Question

In Exercises \(41-48,\) find the equation of the line tangent to the curve at the point defined by the given value of \(t\) . $$x=\sec t, \quad y=\tan t, \quad t=\pi / 6$$

Short answer

The equation of the tangent line is \(y = x \sqrt {3} - 1\)

Step-by-step solution

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

Find the derivative of \(y\) with respect to \(x\) using the chain rule: \(\frac{dy}{dt} / \frac{dx}{dt}\). Here, \(\frac{dy}{dt} = \sec^2(t)\) and \(\frac{dx}{dt} = \sec(t) \tan(t)\). Hence, \(\frac{dy}{dx} = \frac{\sec^2(t)}{\sec(t) \tan(t)}\).

Evaluate \(\frac{dy}{dx}\) at \(t = \pi / 6\)

Substitute \(t = \pi/6\) into \(\frac{dy}{dx}\) to find the slope of the tangent line: \(\frac{dy}{dx}(\pi / 6) = \frac{\sec^2(\pi / 6)}{\sec(\pi / 6) \tan(\pi / 6)} = 3 / \sqrt {3}\).

Find the Point of Tangency

Substitute \(t = \pi/6\) into the parametric equations to find the coordinates of the point of tangency: \(x = \sec(\pi/6) = 2/\sqrt{3}\), \(y = \tan(\pi/6) = \sqrt{3} / 3\).

Write the Equation of the Tangent Line

Use the point-slope form of a line to write the equation of the tangent line: \(y - (\sqrt{3} / 3) = 3 / \sqrt {3} [x - (2/\sqrt{3})\]. Simplify the equation to its simplest form: \(y = x \sqrt {3} - 1\).

Key concepts

Chain Rule Differentiation

Understanding the chain rule is crucial for calculating derivatives in calculus, especially when dealing with composite functions. In essence, the chain rule is a formula for computing the derivative of a composition of two or more functions. For example, if a function y can be expressed as another function of u, which itself is a function of x, that is, if we have y = f(u) and u = g(x), then the derivative of y with respect to x is given by the product of the derivatives of y with respect to u and u with respect to x. In mathematical terms, this is expressed as \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

In the given exercise, we apply the chain rule to differentiate parametric equations where y and x are both functions of a third variable, t. This allows us to find \( \frac{dy}{dx} \) by dividing \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \), each of which is found using straightforward differentiation techniques.

Parametric Equations

Parametric equations are a set of equations that express a set of quantities as explicit functions of a number of independent variables, known as parameters. In the context of calculus, parametric equations allow us to represent curves in the plane by defining both x and y coordinates in terms of a third variable, often denoted as t. The beauty of parametric equations lies in their ability to describe complex curves and motions that are difficult to express with a single equation in x and y.

For instance, the parametric equations given in the exercise, \(x = \sec t\) and \(y = \tan t\), define a curve in terms of trigonometric functions of the parameter t. By substituting a specific value of t, we can find a specific point on the curve, which is necessary when determining the equation of a tangent line at that point.

Slope of a Tangent

The slope of a tangent line to a curve at a given point is a measure of how steep the line is at that point. It is the rate at which y changes with respect to x along the curve and is defined as the derivative of y with respect to x at that point. For curves described by parametric equations, we calculate the slope of the tangent as the ratio of the derivatives of y and x with respect to the parameter, as seen in the exercise by \( \frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt} \).

Once we have the slope and a specific point from the parametric equations, we can fully determine the tangent line's equation. In the exercise, evaluating \( \frac{dy}{dx} \) at \( t = \frac{\pi}{6} \) yields the slope of the tangent at the point of tangency to the curve.

Point-Slope Form

The point-slope form is a straightforward way of writing the equation of a line when you know the slope and a point on the line. It's given by the formula \( y - y_1 = m(x - x_1) \) where \( m \) is the slope of the line and \( (x_1, y_1) \) is the point on the line. This form is particularly useful when writing the equation of the tangent line to a curve at a specific point.

In the context of our problem, after finding the slope of the tangent line and the coordinates of the tangency point using the parametric equations, we plug these values into the point-slope formula to arrive at the tangent line equation. By simplifying, we obtain the line's equation in a more familiar form, typically y = mx + b, which represents the same line in slope-intercept form.