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Find the points at which the following polar curves have a horizontal or a vertical tangent line. $$r=\sin 2 \theta$$

Short Answer

Expert verified
Question: Determine the points where the polar curve $r = \sin(2\theta)$ has horizontal or vertical tangent lines. Answer: The polar curve has no horizontal tangent points and vertical tangent points at $r = \sin(2(\pm\frac{\pi}{4}+\frac{n\pi}{2}))$, where $\theta = \pm\frac{\pi}{4}+\frac{n\pi}{2}$ and $n = 0, \pm 1, \pm 2, \dots$.

Step by step solution

01

1. Differentiate the polar equation with respect to theta

First, start by finding the derivative of the given polar equation with respect to \(\theta\). The given equation is \(r = \sin(2 \theta)\). To find \(dr/d\theta\), differentiate the polar equation with respect to \(\theta\): $$\frac{dr}{d\theta} = \frac{d}{d\theta}(\sin 2\theta) = 2\cos 2\theta$$
02

2. Determine the horizontal tangent points

To find the points where we have a horizontal tangent, we must find the change of the angle (\(\frac{d\theta}{dr}\)) with respect to the radius. From the given equation, \(r = \sin(2\theta)\), we can express the angle function as \(\theta = \frac{1}{2}\arcsin r\). Now differentiate the angle function with respect to the radius, \(r\): $$\frac{d\theta}{dr} = \frac{1}{2}\frac{d}{dr}(\arcsin r) = \frac{1}{2}\frac{1}{\sqrt{1-r^2}}$$ Set this expression equal to zero to find points where we have a horizontal tangent. Notice that the expression is never equals to zero. Therefore, the polar curve does not have any horizontal tangent points.
03

3. Determine the vertical tangent points

To find the points where we have a vertical tangent, we must look for the points where the change in radius with respect to the angle is zero. From step 1, we found \(dr/d\theta\): $$\frac{dr}{d\theta} = 2\cos 2\theta$$ Set it equal to zero to find the vertical tangent points: $$2\cos 2\theta = 0$$ Solve for \(\theta\): $$2\theta = \pm\frac{\pi}{2} + n\pi \quad \text{where } n = 0, \pm 1, \pm 2, \dots$$ $$\theta = \pm \frac{\pi}{4} + \frac{n\pi}{2} \quad \text{where } n = 0, \pm 1, \pm 2, \dots$$ Now replace these values of \(\theta\) into the given function \(r = \sin(2\theta)\) to find the corresponding radius for each vertical tangent point: $$r = \sin\left(2\left(\pm \frac{\pi}{4} + \frac{n\pi}{2}\right)\right)$$ So the vertical tangent points are at \(r = \sin\left(2\left(\pm \frac{\pi}{4} + \frac{n\pi}{2}\right)\right)\), where \(n = 0, \pm 1, \pm 2, \dots\).
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4. Conclusion

The polar curve \(r = \sin(2\theta)\) has no horizontal tangent points and vertical tangent points at \(r = \sin\left(2\left(\pm \frac{\pi}{4} + \frac{n\pi}{2}\right)\right)\), where \(\theta = \pm \frac{\pi}{4} + \frac{n\pi}{2}\) and \(n = 0, \pm 1, \pm 2, \dots\).

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