Polar Coordinates and Differentiation - AP Calculus BC

1. Let $C$ be the polar curve $r = 3\sin(2\theta)$. What is the rate of change of the radius $r$ with respect to the angle $\theta$ when $\theta = \frac{\pi}{4}$?
   Answer: $0$ The rate of change of the radius $r$ with respect to the angle $\theta$ is $\frac{dr}{d\theta}$. For $r = 3\sin(2\theta)$, the derivative is $\frac{dr}{d\theta} = 3\cos(2\theta) \cdot 2 = 6\cos(2\theta)$. Evaluating at $\theta = \frac{\pi}{4}$ gives $\frac{dr}{d\theta} = 6\cos(2 \cdot \frac{\pi}{4}) = 6\cos(\frac{\pi}{2}) = 6(0) = 0$.
2. A curve is described by $r=2\sin\theta$; what is $\dfrac{dy}{dx}$ at $\theta=\dfrac{\pi}{2}$?
   Answer: $0$ This problem requires the skill of differentiation in polar coordinates to find the slope of the tangent line in Cartesian coordinates. To find dy/dx for a polar curve r = f(θ), use the formula dy/dx = [f'(θ) sin θ + f(θ) cos θ] / [f'(θ) cos θ - f(θ) sin θ]. For r = 2 sin θ, f'(θ) = 2 cos θ, so at θ = π/2, f'(π/2) = 0 and f(π/2) = 2. Plugging in, the numerator is 0·1 + 2·0 = 0 and the denominator is 0·0 - 2·1 = -2, yielding dy/dx = 0/(-2) = 0. A tempting distractor is choice B, dr/dθ at π/2 which is 0, but this fails because while it matches numerically here, it generally ignores the transformation and would not work for other points. When computing polar derivatives, always remember to use the complete formula to account for the coordinate transformation, a strategy applicable to any polar curve.
3. Which of the following is the slope of a tangent line to the graph of the polar curve $r = 1 - 2\cos\theta$ at the pole?
   Answer: $\sqrt{3}$ The curve passes through the pole when $r=0$. So, $1 - 2\cos\theta = 0$, which gives $\cos\theta = \frac{1}{2}$. This occurs at $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$. The slope of the tangent line at the pole is given by $\tan\theta$, provided $\frac{dr}{d\theta} \neq 0$. Here, $\frac{dr}{d\theta} = 2\sin\theta$. At $\theta = \frac{\pi}{3}$, $\frac{dr}{d\theta} = 2\sin(\frac{\pi}{3}) = \sqrt{3} \neq 0$. The slope is $\tan(\frac{\pi}{3}) = \sqrt{3}$. At $\theta = \frac{5\pi}{3}$, the slope is $\tan(\frac{5\pi}{3}) = -\sqrt{3}$. One of these values is $\sqrt{3}$.
4. For the polar curve $r=3-\cos\theta$, what is $\dfrac{dy}{dx}$ at $\theta=0$?
   Answer: undefined This problem involves finding dy/dx for r = 3 - cos θ using polar differentiation. The formula is dy/dx = (dr/dθ·sin θ + r·cos θ)/(dr/dθ·cos θ - r·sin θ). With dr/dθ = sin θ, at θ = 0: r = 3 - 1 = 2, dr/dθ = 0, sin(0) = 0, cos(0) = 1. Substituting: dy/dx = (0·0 + 2·1)/(0·1 - 2·0) = 2/0, which is undefined. A common error is assuming dy/dx = 0 when the numerator is finite but the denominator is zero. When the denominator equals zero, dy/dx is undefined, indicating a vertical tangent line at that point.
