### Topics

## Introduction to Points, Vectors, And Functions - At A Glance:

We can describe trajectories of things with parametric equations.

Before we go forward, we should review parabolas via Super Mario Brothers. When Mario jumps, we can parameterize his trajectory *d*(*x,y*) = as a function of time *t* by writing *d*(*t*) = <*x*(*t*),*y*(*t*)>.

To graph this parametric equation, we graph a bunch of individual points. To find the coordinates of a point we take a *t* value and find the corresponding *x* and *y* values.

How do we know if Mario landed safely on the other side of the canyon? We need to know if point (x_{1},y_{1}) is on the trajectory.

We are going to reverse the graphing process. We'll start with point (x_{1},y_{1}) and find the *t*-value that produced it, if there is one.

To determine if a particular point (*x*,*y*) is on a parametric graph we need to find a single value of the parameter *t* that produces the correct *x*- and *y*-values at the same time.

#### Example 1

Determine if the point (5,-5) is on the graph of the parametric equations * x* = 4*t* + 1
*y* = 2*t*-7.
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If *x* = 5 then
5 = 4*t* + 1
therefore *t* must equal 1. Check to see if t = 1 produces the the correct *y* value. When t = 1,
y = 2(1)-7 = -5.
This is the correct *y*-value. We conclude that the point (5,-5) is on the graph and occurs when t = 1.
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#### Example 2

Determine if the point (17,0) is on the graph of the parametric equations
x = 4t + 1
y = 2t-7. | |

If x = 17 then 17 = 4t + 1 therefore t = 4. See what we find from the *y* equation when t = 4. We find y = 2(4)-7 = 1. Since t = 4 is the only value of the parameter that makes x = 17, and at this time y ≠ 0, the point (17,0) is not on the graph. There's no value of *t* that makes x = 17 and y = 0 at the same time. When the parametric equations aren't linear, there may be more than one value of *t* that could generate the point in question. | |

#### Example 3

Determine if the point (4,-8) is on the graph of the parametric equations x = t^{2}
y = t^{3}. | |

When x = 4 we can have either t = 2 or t = -2. We need to see if either of these *t*-values produces y = -8. When t = 2 we find y = (2)^{3} = 8, which isn't what we want. When t = -2 we find y = (-2)^{3} = -8, which is what we want. The point (4,-8) is on the graph of the parametric equations and occurs when t = -2. If the equation for *y* is nicer, or has few solutions, than the equation for *x*, we might use the *y*-equation first to find the value of *t*. | |

#### Example 4

Determine if the point (4,3) is on the graph of the parametric equations x = t^{2} y = t + 2 | |

If we solve the *x* equation for *t* we'll find two possible *t*-values to check. If we solve the *y* equation we only find one.
When y = 3 we have t = 1. Then
x = (1)^{2} = 1.
Since the only value of *t* that makes y = 3 doesn't make x = 4, the point (4,3) isn't on the graph.
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#### Exercise 1

- Determine whether each point is on the graph of the parametric equations

x = t^{3}y = t^{4}

Answer

- Since the equation x = t
^{3} can have only one solution while the equation y = t^{4} can have two, we'll start with the *x* equation.

- When x = 8 we must have t = 2. Theny = (2)
^{4} = 16.The point (8,16) is on the graph and occurs when t = 2. - When x = 8 we have t = 2. Then y = +16 not -16. Since the only value of
*t* that makes x = 8 doesn't make y = -16, the point(8,-16) isn't on the graph. - When x = -27 we have t = -3. Theny = (-3)
^{4} = 81.The point (-27, 81) is on the graph when t = -3.

#### Exercise 2

- Show that the point (5,-6) is on the graph of the parametric equations

x = t^{2} + 1y = 3t

Answer

Determine if (5,-6) is on the graph. We know the answer should be "yes". We'll use the *y*-equation first since we'll find only one *t*-value that way.
When y = -6 we must have t = -2. Then
x = (-2)^{2} + 1 = 5.
The point (5,-6) occurs on the graph when t = -2.

#### Exercise 3

- Show that the point (11,9) is not on the graph of the parametric equations

x = t^{2} + 1y = 3t

Answer

- Determine if (11,9) is on the graph. We know the answer should be
*no*. Using the *y*-equation first, if y = 9 then t = 3. Then
x = (3)^{2} + 1 = 10.
Since x≠ 11 at the only *t*-value that makes y = 9, the point (11,9) is not on the graph.
- It's also possible to determine the points at which parametric equations intersect. This is how we'd know if Mario ran into a red koopa flying loftily over the canyon during his jump. One example is given here.