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## Parametric Equations

In this section, you will:

- Parameterize a curve.
- Eliminate the parameter.
- Find a rectangular equation for a curve defined parametrically.
- Find parametric equations for curves defined by rectangular equations.

Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in [link] . At any moment, the moon is located at a particular spot relative to the planet. But how do we write and solve the equation for the position of the moon when the distance from the planet, the speed of the moon’s orbit around the planet, and the speed of rotation around the sun are all unknowns? We can solve only for one variable at a time.

In this section, we will consider sets of equations given by x ( t )

and y ( t )

is the independent variable of time. We can use these parametric equations in a number of applications when we are looking for not only a particular position but also the direction of the movement. As we trace out successive values of t ,

the orientation of the curve becomes clear. This is one of the primary advantages of using parametric equations : we are able to trace the movement of an object along a path according to time. We begin this section with a look at the basic components of parametric equations and what it means to parameterize a curve. Then we will learn how to eliminate the parameter, translate the equations of a curve defined parametrically into rectangular equations, and find the parametric equations for curves defined by rectangular equations.

## Parameterizing a Curve

When an object moves along a curve—or curvilinear path —in a given direction and in a given amount of time, the position of the object in the plane is given by the x- coordinate and the y- coordinate. However, both x

vary over time and so are functions of time. For this reason, we add another variable, the parameter , upon which both x

are dependent functions. In the example in the section opener, the parameter is time, t .

position of the moon at time, t ,

is represented as the function x ( t ) ,

and the y

is represented as the function y ( t ) .

Together, x ( t )

are called parametric equations, and generate an ordered pair ( x ( t ) , y ( t ) ) .

Parametric equations primarily describe motion and direction.

When we parameterize a curve, we are translating a single equation in two variables, such as x

into an equivalent pair of equations in three variables, x , y ,

and t .

One of the reasons we parameterize a curve is because the parametric equations yield more information: specifically, the direction of the object’s motion over time.

When we graph parametric equations, we can observe the individual behaviors of x

and of y .

There are a number of shapes that cannot be represented in the form y = f ( x ) ,

meaning that they are not functions. For example, consider the graph of a circle, given as r 2 = x 2 + y 2 .

Solving for y

gives y = ± r 2 − x 2 ,

or two equations: y 1 = r 2 − x 2

and y 2 = − r 2 − x 2 .

If we graph y 1

and y 2

together, the graph will not pass the vertical line test, as shown in [link] . Thus, the equation for the graph of a circle is not a function.

However, if we were to graph each equation on its own, each one would pass the vertical line test and therefore would represent a function. In some instances, the concept of breaking up the equation for a circle into two functions is similar to the concept of creating parametric equations, as we use two functions to produce a non-function. This will become clearer as we move forward.

Suppose t

is a number on an interval, I .

The set of ordered pairs, ( x ( t ) , y ( t ) ) ,

where x = f ( t )

and y = g ( t ) ,

forms a plane curve based on the parameter t .

The equations x = f ( t )

and y = g ( t )

are the parametric equations.

Parameterize the curve y = x 2 − 1

letting x ( t ) = t .

Graph both equations.

If x ( t ) = t ,

then to find y ( t )

we replace the variable x

with the expression given in x ( t ) .

In other words, y ( t ) = t 2 − 1.

Make a table of values similar to [link] , and sketch the graph.

See the graphs in [link] . It may be helpful to use the TRACE feature of a graphing calculator to see how the points are generated as t

The arrows indicate the direction in which the curve is generated. Notice the curve is identical to the curve of y = x 2 − 1.

Construct a table of values and plot the parametric equations: x ( t ) = t − 3 , y ( t ) = 2 t + 4 ; − 1 ≤ t ≤ 2.

Find a pair of parametric equations that models the graph of y = 1 − x 2 ,

using the parameter x ( t ) = t .

Plot some points and sketch the graph.

If x ( t ) = t

and we substitute t

into the y

equation, then y ( t ) = 1 − t 2 .

Our pair of parametric equations is

To graph the equations, first we construct a table of values like that in [link] . We can choose values around t = 0 ,

from t = − 3

to t = 3.

The values in the x ( t )

column will be the same as those in the t

column because x ( t ) = t .

Calculate values for the column y ( t ) .

