Question

Find the slope of the tangent line to the given polar curve at the point given by the value of \(\theta\).[T] Use technology: \(r=2+4 \cos \theta\) at \(\theta=\frac{\pi}{6}\)

Short answer

The slope of the tangent is \( \sqrt{3} - 1 \).

Step-by-step solution

Convert polar to Cartesian coordinates

First, remember the polar to the Cartesian conversion formulas: \[ x = r \cos \theta \quad \text{and} \quad y = r \sin \theta \]At \( \theta = \frac{\pi}{6} \), calculate \( r \) using the polar equation \( r = 2 + 4 \cos \theta \).Calculating, we have:\[ \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]Substituting this into the equation for \( r \):\[ r = 2 + 4 \left(\frac{\sqrt{3}}{2}\right) = 2 + 2\sqrt{3} \]Now, substitute into the Cartesian coordinates:\[ x = (2 + 2\sqrt{3}) \cdot \frac{\sqrt{3}}{2} \]\[ y = (2 + 2\sqrt{3}) \cdot \frac{1}{2} \]Calculate to find \( x \approx 3.464 \) and \( y \approx 2.732 \).

Find derivatives in polar coordinates

For a polar curve, the slope of the tangent is given by \( \frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta} \).First, find \( \frac{dr}{d\theta} \) using the derivative of \( r = 2 + 4 \cos \theta \):\[ \frac{dr}{d\theta} = -4 \sin \theta \]At \( \theta = \frac{\pi}{6} \), substitute \( \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \) into the derivative:\[ \frac{dr}{d\theta} = -4 \cdot \frac{1}{2} = -2 \]

Calculate the tangent slope

Substitute \( \theta = \frac{\pi}{6} \), \( r = 2 + 2\sqrt{3} \), and \( \frac{dr}{d\theta} = -2 \) into the slope formula:\[ \frac{dy}{dx} = \frac{(-2)\cdot\frac{1}{2} + (2 + 2\sqrt{3})\cdot\frac{\sqrt{3}}{2}}{(-2)\cdot\frac{\sqrt{3}}{2} - (2 + 2\sqrt{3})\cdot\frac{1}{2}} \]Time to simplify the expression:\[ \frac{dy}{dx} = \frac{-1 + (2 + 2\sqrt{3})\frac{\sqrt{3}}{2}}{-\sqrt{3} - (2 + 2\sqrt{3})\frac{1}{2}} \]After further simplification, we get:\[ \frac{dy}{dx} \approx -\frac{2 + \sqrt{3}}{4} \]

Simplify result

Further simplify the expression to get the exact value:\[ \frac{dy}{dx} = -1 + \sqrt{3} \]Thus, the slope of the tangent line at \( \theta = \frac{\pi}{6} \) is \( \sqrt{3} - 1 \).

Key concepts

Polar to Cartesian Conversion

Polar coordinates describe a point in the plane using two values: the distance from the origin, called the radial coordinate \(r\), and the angle \(\theta\), often from the positive x-axis. To work with these in many applications, converting polar coordinates to Cartesian coordinates, which use \(x\) and \(y\) values, is crucial. This is done using the formulas:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)

These formulas allow us to determine the precise location of a point in the standard Cartesian coordinate plane. For our given polar curve, where \(r = 2 + 4 \cos \theta\) at \(\theta = \frac{\pi}{6}\), we found \(r\) by calculating \(2 + 2\sqrt{3}\). Substituting into our conversion formulas:

• \(x = (2 + 2\sqrt{3}) \cdot \frac{\sqrt{3}}{2}\) leading to \(x \approx 3.464\)
• \(y = (2 + 2\sqrt{3}) \cdot \frac{1}{2}\) resulting in \(y \approx 2.732\)

These values help in determining the exact location on the Cartesian plane, which is essential for further analysis, such as finding slopes of tangent lines.

Derivative in Polar Coordinates

Derivatives provide a way to measure change and are crucial in understanding curves and dynamic systems. In polar coordinates, we often need to find the derivative of \(r\) with respect to \(\theta\) to compute slopes and other properties. For our curve, \(r = 2 + 4 \cos \theta\), we found the derivative:

• \(\frac{dr}{d\theta} = -4 \sin \theta\)

At \(\theta = \frac{\pi}{6}\), this becomes \(-4 \cdot \frac{1}{2} = -2\). This derivative is essential because it gives us the rate of change of the distance \(r\) as \(\theta\) changes. This understanding allows us to further analyze the curve and calculate the slope of the tangent line using polar information. Knowing how these derivatives behave helps us understand the dynamic nature of polar graphs.

Slope of Tangent Line

Finding the slope of a tangent line to a polar curve at a specific point is a key task in calculus. To acquire this, we must use both the derivatives of \(r\) and the trigonometric components \(\cos \theta\) and \(\sin \theta\) to convert polar derivatives to Cartesian derivatives. The formula used is:\[\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}\]For the given polar equation, at \(\theta = \frac{\pi}{6}\), and with the calculated \(r\) and \(\frac{dr}{d\theta}\), we arrive at:

• Substituting the values results in \(\frac{dy}{dx} \approx -\frac{2 + \sqrt{3}}{4}\)
• Which simplifies to \(\sqrt{3} - 1\)

This slope represents how steep the tangent is to the curve at that particular angle, reflecting the rate of change in the y-direction compare to the x-direction, using the polar values.

Polar Curves Analysis

Analyzing polar curves combines several calculus concepts including differentiation, trigonometry, and coordinate conversion. Such analysis offers insights into the behavior of curves defined in terms of angles and radii rather than Cartesian coordinates. For the polar curve given by \(r = 2 + 4 \cos \theta\), at \(\theta = \frac{\pi}{6}\):

• Conversion to Cartesian coordinates makes graphing and geometric interpretation possible.
• Finding derivatives allows us to understand change and analyze features like tangents and normals.
• The calculation of tangent slopes provides information on direction and behavior of the curve at precise points.

These steps showcase how polar graphs can be critically assessed using these mathematical methods, allowing a deeper comprehension and visualization of the curve's properties and potentials.