Question

, find the equation of the tangent line to the given curve at the given value of \(t\) without eliminating the parameter. Make a sketch. $$ x=2 \sec t, y=2 \tan t ; t=-\frac{\pi}{6} $$

Short answer

The equation of the tangent line is \( y + \frac{2}{\sqrt{3}} = -2(x - \frac{4\sqrt{3}}{3}) \).

Step-by-step solution

Parameterize and Differentiate

The given parametric equations are \( x = 2 \sec t \) and \( y = 2 \tan t \). To find the slope of the tangent line, we need \( \frac{dy}{dx} \). First, find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Using derivatives of trigonometric functions: \( \frac{dx}{dt} = 2 \sec t \tan t \), \( \frac{dy}{dt} = 2 \sec^2 t \).

Find \( \frac{dy}{dx} \)

The slope of the tangent line is \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2 \sec^2 t}{2 \sec t \tan t} = \frac{\sec t}{\tan t} = \frac{1}{\sin t} \).

Evaluate Derivative at \( t = -\frac{\pi}{6} \)

Substitute \( t = -\frac{\pi}{6} \) into \( \frac{1}{\sin t} \). We have \( \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \). Thus, \( \frac{1}{\sin t} = -2 \).

Calculate Point on the Curve

Using \( t = -\frac{\pi}{6} \), substitute into the original equations: \( x = 2 \sec\left(-\frac{\pi}{6}\right) = 2 \cdot \frac{2}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \) and \( y = 2 \tan\left(-\frac{\pi}{6}\right) = 2 \left(-\frac{1}{\sqrt{3}}\right) = -\frac{2}{\sqrt{3}} \).

Write the Equation of the Tangent Line

The equation of the tangent line can be written in the form \( y - y_0 = m(x - x_0) \), where \((x_0, y_0)\) is the point \( \left(\frac{4\sqrt{3}}{3}, -\frac{2}{\sqrt{3}}\right) \) and \( m = -2 \) is the slope. Substituting these in, the equation becomes \( y + \frac{2}{\sqrt{3}} = -2\left(x - \frac{4\sqrt{3}}{3}\right) \).

Key concepts

Parametric Equations

Parametric equations are a powerful mathematical tool that express the coordinates of the points making up a geometric figure as functions of a variable, often denoted as \( t \). This approach provides flexibility, especially with complex curves that aren't easily represented by a simple equation in \( x \) and \( y \).

In the given problem, the curve is defined by the parametric equations \( x = 2 \sec t \) and \( y = 2 \tan t \). Here, \( t \) is the parameter that influences both \( x \) and \( y \), allowing us to trace the curve as \( t \) changes. This representation is particularly useful when dealing with curves influenced by circular functions, like those involving secant and tangent, due to their periodic nature.

By not eliminating the parameter, we maintain the parametric nature of the expression, which can simplify the problem-solving process, especially for curves where traditional Cartesian forms become too complex. This is why, for many curves, especially those in periodic or cyclical contexts, parametric equations become the preferred method of representation.

Trigonometric Functions

Trigonometric functions, like \( \sec t \) and \( \tan t \), are key components in modeling relationships in cyclical patterns, such as waves and oscillations. These functions relate an angle measure to side lengths in a right triangle and can extend this relationship into the unit circle, providing continuity and periodic behavior.

For example, the secant function \( \sec t \) is the reciprocal of the cosine function \( \cos t \), expressed as \( \sec t = \frac{1}{\cos t} \). Similarly, the tangent function \( \tan t \) describes the ratio of the sine function to the cosine function, or \( \tan t = \frac{\sin t}{\cos t} \).

In the exercise, these trigonometric functions are used to parameterize the curve. This involves crafting an expression where \( x \) and \( y \) are defined with respect to these periodic functions, hence allowing the curve to be explored across the cycles of \( t \). The cyclic properties are crucial for finding intersections and tangents because of their consistent patterns.

Derivative Calculation

When finding the equation of a tangent line to a curve given in parametric form, calculating derivatives is essential. Specifically, you need to determine both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), individually.

In our scenario, these derivatives were computed using rules for trigonometric functions, such as \( \frac{d}{dt}(\sec t) = \sec t \tan t \) and \( \frac{d}{dt}(\tan t) = \sec^2 t \).

Next, obtaining \( \frac{dy}{dx} \), the slope of the tangent line, is done through dividing the two derivatives: \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). This step involves simplifying complex fractions that come from the trigonometric, which reduces to a function of \( t \).

At the specified value of \( t \), which was \( t = -\frac{\pi}{6} \), the slope calculation yielded a straightforward result: \( -2 \). This computation forms the backbone of identifying how sharply the curve's path changes at \( t \), and the eventual tangent line equation leverages this rate of change.