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4.7: Parametric Equations


Learning Objectives

  • Plot a curve described by parametric equations.
  • Convert the parametric equations of a curve into the form (y=f(x)).
  • Recognize the parametric equations of basic curves, such as a line and a circle.
  • Recognize the parametric equations of a cycloid.

In this section we examine parametric equations and their graphs. In the two-dimensional coordinate system, parametric equations are useful for describing curves that are not necessarily functions. The parameter is an independent variable that both (x) and (y) depend on, and as the parameter increases, the values of (x) and (y) trace out a path along a plane curve. For example, if the parameter is (t) (a common choice), then (t) might represent time. Then (x) and (y) are defined as functions of time, and ((x(t),y(t))) can describe the position in the plane of a given object as it moves along a curved path.

Parametric Equations and Their Graphs

Consider the orbit of Earth around the Sun. Our year lasts approximately 365.25 days, but for this discussion we will use 365 days. On January 1 of each year, the physical location of Earth with respect to the Sun is nearly the same, except for leap years, when the lag introduced by the extra (frac{1}{4}) day of orbiting time is built into the calendar. We call January 1 “day 1” of the year. Then, for example, day 31 is January 31, day 59 is February 28, and so on.

The number of the day in a year can be considered a variable that determines Earth’s position in its orbit. As Earth revolves around the Sun, its physical location changes relative to the Sun. After one full year, we are back where we started, and a new year begins. According to Kepler’s laws of planetary motion, the shape of the orbit is elliptical, with the Sun at one focus of the ellipse. We study this idea in more detail in Conic Sections.

Figure ( PageIndex{1}) depicts Earth’s orbit around the Sun during one year. The point labeled (F_2) is one of the foci of the ellipse; the other focus is occupied by the Sun. If we superimpose coordinate axes over this graph, then we can assign ordered pairs to each point on the ellipse (Figure ( PageIndex{2})). Then each (x) value on the graph is a value of position as a function of time, and each (y) value is also a value of position as a function of time. Therefore, each point on the graph corresponds to a value of Earth’s position as a function of time.

We can determine the functions for (x(t)) and (y(t)), thereby parameterizing the orbit of Earth around the Sun. The variable (t) is called an independent parameter and, in this context, represents time relative to the beginning of each year.

A curve in the ((x,y)) plane can be represented parametrically. The equations that are used to define the curve are called parametric equations.

Definition: Parametric Equations

If (x) and (y) are continuous functions of (t) on an interval (I), then the equations

[x=x(t)]

and

[y=y(t)]

are called parametric equations and (t) is called the parameter. The set of points ((x,y)) obtained as (t) varies over the interval (I) is called the graph of the parametric equations. The graph of parametric equations is called a parametric curve or plane curve, and is denoted by (C).

Notice in this definition that (x) and (y) are used in two ways. The first is as functions of the independent variable (t). As (t) varies over the interval (I), the functions (x(t)) and (y(t)) generate a set of ordered pairs ((x,y)). This set of ordered pairs generates the graph of the parametric equations. In this second usage, to designate the ordered pairs, (x) and (y) are variables. It is important to distinguish the variables (x) and (y) from the functions (x(t)) and (y(t)).

Example (PageIndex{1}): Graphing a Parametrically Defined Curve

Sketch the curves described by the following parametric equations:

  1. (x(t)=t−1, quad y(t)=2t+4,quad ext{for }−3≤t≤2)
  2. (x(t)=t^2−3, quad y(t)=2t+1,quad ext{for }−2≤t≤3)
  3. (x(t)=4 cos t, quad y(t)=4 sin t,quad ext{for }0≤t≤2π)

Solution

a. To create a graph of this curve, first set up a table of values. Since the independent variable in both (x(t)) and (y(t)) is (t), let (t) appear in the first column. Then (x(t)) and (y(t)) will appear in the second and third columns of the table.

(t)(x(t))(y(t))
−3−4−2
−2−30
−1−22
0−14
106
218

The second and third columns in this table provide a set of points to be plotted. The graph of these points appears in Figure ( PageIndex{3}). The arrows on the graph indicate the orientation of the graph, that is, the direction that a point moves on the graph as t varies from −3 to 2.

b. To create a graph of this curve, again set up a table of values.

(t)(x(t))(y(t))
−21−3
−1−2−1
0−31
1−23
215
367

The second and third columns in this table give a set of points to be plotted (Figure ( PageIndex{4})). The first point on the graph (corresponding to (t=−2)) has coordinates ((1,−3)), and the last point (corresponding to (t=3)) has coordinates ((6,7)). As (t) progresses from (−2) to (3), the point on the curve travels along a parabola. The direction the point moves is again called the orientation and is indicated on the graph.

c. In this case, use multiples of (π/6) for (t) and create another table of values:

(t)(x(t))(y(t))(t)(x(t))(y(t))
040(frac{7π}{6})(-2sqrt{3}≈−3.5)-2
(frac{π}{6})(2sqrt{3}≈3.5)2(frac{4π}{3})−2(−2sqrt{3}≈−3.5)
(frac{π}{3})2(2sqrt{3}≈3.5)(frac{3π}{2})0−4
(frac{π}{2})04(frac{5π}{3})2(−2sqrt{3}≈−3.5)
(frac{2π}{3})−2(2sqrt{3}≈3.5)(frac{11π}{6})(2sqrt{3}≈3.5)-2
(frac{5π}{6})(−2sqrt{3}≈−3.5)2(2π)40
(π)−40

The graph of this plane curve appears in the following graph.

This is the graph of a circle with radius (4) centered at the origin, with a counterclockwise orientation. The starting point and ending points of the curve both have coordinates ((4,0)).

Exercise (PageIndex{1})

Sketch the curve described by the parametric equations

[ x(t)=3t+2,quad y(t)=t^2−1,quad ext{for }−3≤t≤2. onumber]

Hint

Make a table of values for (x(t)) and (y(t)) using (t) values from (−3) to (2).

Answer

Eliminating the Parameter

To better understand the graph of a curve represented parametrically, it is useful to rewrite the two equations as a single equation relating the variables (x) and (y). Then we can apply any previous knowledge of equations of curves in the plane to identify the curve. For example, the equations describing the plane curve in Example (PageIndex{1b}) are

[egin{align} x(t) &=t^2−3 label{x1} [4pt] y(t) &=2t+1 label{y1} end{align}]

over the region (-2 le t le 3.)

Solving Equation ef{y1} for (t) gives

[t=dfrac{y−1}{2}. onumber]

This can be substituted into Equation ef{x1}:

[egin{align} x &=left(dfrac{y−1}{2} ight)^2−3 [4pt] &=dfrac{y^2−2y+1}{4}−3 [4pt] &=dfrac{y^2−2y−11}{4}. label{y2}end{align}]

Equation ef{y2} describes (x) as a function of (y). These steps give an example of eliminating the parameter. The graph of this function is a parabola opening to the right (Figure (PageIndex{4})). Recall that the plane curve started at ((1,−3)) and ended at ((6,7)). These terminations were due to the restriction on the parameter (t).

Example (PageIndex{2}): Eliminating the Parameter

Eliminate the parameter for each of the plane curves described by the following parametric equations and describe the resulting graph.

