A circle with diameter |a| centered on the x-axis, passing through the origin.

A circle with diameter |a| centered on the y-axis, passing through the origin.

A limacon with an inner loop.

A cardioid.

A convex limacon (no loop or indentation).

The sign of 'a' determines which quadrant the first petal of the rose curve lies in.

A spiral that starts at the pole and winds outwards as θ increases.

A four-petal rose centered at the origin.

A three-petal rose centered at the origin.

$x = r \cos\theta$

$y = r \sin\theta$

$r = \sqrt{x^2 + y^2}$

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{r\cos\theta + \frac{dr}{d\theta}(\sin\theta)}{-r\sin\theta + \frac{dr}{d\theta}(\cos\theta)}$

$\frac{dx}{d\theta} = \frac{d}{d\theta}(r\cos\theta)$

$\frac{dy}{d\theta} = \frac{d}{d\theta}(r\sin\theta)$

$\frac{dr}{d\theta} = f'(\theta)$

$x = r \cos(\theta)$

$y = r \sin(\theta)$

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

Multiply both sides by r: $r^2 = 2r\sin\theta$. Substitute $r^2 = x^2 + y^2$ and $y = r\sin\theta$: $x^2 + y^2 = 2y$. Complete the square: $x^2 + (y-1)^2 = 1$.

Find x and y: $x = r\cos\theta = \theta\cos\theta$, $y = r\sin\theta = \theta\sin\theta$. Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$: $\frac{dx}{d\theta} = \cos\theta - \theta\sin\theta$, $\frac{dy}{d\theta} = \sin\theta + \theta\cos\theta$. Then, $\frac{dy}{dx} = \frac{\sin\theta + \theta\cos\theta}{\cos\theta - \theta\sin\theta}$.

Find $\frac{dr}{d\theta} = -\sin\theta$. Set $- \sin\theta = 0$, so $\theta = 0, \pi$. Evaluate r at these points: $r(0) = 2$, $r(\pi) = 0$. The closest point is 0, the furthest is 2.

1. Find x and y in terms of θ: $x = r\cos(\theta) = \sin(2\theta)\cos(\theta)$, $y = r\sin(\theta) = \sin(2\theta)\sin(\theta)$. 2. Compute $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$. 3. Calculate $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ at $\theta = \frac{\pi}{4}$. 4. Find the (x, y) coordinates at $\theta = \frac{\pi}{4}$. 5. Use the point-slope form to write the equation of the tangent line.

1. Recall that $r^2 = x^2 + y^2$. 2. Substitute $r^2$ for $x^2 + y^2$ in the given equation. 3. The polar form is $r^2 = 9$, which simplifies to $r = 3$.

1. Find x and y: $x = 2\cos^2(\theta)$, $y = 2\cos(\theta)\sin(\theta) = \sin(2\theta)$. 2. Find $\frac{dx}{d\theta} = -4\cos(\theta)\sin(\theta) = -2\sin(2\theta)$ and $\frac{dy}{d\theta} = 2\cos(2\theta)$. 3. Calculate $\frac{dy}{dx} = \frac{2\cos(2\theta)}{-2\sin(2\theta)} = -\cot(2\theta)$. 4. Evaluate at $\theta = \frac{\pi}{3}$: $\frac{dy}{dx} = -\cot(\frac{2\pi}{3}) = \frac{\sqrt{3}}{3}$.

1. Calculate r: $r = 4\sin(\frac{\pi}{6}) = 4 * \frac{1}{2} = 2$. 2. Use the formula $x = r\cos(\theta)$: $x = 2\cos(\frac{\pi}{6}) = 2 * \frac{\sqrt{3}}{2} = \sqrt{3}$.

1. Calculate r: $r = 2 + \cos(\frac{\pi}{2}) = 2 + 0 = 2$. 2. Use the formula $y = r\sin(\theta)$: $y = 2\sin(\frac{\pi}{2}) = 2 * 1 = 2$.

1. The x-axis corresponds to $y = 0$. 2. Set $y = r\sin(\theta) = 0$. 3. This implies $\sin(\theta) = 0$ (since r is not always zero). 4. Solve for $\theta$: $\theta = n\pi$, where n is an integer.

1. The y-axis corresponds to $x = 0$. 2. Set $x = r\cos(\theta) = 0$. 3. This implies $\cos(\theta) = 0$ (since r is not always zero). 4. Solve for $\theta$: $\theta = \frac{\pi}{2} + n\pi$, where n is an integer.