10.7 Parametric equations: graphs

Algebra and trigonometry Page 1 / 4

In this section you will:

Graph plane curves described by parametric equations by plotting points.
Graph parametric equations.

It is the bottom of the ninth inning, with two outs and two men on base. The home team is losing by two runs. The batter swings and hits the baseball at 140 feet per second and at an angle of approximately $45°$ to the horizontal. How far will the ball travel? Will it clear the fence for a game-winning home run? The outcome may depend partly on other factors (for example, the wind), but mathematicians can model the path of a projectile and predict approximately how far it will travel using parametric equations . In this section, we’ll discuss parametric equations and some common applications, such as projectile motion problems.

Photo of a baseball batter swinging. — Parametric equations can model the path of a projectile. (credit: Paul Kreher, Flickr)

Graphing parametric equations by plotting points

In lieu of a graphing calculator or a computer graphing program, plotting points to represent the graph of an equation is the standard method. As long as we are careful in calculating the values, point-plotting is highly dependable.

How To

Given a pair of parametric equations, sketch a graph by plotting points.

Construct a table with three columns: $t, x (t), and y (t) .$
Evaluate $x$ and $y$ for values of $t$ over the interval for which the functions are defined.
Plot the resulting pairs $(x, y) .$

Sketching the graph of a pair of parametric equations by plotting points

Sketch the graph of the parametric equations $x (t) = t^{2} + 1, y (t) = 2 + t .$

Construct a table of values for $t, x (t),$ and $y (t),$ as in [link] , and plot the points in a plane.

$t$	$x (t) = t^{2} + 1$	$y (t) = 2 + t$
$- 5$	$26$	$- 3$
$- 4$	$17$	$- 2$
$- 3$	$10$	$- 1$
$- 2$	$5$	$0$
$- 1$	$2$	$1$
$0$	$1$	$2$
$1$	$2$	$3$
$2$	$5$	$4$
$3$	$10$	$5$
$4$	$17$	$6$
$5$	$26$	$7$

The graph is a parabola with vertex at the point $(1, 2),$ opening to the right. See [link] .

Graph of the given parabola opening to the right.

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Sketch the graph of the parametric equations $x = \sqrt{t}, y = 2 t + 3, 0 \leq t \leq 3.$

Graph of the given parametric equations with the restricted domain - it looks like the right half of an upward opening parabola.

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Sketching the graph of trigonometric parametric equations

Construct a table of values for the given parametric equations and sketch the graph:

\begin{array}{l} \begin{array}{l} x = 2 \cos t \\ y = 4 \sin t \end{array} \end{array}

Construct a table like that in [link] using angle measure in radians as inputs for $t,$ and evaluating $x$ and $y .$ Using angles with known sine and cosine values for $t$ makes calculations easier.

$t$	$x = 2 \cos t$	$y = 4 \sin t$
0	$x = 2 \cos (0) = 2$	$y = 4 \sin (0) = 0$
$\frac{π}{6}$	$x = 2 \cos (\frac{π}{6}) = \sqrt{3}$	$y = 4 \sin (\frac{π}{6}) = 2$
$\frac{π}{3}$	$x = 2 \cos (\frac{π}{3}) = 1$	$y = 4 \sin (\frac{π}{3}) = 2 \sqrt{3}$
$\frac{π}{2}$	$x = 2 \cos (\frac{π}{2}) = 0$	$y = 4 \sin (\frac{π}{2}) = 4$
$\frac{2 π}{3}$	$x = 2 \cos (\frac{2 π}{3}) = - 1$	$y = 4 \sin (\frac{2 π}{3}) = 2 \sqrt{3}$
$\frac{5 π}{6}$	$x = 2 \cos (\frac{5 π}{6}) = - \sqrt{3}$	$y = 4 \sin (\frac{5 π}{6}) = 2$
$π$	$x = 2 \cos (π) = - 2$	$y = 4 \sin (π) = 0$
$\frac{7 π}{6}$	$x = 2 \cos (\frac{7 π}{6}) = - \sqrt{3}$	$y = 4 \sin (\frac{7 π}{6}) = - 2$
$\frac{4 π}{3}$	$x = 2 \cos (\frac{4 π}{3}) = - 1$	$y = 4 \sin (\frac{4 π}{3}) = - 2 \sqrt{3}$
$\frac{3 π}{2}$	$x = 2 \cos (\frac{3 π}{2}) = 0$	$y = 4 \sin (\frac{3 π}{2}) = - 4$
$\frac{5 π}{3}$	$x = 2 \cos (\frac{5 π}{3}) = 1$	$y = 4 \sin (\frac{5 π}{3}) = - 2 \sqrt{3}$
$\frac{11 π}{6}$	$x = 2 \cos (\frac{11 π}{6}) = \sqrt{3}$	$y = 4 \sin (\frac{11 π}{6}) = - 2$
$2 π$	$x = 2 \cos (2 π) = 2$	$y = 4 \sin (2 π) = 0$

[link] shows the graph.

