Q 55. - Calculus

Question

Prove that when n is a positive odd integer, the polar rose $r = \cos n\theta or r = \sin n\theta$ is traced twice on the interval $[0, 2\pi ]$ and thus has exactly $n$ petals.

Step-by-Step Solution

Answer

The graph is traced twice in the interval and has exactly $n$ petals.

Step 1: Given information

The polar roses $r=\cos n\theta$ and $r=\sin n\theta$ where $n$ is a odd positive integer.

Step 2: Concept

A polar curve is a shape constructed using the polar coordinate system.

Step 3: Calculation

The goal is to show that $r=\cos n\theta$ and $r=\sin n\theta$ have the same number of petals.

Let $(r,\theta )$ by any point on the graph,

$f\left(\theta \right)=\cos n\theta$

Thus, $r=\cos n\theta$ where $n$ is a odd positive integer.

If $n$an any odd positive integer then by angle sum identity of cos

$$\begin{gathered} \cos n\theta =\cos n(\pi +\theta ) \\ =-\cos n\theta \end{gathered}$$

Therefore, the point $(-r,\cos n(\pi +\theta \left)\right)=(-r,-\cos n\theta )$ has the same location as $(r,\theta )$

As a result, the graph drawn on the interval $[\pi ,2\pi ]$ is identical to the graph shown on the interval $[0,\pi ]$

As a result, the graph has exactly $n$ petals and is traced twice in the interval.

Step 4: Calculation

Let $(r,\theta )$ by any point on the graph,

$f\left(\theta \right)=\sin n\theta$

Thus, $f\left(\theta \right)=\sin n\theta$ where $n$ is a odd positive integer.

If $n$ is any odd positive integer then by angle sum identity of cos

$$\begin{gathered} \sin n\theta =\sin n(\pi +\theta ) \\ =-\sin n\theta \end{gathered}$$

Therefore, for the point $(-r,\sin n(\pi +\theta \left)\right)=(-r,-\sin n\theta )$ has the same location as $(r, \theta) .$

As a result, the graph is traced the $[\pi ,2\pi ]$ corresponds to the graph drawn on the interval $[0,\pi ]$

Therefore, the graph is traced twice in the interval and has exactly $n$ petals.