The graph of y = 1 − t 2

is a parabola facing downward, as shown in [link] . We have mapped the curve over the interval [ −3 , 3 ] ,

shown as a solid line with arrows indicating the orientation of the curve according to t .

Orientation refers to the path traced along the curve in terms of increasing values of t .

As this parabola is symmetric with respect to the line x = 0 ,

the values of x

are reflected across the y -axis.

Parameterize the curve given by x = y 3 − 2 y .

An object travels at a steady rate along a straight path ( −5 , 3 )

to ( 3 , −1 )

in the same plane in four seconds. The coordinates are measured in meters. Find parametric equations for the position of the object.

The parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. The x -value of the object starts at −5

meters and goes to 3 meters. This means the distance x has changed by 8 meters in 4 seconds, which is a rate of 8 m 4 s ,

or 2 m / s .

We can write the x -coordinate as a linear function with respect to time as x ( t ) = 2 t − 5.

In the linear function template y = m x + b , 2 t = m x

and − 5 = b .

Similarly, the y -value of the object starts at 3 and goes to −1 ,

which is a change in the distance y of −4 meters in 4 seconds, which is a rate of − 4 m 4 s ,

or − 1 m / s .

We can also write the y -coordinate as the linear function y ( t ) = − t + 3.

Together, these are the parametric equations for the position of the object, where x

are expressed in meters and t

represents time:

Using these equations, we can build a table of values for t , x ,

(see [link] ). In this example, we limited values of t

to non-negative numbers. In general, any value of t

can be used.

From this table, we can create three graphs, as shown in [link] .

Again, we see that, in [link] (c), when the parameter represents time, we can indicate the movement of the object along the path with arrows.

## Eliminating the Parameter

In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x

and y .

Eliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameter t

from a set of parametric equations; not every method works for every type of equation. Here we will review the methods for the most common types of equations.

## Eliminating the Parameter from Polynomial, Exponential, and Logarithmic Equations

For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for t .

We substitute the resulting expression for t

into the second equation. This gives one equation in x

Given x ( t ) = t 2 + 1

and y ( t ) = 2 + t ,

eliminate the parameter, and write the parametric equations as a Cartesian equation.

We will begin with the equation for y

because the linear equation is easier to solve for t .

Next, substitute y − 2

in x ( t ) .

The Cartesian form is x = y 2 − 4 y + 5.

This is an equation for a parabola in which, in rectangular terms, x

is dependent on y .

From the curve’s vertex at ( 1 , 2 ) ,

the graph sweeps out to the right. See [link] . In this section, we consider sets of equations given by the functions x ( t )

and y ( t ) ,

where t

is the independent variable of time. Notice, both x

are functions of time; so in general y

is not a function of x .

Given the equations below, eliminate the parameter and write as a rectangular equation for y

as a function * * *

Eliminate the parameter and write as a Cartesian equation: x ( t ) = e − t

and y ( t ) = 3 e t , t > 0.

Isolate e t .

Substitute the expression into y ( t ) .

The Cartesian form is y = 3 x .

The graph of the parametric equation is shown in [link] (a) . The domain is restricted to t > 0.

The Cartesian equation, y = 3 x

is shown in [link] (b) and has only one restriction on the domain, x ≠ 0.

Eliminate the parameter and write as a Cartesian equation: x ( t ) = t + 2

and y ( t ) = log ( t ) .

Solve the first equation for t .

Then, substitute the expression for t

The Cartesian form is y = log ( x − 2 ) 2 .

To be sure that the parametric equations are equivalent to the Cartesian equation, check the domains. The parametric equations restrict the domain on x = t + 2

to t > 0 ;

we restrict the domain on x

to x > 2.

The domain for the parametric equation y = log ( t )

is restricted to t > 0 ;

we limit the domain on y = log ( x − 2 ) 2

to x > 2.

Eliminate the parameter and write as a rectangular equation .

## Eliminating the Parameter from Trigonometric Equations

Eliminating the parameter from trigonometric equations is a straightforward substitution. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem.

First, we use the identities:

Solving for cos t

and sin t ,

Then, use the Pythagorean Theorem:

Substituting gives

Eliminate the parameter from the given pair of trigonometric equations where 0 ≤ t ≤ 2 π

and sketch the graph.

and sin t ,

Next, use the Pythagorean identity and make the substitutions.