  1. (x(t)=sqrt{2t+4}, quad y(t)=2t+1,quad ext{for }−2≤t≤6)
  2. (x(t)=4cos t, quad y(t)=3sin t,quad ext{for }0≤t≤2π)

Solution

a. To eliminate the parameter, we can solve either of the equations for (t). For example, solving the first equation for (t) gives

[egin{align*} x &=sqrt{2t+4} [4pt] x^2 &=2t+4 [4pt] x^2−4 &=2t [4pt] t &=dfrac{x^2−4}{2}. end{align*}]

Note that when we square both sides it is important to observe that (x≥0). Substituting (t=dfrac{x^2−4}{2}) into (y(t)) yields

[ y(t)=2t+1]

[ y=2left(dfrac{x^2−4}{2} ight)+1]

[ y=x^2−4+1]

[ y=x^2−3.]

This is the equation of a parabola opening upward. There is, however, a domain restriction because of the limits on the parameter (t). When (t=−2), (x=sqrt{2(−2)+4}=0), and when (t=6), (x=sqrt{2(6)+4}=4). The graph of this plane curve follows.

b. Sometimes it is necessary to be a bit creative in eliminating the parameter. The parametric equations for this example are

[ x(t)=4 cos t onumber]

and

[ y(t)=3 sin t onumber]

Solving either equation for (t) directly is not advisable because sine and cosine are not one-to-one functions. However, dividing the first equation by (4) and the second equation by (3) (and suppressing the (t)) gives us

[ cos t=dfrac{x}{4} onumber]

and

[ sin t=dfrac{y}{3}. onumber]

Now use the Pythagorean identity (cos^2t+sin^2t=1) and replace the expressions for (sin t) and (cos t) with the equivalent expressions in terms of (x) and (y). This gives

[ left(dfrac{x}{4} ight)^2+left(dfrac{y}{3} ight)^2=1 onumber]

[ dfrac{x^2}{16}+dfrac{y^2}{9}=1. onumber]

This is the equation of a horizontal ellipse centered at the origin, with semi-major axis (4) and semi-minor axis (3) as shown in the following graph.

As t progresses from (0) to (2π), a point on the curve traverses the ellipse once, in a counterclockwise direction. Recall from the section opener that the orbit of Earth around the Sun is also elliptical. This is a perfect example of using parameterized curves to model a real-world phenomenon.

Exercise (PageIndex{2})

Eliminate the parameter for the plane curve defined by the following parametric equations and describe the resulting graph.

[ x(t)=2+dfrac{3}{t}, quad y(t)=t−1, quad ext{for }2≤t≤6 onumber]

Hint

Solve one of the equations for (t) and substitute into the other equation.

Answer

(x=2+frac{3}{y+1},) or (y=−1+frac{3}{x−2}). This equation describes a portion of a rectangular hyperbola centered at ((2,−1)).

So far we have seen the method of eliminating the parameter, assuming we know a set of parametric equations that describe a plane curve. What if we would like to start with the equation of a curve and determine a pair of parametric equations for that curve? This is certainly possible, and in fact it is possible to do so in many different ways for a given curve. The process is known as parameterization of a curve.

Example (PageIndex{3}): Parameterizing a Curve

Find two different pairs of parametric equations to represent the graph of (y=2x^2−3).

Solution

First, it is always possible to parameterize a curve by defining (x(t)=t), then replacing (x) with (t) in the equation for (y(t)). This gives the parameterization

[ x(t)=t, quad y(t)=2t^2−3. onumber]

Since there is no restriction on the domain in the original graph, there is no restriction on the values of (t).

We have complete freedom in the choice for the second parameterization. For example, we can choose (x(t)=3t−2). The only thing we need to check is that there are no restrictions imposed on (x); that is, the range of (x(t)) is all real numbers. This is the case for (x(t)=3t−2). Now since (y=2x^2−3), we can substitute (x(t)=3t−2) for (x). This gives

[ y(t)=2(3t−2)^2−2=2(9t^2−12t+4)−2=18t^2−24t+8−2=18t^2−24t+6. onumber]

Therefore, a second parameterization of the curve can be written as

( x(t)=3t−2) and ( y(t)=18t^2−24t+6.)

Exercise (PageIndex{3})

Find two different sets of parametric equations to represent the graph of (y=x^2+2x).

Hint

Follow the steps in Example (PageIndex{3}). Remember we have freedom in choosing the parameterization for (x(t)).

Answer

One possibility is (x(t)=t, quad y(t)=t^2+2t.) Another possibility is (x(t)=2t−3, quad y(t)=(2t−3)^2+2(2t−3)=4t^2−8t+3.) There are, in fact, an infinite number of possibilities.

Cycloids and Other Parametric Curves

Imagine going on a bicycle ride through the country. The tires stay in contact with the road and rotate in a predictable pattern. Now suppose a very determined ant is tired after a long day and wants to get home. So he hangs onto the side of the tire and gets a free ride. The path that this ant travels down a straight road is called a cycloid (Figure ( PageIndex{8})). A cycloid generated by a circle (or bicycle wheel) of radius a is given by the parametric equations

[x(t)=a(t−sin t), quad y(t)=a(1−cos t). onumber]

To see why this is true, consider the path that the center of the wheel takes. The center moves along the (x)-axis at a constant height equal to the radius of the wheel. If the radius is (a), then the coordinates of the center can be given by the equations

[x(t)=at,quad y(t)=a onumber]

for any value of (t). Next, consider the ant, which rotates around the center along a circular path. If the bicycle is moving from left to right then the wheels are rotating in a clockwise direction. A possible parameterization of the circular motion of the ant (relative to the center of the wheel) is given by

[egin{align*} x(t) &=−a sin t [4pt] y(t) &=−acos t.end{align*}]

(The negative sign is needed to reverse the orientation of the curve. If the negative sign were not there, we would have to imagine the wheel rotating counterclockwise.) Adding these equations together gives the equations for the cycloid.

[egin{align*} x(t) &=a(t−sin t) [4pt] y(t) &=a(1−cos t ) end{align*}]

Now suppose that the bicycle wheel doesn’t travel along a straight road but instead moves along the inside of a larger wheel, as in Figure ( PageIndex{9}). In this graph, the green circle is traveling around the blue circle in a counterclockwise direction. A point on the edge of the green circle traces out the red graph, which is called a hypocycloid.

The general parametric equations for a hypocycloid are

[x(t)=(a−b) cos t+b cos (dfrac{a−b}{b})t onumber]

[y(t)=(a−b) sin t−b sin (dfrac{a−b}{b})t. onumber]

These equations are a bit more complicated, but the derivation is somewhat similar to the equations for the cycloid. In this case we assume the radius of the larger circle is (a) and the radius of the smaller circle is (b). Then the center of the wheel travels along a circle of radius (a−b.) This fact explains the first term in each equation above. The period of the second trigonometric function in both (x(t)) and (y(t)) is equal to (dfrac{2πb}{a−b}).

The ratio (dfrac{a}{b}) is related to the number of cusps on the graph (cusps are the corners or pointed ends of the graph), as illustrated in Figure ( PageIndex{10}). This ratio can lead to some very interesting graphs, depending on whether or not the ratio is rational. Figure (PageIndex{9}) corresponds to (a=4) and (b=1). The result is a hypocycloid with four cusps. Figure (PageIndex{10}) shows some other possibilities. The last two hypocycloids have irrational values for (dfrac{a}{b}). In these cases the hypocycloids have an infinite number of cusps, so they never return to their starting point. These are examples of what are known as space-filling curves.