5. For the polar curve $r=\sec\theta$, determine $\dfrac{dy}{dx}$ at $\theta=0$.
   Answer: undefined This problem requires the skill of differentiation in polar coordinates to find the slope of the tangent line in Cartesian coordinates. To find $dy/dx$ for a polar curve $r = f(\theta)$, use the formula $\dfrac{dy}{dx} = \dfrac{f'(\theta) \sin \theta + f(\theta) \cos \theta}{f'(\theta) \cos \theta - f(\theta) \sin \theta}$. For $r = \sec \theta$, $f'(\theta) = \sec \theta \tan \theta$, so at $\theta = 0$, $f'(0) = 0$ and $f(0) = 1$. Plugging in, the numerator is $0 \cdot 0 + 1 \cdot 1 = 1$ and the denominator is $0 \cdot 1 - 1 \cdot 0 = 0$, yielding $dy/dx$ undefined. A tempting distractor is choice C, $\dfrac{dr}{d\theta}$ at 0 which is 0, but this fails because it does not reveal the division by zero in the denominator, which indicates a vertical tangent. When computing polar derivatives, always remember to use the complete formula to account for the coordinate transformation, a strategy applicable to any polar curve.
6. At $\theta=0$, the polar curve $r=3\cos(2\theta)$ has a tangent line that is
   Answer: vertical. For $r=3\cos(2\theta)$, we have $\frac{dr}{d\theta}=-6\sin(2\theta)$. We evaluate the derivatives of x and y at $\theta=0$. At $\theta=0$, $r=3\cos(0)=3$ and $\frac{dr}{d\theta}=-6\sin(0)=0$. Then $\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta+r\cos\theta = (0)\sin(0)+(3)\cos(0)=3$. And $\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta-r\sin\theta = (0)\cos(0)-(3)\sin(0)=0$. Since $\frac{dx}{d\theta}=0$ and $\frac{dy}{d\theta}\neq 0$, the tangent line is vertical.
7. For $r=\theta^2$, what is $\dfrac{dy}{dx}$ at $\theta=1$?
   Answer: $\dfrac{2\sin 1+\cos 1}{2\cos 1-\sin 1}$ This problem requires finding dy/dx for the polar curve r = θ² using polar differentiation techniques. For polar curves, dy/dx = (dy/dθ)/(dx/dθ) where x = r cos θ and y = r sin θ. With r = θ² and r' = 2θ, at θ = 1 we have r = 1 and r' = 2. Computing dx/dθ = -r sin θ + r' cos θ = -1 sin 1 + 2 cos 1 = 2 cos 1 - sin 1 and dy/dθ = r cos θ + r' sin θ = 1 cos 1 + 2 sin 1 = cos 1 + 2 sin 1. Therefore dy/dx = (cos 1 + 2 sin 1)/(2 cos 1 - sin 1) = (2 sin 1 + cos 1)/(2 cos 1 - sin 1). Students might compute dr/dθ = 2 or try tan 1, but these miss the parametric nature of polar curves. Always use the full parametric derivative formula for polar curves, not shortcuts.
8. What is the slope of the line tangent to the graph of the polar curve $r = 4\cos\theta$ at the point where $\theta = \frac{\pi}{3}$?
   Answer: $\frac{1}{\sqrt{3}}$ We use the formula for the slope in polar coordinates: $\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}$. For $r = 4\cos\theta$, we have $\frac{dr}{d\theta} = -4\sin\theta$. Substituting these into the formula gives $\frac{dy}{dx} = \frac{(-4\sin\theta)\sin\theta + (4\cos\theta)\cos\theta}{(-4\sin\theta)\cos\theta - (4\cos\theta)\sin\theta} = \frac{4(\cos^2\theta - \sin^2\theta)}{-8\sin\theta\cos\theta} = \frac{4\cos(2\theta)}{-4\sin(2\theta)} = -\cot(2\theta)$. At $\theta = \frac{\pi}{3}$, the slope is $- \cot(\frac{2\pi}{3}) = -(-\frac{1}{\sqrt{3}}) = \frac{1}{\sqrt{3}}$.
