Polar Coordinates & Differentiation - Calculus 2

1. A polar curve is defined implicitly by the equation $r^2 = \cos(2\theta)$. At the point where $\theta = \frac{\pi}{6}$ and $r > 0$, what is the slope of the tangent line?
   Answer: $-\frac{1}{\sqrt{3}}$ From $r^2 = \cos(2\theta)$, we get $r = \sqrt{\cos(2\theta)}$ (taking positive square root). At $\theta = \frac{\pi}{6}$: $r^2 = \cos(\frac{\pi}{3}) = \frac{1}{2}$, so $r = \frac{1}{\sqrt{2}}$. To find $\frac{dr}{d\theta}$, we differentiate implicitly: $2r\frac{dr}{d\theta} = -2\sin(2\theta)$, giving $\frac{dr}{d\theta} = -\frac{\sin(2\theta)}{r}$. At $\theta = \frac{\pi}{6}$: $\frac{dr}{d\theta} = -\frac{\sin(\frac{\pi}{3})}{\frac{1}{\sqrt{2}}} = -\frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}} = -\frac{\sqrt{3}\sqrt{2}}{2} = -\frac{\sqrt{6}}{2}$. Using $\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}$ with $\sin(\frac{\pi}{6}) = \frac{1}{2}$ and $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$, we get $\frac{dy}{dx} = -\frac{1}{\sqrt{3}}$. The other choices result from sign errors or incorrect trigonometric calculations.
2. Consider the polar curve $r = \sin(\theta) + \cos(\theta)$. At the point where this curve intersects the positive $x$-axis, what is the value of $\frac{dy}{dx}$?
   Answer: $\frac{dy}{dx} = 1$ The curve intersects the positive $x$-axis when $y = 0$ and $x > 0$. Since $y = r\sin\theta$, we need $r\sin\theta = 0$. Either $r = 0$ or $\sin\theta = 0$. If $\sin\theta = 0$, then $\theta = 0$ or $\pi$. At $\theta = 0$: $r = \sin(0) + \cos(0) = 1 > 0$, giving point $(1,0)$. At $\theta = \pi$: $r = \sin(\pi) + \cos(\pi) = -1 < 0$, giving point $(1,0)$ as well (since negative $r$ reverses direction). Using $\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}$ with $\frac{dr}{d\theta} = \cos\theta - \sin\theta$, at $\theta = 0$: $\frac{dy}{dx} = \frac{(1)(0) + (1)(1)}{(1)(1) - (1)(0)} = 1$. The other choices result from computational errors or using the wrong intersection point.
3. A curve is defined by the polar equation $r = 3 + 2\cos(\theta)$. At the point where $\theta = \frac{\pi}{3}$, what is the slope of the tangent line to the curve?
   Answer: $-\frac{2\sqrt{3}}{7}$ To find the slope, we use $\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}$. Here $r = 3 + 2\cos\theta$ and $\frac{dr}{d\theta} = -2\sin\theta$. At $\theta = \frac{\pi}{3}$: $r = 3 + 2(\frac{1}{2}) = 4$, $\frac{dr}{d\theta} = -2(\frac{\sqrt{3}}{2}) = -\sqrt{3}$. So $\frac{dy}{dx} = \frac{-\sqrt{3} \cdot \frac{\sqrt{3}}{2} + 4 \cdot \frac{1}{2}}{-\sqrt{3} \cdot \frac{1}{2} - 4 \cdot \frac{\sqrt{3}}{2}} = \frac{-\frac{3}{2} + 2}{-\frac{\sqrt{3}}{2} - 2\sqrt{3}} = \frac{\frac{1}{2}}{-\frac{7\sqrt{3}}{2}} = -\frac{2\sqrt{3}}{7}$. Choice A is the negative reciprocal error. Choice C omits the negative sign. Choice D doubles the numerator incorrectly.
4. A polar curve is given by $r = f(\theta)$ where $f$ is differentiable. At a point where $r = 3$, $\frac{dr}{d\theta} = -2$, and $\theta = \frac{\pi}{4}$, what is the value of $\frac{dx}{d\theta}$?
   Answer: $-\frac{5\sqrt{2}}{2}$ We use $\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta$. Given $r = 3$, $\frac{dr}{d\theta} = -2$, and $\theta = \frac{\pi}{4}$, we have $\cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$. Therefore, $\frac{dx}{d\theta} = (-2)\left(\frac{\sqrt{2}}{2}\right) - (3)\left(\frac{\sqrt{2}}{2}\right) = -\sqrt{2} - \frac{3\sqrt{2}}{2} = -\frac{2\sqrt{2} + 3\sqrt{2}}{2} = -\frac{5\sqrt{2}}{2}$. Choice B omits the negative signs. Choice C only includes the first term. Choice D omits the negative sign from the final result.

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