy/dt. This means that to find the second derivative, you must first find the first derivative with respect to x, then take the derivative of the first derivative (usually using the quotient rule), then set all of that over the first derivative of x(t). As you can see, there are quite a few steps involved, but with some practice, you will master second derivatives in no time.

9.3 Finding Arc Lengths of Curves Given by Parametric Equations

The arc length of a function is a measure of the distance along a curve defined by the function. More specifically, it is the length of the curve between two points. Remember that for Cartesian equations, the formula for the arc length of a curve was:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-DVCzVsQx6SbX.PNG?alt=media&token=b4d43d87-7965-4884-b15e-cb8f592a29d3

The same logic still applies to , but the formula looks a bit different since x and y are both dependent variables. This is the formula for the arc length of a parametric equation:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-DaamV9mw7Zmx.PNG?alt=media&token=eb257600-31d8-4375-a9b5-b67eaceed829

In this formula, you are still squaring the derivative of the equation, but since there are two dependent variables, we square the derivatives of both variables. Remember to still take the square root of the sum of the two derivatives and take the integral across your interval. Instead of being "the integral from x=a to x=b," we now say "the " because t is now the independent variable, instead of x like we are used to.

9.4 Defining and Differentiating Vector-Valued Functions

A vector-valued function is a function that maps a real number to a vector in a vector space. It is written in the form

r(t) = <f(t), g(t)> or r(t) = f(t)i + g(t)j

where f(t) and g(t) are real-valued functions and i and j are the in the x and y direction respectively. This can be represented geometrically as a point in space moving in the xy-plane.

For , vector-valued functions are used to represent the position, velocity and acceleration of an object moving in space. All these functions are related to each other in the way that velocity is the first derivative of position and acceleration is the second derivative of position.

In order to , you simply differentiate each of its components individually. For example, to go from this vector-valued function for position:

s(t) = <3x+2, ln(x+9)>

to the vector-valued function for velocity, which is the derivative of the position function, you would simply take the derivative of the x-component and the y-component, but you don't need to combine them:

v(t) = s'(t) = <3, 1/(x+9)>

Remember, the derivative rule for the natural log of a function is u'/u where u is the expression inside the parentheses.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1166).png?alt=media&token=4cf35df2-b81c-4a42-9cac-d9e914545f71

9.5 Integrating Vector-Valued Functions

Integration of vector-valued functions is the process of finding an with respect to a scalar variable. is very similar to differentiating vector-valued functions. You simply integrate each component of the function individually.

For example, let's take the integral of a velocity function that is vector-valued to find the :

v(t) = <2x, 3x^2>

To take this integral, we will integrate the x and y components separately, like this:

s(t) = ∫v(t) = <∫ 2x dx, ∫ 3x^2 dx>

Answer: s(t) = ∫v(t) = <x^2, x^3>

9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions

To solve a motion problem using , we first need to identify the that describe the position of the object. Once we have the , we can use them to find the velocity and acceleration of the object. The is represented by the first derivative of the position vector-valued function with respect to t. The is represented by the second derivative of the position vector-valued function with respect to t.

Let's solve the following problem step by step:

A particle moves in the xy-plane according to the :

x = t^3 - 6t^2 y = 2t^2 - 4t

where t is measured in seconds. Find the position and velocity of the particle at t=2 seconds.

To solve this problem using parametric vector-valued functions, we first need to find the position vector-valued function:

r(t) = <x(t), y(t)> = <t^3 - 6t^2, 2t^2 - 4t>

Next, we need to find the , which is the derivative of the position vector-valued function with respect to t:

v(t) = dr/dt = <dx/dt, dy/dt> = <3t^2 - 12t, 4t - 4>

To find the position at time t = 2 seconds, we substitute t = 2 into the position vector-valued function:

r(2) = <2^3 - 6(2)^2, 2(2)^2 - 4(2)> = <-4, 4>

This means the particle is at x=-4 units and y=4 units at t=2 seconds. Similarly, we can find the velocity of the particle at any given time by substituting the time into the . For example, the velocity of the particle at time t = 2 seconds is:

v(2) = <3(2)^2 - 12(2), 4(2) - 4> = <4, 0>

This means that the particle has a velocity of 4 units/s in the x-direction and 0 units/s in the y-direction at time t = 2 seconds.

