Question

In Exercises \(15-20,\) find an equation in rectangular coordinates for the equation given in cylindrical coordinates, and sketch its graph. $$ r=2 \sin \theta $$

Short answer

The rectangular equation equivalent to the given cylindrical coordinate system equation \(r = 2\sin\theta\) is \(x^2 = 3y^2\). It represents a hyperbola in the xy-plane with foci at (0, √3) and (0, -√3).

Step-by-step solution

Understand and Write down the Conversion Formulas

The conversion formulas from cylindrical to rectangular coordinates are: \(x = r\cos\theta\), \(y = r\sin\theta\), and \(z = z\). So, we need to use these to convert the given equation.

Substitute the Conversion Equations into the Given Equation

Given the equation, \(r = 2\sin\theta\). The fact that we want to convert it into rectangular coordinates implies that we want it in terms of \(x\) and \(y\) not \(r\) and \(\theta\).So, here, \(r\) is actually \(\sqrt{x^2 + y^2}\) and \(\sin\theta\) can be read as \(y/r\). So, substitute these equations into the given equation to get: \( \sqrt{x^2 + y^2} = 2(y/ \sqrt{x^2 + y^2}) \)

Simplify to Get Rectangular Coordinates

Next, we simplify the equation to find \(x\) and \(y\). To do that, we can square both sides then simplify to get \(x^2 + y^2 = 4y^2\). Then if you subtract \(3y^2\) from both sides we get: \(x^2 = 3y^2\). This is our equation in rectangular form.

Sketch the Graph

To sketch its graph, recognize that the equation \(x^2 = 3y^2\) is that of a hyperbola because it's a quadratic equation in two variables where the variables terms have opposite signs. It's a hyperbola that opens to the right and left along the x-axis. Given the form of the equation, we can easily sketch the graph of the equation in the xy-plane. Its foci are located at \((0, \sqrt{3})\) and \((0, -\sqrt{3})\).