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Precalculus : Parametric Equations

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Example Questions

Example Question #31 : Algebraic Vectors And Parametric Equations

Write a vector equation describing the line passing through P₁ (1, 4) and parallel to the vector $\vec{a}$ = (3, 4).

Possible Answers:

(x+3,y+4)=t(1,4)

(x-1, y-4)=t(3, 4)

(x-3,y-4)=t(1,4)

(x+1,y+4)=t(3,4)

Correct answer:

(x-1, y-4)=t(3, 4)

Explanation:

First, draw the vector $\vec{a}$ = (3, 4); this is represented in red below. Then, plot the point P₁ (1, 4), and draw a line (represented in blue) through it that is parallel to the vector $\vec{a}$ .

Screen shot 2020 05 29 at 11.28.09 am

We must find the equation of line . For any point P₂ (x, y) on , $\vec{P1P2}=(x-1,y-4)$ . Since $\vec{P1P2}$ is on line and is parallel to $\vec{a}$ , $\vec{P1P2}=t\vec{a}$ for some value of t. By substitution, we have (x-1, y-4)=t(3, 4) . Therefore, the equation is a vector equation describing all of the points (x, y) on line parallel to $\vec{a}$ through P₁ (1, 4).

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Example Question #32 : Algebraic Vectors And Parametric Equations

True or false: A line through P₁ (x₁, y₁) that is parallel to the vector $\vec{a}=(a_1,a_2)$ is defined by the set of points P_2(x,y) such that $\vec{P_1P_2}=t\vec{a}$ for some real number t. Therefore, (x-x_1, y-y_1)=t(a_1,a_2) .

Possible Answers:

True

False

Correct answer:

True

Explanation:

This is true. The independent variable in this equation is called a parameter.

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Example Question #1 : Parametric Equations

Find the parametric equations for a line parallel to $\vec{a}=(3,2)$ and passing through the point (0, 5).

Possible Answers:

x = 3t

y = 5 + 2t

x = 3 + 2t

y = 5t

x = 5t

y = 3 + 2t

x = 3

y = 2 + 5t

Correct answer:

x = 3t

y = 5 + 2t

Explanation:

A line through a point (x₁,y₁) that is parallel to the vector $\vec a$ = (a₁, a₂) has the following parametric equations, where t is any real number.

x=x_1 + ta_1

y=y_1 + ta_2

Using the given vector and point, we get the following:

x = 3t

y = 5 + 2t

Each value of t creates a distinct (x, y) ordered pair. You can think of these points as representing positions of an object, and of t as representing time in seconds. Evaluating the parametric equations for a value of t gives us the coordinates of the position of the object after t seconds have passed.

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Example Question #2 : Parametric Equations

Find the parametric equations for a line parallel to $\vec{a}=(-7,3.5)$ and passing through the point (4, -3).

Possible Answers:

x = -3 - 3.5t

y = 4 + 7t

x = -7 + 4t

y = 3.5 - 3t

x = -3 + 3.5t

y = 4 - 7t

x = 4 - 7t

y = -3 + 3.5t

Correct answer:

x = 4 - 7t

y = -3 + 3.5t

Explanation:

A line through a point (x₁,y₁) that is parallel to the vector $\vec a$ = (a₁, a₂) has the following parametric equations, where t is any real number.

x=x_1 + ta_1

y=y_1 + ta_2

Using the given vector and point, we get the following:

x = 4 - 7t

y = -3 + 3.5t

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Example Question #2 : Parametric Equations

Write the parametric equation for the line y = -3x +1.5

Possible Answers:

x = t

y = -3t +1.5

x = -3t +1.5

y = t

x = -3t +1.5

y = -3t +1.5

x = t

y = 1.5t - 3

Correct answer:

x = t

y = -3t +1.5

Explanation:

In the equation y = -3x +1.5, x is the independent variable and y is the dependent variable. In a parametric equation, t is the independent variable, and x and y are both dependent variables.

Start by setting the independent variables x and t equal to one another, and then you can write two parametric equations in terms of t:

x = t

y = -3t +1.5

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Example Question #3 : Parametric Equations

Write the parametric equation for the line y = 5x - 3.

Possible Answers:

x = 5t - 3

y = 5t - 3

x = 5t - 3

y = t

x = t

y = t

x = t

y = 5t - 3

Correct answer:

x = t

y = 5t - 3

Explanation:

In the equation y = 5x - 3, x is the independent variable and y is the dependent variable. In a parametric equation, t is the independent variable, and x and y are both dependent variables.

Start by setting the independent variables x and t equal to one another, and then you can write two parametric equations in terms of t:

x = t

y = 5t - 3

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Example Question #4 : Parametric Equations

Write an equation in slope-intercept form of the line with the given parametric equations:

x = -4t+7

y=t-12

Possible Answers:

y=t-12

$y=\frac{-x+55}{4}$

y=-4t-84

$y=\frac{-x+7}{4}$

Correct answer:

$y=\frac{-x+55}{4}$

Explanation:

Start by solving each parametric equation for t:

x = -4t+7

x-7=-4t

$\frac{-x+7}{4}=t$

y=t-12

y-12=t

Next, write an equation involving the expressions for t; since both are equal to t, we can set them equal to one another:

$\frac{-x+7}{4}=y-12$

Multiply both sides by the LCD, 4:

-x+7=4y-48

Get y by itself to represent this as an equation in slope-intercept (y = mx + b) form:

4y = -x+55

$y=\frac{-x+55}{4}$

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Example Question #5 : Parametric Equations

Write an equation in slope-intercept form of the line with the given parametric equations:

x = 3t-5

y=-2t+7

Possible Answers:

