Welcome back, everyone. So in this example, we're gonna take a look at this equation, y=2x3, and we're gonna write 2 sets of parametric equations that describe this rectangular equation. So in other words, they want us to parameterize this y=2x3 equation. So let's take a look at the steps of how to do this. The first thing you're gonna do in these problems is you're gonna have to define t unless it's given to you already. In this case, it's not. So in this case, what we're gonna have to do is choose a t to parameterize our equation, and there's always some guidance to this. In this case, you can always just try t equals x or t equals whatever the thing in the parentheses is if you're given some kind of a parentheses in the equation. And in this problem, we're not told that we can't use x=t or t=x. So this is always gonna be a viable solution. So option 1 is that we just choose x=t or t=x. Right? So if we let t=x, then the second step is we're gonna have to solve for x(t). So in other words, we're gonna have to get x in terms of t. And if t=x, so you just flip the equation around. In other words, x(t)=t itself. That's the first equation, the first parameterized equation. Now to get the y equation, we're just gonna have to plug that x(t) into the original equation for x. In other words, the y=2x3. And let's go ahead and do that. So if x(t)=t itself, that means that y is equal to 2. And instead of now plugging in x cubes, we're really just gonna replace this with t3 because that's what x is equal to. Right? So in other words, this really just becomes 2t3. Alright? So this is your x(t) equation, and this is your y(t) equation. So again, it's always kind of silly how these problems work out because if you're not told that you can't use this option or this definition of t, then this is always a viable set or a solution to your parametric equations. You can always just basically replace x with t, and then your y of equation just becomes the original equation, but you've replaced x with t. Alright? So that's always one of your options. So that's one of your sets of parametric equations. Or what we can also do is we're gonna have to pick another sort of definition for t and sort of parameterize the equation a different way. Remember, there's always a ton of different ways that you can do this. There's no really correct answer, but there are some easier ones, for defining t. Alright? So let's go ahead and take a look at the second one here. You're going to define t unless it's given to us. In this case, we can't use t=x, obviously, because we just use that in the first one. So let's just try to pick a different definition for t. What we can do here is whenever you're just exhausted with you know, t equals x or whatever the parenthesis is, you can start to just try to set t in terms of powers of x if you have t if you have powers of x. And so, again, so what you can do here is you can set t=x3. You always wanna avoid even powers of x, so like x2 or the square root of x or something like that, because, remember that you have to sort of avoid those domain restrictions. But t=x3 is a perfectly valid solution because t and x are both defined for all numbers, positive and negative. So t=x3 is perfectly fine here. So now what we do is we've defined t, and we just go ahead and solve for x(t). So now in this case, if we want x in terms of t, I'm going to have to get x by itself. And in this case, what happens is if t=x3, that means that x is actually just going to be the cube root of t. I'm just solving for the other variable, the x. Alright? And, again, this is perfectly fine here because I'm gonna have I'm gonna get negative and positive numbers, and those are perfectly fine. So now, I'm just gonna plug that x(t) into the original equation for y. So in other words, I'm gonna get y=2. And instead of x, now I'm gonna replace it with the function that I have, which is just cube root of t. So the original equation was 2x3. So now if I plug in the x of t expression into this, I'm gonna have the cube root of t, but then I'm gonna have to cube it. So if we simplify this, what we're gonna see here is that this is really just becomes 2, and then the cube root of y cubed, that actually just sort of undoes the cubing. So this actually just ends up becoming 2t. Alright? So your parameterized equations, your second set of parametric equations ends up being x(t)=t and y=2t itself. If you picked a different set of equations, you may have gotten different answers than me, but this is one possible solution to this parameterized equations. Alright. That's it for this one, folks. Let me know if you have any questions.
Table of contents
- 0. Review of College Algebra4h 43m
- 1. Measuring Angles39m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
10. Parametric Equations
Writing Parametric Equations
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Writing Parametric Equations practice set
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