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Problem 705

Find the conditions which the equation of a polar figure must satisfy for the figure to be (a) symmetric about the polar axis, (b) symmetric about the $$90^{\circ}$$ -axis, (c) symmetric about the pole.

Short Answer

Expert verified
A polar equation $$r = f(θ)$$ is symmetric about (a) the polar axis if $$r = f(-θ)$$, (b) the $$90^{\circ}$$-axis if $$r = f(\frac{π}{2} - θ)$$, and (c) the pole if $$-r = f(θ + π)$$ or $$r = f(θ + π)$$.
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Step 1: Symmetric about the polar axis

To verify if a polar equation is symmetric about the polar axis, we must check if the equation remains unchanged when we replace $$θ$$ by $$-θ$$. So, given a polar equation $$r = f(θ)$$, if $$r = f(-θ)$$ holds true, then the polar equation is symmetric about the polar axis. (a) For example, let's check if the polar equation $$r = 1 + \cos(θ)$$ is symmetric about the polar axis: Replace $$θ$$ with $$-θ$$ in the expression: $$r = 1 + \cos(-θ)$$. Since $$\cos(-θ) = \cos(θ)$$, the new expression becomes $$r = 1 + \cos(θ)$$, which is the original equation. Therefore, the polar equation $$r = 1 + \cos(θ)$$ is symmetric about the polar axis.

Step 2: Symmetric about the $$90^{\circ}$$-axis

To verify if a polar equation is symmetric about the $$90^{\circ}$$-axis, we must check if the equation remains unchanged when we replace $$θ$$ by $$\frac{π}{2} - θ$$. So, given a polar equation $$r = f(θ)$$, if $$r = f(\frac{π}{2} - θ)$$ holds true, then the polar equation is symmetric about the $$90^{\circ}$$-axis. (b) For example, let's check if the polar equation $$r = \sin(2θ)$$ is symmetric about the $$90^{\circ}$$-axis: Replace $$θ$$ with $$\frac{π}{2} - θ$$ in the expression: $$r = \sin\left(2\left(\frac{π}{2} - θ\right)\right)$$. Using the double angle identity, we get $$r = \sin(π - 4θ)$$. Since $$\sin(π - 4θ) = \sin(4θ)$$, the new expression becomes $$r = \sin(4θ)$$, which is the original equation. Therefore, the polar equation $$r = \sin(4θ)$$ is symmetric about the $$90^{\circ}$$-axis.

Step 3: Symmetric about the pole

To verify if a polar equation is symmetric about the pole, we must check if the equation remains unchanged when we replace $$r$$ by $$-r$$, and $$θ$$ by $$θ + π$$. So, given a polar equation $$r = f(θ)$$, if $$-r = f(θ + π)$$ or $$r = f(θ + π)$$ holds true, then the polar equation is symmetric about the pole. (c) For example, let's check if the polar equation $$r = 2\cos(θ)$$ is symmetric about the pole: Replace $$θ$$ with $$θ + π$$ in the expression: $$r = 2\cos(θ + π)$$. Using the cosine addition formula, we get $$r = 2(-\cosθ) = -2\cos(θ)$$. The equation $$r =-2\cos(θ)$$ is equivalent to $$-r = 2\cos(θ)$$. Therefore, the polar equation $$r = 2\cos(θ)$$ is indeed symmetric about the pole.

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