Homework: Parametric Equations, Tangent Lines, and Area Enclosed by Curves

Parametric Equations and Tangent Lines

Understanding Parametric Equations

Parametric equations express the coordinates of the points that make up a geometric object as functions of a variable, usually denoted as t (the parameter). This approach is especially useful for describing curves that are not functions in the traditional sense (i.e., they may fail the vertical line test).

• Parametric Form: $x = f(t),\ y = g(t)$, where $t$ varies over an interval.

• Example: $x = t^2 - 4$, $y = t^3 - 2t$ for $t = -2$.

Finding the Equation of the Tangent Line

The tangent line to a parametric curve at a given value of $t$ can be found by computing the derivatives of $x$ and $y$ with respect to $t$ and then using the point-slope form of a line.

• Step 1: Compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

• Step 2: The slope of the tangent line is $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ (provided $\frac{dx}{dt} \neq 0$).

• Step 3: Find the coordinates $(x_0, y_0)$ at the given $t$ value.

• Step 4: Use the point-slope form: $y - y_0 = m(x - x_0)$, where $m = \frac{dy}{dx}$.

Example: For $x = t^2 - 4$, $y = t^3 - 2t$ at $t = -2$:

• $x(-2) = (-2)^2 - 4 = 0$

• $y(-2) = (-2)^3 - 2(-2) = -8 + 4 = -4$

• $\frac{dx}{dt} = 2t$, $\frac{dy}{dt} = 3t^2 - 2$

• At $t = -2$: $\frac{dx}{dt} = -4$, $\frac{dy}{dt} = 3(4) - 2 = 10$

• Slope: $\frac{dy}{dx} = \frac{10}{-4} = -\frac{5}{2}$

• Tangent line: $y + 4 = -\frac{5}{2}(x - 0)$

Vertical and Horizontal Tangents

Points where the tangent is vertical or horizontal are found by analyzing the derivatives:

• Horizontal Tangent: Occurs when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$.

• Vertical Tangent: Occurs when $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$.

Procedure:

1. Solve $\frac{dy}{dt} = 0$ for $t$ to find horizontal tangents.

2. Solve $\frac{dx}{dt} = 0$ for $t$ to find vertical tangents.

3. Substitute these $t$ values into $x(t)$ and $y(t)$ to find the corresponding points.

Example: For $x = t^2 - 4$, $y = t^3 - 2t$:

• $\frac{dx}{dt} = 2t$; vertical tangent when $t = 0$

• $\frac{dy}{dt} = 3t^2 - 2$; horizontal tangent when $t = \pm\sqrt{2/3}$

Area Enclosed by a Parametric Curve

Formula for Area

The area enclosed by a parametric curve and the x-axis can be found using the following integral:

• $A = \int_{a}^{b} y(t) \frac{dx}{dt} dt$

Where $t$ ranges over the interval that traces the closed curve once.

Example Calculation

Given $x = t^2 - 4$, $y = t^3 - 2t$, the area between the curve and the x-axis from $t = a$ to $t = b$ is:

• $A = \int_{a}^{b} (t^3 - 2t) \cdot 2t\ dt$

• Simplify the integrand: $2t^4 - 4t^2$

• $A = \int_{a}^{b} (2t^4 - 4t^2)\ dt$

Evaluate the definite integral over the appropriate interval to find the area.

Summary Table: Tangent Line Conditions

Condition	Mathematical Statement	Type of Tangent
Horizontal Tangent	$\frac{dy}{dt} = 0$, $\frac{dx}{dt} \neq 0$	Horizontal
Vertical Tangent	$\frac{dx}{dt} = 0$, $\frac{dy}{dt} \neq 0$	Vertical

Additional info: These concepts are foundational for understanding parametric curves, their tangents, and areas, which are key topics in Calculus II (Ch. 10 - Parametric Equations and Polar Coordinates).