9. What is the slope of the line tangent to the polar curve $r = 3\theta$ at the point where $\theta = \frac{\pi}{2}$?
   Answer: $-\frac{2}{\pi}$ To find the slope of the tangent line, we compute $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$. The Cartesian coordinates are $x = r\cos\theta = 3\theta\cos\theta$ and $y = r\sin\theta = 3\theta\sin\theta$. Using the product rule, we find the derivatives with respect to $\theta$: $\frac{dx}{d\theta} = 3\cos\theta - 3\theta\sin\theta$ and $\frac{dy}{d\theta} = 3\sin\theta + 3\theta\cos\theta$. Evaluating at $\theta = \frac{\pi}{2}$ gives $\frac{dx}{d\theta} = 3(0) - 3(\frac{\pi}{2})(1) = -\frac{3\pi}{2}$ and $\frac{dy}{d\theta} = 3(1) + 3(\frac{\pi}{2})(0) = 3$. The slope is $\frac{dy}{dx} = \frac{3}{-3\pi/2} = -\frac{2}{\pi}$.
10. Consider the polar curve $r = 2\sin\theta$. What is the value of $\frac{d^2y}{dx^2}$ at $\theta = \frac{\pi}{2}$?
    Answer: $-1$ The curve $r=2\sin\theta$ is a circle of radius 1 centered at (0,1). In Cartesian coordinates, $x^2 + (y-1)^2 = 1$. The point at $\theta=\frac{\pi}{2}$ is $(x,y)=(0,2)$, which is the top of the circle. At this point, the curve has a horizontal tangent, so $\frac{dy}{dx}=0$. Differentiating $x^2 + y^2 - 2y = 0$ implicitly gives $2x + 2y\frac{dy}{dx} - 2\frac{dy}{dx} = 0$. Differentiating again gives $2 + 2(\frac{dy}{dx})^2 + 2y\frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} = 0$. At $(0,2)$, we know $\frac{dy}{dx}=0$. Substituting these values gives $2 + 2(0)^2 + 2(2)\frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} = 0$, which simplifies to $2 + 2\frac{d^2y}{dx^2} = 0$. Thus, $\frac{d^2y}{dx^2} = -1$.
11. For the polar curve $r=\theta$, what is $\dfrac{dy}{dx}$ at $\theta=\dfrac{\pi}{2}$?
    Answer: $-\dfrac{2}{\pi}$ This problem involves finding dy/dx for the polar spiral r = θ using polar differentiation. For polar curves, we apply dy/dx = (dy/dθ)/(dx/dθ) where x = r cos θ and y = r sin θ. With r = θ and r' = 1, we get dx/dθ = -θ sin θ + cos θ and dy/dθ = θ cos θ + sin θ. At θ = π/2, we have dx/dθ = -(π/2)(1) + 0 = -π/2 and dy/dθ = (π/2)(0) + 1 = 1. Therefore dy/dx = 1/(-π/2) = -2/π. Students often confuse the reciprocal relationship and write π/2 instead of -2/π. Remember that polar differentiation requires careful application of the product rule to both x(θ) and y(θ).
12. For the polar curve $r=2-\sin\theta$, what is $\frac{dy}{dx}$ at $\theta=\frac{\pi}{2}$?
    Answer: $0$ This problem asks for dy/dx of r = 2 - sin θ using polar differentiation. We use dy/dx = (dr/dθ·sin θ + r·cos θ)/(dr/dθ·cos θ - r·sin θ). Given r = 2 - sin θ, we find dr/dθ = -cos θ. At θ = π/2: r = 2 - sin(π/2) = 2 - 1 = 1, dr/dθ = -cos(π/2) = 0, and dy/dx = (0·1 + 1·0)/(0·0 - 1·1) = 0/(-1) = 0. A common mistake is computing dr/dθ incorrectly or mixing up signs in the formula. For polar differentiation, carefully evaluate each component and maintain proper signs throughout the calculation.
13. For which value of $\theta$ in the interval $[0, 2\pi)$ does the polar curve $r = 1 - \sin\theta$ have a horizontal tangent line?
