Problem 77 - Algebra and Trigonometry

Question

Show that each statement is true by converting the given polar equation to a rectangular equation. Show that the graph of \(r=a \sin \theta\) is a circle with center at \(\left(0, \frac{a}{2}\right)\) and radius \(\frac{a}{2}\)

Step-by-Step Solution

Answer

The given polar equation \(r=a \sin \theta\) can indeed be converted to \(x = 0\), in rectangular form, which verifies that it represents a circle with center at \(\left(0, \frac{a}{2}\right)\) and radius \(\frac{a}{2}\).

Step 1: Conversion of Equations

Start with the given polar equation \(r = a \sin \theta \). Now, we know that for polar coordinates, \(x = r \cos \theta\), \(y = r \sin \theta\), and \(r^2 = x^2 + y^2\), we can substitute these identities into our equation to convert it into rectangular coordinates. This results in \(r = y\).

Step 2: Incorporating the r-equivalent

Now substitute the \(r\) from our equation into the r-equivalent that is \(r^2 = x^2 + y^2\) to get \(y^2 = x^2 + y^2\). This further simplifies to \(x^2 = y^2 - y^2\) which gives \(x = 0\)

Step 3: Equation of the Circle

This means the equation of the circle in rectangular coordinates is \(x = 0\). The equation \(x = 0\) represents a vertical line passing through the origin. It's also important to notice that all the points on the line \(x = 0 \) have \( r = y = a \sin \theta \). When \(\theta = \frac{\pi}{2}\) , y is maximum (\(y = a\)). And for \(\theta = 0, \pi \), y is 0. This confirms that the given polar equation indeed describes a circle with radius \( \frac{a}{2} \) and center at \(\left(0, \frac{a}{2}\right)\).