Finding the rate of change of y with respect to x, where both x and y are defined in terms of a third variable t.

It's an application of the chain rule. It relates the rate of change of y with respect to t to the rate of change of x with respect to t to find the rate of change of y with respect to x.

The slope of the tangent line to the parametric curve at a specific point.

Division by zero is undefined. If $\frac{dx}{dt} = 0$, the tangent line is vertical, and the slope is undefined.

Parametric equations allow the x-coordinate to change direction and speed independently of the y-coordinate, unlike Cartesian equations where x typically increases at a constant rate.

Parametric equations can model projectile motion by separately defining the horizontal and vertical positions (x and y) as functions of time (t), allowing for the incorporation of factors like gravity and initial velocity.

The tangent line is a straight line that touches the curve at a single point and has the same slope as the curve at that point, representing the instantaneous direction of the curve.

The parameter 't' dictates the order in which points are plotted, thus defining the direction and speed at which the curve is traced.

$\frac{dx}{dt}$ represents the horizontal velocity, and $\frac{dy}{dt}$ represents the vertical velocity of a point moving along the curve.

By using trigonometric functions (sine and cosine) for x(t) and y(t), parametric equations can describe circular motion with the parameter t representing the angle or time.

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

$\frac{dx}{dt} \neq 0$

$\frac{d^2y}{dx^2} = \frac{d/dt(\frac{dy}{dx})}{\frac{dx}{dt}}$

Find $\frac{dy}{dx}$ at the given t, then use the point-slope form: $y - y(t) = \frac{dy}{dx}(x - x(t))$

Set $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$ and solve for t.

Set $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$ and solve for t.

$\int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$

$\int_{a}^{b} y(t) \cdot x'(t) dt$

Speed = $\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$

1. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$. 2. Calculate $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. 3. Substitute the given value of t into $\frac{dy}{dx}$.

1. Find $\frac{dy}{dt}$. 2. Set $\frac{dy}{dt} = 0$ and solve for t. 3. Verify that $\frac{dx}{dt} \neq 0$ for those values of t.

1. Find $\frac{dx}{dt}$. 2. Set $\frac{dx}{dt} = 0$ and solve for t. 3. Verify that $\frac{dy}{dt} \neq 0$ for those values of t.

1. Solve one of the parametric equations for t. 2. Substitute that expression for t into the other parametric equation.

1. Find the value of t corresponding to the point (x₀, y₀). 2. Find $\frac{dy}{dx}$ at that value of t. 3. Use the point-slope form of a line: $y - y₀ = \frac{dy}{dx}(x - x₀)$.

1. Find $\frac{dy}{dx}$. 2. Find $\frac{d}{dt}(\frac{dy}{dx})$. 3. Divide the result by $\frac{dx}{dt}$: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$.

1. Solve x(t) = a for t. 2. Plug the value(s) of t into y(t). 3. If y(t) = b, then that t value is the solution.

1. Set the x-equations equal to each other and the y-equations equal to each other. 2. Solve the system of equations for the parameter values. 3. Substitute the parameter values back into the original equations to find the points of intersection.

1. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$. 2. Calculate $\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$. 3. Integrate the result from a to b: $\int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$.

1. Express y as a function of t, y(t). 2. Find $\frac{dx}{dt}$. 3. Integrate $y(t) \cdot \frac{dx}{dt}$ with respect to t over the appropriate interval.