The graph for the equation is shown in [link] .

Applying the general equations for conic sections (introduced in Analytic Geometry , we can identify x 2 16 + y 2 9 = 1

as an ellipse centered at ( 0 , 0 ) .

Notice that when t = 0

the coordinates are ( 4 , 0 ) ,

and when t = π 2

the coordinates are ( 0 , 3 ) .

This shows the orientation of the curve with increasing values of t .

Eliminate the parameter from the given pair of parametric equations and write as a Cartesian equation: x ( t ) = 2 cos t

and y ( t ) = 3 sin t .

## Finding Cartesian Equations from Curves Defined Parametrically

When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially “eliminating the parameter.” However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such as x ( t ) = t .

In this case, y ( t )

can be any expression. For example, consider the following pair of equations.

Rewriting this set of parametric equations is a matter of substituting x

for t .

Thus, the Cartesian equation is y = x 2 − 3.

Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations.

Method 1 . First, let’s solve the x

equation for t .

Then we can substitute the result into the y

Now substitute the expression for t

Method 2 . Solve the y

equation for t

and substitute this expression in the x

Make the substitution and then solve for y .

Write the given parametric equations as a Cartesian equation: x ( t ) = t 3

and y ( t ) = t 6 .

## Finding Parametric Equations for Curves Defined by Rectangular Equations

Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy we may use to find the parametric equations is valid if it produces equivalency. In other words, if we choose an expression to represent x ,

and then substitute it into the y

equation, and it produces the same graph over the same domain as the rectangular equation, then the set of parametric equations is valid. If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values for x

as the domain of the rectangular equation, then the graphs will be different.

Find a set of equivalent parametric equations for y = ( x + 3 ) 2 + 1.

An obvious choice would be to let x ( t ) = t .

Then y ( t ) = ( t + 3 ) 2 + 1.

But let’s try something more interesting. What if we let x = t + 3 ?

Then we have

The set of parametric equations is

See [link] .

Access these online resources for additional instruction and practice with parametric equations.

- Introduction to Parametric Equations
- Converting Parametric Equations to Rectangular Form

## Key Concepts

into two equations in three variables, x , y , and t . Often, more information is obtained from a set of parametric equations. See [link] , [link] , and [link] .

an equation in

is the result.

solve one of the equations for

and substitute the expression into the second equation. See [link] , [link] , [link] , and [link] .

in one of the equations, and substitute the expression into the second equation. See [link] .

- There are an infinite number of ways to choose a set of parametric equations for a curve defined as a rectangular equation.

such that the domain of the set of parametric equations remains the same as the original rectangular equation. See [link] .

## Section Exercises

What is a system of parametric equations?

A pair of functions that is dependent on an external factor. The two functions are written in terms of the same parameter. For example, x = f ( t )

and y = f ( t ) .

Some examples of a third parameter are time, length, speed, and scale. Explain when time is used as a parameter.

Explain how to eliminate a parameter given a set of parametric equations.

Choose one equation to solve for t ,

substitute into the other equation and simplify.

What is a benefit of writing a system of parametric equations as a Cartesian equation?

What is a benefit of using parametric equations?

Some equations cannot be written as functions, like a circle. However, when written as two parametric equations, separately the equations are functions.

Why are there many sets of parametric equations to represent on Cartesian function?

For the following exercises, eliminate the parameter t

to rewrite the parametric equation as a Cartesian equation.

or y = 1 − 5 l n ( x 2 )

For the following exercises, rewrite the parametric equation as a Cartesian equation by building an x - y

For the following exercises, parameterize (write parametric equations for) each Cartesian equation by setting x ( t ) = t

or by setting y ( t ) = t .

For the following exercises, parameterize (write parametric equations for) each Cartesian equation by using x ( t ) = a cos t

and y ( t ) = b sin t .

Identify the curve.

Parameterize the line from ( 3 , 0 )

to ( −2 , −5 )

so that the line is at ( 3 , 0 )

at t = 0 ,

and at ( −2 , −5 )

at t = 1.