The Witch of Agnesi

Many plane curves in mathematics are named after the people who first investigated them, like the folium of Descartes or the spiral of Archimedes. However, perhaps the strangest name for a curve is the witch of Agnesi. Why a witch?

Maria Gaetana Agnesi (1718–1799) was one of the few recognized women mathematicians of eighteenth-century Italy. She wrote a popular book on analytic geometry, published in 1748, which included an interesting curve that had been studied by Fermat in 1630. The mathematician Guido Grandi showed in 1703 how to construct this curve, which he later called the “versoria,” a Latin term for a rope used in sailing. Agnesi used the Italian term for this rope, “versiera,” but in Latin, this same word means a “female goblin.” When Agnesi’s book was translated into English in 1801, the translator used the term “witch” for the curve, instead of rope. The name “witch of Agnesi” has stuck ever since.

The witch of Agnesi is a curve defined as follows: Start with a circle of radius a so that the points ((0,0)) and ((0,2a)) are points on the circle (Figure ( PageIndex{11})). Let O denote the origin. Choose any other point A on the circle, and draw the secant line OA. Let B denote the point at which the line OA intersects the horizontal line through ((0,2a)). The vertical line through B intersects the horizontal line through A at the point P. As the point A varies, the path that the point P travels is the witch of Agnesi curve for the given circle.

Witch of Agnesi curves have applications in physics, including modeling water waves and distributions of spectral lines. In probability theory, the curve describes the probability density function of the Cauchy distribution. In this project you will parameterize these curves.

1. On the figure, label the following points, lengths, and angle:

a. (C) is the point on the (x)-axis with the same (x)-coordinate as (A).

b. (x) is the (x)-coordinate of (P), and (y) is the (y)-coordinate of (P).

c. (E) is the point ((0,a)).

d. (F) is the point on the line segment (OA) such that the line segment (EF) is perpendicular to the line segment (OA).

e. (b) is the distance from (O) to (F).

f. (c) is the distance from (F) to (A).

g. (d) is the distance from (O) to (C).

h. (θ) is the measure of angle (∠COA).

The goal of this project is to parameterize the witch using (θ) as a parameter. To do this, write equations for (x) and (y) in terms of only (θ).

2. Show that (d=dfrac{2a}{sin θ}).

3. Note that (x=dcos θ). Show that (x=2acot θ). When you do this, you will have parameterized the (x)-coordinate of the curve with respect to (θ). If you can get a similar equation for (y), you will have parameterized the curve.

4. In terms of (θ), what is the angle (∠EOA)?

5. Show that (b+c=2acosleft(frac{π}{2}−θ ight)).

6. Show that (y=2acosleft(frac{π}{2}−θ ight)sin θ).

7. Show that (y=2asin^2θ). You have now parameterized the (y)-coordinate of the curve with respect to (θ).

8. Conclude that a parameterization of the given witch curve is

[x=2acot θ, quad y=2a sin^2θ, quad ext{for }−∞<θ<∞.]

9. Use your parameterization to show that the given witch curve is the graph of the function (f(x)=dfrac{8a^3}{x^2+4a^2}).

Travels with My Ant: The Curtate and Prolate Cycloids

Earlier in this section, we looked at the parametric equations for a cycloid, which is the path a point on the edge of a wheel traces as the wheel rolls along a straight path. In this project we look at two different variations of the cycloid, called the curtate and prolate cycloids.

First, let’s revisit the derivation of the parametric equations for a cycloid. Recall that we considered a tenacious ant trying to get home by hanging onto the edge of a bicycle tire. We have assumed the ant climbed onto the tire at the very edge, where the tire touches the ground. As the wheel rolls, the ant moves with the edge of the tire (Figure (PageIndex{12})).

As we have discussed, we have a lot of flexibility when parameterizing a curve. In this case we let our parameter t represent the angle the tire has rotated through. Looking at Figure ( PageIndex{12}), we see that after the tire has rotated through an angle of (t), the position of the center of the wheel, (C=(x_C,y_C)), is given by

(x_C=at) and (y_C=a).

Furthermore, letting (A=(x_A,y_A)) denote the position of the ant, we note that

(x_C−x_A=asin t) and (y_C−y_A=a cos t)

Then

[x_A=x_C−asin t=at−asin t=a(t−sin t)]

[y_A=y_C−acos t=a−acos t=a(1−cos t).]

Note that these are the same parametric representations we had before, but we have now assigned a physical meaning to the parametric variable (t).

After a while the ant is getting dizzy from going round and round on the edge of the tire. So he climbs up one of the spokes toward the center of the wheel. By climbing toward the center of the wheel, the ant has changed his path of motion. The new path has less up-and-down motion and is called a curtate cycloid (Figure ( PageIndex{13})). As shown in the figure, we let b denote the distance along the spoke from the center of the wheel to the ant. As before, we let t represent the angle the tire has rotated through. Additionally, we let (C=(x_C,y_C)) represent the position of the center of the wheel and (A=(x_A,y_A)) represent the position of the ant.

1. What is the position of the center of the wheel after the tire has rotated through an angle of (t)?

2. Use geometry to find expressions for (x_C−x_A) and for (y_C−y_A).

3. On the basis of your answers to parts 1 and 2, what are the parametric equations representing the curtate cycloid?

Once the ant’s head clears, he realizes that the bicyclist has made a turn, and is now traveling away from his home. So he drops off the bicycle tire and looks around. Fortunately, there is a set of train tracks nearby, headed back in the right direction. So the ant heads over to the train tracks to wait. After a while, a train goes by, heading in the right direction, and he manages to jump up and just catch the edge of the train wheel (without getting squished!).

The ant is still worried about getting dizzy, but the train wheel is slippery and has no spokes to climb, so he decides to just hang on to the edge of the wheel and hope for the best. Now, train wheels have a flange to keep the wheel running on the tracks. So, in this case, since the ant is hanging on to the very edge of the flange, the distance from the center of the wheel to the ant is actually greater than the radius of the wheel (Figure (PageIndex{14})).

The setup here is essentially the same as when the ant climbed up the spoke on the bicycle wheel. We let b denote the distance from the center of the wheel to the ant, and we let t represent the angle the tire has rotated through. Additionally, we let (C=(x_C,y_C)) represent the position of the center of the wheel and (A=(x_A,y_A)) represent the position of the ant (Figure ( PageIndex{14})).

When the distance from the center of the wheel to the ant is greater than the radius of the wheel, his path of motion is called a prolate cycloid. A graph of a prolate cycloid is shown in the figure.

4. Using the same approach you used in parts 1– 3, find the parametric equations for the path of motion of the ant.

5. What do you notice about your answer to part 3 and your answer to part 4?

Notice that the ant is actually traveling backward at times (the “loops” in the graph), even though the train continues to move forward. He is probably going to be really dizzy by the time he gets home!

Key Concepts

  • Parametric equations provide a convenient way to describe a curve. A parameter can represent time or some other meaningful quantity.
  • It is often possible to eliminate the parameter in a parameterized curve to obtain a function or relation describing that curve.
  • There is always more than one way to parameterize a curve.
  • Parametric equations can describe complicated curves that are difficult or perhaps impossible to describe using rectangular coordinates.

