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Question: How do you find the area of one petal of \(r=\cos 5\theta \)?...

How do you find the area of one petal of r=cos5θr=\cos 5\theta ?

Explanation

Solution

We first need to consider the graph of the equation r=cos5θr=\cos 5\theta . Then we have to put r=0r=0 in the given equation to get cos5θ=0\cos 5\theta =0. The two consecutive solutions of this equation will give the θ\theta coordinate of the endpoints of one petal, as θ1{{\theta }_{1}} and θ2{{\theta }_{2}}. Then using the formula for the area in polar coordinates, which is given by A=12θ1θ2r2dθA=\dfrac{1}{2}\int_{{{\theta }_{1}}}^{{{\theta }_{2}}}{{{r}^{2}}d\theta } we can determine the required area of one petal.

Complete step by step answer:
The equation given is
r=cos5θ\Rightarrow r=\cos 5\theta
As we can see that the above equation is a relation between the variables rr and θ\theta , which are the polar variables. The graph of the above equation is as shown in the below diagram.

We can clearly see that the graph of the equation r=cos5θr=\cos 5\theta consists of five identical petals. Let us consider the horizontal petal and try to evaluate its area. For this, we need to determine the polar coordinates of the end points of the horizontal petal.
We can see that the petal starts from and ends at the origin. So the rr coordinate for both of its end points is equal to 00. Therefore, putting r=0r=0 in the given equation we get
0=cos5θ cos5θ=0 \begin{aligned} & \Rightarrow 0=\cos 5\theta \\\ & \Rightarrow \cos 5\theta =0 \\\ \end{aligned}
Now, we know that the solution of the equation cosx=0\cos x=0 is x=(2n+1)π2x=\left( 2n+1 \right)\dfrac{\pi }{2}. So the solution of the above equation is given by
5θ=(2n+1)π2 θ=(2n+1)π10 \begin{aligned} & \Rightarrow 5\theta =\left( 2n+1 \right)\dfrac{\pi }{2} \\\ & \Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{10} \\\ \end{aligned}
Since the horizontal petal lies in the first and the fourth quadrant, we get θ=π10\theta =-\dfrac{\pi }{10} and θ=π10\theta =\dfrac{\pi }{10}.
Now, we know that the area in terms of the polar coordinates is given by
A=12θ1θ2r2dθ\Rightarrow A=\dfrac{1}{2}\int_{{{\theta }_{1}}}^{{{\theta }_{2}}}{{{r}^{2}}d\theta }
Substituting r=cos5θr=\cos 5\theta , θ1=π10{{\theta }_{1}}=-\dfrac{\pi }{10} and θ2=π10{{\theta }_{2}}=\dfrac{\pi }{10}, we get the area of the horizontal petal as
A=12π10π10(cos25θ)dθ\Rightarrow A=\dfrac{1}{2}\int_{-\dfrac{\pi }{10}}^{\dfrac{\pi }{10}}{\left( {{\cos }^{2}}5\theta \right)d\theta }
We know that 2cos2x=cos2x+12{{\cos }^{2}}x=\cos 2x+1. Substituting x=5θx=5\theta , we get

& \Rightarrow 2{{\cos }^{2}}5\theta =\cos \left( 10\theta \right)+1 \\\ & \Rightarrow {{\cos }^{2}}5\theta =\dfrac{1}{2}\left( \cos \left( 10\theta \right)+1 \right) \\\ \end{aligned}$$. So we have $$\begin{aligned} & \Rightarrow A=\dfrac{1}{2}\int_{-\dfrac{\pi }{10}}^{\dfrac{\pi }{10}}{\dfrac{1}{2}\left( \cos \left( 10\theta \right)+1 \right)d\theta } \\\ & \Rightarrow A=\dfrac{1}{4}\left( \int_{-\dfrac{\pi }{10}}^{\dfrac{\pi }{10}}{\cos \left( 10\theta \right)d\theta }+\int_{-\dfrac{\pi }{10}}^{\dfrac{\pi }{10}}{d\theta } \right) \\\ & \Rightarrow A=\dfrac{1}{4}\left\\{ \left[ \dfrac{\sin \left( 10\theta \right)}{10} \right]_{-\dfrac{\pi }{10}}^{\dfrac{\pi }{10}}+\left[ \theta \right]_{-\dfrac{\pi }{10}}^{\dfrac{\pi }{10}} \right\\} \\\ & \Rightarrow A=\dfrac{1}{4}\left\\{ \left[ \dfrac{\sin \left( 10\times \dfrac{\pi }{10} \right)}{10}-\dfrac{\sin \left( -10\times \dfrac{\pi }{10} \right)}{10} \right]+\left[ \dfrac{\pi }{10}-\left( -\dfrac{\pi }{10} \right) \right] \right\\} \\\ & \Rightarrow A=\dfrac{1}{4}\left\\{ \left[ \dfrac{\sin \left( \pi \right)}{10}-\dfrac{\sin \left( -\pi \right)}{10} \right]+\left[ \dfrac{\pi }{10}+\dfrac{\pi }{10} \right] \right\\} \\\ \end{aligned}$$ Now, we know that $$\sin \pi =0$$. So we get $$\begin{aligned} & \Rightarrow A=\dfrac{1}{4}\left\\{ \left[ 0-0 \right]+\left[ \dfrac{\pi }{10}+\dfrac{\pi }{10} \right] \right\\} \\\ & \Rightarrow A=\dfrac{1}{4}\left( \dfrac{2\pi }{10} \right) \\\ & \Rightarrow A=\dfrac{\pi }{20} \\\ \end{aligned}$$ **Hence, the area of one petal of $r=\cos 5\theta $ is equal to $$\dfrac{\pi }{20}$$.** **Note:** The graph of the given equation $r=\cos 5\theta $ is not easy to be sketched. But we can solve this question without using the graph as well. We can see in the graph that each petal starts from and ends at the origin. So by considering any consecutive roots of the equation $\cos 5\theta =0$ as the limits of the integration, we can determine the area as done in the above solution.