37–48. Polar-to-Cartesian coordinates Convert the following eq...

Question

37–48. Polar-to-Cartesian coordinates Convert the following equations to Cartesian coordinates. Describe the resulting curve.

r = sin θ sec² θ

Accepted Answer

Welcome back, everyone. Convert the polar equation R equals cosine of the multiplied by cos squared of theta to Cartesian coordinates and describe the resulting curve. For this problem, let's begin by writing the original form, R equals cosine of theta multiplied by co-sequence squared of theta. We can express it as cosine of the multiplied by Cosequent is a 1 divided by sine of theta, and we're basically squaring it. So we can just square sign, right? We get cosine of data multiplied by a 1 divided by sine squared of theta. Now, what we're going to do is simply rewrite it as a single fraction. We got cosine of the divided by sin squared of data. This is the expression of R in terms of sine and cosine. From here, we can multiply both sides by sin squared. Theta and we get R multiplied by sin square of theta equals. Cosine theta. And now we can substitute the polar to Cartesian formulas into this equation. Let's remember that cosine theta can be expressed as X divided by R, and sine of theta is Y divided by R. So we get R multiplied by Y divided by R squared is equal to X divided by R. Now, let's simplify. We get R multiplied by Y2 divided by R2 is equal to x divided by R. We can cancel out. R on the left hand side and our denominator becomes R, right? So we get Y2 divided by R equals x divided by R and multiplying both sides by r we can show that Y2 is equal to X, which is our equation for this problem. And then we want to describe the curve itself. So for this problem. Let's recognize that this is a parabola. In particular, we have to understand that Y is equal to X, which essentially means that X must be greater than or equal to 0 to begin with, right, because a square of a real number has a minimum value of 0, it cannot be negative. So when we think about this parabola. We can solve for a while, right? Y is going to be equal to + or minus square root of X because taking square root of a2 gives us two possible solutions. We know that X begins at 0. This is the minimum value. With an increase in X, we get an increase in square root. So we have a positive branch and a symmetric negative branch, right? So we Used a simple sketch to demonstrate that this is a parabola. That Ops To the right Because we have non-negative. Why, I'm sorry, non-negative X values, right? Y can be both positive and negative. So we have a parabola that opens to the right. With a vertex. At The origin. This is how we can describe this curve. Let's label our final answer and thank you for watching.

37–48. Polar-to-Cartesian coordinates Convert the following equations to Cartesian coordinates. Describe the resulting curve.

r = sin θ sec² θ

Key Concepts

Polar and Cartesian Coordinate Systems

Trigonometric Identities

Equation Conversion and Curve Description

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