Glossary

cycloid
the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slippage
cusp
a pointed end or part where two curves meet
orientation
the direction that a point moves on a graph as the parameter increases
parameter
an independent variable that both (x) and (y) depend on in a parametric curve; usually represented by the variable (t)
parametric curve
the graph of the parametric equations (x(t)) and (y(t)) over an interval (a≤t≤b) combined with the equations
parametric equations
the equations (x=x(t)) and (y=y(t)) that define a parametric curve
parameterization of a curve
rewriting the equation of a curve defined by a function (y=f(x)) as parametric equations

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Solutions for Chapter 4-7: Trigonometric Function Properties, Identities, and Parametric Functions

Solutions for Chapter 4-7: Trigonometric Function Properties, Identities, and Parametric Functions

  • 4-7.R.0: Update your journal with what you have learned in this chapter. Inc.
  • 4-7.R.1: Figure 4-7a shows a unit circle and an angle in standard position. .
  • 4-7.R.2: a. Write equations expressing tan x and cot x in terms of sin x and.
  • 4-7.R.3: a. Transform tan A sin A + cos A to sec A. What values of A are exc.
  • 4-7.R.4: a. Find the general solution for = arcsin 0. b. Solve 1 + tan 2 (x .
  • 4-7.R.5: a. Plot the graph of this parametric function on your grapher. Sket.
  • 4-7.R.6: a. Using parametric mode on your grapher, duplicate the graph of th.
  • 4-7.C.1: Pendulum Problem: Figure 4-7d shows a pendulum hanging from the cei.
  • 4-7.C.2: Prove that each of these equations is an identity. a. b. C
  • 4-7.C.3: Square of a Sinusoid Problem: Figure 4-7e shows the graphs of y1 = .
  • 4-7.T.1: Write the Pythagorean property for cosine and sine. T
  • 4-7.T.2: Write a quotient property involving cosine and sine. T
  • 4-7.T.3: Write the reciprocal property for cotangent. T
  • 4-7.T.4: Write the reciprocal property for secant. T
  • 4-7.T.5: The value of sin1 0.5 is 30. Write the general solution for = arcsi.
  • 4-7.T.6: The value of tan1 is . Write the general solution for x = arctan . T
  • 4-7.T.7: Explain why the range of y = cos1 x is [0, ] but the range of y = s.
  • 4-7.T.8: Find geometrically the exact value of cos (tan1 2). T
  • 4-7.T.9: Transform (1 + sin A)(1 sin A) to cos2 A. What values of A are excl.
  • 4-7.T.10: Prove that tan B + cot B = csc B sec B is an identity. What values .
  • 4-7.T.11: Multiply the numerator and denominator of by the conjugate of the d.
  • 4-7.T.12: Plot the graphs of both expressions in T11 to confirm that the two .
  • 4-7.T.13: With your calculator in degree mode, find the value of cos1 0. Show.
  • 4-7.T.14: Find another angle between 0 and 360 whose cosine is 0. Show it on .
  • 4-7.T.15: Write the general solution for the inverse trigonometric relation =.
  • 4-7.T.16: Find the fifth positive value of for which cos = 0. How many revolu.
  • 4-7.T.17: Find algebraically the general solution for 4 tan ( 25) = 7 T1
  • 4-7.T.18: Write parametric equations for this ellipse (Figure 4-7f). Figure 4.
  • 4-7.T.19: Write the parametric equations you use to plot y = arctan x (Figure.
  • 4-7.T.20: What did you learn as a result of working this test that you did no.
Textbook: Precalculus with Trigonometry: Concepts and Applications
Edition: 1
Author: Foerster
ISBN: 9781559533911

Chapter 4-7: Trigonometric Function Properties, Identities, and Parametric Functions includes 30 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Precalculus with Trigonometry: Concepts and Applications, edition: 1. Precalculus with Trigonometry: Concepts and Applications was written by and is associated to the ISBN: 9781559533911. Since 30 problems in chapter 4-7: Trigonometric Function Properties, Identities, and Parametric Functions have been answered, more than 35362 students have viewed full step-by-step solutions from this chapter.

Union of two rays with a common endpoint (the vertex). The beginning ray (the initial side) can be rotated about its endpoint to obtain the final position (the terminal side)

A graph that displays a five-number summary

(a + bi) - (c + di) = (a - c) + (b - d)i

A statement of equality between two expressions.

Arrows that have the same magnitude and direction.

A procedure for fitting an exponential function to a set of data.

The net amount of money returned from an annuity.

A visible representation of a numerical or algebraic model.

The line is a horizontal asymptote of the graph of a function ƒ if lim x:- q ƒ(x) = or lim x: q ƒ(x) = b

A function ƒ is increasing on an interval I if, for any two points in I, a positive change in x results in a positive change in.

Length of 1 minute of arc along the Earth’s equator.

A complex number v such that vn = z

If bn = a, then b is an nth root of a. If bn = a and a and b have the same sign, b is the principal nth root of a (see Radical), p. 508.

The difference y1 - (ax 1 + b), where (x1, y1)is a point in a scatter plot and y = ax + b is a line that fits the set of data.


Parametric Equation of a Circle

The equations x = a cosθ , y = a sinθ are called parametric equations of the circle x 2 + y 2 = a 2 and θ is called a parameter.

The point (a cos θ, a sinθ) is also referred to as point θ

The parametric coordinates of any point on the circle

(x – h) 2 + (y – k) 2 =a 2 are given by (h + a cosθ, k + a sinθ) with 0 ≤ θ < 2π

Example : If a straight line through C(-√8, √8), making an angle of 135° with the x-axis, cuts the circle x = 5 cosθ , y = 5 sinθ , in points A and B, find the length of the segment AB.

Solution: The given circle is x 2 + y 2 = 25 . . .(1)

Equation of line through C is (x + √8)/cos135° = (y − √8)/sin135° = r

Any point on this line is ( − √8 − r/√2 , √8 + r/√2)

If the point lies on the circle x 2 + y 2 = 25, the

Example : (i) Find the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6).

(ii) Find the equation of a circle passing through (1, 1), (2, − 1) and (3, 2).

(i) The radius of the circle is

⇒ r = 5
Hence the equation of the circle is (x − 1) 2 + (y − 2) 2 = 25

Substituting for the three points, we get

Solving the above three equations, we obtain:

Hence the equation of the circle is

Example : Find the equation of the circle whose centre is (3, 4) and which touches the line 5x + 12y = 1.

Solution: Let r be the radius of the circle. Then

r = distance of the centre i.e. point (3 , 4) from the line 5x + 12y = 1

Hence the equation of the required circle is (x − 3) 2 + (y − 4) 2 = (62/13) 2

=> x 2 + y 2 − 6x − 8y + 381/169 = 0

Example : Find the equation of the circle whose diameter is the line joining the points ( −4, 3) and (12, −1). Find also the intercept made by it on the y-axis.

Solution: The equation of the required circle is

Intercept made on the y-axis = 2 √(f 2 − c)

Example : A circle has radius equal to 3 units and its centre lies on the line y = x − 1. Find the equation of the circle if it passes through (7 , 3)

Solution: Let the centre of the circle be (α , β).

It lies on the line y = x − 1

Hence the centre is (α , α − 1)

=> The equation of the circle is

Hence the required equations are x 2 + y 2 − 8x − 6y + 16 = 0 and x 2 + y 2 − 14x − 12y + 76 = 0.


