Algebraic Vectors and Parametric Equations - Pre-Calculus

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 = 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

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

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:

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:

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.

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:

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

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:

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.

It will be helpful to first draw the vector so we can see what quadrant the angle is in:

Since the vector is pointing up and to the right, it is in the first quadrant. To determine the angle, set up a trig equation with tangent, since the component 5 is opposite and the component 4 is adjacent to the angle we are looking for:

to solve for theta, take the inverse tangent of both sides:

Now we have the direction, and we can solve for the magnitude using Pythagorean Theorem:

take the square root of both sides

The vector in polar form is

To convert to polar form, we need to find the magnitude of the vector, , and the angle it forms with the positive -axis going counterclockwise, or . This is shown in the figure below.

We find the angle using trigonometric identities:

Using a calculator,

To find the magnitude of a vector, we add up the squares of each component and take the square root:

.

So, our vector written in polar form is

To convert a point or a vector to its polar form, use the following equations to determine the magnitude and the direction.

Substitute the vector to the equations to find the magnitude and the direction.

The polar form is:

The polar form of the vector is:

Find .

Find the angle.

We know that converting into polar form requires using the formulas : and .

Solving for r will give us the equation:

We can then solve this equation for theta thusly:

We substitute the values of x and y found in the vector equation to get the angle measure:

Since we have already solved for the radius in terms of x and y and the angle, we substitute the proper values into the equation to get the radius.

Therefore, the vector expressed in polar form is:

To find the polar form of , two formula will be needed since the polar form of a vector is defined as .

However, the direction of is not in the first quadrant, but lies in the third quadrant. It is mandatory to add 180 degrees so that the angle corresponds to the correct quadrant.

Therefore, the answer is:

To figure out the horizontal component, set up an equation involving cosine, since that side of the implied triangle is adjacent to the 48-degree angle:

To solve for x, first find the cosine of 48, then multiply by 11:

To figure out the vertical component, set up an equation involving sine, since that side of the implied triangle is opposite the 48-degree angle:

to solve for y, just like x, first find the sine of 48, then multiply by 11:

Putting this in component form results in the vector

First, it could be helpful to draw the vector so that we can get a sense of what it looks like. The component form means from the start to the end, it moves forward 2 and down 7:

We can now use the Pythagorean Theorem to solve for the magnitude:

note that if you had used -7 that would be perfect as well, since that would give you the exact same answer.

take the square root of both sides

The magnitude is .

Now to find the angle we should use trigonometric ratios. We can consider the angle being formed by the vector and the component 2, then we can place it in the right quadrant later on. We know that the tangent of that angle is :

now we can take of both sides to determine theta:

We can see that the angle for this particular vector is pointing down and to the right, so the angle we want is in the 4th quadrant. This angle would be

To find vector , the point A is the terminal point and point B is the starting point.

The directional vector can be determined by subtracting the start from the terminal point.

The correct vector is given by the subtraction of the two points: .

Since the subtraction here is component-wise, it is given by the formula: .

This results in the vector .

The vector is also correct as it is a scalar multiple of the vector marked as correct, it is found by reversing the order of the subtraction of the two points.