AP Calculus BC Practice Quiz: Defining Polar Coordinates and Differentiating in Polar Form

Correct Answer: A

To find the derivative of a polar function, we first express x and y in terms of the parameter $\theta$: $x = r\cos\theta = f(\theta)\cos\theta$ and $y = r\sin\theta = f(\theta)\sin\theta$. Then, we use the formula for the derivative of parametric equations, $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$. Using the product rule, we find $\frac{dy}{d\theta} = f'(\theta)\sin\theta + f(\theta)\cos\theta$ and $\frac{dx}{d\theta} = f'(\theta)\cos\theta - f(\theta)\sin\theta$. The ratio of these two expressions gives the correct formula.

Correct Answer: C

First, find the expressions for $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$. Given $r = 4\sin\theta$, we have $\frac{dr}{d\theta} = 4\cos\theta$. Using the formulas $\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta$ and $\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta$, we get $\frac{dy}{d\theta} = (4\cos\theta)\sin\theta + (4\sin\theta)\cos\theta = 8\sin\theta\cos\theta = 4\sin(2\theta)$ and $\frac{dx}{d\theta} = (4\cos\theta)\cos\theta - (4\sin\theta)\sin\theta = 4(\cos^2\theta - \sin^2\theta) = 4\cos(2\theta)$. The slope is $\frac{dy}{dx} = \frac{4\sin(2\theta)}{4\cos(2\theta)} = \tan(2\theta)$. At $\theta = \frac{\pi}{6}$, the slope is $\tan(2 \cdot \frac{\pi}{6}) = \tan(\frac{\pi}{3}) = \sqrt{3}$.

Correct Answer: C

A horizontal tangent occurs when $\frac{dy}{d\theta} = 0$ and $\frac{dx}{d\theta} \neq 0$. For $r = 2 + 2\cos\theta$, $\frac{dr}{d\theta} = -2\sin\theta$. We calculate $\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta = (-2\sin\theta)\sin\theta + (2+2\cos\theta)\cos\theta = -2\sin^2\theta + 2\cos\theta + 2\cos^2\theta = 2(2\cos^2\theta + \cos\theta - 1) = 2(2\cos\theta - 1)(\cos\theta + 1)$. Setting $\frac{dy}{d\theta}=0$ gives $\cos\theta = 1/2$ or $\cos\theta = -1$. This corresponds to $\theta = \pi/3$ and $\theta = \pi$. We must check that $\frac{dx}{d\theta}$ is not zero. At $\theta = \pi$, $\frac{dx}{d\theta}=0$, which indicates a cusp. At $\theta = \pi/3$, $\frac{dx}{d\theta} \neq 0$. Thus, a horizontal tangent exists at $\theta=\pi/3$.

Correct Answer: C

A vertical tangent occurs when $\frac{dx}{d\theta} = 0$ and $\frac{dy}{d\theta} \neq 0$. For $r=2\cos\theta$, we have $\frac{dr}{d\theta} = -2\sin\theta$. Then $\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta = (-2\sin\theta)\cos\theta - (2\cos\theta)\sin\theta = -4\sin\theta\cos\theta = -2\sin(2\theta)$. Setting this to zero gives $\sin(2\theta)=0$, so $2\theta = k\pi$ for integer k. Possible values for $\theta$ are $0, \pi/2, \pi, ...$. We check $\frac{dy}{d\theta} = 2\cos(2\theta)$. At $\theta=\pi/2$, $\frac{dy}{d\theta} = 2\cos(\pi) = -2 \neq 0$. Therefore, there is a vertical tangent at $\theta=\pi/2$.

Correct Answer: B

The variable $r$ in a polar equation represents the directed distance from the origin (the pole). Therefore, its derivative with respect to the angle $\theta$, which is $\frac{dr}{d\theta}$, represents the rate at which this distance is changing as the angle $\theta$ changes. It tells us if the curve is moving away from or towards the pole as $\theta$ increases.

Correct Answer: B

The slope is given by $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$. For $r=2\theta$, we have $\frac{dr}{d\theta}=2$. Then $\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta = 2\sin\theta + 2\theta\cos\theta$ and $\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta = 2\cos\theta - 2\theta\sin\theta$. Evaluating at $\theta=\pi/2$ gives $\frac{dy}{d\theta} = 2\sin(\pi/2) + 2(\pi/2)\cos(\pi/2) = 2(1) + \pi(0) = 2$. And $\frac{dx}{d\theta} = 2\cos(\pi/2) - 2(\pi/2)\sin(\pi/2) = 2(0) - \pi(1) = -\pi$. The slope is $\frac{2}{-\pi}$.

Correct Answer: B

Since a polar curve can be treated as a set of parametric equations with parameter $\theta$, the second derivative $\frac{d^2y}{dx^2}$ is found using the formula for parametric derivatives: $\frac{d^2y}{dx^2} = \frac{\frac{d}{d\theta}\left(\frac{dy}{dx}\right)}{\frac{dx}{d\theta}}$. This process involves finding the derivative of the first derivative ($\frac{dy}{dx}$) with respect to the parameter $\theta$, and then dividing the result by $\frac{dx}{d\theta}$.

Correct Answer: D

The curve passes through the pole when $r = 0$. For a polar curve $r=f(\theta)$, the slope of the tangent line at the pole (where $r=0$ at $\theta = \theta_0$) is given by $\tan(\theta_0)$, provided $\frac{dr}{d\theta}$ is not zero at $\theta_0$. Here, $r=0$ when $\cos\theta = -1/2$. One such angle is $\theta = 2\pi/3$. We check $\frac{dr}{d\theta} = -2\sin\theta$. At $\theta=2\pi/3$, $\frac{dr}{d\theta} = -2\sin(2\pi/3) = -\sqrt{3} \neq 0$. Therefore, the slope is $\tan(2\pi/3) = -\sqrt{3}$.

Correct Answer: B

We need to find the signs of $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ at $\theta = \pi/2$. For $r=1+\cos\theta$, we have $\frac{dr}{d\theta}=-\sin\theta$. At $\theta=\pi/2$, $r=1+\cos(\pi/2)=1$ and $\frac{dr}{d\theta}=-\sin(\pi/2)=-1$. Using the conversion formulas, $\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta$ and $\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta$. At $\theta=\pi/2$, we substitute the values: $\frac{dx}{d\theta} = (-1)\cos(\pi/2) - (1)\sin(\pi/2) = (-1)(0) - (1)(1) = -1$. And $\frac{dy}{d\theta} = (-1)\sin(\pi/2) + (1)\cos(\pi/2) = (-1)(1) + (1)(0) = -1$. Since both derivatives are negative, both x and y are decreasing with respect to $\theta$ at this point.