4.7: Parametric Equations

State which of the following equations define lines and which define planes. Explain how you made your decision.

a) vec = (1, 2, 3) + s(1, 1, 0) + t(3, 4, -6), s, t in mathbb

b) vec = (-2, 3, 0) + m(3, 4, 7), m in mathbb

c) x = -3 - t, y = 5, z= 4 + t, t in mathbb

d) vec = m(4, -1, 2) + t(4, -1, 5), m, t in mathbb

A plane has a vector equation vec = (2, 1, 3) + s(displaystyle<3>>, -2, displaystyle<4>>) + t(6, -12, 30), s, t in mathbb.

a) Express the first direction vector with only integers.

b) Reduce the second direction vector.

c) Write a new equation for the plane using the calculations from part a. and b.

A plane has x = 2m, y = -3m + 5n, z = - 1 - 3m - 2n, m, n in mathbb as its parametric equations.

a) By inspection, identify the coordinates of a point that is on this plane.

b) What are the direction vectors for this plane?

c) What point corresponds to the parameter values of m = -1 and n = -4 ?

d) What are the parametric values corresponding to the point A(0, 15, -7) ?

e) Using your answer for part d., explain why the point B(0, 15, -8) cannot be on this plane.

A plane passes through the points P(-2, ,3, 1) , Q(-2, 3, 2) , and R(1, 0 , 1) .

a) Using vec and vec as direction vectors, write a vector equation for this plane.

b) Using vec and one other direction vector, write a second vector equation for this plane.

Explain why the equation vec = (-1, 0, -1) + s(2, 3, -4) + t(4, 6, -8), s, t, in mathbb , does not represent the equation of a plane. What does this equation represent?

Determine vector equations and the corresponding parametric equations of the plane.

Determine vector equations and the corresponding parametric equations of the plane.

Determine vector equations and the corresponding parametric equations of the plane.

a) Determine parameters corresponding to the point P(5, 3, 2) , where P is a point on the plane with equation

pi: vec = (2, 0 , 1) + s(4, 2, -1) + t(-1, 1, 2), s, t, in mathbb

Show that A does not lie on pi . displaystyleeginpi: vec=(2,0,1)+s(4,2,-1)+t(-1,1,2), s, t in mathbfA(0,5,-4) end

A plane has vec = (-3, 5, 6) + s(-1, 1, 2) + v(2, 1, -3), s, v in mathbb as its equation.

a) Give the equations of two intersecting lines that lie on this plane.

b) What point do these two lines have in common?

Determine the coordinates of the point where the plane with equation vec = (4, 1, 6) + s(11, -1, 3) + t(-7, 2, -2), s, t in mathbb , crosses the z-axis.

Determine the equation of the plane that contains the point P(-1, 2, 1) and the line vec= (2, 1, 3) + s(4, 1, 5), sin mathbb.

Determine the equation of the plane that contains the point A(-2, 2, 3) and the line vec = m(2, -1, 7), m inmathbb.

a) Determine two pairs of direction vectors that can be used to represent the xy-plane in mathbb^3

b) Write a vector and parametric equations for the xy-plane in mathbb^3

Show that the following equations represent the same plane:

a) vec = u(-2, 2, 4) + v(-4, 7, 1), u, v in mathbb and

b) vec = s(-1, 5, -3) + t(-1, -5, 7), s, t in mathbb

(Hint: Express each direction vector in the first equation as a linear combination of the direction vectors in the second equation.)


Contents

Analogue parametric design Edit

One of the earliest examples of parametric design was the upside down model of churches by Antonio Gaudi. In his design for the Church of Colònia Güell he created a model of strings weighted down with birdshot to create complex vaulted ceilings and arches. By adjusting the position of the weights or the length of the strings he could alter the shape of each arch and also see how this change influenced the arches connected to it. He placed a mirror on the bottom of the model to see how it should look upside-down.

Features of Gaudi's method Edit

Gaudi's analogue method includes the main features of a computational of a parametric model (input parameters, equation, output):

  • The string length, birdshot weight and anchor point location all form independent input parameters
  • The vertex locations of the points on the strings being the outcomes of the model
  • The outcomes are derived by explicit functions, in this case gravity or Newtons law of motion.

By modifying individual parameters of these models Gaudi could generate different versions of his model while being certain the resulting structure would stand in pure compression. Instead of having to manually calculate the results of parametric equations he could automatically derive the shape of the catenary curves through the force of gravity acting on the strings. [5]

Luigi Moretti's Parametric Architecture Edit

We owe to Luigi Moretti the first definition of parametric architecture, the result of studies on the history of architecture, art, and the application of operation research to architectural design. [6] His researches on parametric architecture methods were developed within the Institute for Operations Research and Applied Mathematics Urbanism. On the occasion of an exhibition at the Milan Triennale, he will present various models of stadiums designed according to the principles of parametric architecture.

Principles of Luigi Moretti's parametric architecture Edit

  • Rejection of empirical decisions.
  • Assessment of traditional phenomena as objective facts based on the interdependence of expressive, social and technical values.
  • Exact and complete definition of architectural themes.
  • Objective observation of all the conditioning elements (parameters) related to the architectural theme and identification of their quantitative values.
  • Definition of the relationships between the values of the parameters.
  • Indispensability of different skills and scientific methodologies according to the criteria of operational research to define conditioning elements and their quantities.
  • Affirmation of the Architect's freedom in decision and expression, only if it does not affect the characteristics determined by the analytical investigations.
  • Research of architectural forms towards a maximum, therefore definitive, exactness of relationships in their general "structure".

Nature has often served as inspiration for architects and designers. [ citation needed ] Computer technology has given designers and architects the tools to analyse and simulate the complexity observed in nature and apply it to structural building shapes and urban organizational patterns. In the 1980s architects and designers started using computers running software developed for the aerospace and moving picture industries to "animate form". [7]

One of the first architects and theorists that used computers to generate architecture was Greg Lynn. His blob and fold architecture is some of the early examples of computer generated architecture. Shenzhen Bao'an International Airport's new Terminal 3, finished in 2013, designed by Italian architect Massimiliano Fuksas, with parametric design support by the engineering firm Knippers Helbig, is an example for the use of parametric design and production technologies in a large scale building.

Parametric urbanism is concerned with the study and prediction of settlement patterns. Architect Frei Otto distinguishes occupying and connecting as the two fundamental processes that are involved with all urbanisation. [8] Studies look at producing solutions that reduce overall path length in systems while maintaining low average detour factor or facade differentiation [ clarification needed ] .

Power Surfacing Edit

Power Surfacing is a SolidWorks application for industrial design / freeform organic surface / solids modeling. Tightly integrated with SolidWorks, it works with all SolidWorks commands. Reverse Engineer scanned meshes with Power Surfacing RE.

Catia Edit

CATIA (Computer Aided three-dimensional Interactive Application) was used by architect Frank Gehry to design some of his award-winning curvilinear buildings such as the Guggenheim Museum Bilbao. [9] Gehry Technologies, the technology arm of his firm, have since created Digital Project, their own parametric design software based on their experience with CATIA.

Autodesk 3DS Max Edit

Autodesk 3ds Max is a parametric 3D modeling software which provides modeling, animation, simulation, and rendering functions for games, film, and motion graphics. 3ds Max uses the concept of modifiers and wired parameters to control its geometry and gives the user the ability to script its functionality. Max Creation Graph is a visual programming node-based tool creation environment in 3ds Max 2016 that is similar to Grasshopper and Dynamo.

