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 = t² + 3, y = 6 − t³; 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 U to the power of five plus three and Y equals six minus U to the seventh power where U is equal to one for our answer choices. A is the coordinate 54, B is 45, C is negative 25 and D is 47. Now, what do we already know? Well, we know that we're talking about parametric equations and that means it's a way to link two equations to make some fancy curve that's linked by just one simple parameter. In this case, the parameter you. So if we can substitute that value of U into both of our equations for X and Y, then we'll find the coordinates. So I'm going to call X equals U to the fifth power plus three equation one. And I'm going to call Y equals six minus U to the seventh power equation two. So let's start with X solving for X in our first equation that means X will be equal to U to the fifth. Now, remember U is one U is equal to one. So that's going to be one to the fifth power plus three, one raise to any power is still one. So that would be one plus three, X equals four in our second equation solving for Y Y equals six minus U to the seventh. So that would be one to the seventh power, no one to the seventh power is equal to one. So that would be six minus one, which equals five. Therefore, X equals four, Y equals five, which means that our answer is B 45. Thanks for watching this video. I hope it helped.

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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 = t² + 3, y = 6 − t³; t = 2