9.7 Defining Polar Coordinates and Differentiating in Polar Form

A is a two-dimensional coordinate system in which the position of a point is determined by the distance from the origin (r) and the angle (theta θ) between the positive x-axis and the line connecting the point to the origin, counterclockwise. The of a point (r, θ) in the are represented by an ordered pair of real numbers, where r is the distance from the origin and theta is the angle measured in radians.

A polar function is a function of the form:

r = f(θ)

where r is the distance from the origin to a point on the , and theta is the angle between the positive x-axis and the line connecting the point to the origin.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1185).png?alt=media&token=3ffc2c3a-5eab-4d11-9ded-7e363db131cb

To find the derivative of a polar function, we can use the chain rule to derive a formula. It is helpful to memorize the formula, but you can also derive it during the test.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1199).png?alt=media&token=11a80644-00e0-4f67-881f-e9f62f53e624

You can also convert between a polar function and a Cartesian function. To go from polar to Cartesian, use the first two formulas, and to go from Cartesian to polar, use the third formula.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1183).png?alt=media&token=c9db3c65-7ad8-4c6d-baaa-96529a43e114

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

The area of the region enclosed by a polar curve is given by the definite integral:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1205).png?alt=media&token=91d91005-db7b-41f2-a8d4-3b8be30c9661

Where a and b are the and r is the polar function. This integral is calculated by taking the product of 1/2 and the square of the polar function, and then integrating this expression with respect to theta from a to b.

It's important to note that this method of finding the area under a polar curve is only valid for , meaning the curve starts and ends at the same point. If it's not a closed curve, we have to find the area enclosed by the curve and a line connecting the start and the end of the curve.

9.9 Finding the Area of the Region Bounded by Two Polar Curves

The area of the region enclosed by two polar curves is given by the definite integral:

A = (1/2) ∫(a,b) (R^2 - r^2) dθ

Where a and b are the , R is the equation of the and r is the equation of the . This integral is calculated by taking the difference of the square of the and the square of the and then integrating this expression with respect to theta from a to b.

This is similar to finding the integral between two curves in the . Where you subtracted the bottom curve from the top curve, you'll now subtract the from the .

https://i.stack.imgur.com/Ds9oG.png

Image courtesy of Math Stack Exchange.

Key Terms to Review (21)

Acceleration vector-valued function

: An acceleration vector-valued function describes how quickly an object's velocity changes over time. It includes both magnitude and direction.

Antiderivative of a Vector-Valued Function

: Finding an antiderivative (or integral) for each component of a vector-valued function independently yields an antiderivative for that entire function. It allows us to find the original vector-valued function given its derivative.

Area of the Region Bounded by Two Polar Curves

: The area of the region bounded by two polar curves is the total amount of space enclosed between these curves. It represents the area within a specific range of angles and radii in a polar coordinate system.

Cartesian plane

: The Cartesian plane, also known as ℝ^2 or the xy-plane, is a two-dimensional coordinate system that uses two perpendicular number lines called axes to locate points. The horizontal axis is called the x-axis and the vertical axis is called the y-axis.

Closed Curves

: Closed curves are continuous curves that form closed loops without any endpoints. They start and end at the same point, creating shapes like circles, ellipses, or closed polygons.

Derivative of a parametric function

: The derivative of a parametric function represents the rate at which the y-coordinate changes with respect to the x-coordinate. It measures how fast the curve is changing at any given point.

Differentiate a vector-valued function

: Differentiating a vector-valued function involves finding the derivative of each component of the vector separately. It measures how the vector changes with respect to its independent variable.

Displacement of the object

: The displacement of an object refers to the change in its position from one point to another. It is a vector quantity that includes both magnitude (distance) and direction.

Inner curve

: The inner curve refers to the boundary or edge of a shape or graph that is closest to the center. It represents the minimum values or points on a graph.

Integral from t=a to t=b

: The integral from t=a to t=b represents the area under a curve between two points on the x-axis. It calculates the total accumulation of a quantity over a given interval.

Integrating Vector-Valued Functions

: Integrating a vector-valued function involves finding the antiderivative (or integral) of each component of the vector separately. It calculates the area under the curve represented by the vector.

Key Term: r = f(θ)

: Definition: In polar coordinates, r = f(θ) represents a relationship between the distance from the origin (r) and the angle (θ). It describes a curve in the polar plane.

Key Term: y = g(t)

: Definition: In calculus, y = g(t) represents a function where the value of y depends on the value of t. It is a way to express how one quantity changes with respect to another.

Limits of Integration

: The limits of integration are the values that determine the range over which an integral is evaluated. They specify the starting and ending points on the x-axis for finding the area or calculating other quantities using integration.