Autodesk Maya Edit

Autodesk Maya is a 3D computer graphics software originally developed by Alias Systems Corporation (formerly Alias|Wavefront) and currently owned and developed by Autodesk, Inc. It is used to create interactive 3D applications, including video games, animated film, TV series, or visual effects. Maya exposes a node graph architecture. Scene elements are node-based, each node having its own attributes and customization. As a result, the visual representation of a scene is based on a network of interconnecting nodes, depending on each other's information. Maya is equipped with a cross-platform scripting language, called Maya Embedded Language. MEL is provided for scripting and a means to customize the core functionality of the software, since many of the tools and commands used are written in it. MEL or Python can be used to engineer modifications, plug-ins or be injected into runtime. User interaction is recorded in MEL, allowing novice users to implement subroutines.

Grasshopper 3D Edit

Grasshopper 3d (originally Explicit History) is a plug-in for Rhinoceros 3D that presents the users with a visual programming language interface to create and edit geometry. [10]

Components or nodes are dragged onto a canvas in order to build a grasshopper definition. Grasshopper is based on graphs (see Graph (discrete mathematics)) that map the flow of relations from parameters through user-defined functions (nodes), resulting in the generation of geometry. Changing parameters or geometry causes the changes to propagate throughout all functions, and the geometry to be redrawn. [5]

Autodesk Revit Edit

Autodesk Revit is building information modeling (BIM) software used by architects and other building professionals. Revit was developed in response to the need for software that could create three-dimensional parametric models that include both geometry and non-geometric design and construction information. Every change made to an element in Revit is automatically propagated through the model to keep all components, views and annotations consistent. This eases collaboration between teams and ensures that all information (floor areas, schedules, etc.) are updated dynamically when changes in the model are made.

Autodesk Dynamo Edit

Dynamo is an open source graphical programming environment for design. Dynamo extends building information modeling with the data and logic environment of a graphical algorithm editor.

GenerativeComponents Edit

GenerativeComponents, parametric CAD software developed by Bentley Systems, [11] was first introduced in 2003, became increasingly used in practice (especially by the London architectural community) by early 2005, and was commercially released in November 2007. GenerativeComponents has a strong traditional base of users in academia and at technologically advanced design firms. [ citation needed ] GenerativeComponents is often referred to by the nickname of 'GC'. GC epitomizes the quest to bring parametric modeling capabilities of 3D solid modeling into architectural design, seeking to provide greater fluidity and fluency than mechanical 3D solid modeling. [ citation needed ]

Users can interact with the software by either dynamically modeling and directly manipulating geometry, or by applying rules and capturing relationships among model elements, or by defining complex forms and systems through concisely expressed algorithms. The software supports many industry standard file input and outputs including DGN by Bentley Systems, DWG by Autodesk, STL (Stereo Lithography), Rhino, and others. The software can also integrate with Building Information Modeling systems.

The software has a published API and uses a simple scripting language, both allowing the integration with many different software tools, and the creation of custom programs by users.

This software is primarily used by architects and engineers in the design of buildings, but has also been used to model natural and biological structures and mathematical systems.

Generative Components runs exclusively on Microsoft Windows operating systems.

VIKTOR Edit

VIKTOR is an application development platform that enables engineers and other domain experts to rapidly build their own online applications using Python. It is used to create parametric design models and integrates with many software packages. [12] It enables users to make intuitive user interfaces (GUI), which include different form of visualizing results like 3D models, drawings, map or satellite views, and interactive graphs. This makes it possible to make the applications available to persons without programming affinity.

Applications made with VIKTOR are online, meaning data is update automatically and everyone works with the same information and the latest models. It includes a user management system, allowing to give different rights to users.

Marionette Edit

Marionette is an open source [ citation needed ] graphical scripting tool (or visual programming environment) for the architecture, engineering, construction, landscape, and entertainment design industries that is built into the Mac and Windows versions of Vectorworks software. The tool was first made available in the Vectorworks 2016 line of software products. Marionette enables designers to create custom application algorithms that build interactive parametric objects and streamline complex workflows, as well as build automated 2D drawing, 3D modeling, and BIM workflows within Vectorworks software.

Built in the Python programming language, everything in Marionette consists of nodes which are linked together in a flowchart arrangement. Each node contains a Python script with predefined inputs and outputs that can be accessed and modified with a built-in editor. Nodes are placed directly into the Vectorworks document and then connected to create complex algorithms. Since Marionette is fully integrated into Vectorworks software, it can also be used to create entirely self-contained parametric objects that can be inserted into new and existing designs.

Modelur Edit

Modelur is a parametric urban design software plug-in for Trimble SketchUp, developed by Agilicity d.o.o. (LLC).. Its primary goal is to help the users create conceptual urban massing. In contrast to common CAD applications, where the user designs buildings with usual dimensions such as width, depth and height, Modelur offers design of built environment through key urban parameters such as number of storeys and gross floor area of a building.

Modelur calculates key urban control parameters on the fly (e.g. floor area ratio or required number of parking lots), delivering urban design information while the development is still evolving. This way it helps taking well-informed decision during the earliest stages, when design decisions have the highest impact.

Archimatix Edit

Archimatix is a node-based parametric modeler extension for Unity 3D. It enables visual modeling of 3D models within the Unity 3D editor.


Transform a parametric plane form to the normal form

This transformation is nearly identical to the transformation from the parametric form to the cartesian form. We are given a plane in the parametric form and want to transform it to the normal form . We only need to calculate the normal vector of the plane by using the cross product and then we have all information for the normal form:


Subsection 2.4.1 Homogeneous Systems ¶ permalink

A homogeneous system is just a system of linear equations where all constants on the right side of the equals sign are zero.

A homogeneous system always has the solution

This is called the trivial solution. Any nonzero solution is called nontrivial.

Observation

has a nontrivial solution

Example (No nontrivial solutions)
Observation

When we row reduce the augmented matrix for a homogeneous system of linear equations, the last column will be zero throughout the row reduction process. We saw this in the last example:

So it is not really necessary to write augmented matrices when solving homogeneous systems.

When the homogeneous equation

does have nontrivial solutions, it turns out that the solution set can be conveniently expressed as a span.

Parametric Vector Form (homogeneous case)

Consider the following matrix in reduced row echelon form:

corresponds to the system of equations

We can write the parametric form as follows:

We wrote the redundant equations

in order to turn the above system into a vector equation:

This vector equation is called the parametric vector form of the solution set. Since

are allowed to be anything, this says that the solution set is the set of all linear combinations of

In other words, the solution set is

Here is the general procedure.

Recipe: Parametric vector form (homogeneous case)

matrix. Suppose that the free variables in the homogeneous equation

    Find the reduced row echelon form of

Put equations for all of the

will then be expressed in the form

This is called the parametric vector form of the solution.

In this case, the solution set can be written as

We emphasize the following fact in particular.

The set of solutions to a homogeneous equation

Example (The solution set is a line)

Since there were two variables in the above example, the solution set is a subset of

Since one of the variables was free, the solution set is a line:

In order to actually find a nontrivial solution to

in the above example, it suffices to substitute any nonzero value for the free variable

gives the nontrivial solution

Example (The solution set is a plane)

Since there were three variables in the above example, the solution set is a subset of

Since two of the variables were free, the solution set is a plane.

There is a natural question to ask here: is it possible to write the solution to a homogeneous matrix equation using fewer vectors than the one given in the above recipe? We will see in example in Section 2.5 that the answer is no: the vectors from the recipe are always linearly independent, which means that there is no way to write the solution with fewer vectors.

Another natural question is: are the solution sets for inhomogeneuous equations also spans? As we will see shortly, they are never spans, but they are closely related to spans.