Parameterize the line from ( −1 , 0 )

to ( 3 , −2 )

so that the line is at ( −1 , 0 )

and at ( 3 , −2 )

Parameterize the line from ( −1 , 5 )

to ( 2 , 3 )

so that the line is at ( −1 , 5 )

and at ( 2 , 3 )

Parameterize the line from ( 4 , 1 )

to ( 6 , −2 )

so that the line is at ( 4 , 1 )

and at ( 6 , −2 )

For the following exercises, use the table feature in the graphing calculator to determine whether the graphs intersect.

yes, at t = 2

For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations.

Find two different sets of parametric equations for y = ( x + 1 ) 2 .

answers may vary: { x ( t ) = t − 1 y ( t ) = t 2 and { x ( t ) = t + 1 y ( t ) = ( t + 2 ) 2

Find two different sets of parametric equations for y = 3 x − 2.

Find two different sets of parametric equations for y = x 2 − 4 x + 4.

answers may vary: , { x ( t ) = t y ( t ) = t 2 − 4 t + 4 and { x ( t ) = t + 2 y ( t ) = t 2

are both dependent

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Precalculus by Richard Wright

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This book is available to download as an epub.

For since death came through a man, the resurrection of the dead comes also through a man. 1 Corinthians 15:21 NIV

## 7-06 Parametric Equations

- Graph parametric equations.
- Write parametric equations.

Eliminate the parameter.

SDA NAD Content Standards (2018): PC.5.2, PC.6.7

In physics, projectiles are objects that move only under the influence of gravity. Near the earth, the path of a projectile is a parabola that opens down. While it is possible to write a function to model the motion, it is more useful to have separate functions for the horizontal motion and the vertical motion. Both these functions are based on time.

$$\left\{\begin{align}x &= v_{x0} t\\y &= -9.8 t^2 + v_{y0} t + y_0\end{align}\right.$$

## Parametric Equations

Parametric equations are separate equations for each of the dimensions. The variable in the equations is called the parameter and is often t or θ. To graph parametric equations, make a table of values where you choose values of the parameter and calculate x and y .

## Graph Parametric Equations

- Make a table of values.
- Choose values of t (or θ).
- Calculate x and y from the equations.
- Graph the sets of x and y values as coordinates ( x , y ).
- Draw a curve through the points.

## Example 1: Graph Parametric Equations

Graph the curve given by

$$\left\{\begin{align}x &= t^2 - 1\\ y &= t - 2\end{align}\right.$$

Start by making a table by picking values of t and calculating x and y .

Now graph the points formed from corresponding x and y coordinates.

## Example 2: Graph Parametric Equations

Graph \(\left\{\begin{align}x &= 3\sin θ\\ y &= \cos θ\end{align}\right.\).

Start by making a table by picking values of θ and calculating x and y .

Graph \(\left\{\begin{align}x &= 2 \sec θ\\y &= \tan θ\end{align}\right.\).

## Parametric Equations of Conics

Example 3: write parametric equations.

Write a set of parametric equations for a vertical ellipse with center (2, 3), major axis length 10, and minor axis length of 8.

The equations for a vertical ellipse are \(\left\{\begin{align}x &= h + b \sin t\\y &= k + a \cos t\end{align}\right.\). The center is ( h , k ) = (2, 3) so h = 2 and k = 3. The variable a is half the major axis which is 10, so a = 5. b is half the minor axis length which is 8, so b = 4. Now fill in the equations.

$$\left\{\begin{align}x &= h + b \sin t\\y &= k + a \cos t\end{align}\right.$$

$$\left\{\begin{align}x &= 2 + 4 \sin t\\y &= 3 + 5 \cos t\end{align}\right.$$

## Example 4: Write Parametric Equations

Write a set of parametric equations for y 2 = 4 x − 4.

This is a parabola, so use the equation for a horizontal parabola.

$$\left\{\begin{align}x &= pt^2 + h\\y &= 2pt + k\end{align}\right.$$

To use these p , h , and k need to be known, so put the equation in standard form.

$$y^2 = 4x - 4$$

$$y^2 = 4(x - 1)$$

When compared to ( y − k ) 2 = 4 p ( x − h ), we find that p = 1, h = 1, and k = 0. Now fill in the parametric equations.

$$\left\{\begin{align}x &= t^2 + 1\\y &= 2t\end{align}\right.$$

Another way is to choose something to equal t . Since to solve y 2 = 4( x − 1) for y , the square root would have to be taken, it would be convenient to let t 2 = x − 1. Solve this for x to get

$$x = t^2 + 1$$

Substitute this for x in the equation to get

$$y^2 = 4(t^2 + 1 - 1)$$

Then solve for y.

$$y^2 = 4t^2$$

So the parametric equations are

Write a set of parametric equations for \(\frac{x^2}{4} + \frac{y^2}{16} = 1\).