Outer curve

: The outer curve refers to the boundary or edge of a shape or graph that is farthest away from the center. It represents the maximum values or points on a graph.

Parametric Equations

: Parametric equations are a set of equations that express the coordinates of points on a curve or surface in terms of one or more parameters. They allow us to represent complex shapes and motions by breaking them down into simpler components.

Polar Coordinates

: Polar coordinates are a two-dimensional coordinate system used to locate points in space using radial distance (r) and angular displacement (θ) from a reference point called the pole.

Polar Form

: The polar form is an alternative way to represent complex numbers using their magnitude (r) and argument (θ). It allows for easier multiplication, division, exponentiation, and root extraction of complex numbers.

Polar plane

: The polar plane is a coordinate system that represents points using a distance (r) and an angle (θ) from the origin. It is an alternative to the Cartesian coordinate system.

Unit vectors

: Unit vectors are vectors that have a magnitude of 1. They are used to represent direction and can be added or subtracted to other vectors.

Velocity vector-valued function

: A velocity vector-valued function represents the rate of change of an object's position with respect to time. It provides both the magnitude and direction of the object's velocity at any given time.

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

8 min readjanuary 23, 2023

Sumi Vora

Sumi Vora

Kashvi Panjolia

Kashvi Panjolia

Sumi Vora

Sumi Vora

Kashvi Panjolia

Kashvi Panjolia

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(565).png?alt=media&token=7dc52252-995a-4df2-96bc-e31057973a52

There are many different kinds of functions in math because not everything in the world exists on a plane with two variables. So far, everything we have been doing has been on the Cartesian plane: ℝ^2. This is also known as the xy-plane. However, some functions that model the world around us are better graphed using other types of planes, which we will explore in this unit. This unit makes up 11-12% of the AP Calculus BC Exam.

As you are reading through this guide, pay special attention to the formulas mentioned. This unit is very formula-heavy, and ideally you should have all these formulas memorized, but some of them you can derive on the exam.

9.1 Defining and Differentiating Parametric Equations

Parametric functions are a way to express a relationship between variables in the form of an equation that involves time. We will often use parametric functions to express the position of an object moving in space, or to describe the shape of a curve.

A parametric equation is typically written in the form:

x = f(t) y = g(t)

where x and y are the coordinates of a point on the curve, and t represents time. By changing the value of t, we can trace out the entire curve defined by the . In a parametric function, both the x and y variables are dependent variables, and time is the independent variable.

To find the derivative of a parametric function, we need to find the derivative of x(t) and y(t) and set y'(t) over x'(t). When we do this, the dt's cancel out and we are left with the derivative dy/dx.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1072).png?alt=media&token=6aa39361-ae1a-4311-95ca-d1e78289129b

9.2 Second Derivatives of Parametric Equations

As with equations in the , we can take the second derivative of a parametric function. The process for finding the second derivative is a bit different than the process you are used to. We use the chain rule after finding the first derivative to arrive at this equation for the second :

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1156).png?alt=media&token=b1735297-ce41-4edc-a2be-d557c5f5ef18

Notice how inside the parentheses, the formula states we need to find dy/dx, not dy/dt. This means that to find the second derivative, you must first find the first derivative with respect to x, then take the derivative of the first derivative (usually using the quotient rule), then set all of that over the first derivative of x(t). As you can see, there are quite a few steps involved, but with some practice, you will master second derivatives in no time.

9.3 Finding Arc Lengths of Curves Given by Parametric Equations

The arc length of a function is a measure of the distance along a curve defined by the function. More specifically, it is the length of the curve between two points. Remember that for Cartesian equations, the formula for the arc length of a curve was:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-DVCzVsQx6SbX.PNG?alt=media&token=b4d43d87-7965-4884-b15e-cb8f592a29d3

The same logic still applies to , but the formula looks a bit different since x and y are both dependent variables. This is the formula for the arc length of a parametric equation:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-DaamV9mw7Zmx.PNG?alt=media&token=eb257600-31d8-4375-a9b5-b67eaceed829

In this formula, you are still squaring the derivative of the equation, but since there are two dependent variables, we square the derivatives of both variables. Remember to still take the square root of the sum of the two derivatives and take the integral across your interval. Instead of being "the integral from x=a to x=b," we now say "the " because t is now the independent variable, instead of x like we are used to.