There is a natural relationship between the number of free variables and the “size” of the solution set, as follows.

Dimension of the solution set

The above examples show us the following pattern: when there is one free variable in a consistent matrix equation, the solution set is a line, and when there are two free variables, the solution set is a plane, etc. The number of free variables is called the dimension of the solution set.

We will develop a rigorous definition of dimension in Section 2.7, but for now the dimension will simply mean the number of free variables. Compare with this important note in Section 2.5.

Intuitively, the dimension of a solution set is the number of parameters you need to describe a point in the solution set. For a line only one parameter is needed, and for a plane two parameters are needed. This is similar to how the location of a building on Peachtree Street—which is like a line—is determined by one number and how a street corner in Manhattan—which is like a plane—is specified by two numbers.


APEX Calculus

The previous section defined curves based on parametric equations. In this section we'll employ the techniques of calculus to study these curves.

We are still interested in lines tangent to points on a curve. They describe how the (y)-values are changing with respect to the (x)-values, they are useful in making approximations, and they indicate instantaneous direction of travel.

The slope of the tangent line is still (frac ext<,>) and the Chain Rule allows us to calculate this in the context of parametric equations. If (x=f(t)) and (y=g(t) ext<,>) the Chain Rule states that

Solving for (frac ext<,>) we get

provided that (fp(t) eq 0 ext<.>) This is important so we label it a Key Idea.

Key Idea 9.3.2 . Finding (frac) with Parametric Equations.

Let (x=f(t)) and (y=g(t) ext<,>) where (f) and (g) are differentiable on some open interval (I) and (fp(t) eq 0) on (I ext<.>) Then

We use this to define the tangent line.

Definition 9.3.3 . Tangent and Normal Lines.

Let a curve (C) be parametrized by (x=f(t)) and (y=g(t) ext<,>) where (f) and (g) are differentiable functions on some interval (I) containing (t=t_0 ext<.>) The to (C) at (t=t_0) is the line through (ig(f(t_0),g(t_0)ig)) with slope (m=g'(t_0)/fp(t_0) ext<,>) provided (fp(t_0) eq 0 ext<.>)

The to (C) at (t=t_0) is the line through (ig(f(t_0),g(t_0)ig)) with slope (m=-fp(t_0)/g'(t_0) ext<,>) provided (g'(t_0) eq 0 ext<.>)

The definition leaves two special cases to consider. When the tangent line is horizontal, the normal line is undefined by the above definition as (g'(t_0)=0 ext<.>) Likewise, when the normal line is horizontal, the tangent line is undefined. It seems reasonable that these lines be defined (one can draw a line tangent to the “right side” of a circle, for instance), so we add the following to the above definition.

If the tangent line at (t=t_0) has a slope of 0, the normal line to (C) at (t=t_0) is the line (x=f(t_0) ext<.>)

If the normal line at (t=t_0) has a slope of 0, the tangent line to (C) at (t=t_0) is the line (x=f(t_0) ext<.>)

Example 9.3.4 . Tangent and Normal Lines to Curves.

Let (x=5t^2-6t+4) and (y=t^2+6t-1 ext<,>) and let (C) be the curve defined by these equations.

Find the equations of the tangent and normal lines to (C) at (t=3 ext<.>)

Find where (C) has vertical and horizontal tangent lines.

We start by computing (fp(t) = 10t-6) and (g'(t) =2t+6 ext<.>) Thus

Make note of something that might seem unusual: (frac) is a function of (t ext<,>) not (x ext<.>) Just as points on the curve are found in terms of (t ext<,>) so are the slopes of the tangent lines. The point on (C) at (t=3) is ((31,26) ext<.>) The slope of the tangent line is (m=1/2) and the slope of the normal line is (m=-2 ext<.>) Thus,

the equation of the tangent line is (ds y=frac12(x-31)+26 ext<,>) and

the equation of the normal line is (ds y=-2(x-31)+26 ext<.>)

To find where (C) has a horizontal tangent line, we set (frac=0) and solve for (t ext<.>) In this case, this amounts to setting (g'(t)=0) and solving for (t) (and making sure that (fp(t) eq 0)).

The point on (C) corresponding to (t=-3) is ((67,-10) ext<>) the tangent line at that point is horizontal (hence with equation (y=-10)). To find where (C) has a vertical tangent line, we find where it has a horizontal normal line, and set (-frac=0 ext<.>) This amounts to setting (fp(t)=0) and solving for (t) (and making sure that (g'(t) eq 0)).

The point on (C) corresponding to (t=0.6) is ((2.2,2.96) ext<.>) The tangent line at that point is (x=2.2 ext<.>) The points where the tangent lines are vertical and horizontal are indicated on the graph in Figure 9.3.5.

Example 9.3.6 . Tangent and Normal Lines to a Circle.

Find where the unit circle, defined by (x=cos(t)) and (y=sin(t)) on ([0,2pi] ext<,>) has vertical and horizontal tangent lines.

Find the equation of the normal line at (t=t_0 ext<.>)

We compute the derivative following Key Idea 9.3.2:

The derivative is (0) when (cos(t) = 0 ext<>) that is, when (t=pi/2,, 3pi/2 ext<.>) These are the points ((0,1)) and ((0,-1)) on the circle. The normal line is horizontal (and hence, the tangent line is vertical) when (sin(t) =0 ext<>) that is, when (t= 0,,pi,,2pi ext<,>) corresponding to the points ((-1,0)) and ((0,1)) on the circle. These results should make intuitive sense.

The slope of the normal line at (t=t_0) is (ds m=frac = an(t_0) ext<.>) This normal line goes through the point ((cos(t_0) ,sin(t_0) ) ext<,>) giving the line

as long as (cos(t_0) eq 0 ext<.>) It is an important fact to recognize that the normal lines to a circle pass through its center, as illustrated in Figure 9.3.7. Stated in another way, any line that passes through the center of a circle intersects the circle at right angles.

Example 9.3.8 . Tangent lines when (frac) is not defined.

Find the equation of the tangent line to the astroid (x=cos^3(t) ext<,>) (y=sin^3(t)) at (t=0 ext<,>) shown in Figure 9.3.9.

We start by finding (x'(t)) and (y'(t) ext<:>)

Note that both of these are 0 at (t=0 ext<>) the curve is not smooth at (t=0) forming a cusp on the graph. Evaluating (frac) at this point returns the indeterminate form of “0/0”.

We can, however, examine the slopes of tangent lines near (t=0 ext<,>) and take the limit as (t o 0 ext<.>)

We have accomplished something significant. When the derivative (frac) returns an indeterminate form at (t=t_0 ext<,>) we can define its value by setting it to be (limlimits_)(frac ext<,>) if that limit exists. This allows us to find slopes of tangent lines at cusps, which can be very beneficial.

We found the slope of the tangent line at (t=0) to be 0 therefore the tangent line is (y=0 ext<,>) the (x)-axis.

Subsection 9.3.1 Concavity

We continue to analyze curves in the plane by considering their concavity that is, we are interested in (frac ext<,>) “the second derivative of (y) with respect to (x ext<.>)” To find this, we need to find the derivative of (frac) with respect to (x ext<>) that is,

but recall that (frac) is a function of (t ext<,>) not (x ext<,>) making this computation not straightforward.