- \(\left\{\begin{align}x &= 2 \sin θ\\y &= 4 \cos θ\end{align}\right.\)

## Eliminating the Parameter

It is possible to change parametric equations into regular functions by eliminating the parameter. To do this, solve one of the parametric equations for t and substitute the expression into the other equation. If the equations contain sine and cosine, or tangent and secant, it might be easier to solve each equation for the trigonometric function and use a Pythagorean Identity such as \(\sin^2 θ + \cos^2 θ = 1\) or \(1 + \tan^2 θ = \sec^2 θ\).

## Eliminate the Parameter

If the equations do not have trigonometric functions.

- Solve one equation for t .
- Substitute that expression for t in the other equation.
- Simplify the result.

If the equations do have trigonometric functions.

- Solve each equation for the trigonometric function.
- Substitute those into identities such as Pythagorean Identities to eliminate the parameter.

## Example 5: Eliminate the Parameter

Eliminate the parameter in \(\left\{\begin{align}x &= t^2 - 1\\y &= 3t\end{align}\right.\).

Since these do not have trigonometric functions, start by solving one equation for t .

$$x = t^2 - 1$$

$$x + 1 = t^2$$

$$±\sqrt{x + 1} = t$$

Now substitute this into the other equation.

$$y = 3 \left(± \sqrt{x + 1}\right)$$

Eliminate the ± by squaring both sides.

$$y^2 = 9(x + 1)$$

This is a horizontal parabola.

## Example 6: Eliminate the Parameter

Eliminate the parameter in \(\left\{\begin{align}x &= 2 \tan θ\\y &= 4 \sec θ\end{align}\right.\).

Since these do have trigonometric functions, it will be easier to solve the equations for the trig functions.

$$\left\{\begin{align}x &= 2 \tan θ\\y &= 4 \sec θ\end{align}\right.$$

$$\left\{\begin{align}\frac{x}{2} &= \tan θ\\ \frac{y}{4} &= \sec θ\end{align}\right.$$

An identity that relates secant and tangent is

$$ 1 + \tan^2 θ = \sec^2 θ $$

$$1 + \left(\frac{x}{2}\right)^2 = \left(\frac{y}{4}\right)^2$$

$$1 + \frac{x^2}{4} = \frac{y^2}{16}$$

Rearrange the equation to put it in standard form.

$$\frac{y^2}{16} - \frac{x^2}{4} = 1$$

This is a vertical hyperbola.

- \(\left\{\begin{align}x &= t + 2\\y &= t^2\end{align}\right.\)
- \(\left\{\begin{align}x &= 2 \sin θ\\y &= 3 \cos θ\end{align}\right.\)

\(y = x^2 - 4x + 4\), \(\frac{x^2}{4} + \frac{y^2}{9} = 1\)

## Lesson Summary

Helpful videos about this lesson.

- Mr. Wright Teaches the Lesson
- Introduction to Parametric Equations
- Converting Parametric Equations to Rectangular Form
- Graphing Parametric Equations on the TI-84

## Practice Exercises

- What is a parameter?

Graph the parametric equations.

- \(\left\{\begin{align}x &= 4t^2\\y &= 8t + 2\end{align}\right.\)
- \(\left\{\begin{align}x &= 4t + 1\\y &= 2t^2\end{align}\right.\)
- \(\left\{\begin{align}x &= 3 \sec θ\\y &= 2 \tan θ\end{align}\right.\)

Write a set of parametric equations for the following conditions.

