In Exercises 1–8, parametric equations and a value for the parame...

Question

In Exercises 1–8, parametric equations and a value for the parameter t are given. Find the coordinates of the point on the plane curve described by the parametric equations corresponding to the given value of t.x = (60 cos 30°)t, y = 5 + (60 sin 30°)t − 16t²; t = 2

Accepted Answer

Welcome back. Everyone. In this problem, we want to determine the coordinates of the point on the plane curve represented by the parametric equations. X equals 90 multiplied by the cosine of 60 degrees multiplied by U and Y equals 12 plus 90 multiplied by the sine of 60 degrees multiplied by U minus 25 multiplied by U squared where U equals four. For our answer choices, there are the coordinates A which is 18 and 18 multiplied by the skirt of three minus 388. B is 18 and 18 multiplied by the skirt of three plus 388. C is 181 180 multiplied by the skirt of three plus 388 and D is 181 180 multiplied by the square of three minus 388. Now, what do we already know? Well, we know that both equations are parametric. And what that means is that it can make some fancy graph where the X and Y coordinates are linked by a common parameter U. So if we want to know what those coordinates are, we can substitute U into each equation to solve for the values of X and Y. Now we know that you equals four. So to solve, I'm going to call my first equation for X equals multiplied by the cosine of 60 degrees multiplied by U equation one. And my second equation for Y where Y is plus 90 multiplied by the sin of 60 degrees multiplied by U minus 25 U squared. So first let me solve for eggs. So solving for X in equation one, it just means that no, wherever I see you, I'm going to substitute for. So that tells us that X equals 90 multiplied by the cosine of 60 degrees multiplied by U which is four. Now remember that from our unit circle, recall that the cosine of 60 degrees equals a half. So that means X equals 90 multiplied by a half, all multiplied by four, a half of 90 is 45 and 45 multiplied by four is 180. So that would be the X value of our coordinate. Next, let's see if we can solve for or Y value. So solving for Y or more second equation, no, Y equals 12 plus 90 multiplied by the sign of 60 degrees multiplied by U which is now four minus multiplied by U squared. So that's going to be four squared. Now, if we go ahead and solve, recall again from our unit circle that the sign of degrees equals the square of three divided by two. So that means I can put my value into my equation to help me to solve. Because that not tells me that Y is equal to 12 plus 90 multiplied by square root of three divided by two are multiplied by four minus multiplied by four square. So that's 25 multiplied by 16. No, put back by 12 90 multiplied by the skirt of three, divided by two, all multiplied by four. That would be equal to 180 multiplied by the skirt of three and 25 multiplied by 16 is 400. Now 12, we have two common terms, 12 and negative. 412 minus 400 is negative 388. So my final answer for why is 180 multiplied by the square of three minus 388. Therefore, X equals 180 Y equals multiplied by the square of three minus 388 which tells us that our answer is choice. D thanks a lot for listening. Everyone. I hope this video helped.

In Exercises 1–8, parametric equations and a value for the parameter t are given. Find the coordinates of the point on the plane curve described by the parametric equations corresponding to the given value of t.x = (60 cos 30°)t, y = 5 + (60 sin 30°)t − 16t²; t = 2

Key Concepts

Parametric Equations

Trigonometric Functions

Quadratic Functions

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