    Answer: $\frac{3\pi}{2}$ A horizontal tangent occurs when $\frac{dy}{d\theta} = 0$ and $\frac{dx}{d\theta} \neq 0$. We have $r = 1 - \sin\theta$ and $\frac{dr}{d\theta} = -\cos\theta$. Then $\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta = (-\cos\theta)\sin\theta + (1 - \sin\theta)\cos\theta = \cos\theta(1 - 2\sin\theta) = 0$. This implies $\cos\theta = 0$ (so $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$) or $\sin\theta = \frac{1}{2}$ (so $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$). We also need to check $\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta = (-\cos\theta)\cos\theta - (1 - \sin\theta)\sin\theta = 2\sin^2\theta - \sin\theta - 1$. At $\theta = \frac{3\pi}{2}$, $\frac{dx}{d\theta} = 2(-1)^2 - (-1) - 1 = 2 \neq 0$. At $\theta=\frac{\pi}{2}$, $\frac{dx}{d\theta} = 0$, so the slope is indeterminate. Choices B and C are values where $\frac{dx}{d\theta}=0$, which correspond to vertical tangents. Thus, the only valid option is $\theta = \frac{3\pi}{2}$.
14. For a particle with position given by the polar curve $r = 1 + \cos\theta$, what is the rate of change of the x-coordinate with respect to $\theta$ at $\theta = \frac{\pi}{2}$?
    Answer: $-1$ The x-coordinate is given by $x = r\cos\theta$. Substituting $r=1+\cos\theta$, we get $x = (1+\cos\theta)\cos\theta = \cos\theta + \cos^2\theta$. The rate of change of the x-coordinate with respect to $\theta$ is $\frac{dx}{d\theta}$. We have $\frac{dx}{d\theta} = -\sin\theta + 2\cos\theta(-\sin\theta) = -\sin\theta(1+2\cos\theta)$. Evaluating at $\theta = \frac{\pi}{2}$ gives $\frac{dx}{d\theta} = -\sin(\frac{\pi}{2})(1+2\cos(\frac{\pi}{2})) = -1(1+2(0)) = -1$.
15. What is the equation of the line tangent to the polar curve $r=2$ at $\theta = \frac{\pi}{6}$?
    Answer: $y = -\sqrt{3}x + 4$ For $r=2$, we have $\frac{dr}{d\theta}=0$. The slope is $\frac{dy}{dx} = \frac{0 \cdot \sin\theta + 2\cos\theta}{0 \cdot \cos\theta - 2\sin\theta} = -\cot\theta$. At $\theta=\frac{\pi}{6}$, the slope is $m = -\cot(\frac{\pi}{6}) = -\sqrt{3}$. The point of tangency in Cartesian coordinates is $x=r\cos\theta = 2\cos(\frac{\pi}{6}) = 2(\frac{\sqrt{3}}{2})=\sqrt{3}$ and $y=r\sin\theta = 2\sin(\frac{\pi}{6}) = 2(\frac{1}{2})=1$. The equation of the tangent line is $y-1 = -\sqrt{3}(x-\sqrt{3})$, which simplifies to $y-1 = -\sqrt{3}x+3$, or $y = -\sqrt{3}x+4$.
16. A curve is given by $r=2-\cos\theta$; determine $\dfrac{dy}{dx}$ at $\theta=\pi$.
    Answer: undefined This problem requires the skill of differentiation in polar coordinates to find the slope of the tangent line in Cartesian coordinates. To find $dy/dx$ for a polar curve $r = f(\theta)$, use the formula $\frac{dy}{dx} = \frac{f'(\theta) \sin \theta + f(\theta) \cos \theta}{f'(\theta) \cos \theta - f(\theta) \sin \theta}$. For $r = 2 - \cos \theta$, $f'(\theta) = \sin \theta$, so at $\theta = \pi$, $f'(\pi) = 0$ and $f(\pi) = 3$. Plugging in, the numerator is $0 \cdot 0 + 3 \cdot (-1) = -3$ and the denominator is $0 \cdot (-1) - 3 \cdot 0 = 0$, yielding $dy/dx$ undefined. A tempting distractor is choice B, $\frac{dr}{d\theta}$ at $\pi$ which is 0, but this fails because it misses the zero denominator that signifies a vertical tangent. When computing polar derivatives, always remember to use the complete formula to account for the coordinate transformation, a strategy applicable to any polar curve.