- Parabola with vertex at (−2, −1) and focus (−5, −1).
- Horizontal ellipse with center (0, 0), vertices (±5, 0), and foci (±3, 0).
- \(y = 3x + 1\)
- \(\left(y - 2\right)^2 = -16x\)
- \(\left\{\begin{align}x &= 3t\\y &= \frac{9}{t}\end{align}\right.\)
- \(\left\{\begin{align}x &= 7 \sin θ\\y &= 4 \cos θ\end{align}\right.\)
- \(\left\{\begin{align}x &= 6 \tan θ\\y &= 5 \sec θ\end{align}\right.\)

Problem Solving

- On televised baseball games, sometimes the distance of a home run is given, but it was not measured. The distance was calculated. Let \(\left\{\begin{align}x &= 35t\\y &= -16t^2 + 30t\end{align}\right.\) model the path of the ball after it was hit. How far did the ball go? (Hint: Let y = 0 and find t .)

Mixed Review

- (7-05) Classify the conic and rewrite it in standard form by eliminating the Bxy term. \(x^2 - 2xy + y^2 - x - y = 0\).
- (7-04) Find the standard equation of the hyperbola with vertices (4, ±3) and foci (4, ±5).
- (7-03) Find the center, vertices, covertices, and foci of the ellipse \(\frac{(x + 1)^2}{16} + \frac{y^2}{9} = 1\).
- (7-02) Graph \((x - 1)^2 = 4(y + 2)\).
- (7-01) Find the distance from the point (2, 0) to the line \(2x - y + 4 = 0\).
- The parameter is the value in the equations that is used to calculate x and y .
- \(\left\{\begin{align}x &= -3t^2 - 2\\y &= -6t - 1\end{align}\right.\)
- \(\left\{\begin{align}x &= 5 \cos t\\y &= 4 \sin t\end{align}\right.\)
- \(\left\{\begin{align}x &= \frac{t}{3}\\y &= t + 1\end{align}\right.\)
- \(\left\{\begin{align}x &= -4t^2\\y &= -8t + 2\end{align}\right.\)
- \((y - 2)^2 = 16x\)
- \(xy - 27 = 0\)
- \(\frac{x^2}{49} + \frac{y^2}{16} = 1\)
- \(\frac{x^2}{36} - \frac{y^2}{25} = 1\)
- 65.625 feet
- \(\left\{\begin{align}x &= 4t\\y &= 2t^2\end{align}\right.\)
- Parabola; \(y′^2 = \frac{\sqrt{2}}{2}x′\)
- \(\frac{y^2}{9}-\frac{(x - 4)^2}{16} = 1\)
- C(−1, 0); V(−5, 0), (3, 0); CV(−1, −3), (−1, 3); F\((-1 - \sqrt{7}, 0)\), \((-1 + \sqrt{7}, 0)\)
- \(\frac{8\sqrt{5}}{5}\)

## Collapse menu

Introduction, 1 analytic geometry.

- 2. Distance Between Two Points; Circles
- 3. Functions
- 4. Shifts and Dilations

## 2 Instantaneous Rate of Change: The Derivative

- 1. The slope of a function
- 2. An example
- 4. The Derivative Function
- 5. Properties of Functions

## 3 Rules for Finding Derivatives

- 1. The Power Rule
- 2. Linearity of the Derivative
- 3. The Product Rule
- 4. The Quotient Rule
- 5. The Chain Rule

## 4 Trigonometric Functions

- 1. Trigonometric Functions
- 2. The Derivative of $\sin x$
- 3. A hard limit
- 4. The Derivative of $\sin x$, continued
- 5. Derivatives of the Trigonometric Functions
- 6. Implicit Differentiation
- 7. Limits revisited

## 5 Curve Sketching

- 1. Maxima and Minima
- 2. The first derivative test
- 3. The second derivative test
- 4. Concavity and inflection points
- 5. Asymptotes and Other Things to Look For

## 6 Applications of the Derivative

- 1. Optimization
- 2. Related Rates
- 3. Newton's Method
- 4. Linear Approximations
- 5. The Mean Value Theorem

## 7 Integration

- 1. Two examples
- 2. The Fundamental Theorem of Calculus
- 3. Some Properties of Integrals
- 4. Substitution

## 8 Applications of Integration

- 1. Area between curves
- 2. Distance, Velocity, Acceleration
- 4. Average value of a function

## 9 Transcendental Functions

- 1. Inverse functions
- 2. The natural logarithm
- 3. The exponential function
- 4. Other bases
- 5. Inverse Trigonometric Functions
- 6. Hyperbolic Functions

