Problem 59 - Calculus Early Transcendentals

Question

\(57-62\) Find the slope of the tangent line to the given polar curve at the point specified by the value of \(\theta\) . $$r=1 / \theta, \quad \theta=\pi$$

Step-by-Step Solution

Answer

The slope of the tangent line at \(\theta = \pi\) is \(\pi\).

Step 1: Understand the Problem

We need to find the slope of the tangent line to the polar curve given by the equation \(r = \frac{1}{\theta}\) at the point corresponding to \(\theta = \pi\).

Step 2: Identify the Conversion to Cartesian Coordinates

First, we need to convert the polar coordinates to Cartesian coordinates using the formulas \(x = r\cos\theta\) and \(y = r\sin\theta\). For \(r = \frac{1}{\theta}\), we have \(x = \frac{1}{\theta}\cos(\theta)\) and \(y = \frac{1}{\theta}\sin(\theta)\).

Step 3: Differentiate Parametrically

To find the slope of the tangent, we require \(\frac{dy}{dx}\). Start by computing \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\). Differentiate \(x = \frac{1}{\theta}\cos(\theta)\) and \(y = \frac{1}{\theta}\sin(\theta)\) with respect to \(\theta\).

Step 4: Apply the Product Rule

For \(x = \frac{1}{\theta}\cos(\theta)\), apply the product rule: \(\frac{dx}{d\theta} = \frac{d}{d\theta}(\frac{1}{\theta})\cos(\theta) + \frac{1}{\theta}\frac{d}{d\theta}(\cos(\theta))\). This results in \(\frac{dx}{d\theta} = -\frac{1}{\theta^2}\cos(\theta) - \frac{1}{\theta}\sin(\theta)\).

Step 5: Differentiate Y Component

Similarly for \(y = \frac{1}{\theta}\sin(\theta)\), apply the product rule: \(\frac{dy}{d\theta} = \frac{d}{d\theta}(\frac{1}{\theta})\sin(\theta) + \frac{1}{\theta}\frac{d}{d\theta}(\sin(\theta))\). This results in \(\frac{dy}{d\theta} = -\frac{1}{\theta^2}\sin(\theta) + \frac{1}{\theta}\cos(\theta)\).

Step 6: Find \(\frac{dy}{dx}\)

The slope of the tangent line, \(\frac{dy}{dx}\), is given by \(\frac{dy}{d\theta}\) divided by \(\frac{dx}{d\theta}\). Substitute the expressions from Steps 4 and 5 into the formula: \(\frac{dy}{dx} = \frac{-\frac{1}{\theta^2}\sin(\theta) + \frac{1}{\theta}\cos(\theta)}{-\frac{1}{\theta^2}\cos(\theta) - \frac{1}{\theta}\sin(\theta)}\).

Step 7: Evaluate at \(\theta = \pi\)

Substitute \(\theta = \pi\) into the expression for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{-\frac{1}{\pi^2}\sin(\pi) + \frac{1}{\pi}\cos(\pi)}{-\frac{1}{\pi^2}\cos(\pi) - \frac{1}{\pi}\sin(\pi)}\). Simplifying this, noting that \(\sin(\pi) = 0\) and \(\cos(\pi) = -1\), we get \(\frac{dy}{dx} = \pi\).

Key Concepts

Tangent Line

In mathematics, a tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. Imagine you're drawing a line that just barely skims the surface of the curve, merging with it at one single point. At that point, the tangent line provides a good approximation for the curve itself.

For polar coordinates,

• The concept remains similar; the tangent line at a point on a polar curve gives us the direction the curve is going at that precise point.
• This direction is crucial for understanding how the curve behaves locally—meaning in the neighboring vicinity of the point.

Identifying and calculating the tangent line involves finding the slope at the point of interest. This is where the slope and the method of parametric differentiation, which we will discuss next, come into play.

Parametric Differentiation

Parametric differentiation is a technique used to calculate derivatives when the coordinates of a curve are defined by one or more parameters. In simpler terms, parameters act like sliders that change the position on the curve. For polar equations, the parameter often used is \( \theta \), representing the angle.

When finding the slope of the tangent line to a polar curve:

• We typically convert the polar coordinates to Cartesian coordinates using \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \).
• We then differentiate these expressions parametrically with respect to \( \theta \).

This involves applying the product rule, which is a fundamental principle in calculus for differentiating products of functions. Thus, by finding \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \), we can solve for the slope \( \frac{dy}{dx} \), helping determine the tangent line's inclination.

Slope of a Curve

The slope of a curve at a specific point is a measure of how steep the curve is at that point. It could be thought of as the angle of the steepest hill represented by the curve. In terms of tangent lines, the slope identifies the incline of the line that just touches the curve at that point.

For polar equations,

• The task is to find the slope \( \frac{dy}{dx} \) by taking the derivative of both \( x \) and \( y \) with respect to the parameter \( \theta \), and then dividing them: \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \).
• This slope value tells us how sharply the curve is inclined at the specified angle \( \theta \).

In the original exercise, this entire process was applied to the function \( r = \frac{1}{\theta} \) at \( \theta = \pi \), leading to a slope of \( \pi \). This means that, at \( \theta = \pi \), the tangent line has a slope or steepness of \( \pi \), reflecting the curve's trajectory at that point.