10–12. Parametric curves
a. Eliminate the parameter to ...

Question

10–12. Parametric curves

a. Eliminate the parameter to obtain an equation in x and y.

x = 3cos(-t), y = 3sin(-t) - 1, for 0 ≤ t ≤ π; (0, -4)

Accepted Answer

Welcome back, everyone. Find the Cartesian equation of the curve defined parametrically by X equals 5 cosine of net minus 2 and Y equals 5 of negative T + 4 for T between 0 and 2 pi inclusive by eliminating the parameter T. For this problem, let's rewrite each coordinate. X is equal to 5, cosine of negative T minus 2. We can simplify this expression, remembering that cosine is an even function, so cosine of negative T is equal to cosine of T, meaning X can be written as 5. Cosine of T minus 2. Now let's analyze why. It says 5 of negative T + 4. We can remember that sine is an odd function, so sine of negative T is equal to negative sign of T, meaning we get -5 sign of T + 4. And now that we have Simplified The X and Y expressions, we can eliminate the parameter T. What we can do is simply notice that we have cosine of T and sine of T. In the X and Y coordinates respectively. So we can simply solve for cosine and sign. Since we can apply. The Pythagorean identity, which says that sine squared of T plus sine squared of T is equal to 1, right? Solving for cosine of T, we can show that cosine of T equals X + 2, we're adding 2 to both sides divided by 5, we're dividing by the leading coefficient. Now for a sign of tea. It is going to be equal to or minus Y. Divided by 5, right? We can move -5 sign of T to the left, then move Y to the right. So we get 4 minus Y equals 5 signs of T. And 4 minus Y divided by 5, then the returns sin of T. And now we're going to substitute these expressions into the Pythagorean identity we got. X + 2 divided by. Why Squared plus. 4 minus y divided by 5 squared is equal to 1. Or simply X + 2 squad divided by 25, we're applying the properties. Of exponents when it comes to fractions, right, we can distribute the square. Plus 4 minus y squared divided by. Again, 5 squad gives us 25 equals 1. Notice that this is an equation of a circle, we can multiply both sides by 25. We get X + 2 squad plus. Or minus Y2 equals 25. And now, to get a standard form of an equation of a circle, we can factor out -1. In the expression for minus Y, so we get X + 2 squared plus. Negative in Y minus 42 equals 25. And now let's remember that -1 square this 1, so we can just eliminate that negative sign. Which gives us X + 2 squad. Plus Y minus 42 equals 25. That's our final answer and thank you for watching.

10–12. Parametric curves
a. Eliminate the parameter to obtain an equation in x and y.
x = 3cos(-t), y = 3sin(-t) - 1, for 0 ≤ t ≤ π; (0, -4)

Key Concepts

Parametric Equations

Eliminating the Parameter

Trigonometric Identities

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