To make the upcoming notation a bit simpler, let (h(t) = frac ext<.>) We want (frac[h(t)] ext<>) that is, we want (frac ext<.>) We again appeal to the Chain Rule. Note:

In words, to find (frac ext<,>) we first take the derivative of (frac) with respect to (t), then divide by (x'(t) ext<.>) We restate this as a Key Idea.

Key Idea 9.3.10 . Finding (frac) with Parametric Equations.

Let (x=f(t)) and (y=g(t)) be twice differentiable functions on an open interval (I ext<,>) where (fp(t) eq 0) on (I ext<.>) Then

Examples will help us understand this Key Idea.

Example 9.3.11 . Concavity of Plane Curves.

Let (x=5t^2-6t+4) and (y=t^2+6t-1) as in Example 9.3.4. Determine the (t)-intervals on which the graph is concave up/down.

Concavity is determined by the second derivative of (y) with respect to (x ext<,>) (frac ext<,>) so we compute that here following Key Idea 9.3.10.

The graph of the parametric functions is concave up when (frac gt 0) and concave down when (frac lt 0 ext<.>) We determine the intervals when the second derivative is greater/less than 0 by first finding when it is 0 or undefined.

As the numerator of (ds -frac<9><(5t-3)^3>) is never 0, (frac eq 0) for all (t ext<.>) It is undefined when (5t-3=0 ext<>) that is, when (t= 3/5 ext<.>) Following the work established in Section 3.4, we look at values of (t) greater/less than (3/5) on a number line:

Reviewing Example 9.3.4, we see that when (t=3/5=0.6 ext<,>) the graph of the parametric equations has a vertical tangent line. This point is also a point of inflection for the graph, illustrated in Figure 9.3.12.

The video in Figure 9.3.13 shows how this information can be used to sketch the curve by hand.

Example 9.3.14 . Concavity of Plane Curves.

Find the points of inflection of the graph of the parametric equations (x=sqrt ext<,>) (y=sin(t) ext<,>) for (0leq tleq 16 ext<.>)

We need to compute (frac) and (frac ext<.>)

The points of inflection are found by setting (frac=0 ext<.>) This is not trivial, as equations that mix polynomials and trigonometric functions generally do not have “nice” solutions.

In Figure 9.3.15.(a) we see a plot of the second derivative. It shows that it has zeros at approximately (t=0.5,,3.5,,6.5,,9.5,,12.5) and (16 ext<.>) These approximations are not very good, made only by looking at the graph. Newton's Method provides more accurate approximations. Accurate to 2 decimal places, we have:

The corresponding points have been plotted on the graph of the parametric equations in Figure 9.3.15.(b). Note how most occur near the (x)-axis, but not exactly on the axis.

Subsection 9.3.2 Arc Length

We continue our study of the features of the graphs of parametric equations by computing their arc length.

Recall in Section 7.4 we found the arc length of the graph of a function, from (x=a) to (x=b ext<,>) to be

We can use this equation and convert it to the parametric equation context. Letting (x=f(t)) and (y=g(t) ext<,>) we know that (frac = g'(t)/fp(t) ext<.>) It will also be useful to calculate the differential of (x ext<:>)

Starting with the arc length formula above, consider:

Note the new bounds (no longer “(x)” bounds, but “(t)” bounds). They are found by finding (t_1) and (t_2) such that (a= f(t_1)) and (b=f(t_2) ext<.>) This formula is important, so we restate it as a theorem.

Theorem 9.3.17 . Arc Length of Parametric Curves.

Let (x=f(t)) and (y=g(t)) be parametric equations with (fp) and (g') continuous on ([t_1,t_2] ext<,>) on which the graph traces itself only once. The arc length of the graph, from (t=t_1) to (t=t_2 ext<,>) is

Note: Theorem 9.3.17 makes use of differentiability on closed intervals, just as was done in Section 7.4.

As before, these integrals are often not easy to compute. We start with a simple example, then give another where we approximate the solution.

Example 9.3.18 . Arc Length of a Circle.

Find the arc length of the circle parametrized by (x=3cos(t) ext<,>) (y=3sin(t)) on ([0,3pi/2] ext<.>)

By direct application of Theorem 9.3.17, we have

Apply the Pythagorean Theorem.

egin amp = int_0^ <3pi/2>3 , dt amp = 3tBig|_0^ <3pi/2>= 9pi/2 ext <.>end

This should make sense we know from geometry that the circumference of a circle with radius 3 is (6pi ext<>) since we are finding the arc length of (3/4) of a circle, the arc length is (3/4cdot 6pi = 9pi/2 ext<.>)

Example 9.3.19 . Arc Length of a Parametric Curve.

The graph of the parametric equations (x=t(t^2-1) ext<,>) (y=t^2-1) crosses itself as shown in Figure 9.3.20, forming a “teardrop.” Find the arc length of the teardrop.

We can see by the parametrizations of (x) and (y) that when (t=pm 1 ext<,>) (x=0) and (y=0 ext<.>) This means we'll integrate from (t=-1) to (t=1 ext<.>) Applying Theorem 9.3.17, we have

Unfortunately, the integrand does not have an antiderivative expressible by elementary functions. We turn to numerical integration to approximate its value. Using 4 subintervals, Simpson's Rule approximates the value of the integral as (2.65051 ext<.>) Using a computer, more subintervals are easy to employ, and (n=20) gives a value of (2.71559 ext<.>) Increasing (n) shows that this value is stable and a good approximation of the actual value.

Subsection 9.3.3 Surface Area of a Solid of Revolution

Related to the formula for finding arc length is the formula for finding surface area. We can adapt the formula found in Theorem 7.4.13 from Section 7.4 in a similar way as done to produce the formula for arc length done before.

Theorem 9.3.21 . Surface Area of a Solid of Revolution.

Consider the graph of the parametric equations (x=f(t)) and (y=g(t) ext<,>) where (fp) and (g') are continuous on an open interval (I) containing (t_1) and (t_2) on which the graph does not cross itself.

The surface area of the solid formed by revolving the graph about the (x)-axis is (where (g(t)geq 0) on ([t_1,t_2])):

The surface area of the solid formed by revolving the graph about the (y)-axis is (where (f(t)geq 0) on ([t_1,t_2])):

Example 9.3.22 . Surface Area of a Solid of Revolution.

Consider the teardrop shape formed by the parametric equations (x=t(t^2-1) ext<,>) (y=t^2-1) as seen in Example 9.3.19. Find the surface area if this shape is rotated about the (x)-axis, as shown in Figure 9.3.23.

The teardrop shape is formed between (t=-1) and (t=1 ext<.>) Using Theorem 9.3.21, we see we need for (g(t)geq 0) on ([-1,1] ext<,>) and this is not the case. To fix this, we simplify replace (g(t)) with (-g(t) ext<,>) which flips the whole graph about the (x)-axis (and does not change the surface area of the resulting solid). The surface area is:

Once again we arrive at an integral that we cannot compute in terms of elementary functions. Using Simpson's Rule with (n=20 ext<,>) we find the area to be (S=9.44 ext<.>) Using larger values of (n) shows this is accurate to 2 places after the decimal.

After defining a new way of creating curves in the plane, in this section we have applied calculus techniques to the parametric equation defining these curves to study their properties. In the next section, we define another way of forming curves in the plane. To do so, we create a new coordinate system, called polar coordinates, that identifies points in the plane in a manner different than from measuring distances from the (y)- and (x)- axes.


Watch the video: PLTW IED Activity - Parametric Constraints (December 2021).