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 (x1,y1) is on the trajectory.
We are going to reverse the graphing process. We'll start with point (x1,y1) 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.
Practice:
Determine if the point (5,-5) is on the graph of the parametric equations x = 4t + 1 y = 2t-7.
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If x = 5 then
5 = 4t + 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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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. | |
Determine if the point (4,-8) is on the graph of the parametric equations x = t2
y = t3. | |
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. | |
Determine if the point (4,3) is on the graph of the parametric equations x = t2 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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- Determine whether each point is on the graph of the parametric equations
x = t3y = t4
Answer
- Since the equation x = t3 can have only one solution while the equation y = t4 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.
- Show that the point (5,-6) is on the graph of the parametric equations
x = t2 + 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.
- Show that the point (11,9) is not on the graph of the parametric equations
x = t2 + 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.