9.4 Defining and Differentiating Vector-Valued Functions

A vector-valued function is a function that maps a real number to a vector in a vector space. It is written in the form

r(t) = <f(t), g(t)> or r(t) = f(t)i + g(t)j

where f(t) and g(t) are real-valued functions and i and j are the in the x and y direction respectively. This can be represented geometrically as a point in space moving in the xy-plane.

For , vector-valued functions are used to represent the position, velocity and acceleration of an object moving in space. All these functions are related to each other in the way that velocity is the first derivative of position and acceleration is the second derivative of position.

In order to , you simply differentiate each of its components individually. For example, to go from this vector-valued function for position:

s(t) = <3x+2, ln(x+9)>

to the vector-valued function for velocity, which is the derivative of the position function, you would simply take the derivative of the x-component and the y-component, but you don't need to combine them:

v(t) = s'(t) = <3, 1/(x+9)>

Remember, the derivative rule for the natural log of a function is u'/u where u is the expression inside the parentheses.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1166).png?alt=media&token=4cf35df2-b81c-4a42-9cac-d9e914545f71

9.5 Integrating Vector-Valued Functions

Integration of vector-valued functions is the process of finding an with respect to a scalar variable. is very similar to differentiating vector-valued functions. You simply integrate each component of the function individually.

For example, let's take the integral of a velocity function that is vector-valued to find the :

v(t) = <2x, 3x^2>

To take this integral, we will integrate the x and y components separately, like this:

s(t) = ∫v(t) = <∫ 2x dx, ∫ 3x^2 dx>

Answer: s(t) = ∫v(t) = <x^2, x^3>

9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions

To solve a motion problem using , we first need to identify the that describe the position of the object. Once we have the , we can use them to find the velocity and acceleration of the object. The is represented by the first derivative of the position vector-valued function with respect to t. The is represented by the second derivative of the position vector-valued function with respect to t.

Let's solve the following problem step by step:

A particle moves in the xy-plane according to the :

x = t^3 - 6t^2 y = 2t^2 - 4t

where t is measured in seconds. Find the position and velocity of the particle at t=2 seconds.

To solve this problem using parametric vector-valued functions, we first need to find the position vector-valued function:

r(t) = <x(t), y(t)> = <t^3 - 6t^2, 2t^2 - 4t>

Next, we need to find the , which is the derivative of the position vector-valued function with respect to t:

v(t) = dr/dt = <dx/dt, dy/dt> = <3t^2 - 12t, 4t - 4>

To find the position at time t = 2 seconds, we substitute t = 2 into the position vector-valued function:

r(2) = <2^3 - 6(2)^2, 2(2)^2 - 4(2)> = <-4, 4>

This means the particle is at x=-4 units and y=4 units at t=2 seconds. Similarly, we can find the velocity of the particle at any given time by substituting the time into the . For example, the velocity of the particle at time t = 2 seconds is:

v(2) = <3(2)^2 - 12(2), 4(2) - 4> = <4, 0>

This means that the particle has a velocity of 4 units/s in the x-direction and 0 units/s in the y-direction at time t = 2 seconds.

9.7 Defining Polar Coordinates and Differentiating in Polar Form

A is a two-dimensional coordinate system in which the position of a point is determined by the distance from the origin (r) and the angle (theta θ) between the positive x-axis and the line connecting the point to the origin, counterclockwise. The of a point (r, θ) in the are represented by an ordered pair of real numbers, where r is the distance from the origin and theta is the angle measured in radians.

A polar function is a function of the form:

r = f(θ)

where r is the distance from the origin to a point on the , and theta is the angle between the positive x-axis and the line connecting the point to the origin.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1185).png?alt=media&token=3ffc2c3a-5eab-4d11-9ded-7e363db131cb

To find the derivative of a polar function, we can use the chain rule to derive a formula. It is helpful to memorize the formula, but you can also derive it during the test.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1199).png?alt=media&token=11a80644-00e0-4f67-881f-e9f62f53e624

You can also convert between a polar function and a Cartesian function. To go from polar to Cartesian, use the first two formulas, and to go from Cartesian to polar, use the third formula.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1183).png?alt=media&token=c9db3c65-7ad8-4c6d-baaa-96529a43e114

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

The area of the region enclosed by a polar curve is given by the definite integral:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(1205).png?alt=media&token=91d91005-db7b-41f2-a8d4-3b8be30c9661

Where a and b are the and r is the polar function. This integral is calculated by taking the product of 1/2 and the square of the polar function, and then integrating this expression with respect to theta from a to b.

