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.

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.

Use the relations $x = r\cos\theta$, $y = r\sin\theta$, and $r = \sqrt{x^2 + y^2}$ to eliminate r and θ and express the equation in terms of x and y.

Find $\frac{dr}{d\theta}$, set it equal to zero, solve for $\theta$, and plug the $\theta$ values back into the original equation to find the corresponding r values. Also, check endpoints.

It allows for easier visualization and graphing on a traditional x-y coordinate plane, especially for complex functions.

It represents the radial component of the curve and indicates the instantaneous rate of change of the distance from the origin.

It represents the radial curvature of the curve and indicates the rate of change of the curvature.

Polar equations can be expressed parametrically using $x = r\cos\theta$ and $y = r\sin\theta$, where $\theta$ is the parameter.

It gives the slope of the tangent line to the curve at a given point in Cartesian coordinates, indicating the direction of the curve at that point.

Because y and x are both functions of r and θ, and r is a function of θ, the chain rule is needed to relate the derivatives.

Calculate $\frac{dy}{dx}$ at the given angle, find the x and y coordinates using $x = r\cos\theta$ and $y = r\sin\theta$, then use the point-slope form to find the equation of the tangent line.

Trigonometric identities are used to simplify expressions, convert between polar and Cartesian coordinates, and solve equations involving trigonometric functions of θ.