17. A polar curve is defined by the equation $r = f(\theta)$ where $f$ is a differentiable function. If the curve has a horizontal tangent at $\theta = \frac{\pi}{3}$ and $f(\frac{\pi}{3}) = 1$, what is the value of $f'(\frac{\pi}{3})$?
    Answer: $-\frac{1}{\sqrt{3}}$ A horizontal tangent occurs when $\frac{dy}{d\theta} = 0$. The formula is $\frac{dy}{d\theta} = f'(\theta)\sin\theta + f(\theta)\cos\theta$. At $\theta = \frac{\pi}{3}$, we set this to zero: $f'(\frac{\pi}{3})\sin(\frac{\pi}{3}) + f(\frac{\pi}{3})\cos(\frac{\pi}{3}) = 0$. We are given $f(\frac{\pi}{3}) = 1$, and we know $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$. Substituting these values gives $f'(\frac{\pi}{3})\left(\frac{\sqrt{3}}{2}\right) + (1)\left(\frac{1}{2}\right) = 0$. This simplifies to $f'(\frac{\pi}{3})\frac{\sqrt{3}}{2} = -\frac{1}{2}$. Solving for $f'(\frac{\pi}{3})$ gives $f'(\frac{\pi}{3}) = -\frac{1}{\sqrt{3}}$.
18. The equation of a polar curve is given by $r = f(\theta)$. Which of the following expressions gives $\frac{dy}{dx}$ in terms of $\theta$?
    Answer: $\frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta}$ The Cartesian coordinates are given by $x = r\cos\theta = f(\theta)\cos\theta$ and $y = r\sin\theta = f(\theta)\sin\theta$. Using the product rule to differentiate with respect to $\theta$ gives $\frac{dx}{d\theta} = f'(\theta)\cos\theta - f(\theta)\sin\theta$ and $\frac{dy}{d\theta} = f'(\theta)\sin\theta + f(\theta)\cos\theta$. The slope is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$. The correct expression is $\frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta}$.
19. For a particle whose position is described by the polar equation $r = \theta^2$, what is the rate of change of the y-coordinate with respect to $\theta$ at $\theta = \pi$?
    Answer: $-\pi^2$ The y-coordinate is given by $y = r\sin\theta$. Substituting $r=\theta^2$, we get $y = \theta^2\sin\theta$. The rate of change of the y-coordinate with respect to $\theta$ is $\frac{dy}{d\theta}$. Using the product rule, $\frac{dy}{d\theta} = 2\theta\sin\theta + \theta^2\cos\theta$. Evaluating at $\theta = \pi$ gives $\frac{dy}{d\theta} = 2\pi\sin\pi + \pi^2\cos\pi = 2\pi(0) + \pi^2(-1) = -\pi^2$.
20. The slope of the tangent line to a polar curve $r=f(\theta)$ is given by $\frac{dy}{dx}$. At a point where $f(\theta) \neq 0$ and $f'(\theta) \neq 0$, under what condition is the tangent line vertical?
    Answer: $\frac{dr}{d\theta} \cos\theta - r \sin\theta = 0$ A vertical tangent line occurs when the denominator of $\frac{dy}{dx}$ is zero and the numerator is non-zero. The denominator is $\frac{dx}{d\theta}$. Since $x = r\cos\theta$, we have $\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta$. Setting this equal to zero gives the condition for a vertical tangent.

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