It's important to note that this method of finding the area under a polar curve is only valid for , meaning the curve starts and ends at the same point. If it's not a closed curve, we have to find the area enclosed by the curve and a line connecting the start and the end of the curve.

9.9 Finding the Area of the Region Bounded by Two Polar Curves

The area of the region enclosed by two polar curves is given by the definite integral:

A = (1/2) ∫(a,b) (R^2 - r^2) dθ

Where a and b are the , R is the equation of the and r is the equation of the . This integral is calculated by taking the difference of the square of the and the square of the and then integrating this expression with respect to theta from a to b.

This is similar to finding the integral between two curves in the . Where you subtracted the bottom curve from the top curve, you'll now subtract the from the .

https://i.stack.imgur.com/Ds9oG.png

Image courtesy of Math Stack Exchange.

Key Terms to Review (21)

Acceleration vector-valued function

: An acceleration vector-valued function describes how quickly an object's velocity changes over time. It includes both magnitude and direction.

Antiderivative of a Vector-Valued Function

: Finding an antiderivative (or integral) for each component of a vector-valued function independently yields an antiderivative for that entire function. It allows us to find the original vector-valued function given its derivative.

Area of the Region Bounded by Two Polar Curves

: The area of the region bounded by two polar curves is the total amount of space enclosed between these curves. It represents the area within a specific range of angles and radii in a polar coordinate system.

Cartesian plane

: The Cartesian plane, also known as ℝ^2 or the xy-plane, is a two-dimensional coordinate system that uses two perpendicular number lines called axes to locate points. The horizontal axis is called the x-axis and the vertical axis is called the y-axis.

Closed Curves

: Closed curves are continuous curves that form closed loops without any endpoints. They start and end at the same point, creating shapes like circles, ellipses, or closed polygons.

Derivative of a parametric function

: The derivative of a parametric function represents the rate at which the y-coordinate changes with respect to the x-coordinate. It measures how fast the curve is changing at any given point.

Differentiate a vector-valued function

: Differentiating a vector-valued function involves finding the derivative of each component of the vector separately. It measures how the vector changes with respect to its independent variable.

Displacement of the object

: The displacement of an object refers to the change in its position from one point to another. It is a vector quantity that includes both magnitude (distance) and direction.

Inner curve

: The inner curve refers to the boundary or edge of a shape or graph that is closest to the center. It represents the minimum values or points on a graph.

Integral from t=a to t=b

: The integral from t=a to t=b represents the area under a curve between two points on the x-axis. It calculates the total accumulation of a quantity over a given interval.

Integrating Vector-Valued Functions

: Integrating a vector-valued function involves finding the antiderivative (or integral) of each component of the vector separately. It calculates the area under the curve represented by the vector.

Key Term: r = f(θ)

: Definition: In polar coordinates, r = f(θ) represents a relationship between the distance from the origin (r) and the angle (θ). It describes a curve in the polar plane.

Key Term: y = g(t)

: Definition: In calculus, y = g(t) represents a function where the value of y depends on the value of t. It is a way to express how one quantity changes with respect to another.

Limits of Integration

: The limits of integration are the values that determine the range over which an integral is evaluated. They specify the starting and ending points on the x-axis for finding the area or calculating other quantities using integration.

Outer curve

: The outer curve refers to the boundary or edge of a shape or graph that is farthest away from the center. It represents the maximum values or points on a graph.

Parametric Equations

: Parametric equations are a set of equations that express the coordinates of points on a curve or surface in terms of one or more parameters. They allow us to represent complex shapes and motions by breaking them down into simpler components.

Polar Coordinates

: Polar coordinates are a two-dimensional coordinate system used to locate points in space using radial distance (r) and angular displacement (θ) from a reference point called the pole.

Polar Form

: The polar form is an alternative way to represent complex numbers using their magnitude (r) and argument (θ). It allows for easier multiplication, division, exponentiation, and root extraction of complex numbers.

Polar plane

: The polar plane is a coordinate system that represents points using a distance (r) and an angle (θ) from the origin. It is an alternative to the Cartesian coordinate system.

Unit vectors

: Unit vectors are vectors that have a magnitude of 1. They are used to represent direction and can be added or subtracted to other vectors.

Velocity vector-valued function

: A velocity vector-valued function represents the rate of change of an object's position with respect to time. It provides both the magnitude and direction of the object's velocity at any given time.


© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.


© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.