Graph of the given equations - a vertical ellipse.

By the symmetry shown in the values of $x$ and $y,$ we see that the parametric equations represent an ellipse . The ellipse is mapped in a counterclockwise direction as shown by the arrows indicating increasing $t$ values.

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Graph the parametric equations: $x = 5 \cos t, y = 3 \sin t .$

Graph of the given equations - a horizontal ellipse.

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Graphing parametric equations and rectangular form together

Graph the parametric equations $x = 5 \cos t$ and $y = 2 \sin t .$ First, construct the graph using data points generated from the parametric form . Then graph the rectangular form of the equation. Compare the two graphs.

Construct a table of values like that in [link] .

$t$	$x = 5 \cos t$	$y = 2 \sin t$
$0$	$x = 5 \cos (0) = 5$	$y = 2 \sin (0) = 0$
$1$	$x = 5 \cos (1) \approx 2.7$	$y = 2 \sin (1) \approx 1.7$
$2$	$x = 5 \cos (2) \approx −2.1$	$y = 2 \sin (2) \approx 1.8$
$3$	$x = 5 \cos (3) \approx −4.95$	$y = 2 \sin (3) \approx 0.28$
$4$	$x = 5 \cos (4) \approx −3.3$	$y = 2 \sin (4) \approx −1.5$
$5$	$x = 5 \cos (5) \approx 1.4$	$y = 2 \sin (5) \approx −1.9$
$−1$	$x = 5 \cos (−1) \approx 2.7$	$y = 2 \sin (−1) \approx −1.7$
$−2$	$x = 5 \cos (−2) \approx −2.1$	$y = 2 \sin (−2) \approx −1.8$
$−3$	$x = 5 \cos (−3) \approx −4.95$	$y = 2 \sin (−3) \approx −0.28$
$−4$	$x = 5 \cos (−4) \approx −3.3$	$y = 2 \sin (−4) \approx 1.5$
$−5$	$x = 5 \cos (−5) \approx 1.4$	$y = 2 \sin (−5) \approx 1.9$

Plot the $(x, y)$ values from the table. See [link] .

Graph of the given ellipse in parametric and rectangular coordinates - it is the same thing in both images.

Next, translate the parametric equations to rectangular form. To do this, we solve for $t$ in either $x (t)$ or $y (t),$ and then substitute the expression for $t$ in the other equation. The result will be a function $y (x)$ if solving for $t$ as a function of $x,$ or $x (y)$ if solving for $t$ as a function of $y .$

\begin{array}{l} x = 5 \cos t \\ \frac{x}{5} = \cos t & Solve for \cos t . \\ y = 2 \sin t \begin{matrix}  \end{matrix} & Solve for \sin t . \\ \frac{y}{2} = \sin t \end{array}

Then, use the Pythagorean Theorem .

\begin{array}{r} \cos^{2} t + \sin^{2} t = 1 \\ {(\frac{x}{5})}^{2} + {(\frac{y}{2})}^{2} = 1 \\ \frac{x^{2}}{25} + \frac{y^{2}}{4} = 1 \end{array}

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Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.

Kate Reply

To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.

Muhammed

What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?

Kate Reply

what is the change in momentum of a body?

Eunice Reply

what is a capacitor?

Raymond Reply

Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.

Gautam

A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate

Maria Reply

please solve

Sharon

8m/s²

Aishat

What is Thermodynamics

Muordit

velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.

Mehmet

A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat

Saheed Reply

50 m/s due south east

Someone

which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer

Ramon Reply

I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.

Someone

Scratch that

Someone

temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....

Someone

about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle

Someone

definitely of physics

Haryormhidey Reply

how many start and codon

Esrael Reply

what is field

Felix Reply

physics, biology and chemistry this is my Field

ALIYU

field is a region of space under the influence of some physical properties

Collete

what is ogarnic chemistry

WISDOM Reply

determine the slope giving that 3y+ 2x-14=0

WISDOM

Another formula for Acceleration

Belty Reply

a=v/t. a=f/m a

IHUMA

innocent

Adah

pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?

Nassze Reply

how do lnternal energy measures

Esrael

Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.

JALLAH Reply

No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.

Dlovan

Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?

Robert

like charges repel while unlike charges atttact

Raymond

What is specific heat capacity

Destiny Reply

Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).

AI-Robot

specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin

ROKEEB

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Source: OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6

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