Finding Cartesian from Parametric Equations

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Question

Finding Cartesian from Parametric Equations

Exercises 1–18 give parametric equations and parameter intervals for the motion of a particle in the xy-plane. Identify the particle’s path by finding a Cartesian equation for it. Graph the Cartesian equation. (The graphs will vary with the equation used.) Indicate the portion of the graph traced by the particle and the direction of motion.

x = 2 sinh t, y = 2 cosh t, −∞

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Identify the particle's path by finding a Cartesian equation for it. Graph this equation, indicating both the direction of motion and the portion of the graph traced by the particle for the following parametric equations X is equal to 3 multiplied by hyperbolic cosine of t. Y is equal to 3 multiplied by hyperbolic sin of T. Negative infinity is less than T, and T is less than positive infinity. OK, so it appears for this particular prong we are essentially asked to graph our Cartesian equation based on the information that is given to us by the prom itself. So using our parametric equations, we need to determine what our Cartesian equation is, and this Cartesian equation describes the particle's path of motion. And when we determine the Cartesian equation, we need to graph it, and then we need to indicate the direction of motion of this particular particle. Is it going from top to bottom, left to right, up and down, etc. What is the direction of motion? And now that we know what we're ultimately trying to solve for, so to put it basic, you know, then keep it basic, we just need to determine what the Cartesian equation is, and then we need to graph that equation and then determine based on what we will solve together, what the direction of motion is, given the parametric equations and our range, which is negative infinity is less than T and T is less than positive infinity. So now that we know what we're trying to solve for, we need to know based on our prior knowledge that in order to solve a problem like this, we will need to isolate the hyperbolic functions, and then we will need to utilize the fundamental hyperbolic identity, which is, so let's make a note of this in blue. So once again we are utilizing our fundamental hyperbolic identity that states that hyperbolic cosine squared of T minus hyperbolic sine of T is equal to 1, and this will help us to eliminate the parameter T. And then we will need to identify the specific branch of the resulting hyperbola or which once again this resulting hyperbola. We'll be, we can once again check the specific branch of the resulting hyperbola by checking our range of the X and Y functions to help us determine the direction of motion by testing the behavior of the particle as T increases from both negative to positive infinity. Fantastic. So now that we have an outline for how we're going to approach solving this problem, our first step that we need to take, like we mentioned earlier, is we need to eliminate our parameter. So we are told that x is equal to 3 multiplied by hyperbolic cosine of T, and we're told that Y is equal to 3 multiplied by hyperbolic sine of T. And like we mentioned earlier, we need to isolate the hyperbolic functions, so we can go ahead and write that hyperbolic cosine of T is equal to x divided by 3. And hyperbolic sine of T is equal to Y divided by 3. Fantastic. So with that said, we now need to recall and use our identity that states that hyperbolic cosine squared of t minus hyperbolic sine squared. Of t, I forgot my square up above squareot oft will be equal to 1. So once again, it's very important that we recall this identity. Hyperbolic cosine squared of t minus hyperbolic sine to the 2 oft is equal to 1. And when we use this identity, we can go ahead and write that x divided by 3 to 2 minus y divided by 3 to 2 will be equal to 1, which will mean that x 2 divided by 9 minus y2 divided by 9 is equal to 1. And this is the equation for a hyperbola that opens horizontally. So once again, this hyperbola will be opening horizontally instead of vertically. So with that said, for our second step, we now need to identify the portion of the graph that will be traced. So we must determine which part of our hyperbola the particle will actually visit based on the range of hyperbolic cosine of T and hyperbolic sine of T. And we need to note that since hyperbolic cosine of T is greater than or equal to 1 for all real t, that will mean that the x coordinate X is equal to 3 multiplied by hyperbolic cosine of T must satisfy X is greater than or equal to 3. This therefore, indicates that the particle will only trace the right branch of our hyperbola as a result. So with that said, our 3rd step is we now need to determine the direction of motion, and in order to do that, we need to consider the following. So to determine the direction of motion, we will need to evaluate the limits and specific values of t to see how the particle will move along the branch of our hyperbola. So, with that said, we need to consider the following values of T. So as T approaches negative infinity, and where T is equal to 0, and as T approaches positive infinity. We will need to note going in order so ast approaches negative infinity, we will have that x will be equal to infinity because once again x is equal to 3 multiplied by hyperbolic cosine of t will approach infinity and similarly y will also approach negative infinity. So that means that the particle comes from the lower end of the right branch. And when T is equal to 0, so when we plug in a value of zero into our expression, we will find that X will be equal to 3, and we will find that Y will be equal to 0, which will mean that our particle will pass through the vertex 3.0. And as T approaches infinity, that will mean that x will approach positive infinity, and y will also approach positive infinity, which will mean that the particle will move towards the upper end of the right branch of our hyperbola. And we need to note that since Y increases continuously as T increases, that means that the motion will be upward along the right branch of our hyperbola. So With that said, for our fourth and final step, so this is our last step, we now need to begin to graph our equation. So with that in mind, we will need to plot the right branch of our hyperbola, which is x 2 divided by 9 minus y2 divided by 9 is equal to 1, and we will start at the vertex. Parentheses 3.0. And we will need to extend into the 1st and 4th quadrants of our graph, and then we will need to add arrowheads pointing in the upwards direction to show the motion of our particle. So, putting what we said into words, into writing, let's go back to look at the blank graph that was given to us by the prom itself. So when we take all the information that we've gathered together and we apply it to our graph. I've gone ahead and digitally graphed our curves, so I already digitally graphed our curve here, and you could use Desmos or MATLAB Online to create a digital graph, but as you can see, we have the right curve of our hyperbola. And as we can see, we have arrows of direction pointing upwards in the first quadrant, and then in our fourth quadrant, we have an arrow pointing up, so it's going from the bottom, going up, curving through our vertex of 3.0, and then going up to the right. So, think of it almost like a string being drawn on a bow. So imagine a bow being pulled back. As the shape And as we can see, so we could put a box around this because this is our correct and final answer, and our curve for the hyperbola is represented by this bold black line, and we know that our Cartesian equation once again is x 2 divided by 9 minus y2 divided by 9 is equal to 1. And then we have our vertex, which is once again parentheses 3.0. And to recap, once again, we have our Cartesian equation, and this will be for the portion, and let's make a note of this, this will be for the portion X is greater than or equal to 3. And our particle will move upward along the white, once again, it's gonna be moving upwards along the right branch, the right side of our hyperbola, as we can see. And that's it, that is our correct and final answer. Hooray, we did it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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