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Access the following online resource for additional instruction and practice with graphs of parametric equations.

Key concepts

  • When there is a third variable, a third parameter on which x and y depend, parametric equations can be used.
  • To graph parametric equations by plotting points, make a table with three columns labeled t , x ( t ) , and y ( t ) . Choose values for t in increasing order. Plot the last two columns for x and y . See [link] and [link] .
  • When graphing a parametric curve by plotting points, note the associated t -values and show arrows on the graph indicating the orientation of the curve. See [link] and [link] .
  • Parametric equations allow the direction or the orientation of the curve to be shown on the graph. Equations that are not functions can be graphed and used in many applications involving motion. See [link] .
  • Projectile motion depends on two parametric equations: x = ( v 0 cos θ ) t and y = 16 t 2 + ( v 0 sin θ ) t + h . Initial velocity is symbolized as v 0 . θ represents the initial angle of the object when thrown, and h represents the height at which the object is propelled.

Section exercises

Verbal

What are two methods used to graph parametric equations?

plotting points with the orientation arrow and a graphing calculator

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What is one difference in point-plotting parametric equations compared to Cartesian equations?

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Why are some graphs drawn with arrows?

The arrows show the orientation, the direction of motion according to increasing values of t .

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Name a few common types of graphs of parametric equations.

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Why are parametric graphs important in understanding projectile motion?

The parametric equations show the different vertical and horizontal motions over time.

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Graphical

For the following exercises, graph each set of parametric equations by making a table of values. Include the orientation on the graph.

{ x ( t ) = t y ( t ) = t 2 1

t x y
3
2
1
0
1
2
3
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{ x ( t ) = t 1 y ( t ) = t 2

t 3 2 1 0 1 2
x
y
Graph of the given equations - looks like an upward opening parabola.
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{ x ( t ) = 2 + t y ( t ) = 3 2 t

t 2 1 0 1 2 3
x
y
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{ x ( t ) = 2 2 t y ( t ) = 3 + t

t 3 2 1 0 1
x
y
Graph of the given equations - a line, negative slope.
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{ x ( t ) = t 3 y ( t ) = t + 2

t 2 1 0 1 2
x
y
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{ x ( t ) = t 2 y ( t ) = t + 3

t 2 1 0 1 2
x
y
Graph of the given equations - looks like a sideways parabola, opening to the right.
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For the following exercises, sketch the curve and include the orientation.

{ x ( t ) = t y ( t ) = t

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{ x ( t ) = t y ( t ) = t

Graph of the given equations - looks like the left half of an upward opening parabola.
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{ x ( t ) = 5 | t | y ( t ) = t + 2

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{ x ( t ) = t + 2 y ( t ) = 5 | t |

Graph of the given equations - looks like a downward opening absolute value function.
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{ x ( t ) = 4 sin t y ( t ) = 2 cos t

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{ x ( t ) = 2 sin t y ( t ) = 4 cos t

Graph of the given equations - a vertical ellipse.
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{ x ( t ) = 3 cos 2 t y ( t ) = −3 sin t

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{ x ( t ) = 3 cos 2 t y ( t ) = −3 sin 2 t

Graph of the given equations- line from (0, -3) to (3,0). It is traversed in both directions, positive and negative slope.
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{ x ( t ) = sec t y ( t ) = tan t

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{ x ( t ) = sec t y ( t ) = tan 2 t

Graph of the given equations- looks like an upward opening parabola.
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{ x ( t ) = 1 e 2 t y ( t ) = e t

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For the following exercises, graph the equation and include the orientation. Then, write the Cartesian equation.

{ x ( t ) = t 1 y ( t ) = t 2

Graph of the given equations- looks like a downward opening parabola.
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{ x ( t ) = t 3 y ( t ) = t + 3

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{ x ( t ) = 2 cos t y ( t ) = sin t

Graph of the given equations- horizontal ellipse.

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{ x ( t ) = 7 cos t y ( t ) = 7 sin t

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{ x ( t ) = e 2 t y ( t ) = e t

Graph of the given equations- looks like the lower half of a sideways parabola opening to the right
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For the following exercises, graph the equation and include the orientation.

x = t 2 , y = 3 t , 0 t 5

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x = 2 t , y = t 2 , 5 t 5

Graph of the given equations- looks like an upwards opening parabola
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x = t , y = 25 t 2 , 0 < t 5

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x ( t ) = t , y ( t ) = t , t 0

Graph of the given equations- looks like the upper half of a sideways parabola opening to the left
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x = 2 cos t , y = 6 sin t , 0 t π

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x = sec t , y = tan t , π 2 < t < π 2

Graph of the given equations- the left half of a hyperbola with diagonal asymptotes
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For the following exercises, use the parametric equations for integers a and b :

x ( t ) = a cos ( ( a + b ) t ) y ( t ) = a cos ( ( a b ) t )

Graph on the domain [ π , 0 ] , where a = 2 and b = 1 , and include the orientation.

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Graph on the domain [ π , 0 ] , where a = 3 and b = 2 , and include the orientation.

Graph of the given equations - vertical periodic trajectory
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Graph on the domain [ π , 0 ] , where a = 4 and b = 3 , and include the orientation.

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Questions & Answers

Cos45/sec30+cosec30=
dinesh Reply
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
Miiro Reply
I dnt get dis work well
john Reply
what is one-to-one function
Iwori Reply
what is the procedure in solving quadratic equetion at least 6?
Qhadz Reply
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
wisdom Reply
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
Gautam Reply
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
favour Reply
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
Ken Reply
proof
AUSTINE
sebd me some questions about anything ill solve for yall
Manifoldee Reply
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
Kristof Reply
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
SO THE ANSWER IS X=-8
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
1KI POWER 1/3 PLEASE SOLUTIONS
Prashant Reply
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
Reuben Reply
which of these functions is not uniformly cintinuous on (0, 1)? sinx
Pooja Reply
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1
Basant Reply

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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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