## 10 Techniques of Integration

- 1. Powers of sine and cosine
- 2. Trigonometric Substitutions
- 3. Integration by Parts
- 4. Rational Functions
- 5. Numerical Integration
- 6. Additional exercises

## 11 More Applications of Integration

- 1. Center of Mass
- 2. Kinetic energy; improper integrals
- 3. Probability
- 4. Arc Length
- 5. Surface Area

## 12 Polar Coordinates, Parametric Equations

- 1. Polar Coordinates
- 2. Slopes in polar coordinates
- 3. Areas in polar coordinates
- 4. Parametric Equations
- 5. Calculus with Parametric Equations

## 13 Sequences and Series

- 1. Sequences
- 3. The Integral Test
- 4. Alternating Series
- 5. Comparison Tests
- 6. Absolute Convergence
- 7. The Ratio and Root Tests
- 8. Power Series
- 9. Calculus with Power Series
- 10. Taylor Series
- 11. Taylor's Theorem
- 12. Additional exercises

## 14 Three Dimensions

- 1. The Coordinate System
- 3. The Dot Product
- 4. The Cross Product
- 5. Lines and Planes
- 6. Other Coordinate Systems

## 15 Vector Functions

- 1. Space Curves
- 2. Calculus with vector functions
- 3. Arc length and curvature
- 4. Motion along a curve

## 16 Partial Differentiation

- 1. Functions of Several Variables
- 2. Limits and Continuity
- 3. Partial Differentiation
- 4. The Chain Rule
- 5. Directional Derivatives
- 6. Higher order derivatives
- 7. Maxima and minima
- 8. Lagrange Multipliers

## 17 Multiple Integration

- 1. Volume and Average Height
- 2. Double Integrals in Cylindrical Coordinates
- 3. Moment and Center of Mass
- 4. Surface Area
- 5. Triple Integrals
- 6. Cylindrical and Spherical Coordinates
- 7. Change of Variables

## 18 Vector Calculus

- 1. Vector Fields
- 2. Line Integrals
- 3. The Fundamental Theorem of Line Integrals
- 4. Green's Theorem
- 5. Divergence and Curl
- 6. Vector Functions for Surfaces
- 7. Surface Integrals
- 8. Stokes's Theorem
- 9. The Divergence Theorem

## 19 Differential Equations

- 1. First Order Differential Equations
- 2. First Order Homogeneous Linear Equations
- 3. First Order Linear Equations
- 4. Approximation
- 5. Second Order Homogeneous Equations
- 6. Second Order Linear Equations
- 7. Second Order Linear Equations, take two

## 20 Useful formulas

21 introduction to sage.

- 2. Differentiation
- 3. Integration

When we computed the derivative $dy/dx$ using polar coordinates, we used the expressions $x=f(\theta)\cos\theta$ and $y=f(\theta)\sin\theta$. These two equations completely specify the curve, though the form $r=f(\theta)$ is simpler. The expanded form has the virtue that it can easily be generalized to describe a wider range of curves than can be specified in rectangular or polar coordinates.

Suppose $f(t)$ and $g(t)$ are functions. Then the equations $x=f(t)$ and $y=g(t)$ describe a curve in the plane. In the case of the polar coordinates equations, the variable $t$ is replaced by $\theta$ which has a natural geometric interpretation. But $t$ in general is simply an arbitrary variable, often called in this case a parameter , and this method of specifying a curve is known as parametric equations . One important interpretation of $t$ is time . In this interpretation, the equations $x=f(t)$ and $y=g(t)$ give the position of an object at time $t$.

Example 12.4.1 Describe the path of an object that moves so that its position at time $t$ is given by $x=\cos t$, $\ds y=\cos^2 t$. We see immediately that $\ds y=x^2$, so the path lies on this parabola. The path is not the entire parabola, however, since $x=\cos t$ is always between $-1$ and $1$. It is now easy to see that the object oscillates back and forth on the parabola between the endpoints $(1,1)$ and $(-1,1)$, and is at point $(1,1)$ at time $t=0$. $\square$

It is sometimes quite easy to describe a complicated path in parametric equations when rectangular and polar coordinate expressions are difficult or impossible to devise.

Example 12.4.2 A wheel of radius 1 rolls along a straight line, say the $x$-axis. A point on the rim of the wheel will trace out a curve, called a cycloid. Assume the point starts at the origin; find parametric equations for the curve.

