Problem 83 - Essential Calculus

Question

Sketch the graph of \(r=4 \sin \theta\) over each interval. (a) \(0 \leq \theta \leq \frac{\pi}{2}\) (b) \(\frac{\pi}{2} \leq \theta \leq \pi\) (c) \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\)

Step-by-Step Solution

Answer

The graph of \(r=4 \sin \theta\) over each interval is a segment of the full circle in its corresponding quadrant. For \(0 \leq \theta \leq \frac{\pi}{2}\) and \(\frac{\pi}{2} \leq \theta \leq \pi\), the graphs appear as semi-circles in the first and second quadrant respectively. For\(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\), the graph starts from third quadrant, moves to fourth and ends at first quadrant, resembling the movement of the hour hand of a clock from 6 to 9 to 12.

Step 1: Sketch base graph in Polar coordinates

Initially, we sketch the graph of \(r = 4\sin(\theta)\) in its entirety to visualize the base graph in Polar coordinates. The graph of \(r = 4\sin(\theta)\) is a circle with double radius since \(r\) varies from \(-4\) to \(4\) when \(\theta\) goes from \(0\) to \(\pi\) in one full periodic cycle of \(\sin\) function. The circle is centred on origin.

Step 2: Sketch for \(0 \leq \theta \leq \frac{\pi}{2}\)

We need to graph only the first quarter of the base circle as mentioned in the interval. It is in the first quadrant starting from \(\theta = 0\), where \(r = 0\), to \(\theta = \frac{\pi}{2}\), where \(r = 4\). The shape is a semi-circle as the radius \(r\) swing from minimum to maximum and back to minimum.

Step 3: Sketch for \(\frac{\pi}{2} \leq \theta \leq \pi\)

We need to graph only the second quarter of the base circle because this quarter starts from \(\theta = \frac{\pi}{2}\) where radius \(r = 4\) to \(\theta = \pi\) where \(r = 0\). The shape is again a semi-circle but the radius \(r\) swing from maximum to minimum.

Step 4: Sketch for \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\)

We need to graph from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), that means the graph should start from bottom, move to the right, then to the top, which is like a clock ticking from 6 to 9 to 12. The radius \(r\) swing twice starting from minimum to maximum and then to minimum.