$y=\frac{x+5}{3}$

$y=\frac{-2x+11}{3}$

y=3x-5

y=-2x+7

Correct answer:

$y=\frac{-2x+11}{3}$

Explanation:

Start by solving each parametric equation for t:

x = 3t-5

x+5=3t

$\frac{x+5}{3}=t$

y=-2t+7

y-7=-2t

$\frac{-y+7}{2}=t$

Next, write an equation involving the expressions for t; since both are equal to t, we can set them equal to one another:

$\frac{x+5}{3}=\frac{-y+7}{2}$

Multiply both sides by the LCD, 6:

2(x+5)=3(-y+7)

2x+10=-3y+21

Get y by itself to represent this as an equation in slope-intercept (y = mx + b) form:

-3y = 2x - 11

$y=\frac{-2x+11}{3}$

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Example Question #6 : Parametric Equations

A football punter kicks a ball with an initial velocity of 40 ft/s at an angle of 29^o to the horizontal. After 0.5 seconds, how far has the ball travelled horizontally and vertically?

Possible Answers:

x=15.5

y=6.8

x=17.5

y=5.7

x=18.8

y=7.9

x=21.4

y=9.1

Correct answer:

x=17.5

y=5.7

Explanation:

To solve this problem, we need to know that the path of a projectile can be described with the following equations:

$x=t\left | \vec{v} \right |\cos(\theta)$

$y=t\left | \vec{v} \right |\sin(\theta) +\frac{1}{2}gt^2$

In these equations, t is time and g is the acceleration due to gravity.

First, you need to write the position of the ball as a pair of parametric equations that define the path of the ball at anytime, t, in seconds:

$x=t\left | \vec{v} \right |\cos(\theta)$

$x=t(40)\cos(29\degree)$

$x=40t\cos(29\degree)$

As you set up the equation for y, use the value g = -32.

$y=t\left | \vec{v} \right |\sin(\theta) +\frac{1}{2}gt^2$

$y=t(40)\sin(29\degree)+\frac{1}{2}(-32)t^2$

$y=40t\sin(29\degree)-16t^2$

Finally, find x and y when t = .05:

$x=40(0.5)\cos(29\degree)$

$y=40(0.5)\sin(29\degree)-16(0.5)^2$

Use a calculator to solve, making sure you are in degree mode:

x=17.5

y=5.7

This means that after 0.5 seconds, the ball has travelled 17.5 feet horizontally and 5.7 feet vertically.

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Example Question #7 : Parametric Equations

A pitcher throws a fastball, and the batter swings and connects with the ball. If the ball has an initial velocity of 150 f/s and an angle of 20^o about the horizontal, what parametric equations will model the motion of the ball? What height will the ball be at when it has travelled 400 feet horizontally?

Possible Answers:

$x=150t\cos(20\degree)$

$y=150t\sin(20\degree)-16t^2$

Height of ball when it has travelled 400 ft is 15.52 feet above the ground

$x=150t\cos(20\degree)$

$y=150t\sin(20\degree)-16t^2$

Height of ball when it has travelled 400 ft is 21.59 feet above the ground

$x=150t\cos(20\degree)$

$y=150t\sin(20\degree)-16t^2$

Height of ball when it has travelled 400 ft is 16.65 feet above the ground

$x=150t\cos(20\degree)$

$y=150t\sin(20\degree)-16t^2$

Height of ball when it has travelled 400 ft is 14.38 feet above the ground

Correct answer:

$x=150t\cos(20\degree)$

$y=150t\sin(20\degree)-16t^2$

Height of ball when it has travelled 400 ft is 16.65 feet above the ground

Explanation:

To solve this problem, we need to know that the path of a projectile can be described with the following equations:

$x=t\left | \vec{v} \right |\cos(\theta)$

$y=t\left | \vec{v} \right |\sin(\theta) +\frac{1}{2}gt^2$

In these equations, t is time and g is the acceleration due to gravity.

First, you need to write the position of the ball as a pair of parametric equations that define the path of the ball at anytime, t, in seconds:

$x=t\left | \vec{v} \right |\cos(\theta)$

$x=t(150)\cos(20\degree)$

$x=150t\cos(20\degree)$

As you set up the equation for y, use the value g = -32.

$y=t\left | \vec{v} \right |\sin(\theta) +\frac{1}{2}gt^2$

$y=t(150)\sin(20\degree)+\frac{1}{2}(-32)t^2$

$y=150t\sin(20\degree)-16t^2$

Finally, find x and y when t = .05:

$x=150t\cos(20\degree)$

$y=150t\sin(20\degree)-16t^2$

To find the height that the ball will be at when it has travelled horizontally, plug 400 feet (horizontal distance) in for x.

$400=150t\cos(20\degree)$

$t = \frac{400}{150\cos(20\degree)}=2.84$

This tells us that the ball has travelled 400 feet horizontally when 2.84 seconds have passed. Let's use the value of t = 2.84 and plug it into our equation for y to see how high the ball is at this time:

$y=150t\sin(20\degree)-16t^2=150(2.84)\sin(20\degree)-16(2.84)^2=16.65$

This means that when the ball has travelled 400 feet horizontally, 2.84 seconds have passed, and the ball is 16.65 feet above the ground.

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Precalculus : Parametric Equations

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #31 : Algebraic Vectors And Parametric Equations

Example Question #32 : Algebraic Vectors And Parametric Equations

Example Question #1 : Parametric Equations

Example Question #2 : Parametric Equations

Example Question #2 : Parametric Equations

Example Question #3 : Parametric Equations

Example Question #4 : Parametric Equations

Example Question #5 : Parametric Equations

Example Question #6 : Parametric Equations

Example Question #7 : Parametric Equations

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