Figure 12.4.1 illustrates the generation of the curve (click on the AP link to see an animation). The wheel is shown at its starting point, and again after it has rolled through about 490 degrees. We take as our parameter $t$ the angle through which the wheel has turned, measured as shown clockwise from the line connecting the center of the wheel to the ground. Because the radius is 1, the center of the wheel has coordinates $(t,1)$. We seek to write the coordinates of the point on the rim as $(t+\Delta x,1+\Delta y)$, where $\Delta x$ and $\Delta y$ are as shown in figure 12.4.2 . These values are nearly the sine and cosine of the angle $t$, from the unit circle definition of sine and cosine. However, some care is required because we are measuring $t$ from a nonstandard starting line and in a clockwise direction, as opposed to the usual counterclockwise direction. A bit of thought reveals that $\Delta x=-\sin t$ and $\Delta y=-\cos t$. Thus the parametric equations for the cycloid are $x=t-\sin t$, $y=1-\cos t$. $\square$

Animated Cycloid

## Exercises 12.4

You can plot parametric functions with Sage.

Ex 12.4.1 What curve is described by $\ds x=t^2$, $\ds y=t^4$? If $t$ is interpreted as time, describe how the object moves on the curve.

Ex 12.4.2 What curve is described by $x=3\cos t$, $y=3\sin t$? If $t$ is interpreted as time, describe how the object moves on the curve.

Ex 12.4.3 What curve is described by $x=3\cos t$, $y=2\sin t$? If $t$ is interpreted as time, describe how the object moves on the curve.

Ex 12.4.4 What curve is described by $x=3\sin t$, $y=3\cos t$? If $t$ is interpreted as time, describe how the object moves on the curve.

Ex 12.4.5 Sketch the curve described by $\ds x=t^3-t$, $\ds y=t^2$. If $t$ is interpreted as time, describe how the object moves on the curve.

Ex 12.4.6 A wheel of radius 1 rolls along a straight line, say the $x$-axis. A point $P$ is located halfway between the center of the wheel and the rim; assume $P$ starts at the point $(0,1/2)$. As the wheel rolls, $P$ traces a curve; find parametric equations for the curve. ( answer )

Ex 12.4.7 A wheel of radius 1 rolls around the outside of a circle of radius 3. A point $P$ on the rim of the wheel traces out a curve called a hypercycloid , as indicated in figure 12.4.3 . Assuming $P$ starts at the point $(3,0)$, find parametric equations for the curve. ( answer )

Ex 12.4.8 A wheel of radius 1 rolls around the inside of a circle of radius 3. A point $P$ on the rim of the wheel traces out a curve called a hypocycloid , as indicated in figure 12.4.3 . Assuming $P$ starts at the point $(3,0)$, find parametric equations for the curve. ( answer )

Ex 12.4.9 An involute of a circle is formed as follows: Imagine that a long (that is, infinite) string is wound tightly around a circle, and that you grasp the end of the string and begin to unwind it, keeping the string taut. The end of the string traces out the involute. Find parametric equations for this curve, using a circle of radius 1, and assuming that the string unwinds counter-clockwise and the end of the string is initially at $(1,0)$. Figure 12.4.4 shows part of the curve; the dotted lines represent the string at a few different times. ( answer )

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## Parametric Equations: Eliminating Angle Parameters

x = 5 sin θ, y = 3 cos θ, when − π ≤ θ ≤ − π 6

TRIGONOMETRIC IDENTITIES:

sin 2 θ + cos 2 θ = 1

sec 2 θ − tan 2 θ = 1

csc 2 θ − cot 2 θ = 1

x = 5 sin θ → x 5 = sin θ

y = 3 cos θ → y 3 = cos θ

sin 2 θ + cos 2 θ = 1 → ( x 5 ) 2 + ( y 3 ) 2 = 1

x 2 5 2 + y 2 3 2 = 1

GUIDELINES FOR ELIMINATING THE ANGLE PARAMETER:

1. Rewrite the parametric equations in terms that can be substituted into one of the trigonometric identities. 2. Substitute the resulting expression into the corresponding trigonometric identity.

3. Simplify

To link to this Parametric Equations: Eliminating Angle Parameters page, copy the following code to your site:

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