9.1 Defining and Differentiating Parametric Equations

• In parametric equations, $x$ and $y$ are expressed as functions of the same parameter $t$.
• What do $f(t)$ and $g(t)$ represent in parametric equations?


  Functions of $t$
• As $t$ changes in the equations \begin{cases} x =$t \\ y =$ t^{2} \end{cases}, the point moves along a parabola $y =$$x^{2}$ called its path
• Parametric equations allow for easy representation of complex shapes that are difficult to express as explicit functions.
• Match the curve with its parametric equations:

  Circle ↔️ $x =$$r\cos(t), y =$$r\sin(t)$

  Ellipse ↔️ $x =$$a\cos(t), y =$$b\sin(t)$

  Line ↔️ $x =$$x_{0} +$$at, y =$$y_{0} +$$bt$
• Order the steps to describe the movement of a point along a curve defined by parametric equations as $t$ changes:


  1️⃣ Change the value of $t$

  2️⃣ Calculate the new coordinates $(x, y)$

  3️⃣ Plot the point in the plane

  4️⃣ Repeat the process
• To find $\frac{dy}{dx}$ from parametric equations, we use the formula $\frac{dy}{dx} =$$\frac{dy / dt}{dx / dt}$, which is called the chain rule
• The parametric derivative $\frac{dy}{dx}$ gives the slope of the tangent line to the curve at a specific point.
• The equation of the tangent line to a parametric curve at (x_{0}, y_{0})</latex> is $y - y_{0} =$$\frac{dy}{dx}(t_{0}) (x - x_{0})$, where $\frac{dy}{dx}(t_{0})$ is the slope at the parameter value t_{0}
• The velocity vector in parametric equations is given by $\vec{v} =$$\left( \frac{dx}{dt}, \frac{dy}{dt} \right)$.
• Steps to find the tangent line to the curve defined by $x =$$t^{2}$ and $y =$$2t$ at $t =$$1$:


  1️⃣ Find the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$

  2️⃣ Calculate $\frac{dy}{dx}$

  3️⃣ Find the coordinates $(x_{0}, y_{0})$ at $t =$$1$

  4️⃣ Write the tangent line equation
• At $t =$$1$, the slope $\frac{dy}{dx}(1)$ for the curve defined by $x =$$t^{2}$ and $y =$$2t$ is 1
• Parametric equations define the coordinates of a point in terms of a parameter t