In order to find the horizontal component, set up an equation involving cosine with 7 as the hypotenuse, since the side in the implied triangle that represents the horizontal component is adjacent to the 22-degree angle:

First, find the cosine of 22, then multiply by 7

To find the vertical component, set up an equation involving sine, since the side in the implied triangle that represents the vertical component is opposite the 22-degree angle:

First, find the sine of 22, then multiply by 7

We are almost done, but we need to make a small adjustment. The picture indicates that the vector points up and to the left, so the horizontal component, 6.49, should be negative:

First, draw the vector  = (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 .

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

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

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

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.

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

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

x = 4 - 7t

y = -3 + 3.5t

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.

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

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

Start by solving each parametric equation for t:

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

Multiply both sides by the LCD, 4:

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

Start by solving each parametric equation for t:

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

Multiply both sides by the LCD, 6:

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

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

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:

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

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

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

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