Problem 62 - 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 \(\theta .\) $$r=1+2 \cos \theta, \quad \theta=\pi / 3$$

Step-by-Step Solution

Answer

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

Step 1: Convert Polar to Cartesian Coordinates

First, convert the polar equation to Cartesian coordinates. The formulas for conversion are, \[ x = r \cos\theta \quad \text{and} \quad y = r \sin\theta \]For the equation \( r = 1 + 2\cos\theta \), at \( \theta = \frac{\pi}{3} \),\[ x = (1 + 2\cos\frac{\pi}{3}) \cos\frac{\pi}{3} = (1+2 \cdot \frac{1}{2}) \cdot \frac{1}{2} = 1 \cdot \frac{1}{2} = \frac{1}{2} \]\[ y = (1 + 2\cos\frac{\pi}{3}) \sin\frac{\pi}{3} = 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]Thus, the Cartesian coordinates are \((\frac{1}{2}, \frac{\sqrt{3}}{2})\).

Step 2: Differentiate the Polar Equation

To find \(\frac{dy}{dx}\), differentiate \( r \cos\theta \) and \( r \sin\theta \) with respect to \( \theta \),\[ \frac{dy}{d\theta} = \frac{d}{d\theta}[r \sin\theta] = r' \sin\theta + r \cos\theta \]\[ \frac{dx}{d\theta} = \frac{d}{d\theta}[r \cos\theta] = r' \cos\theta - r \sin\theta \]Where \( r' = \frac{dr}{d\theta} = -2\sin\theta \).

Step 3: Calculate The Derivatives at \(\theta = \pi/3\)

Substitute \( \theta = \frac{\pi}{3} \) into the derivative equations;\[ r' = -2\sin\frac{\pi}{3} = -2 \cdot \frac{\sqrt{3}}{2} = -\sqrt{3} \]\[ \frac{dy}{d\theta} = (-\sqrt{3}) \cdot \frac{\sqrt{3}}{2} + 1 \cdot \frac{1}{2} = -\frac{3}{2} + \frac{1}{2} = -1 \]\[ \frac{dx}{d\theta} = (-\sqrt{3}) \cdot \frac{1}{2} - 1 \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = -\sqrt{3} \]

Step 4: Find the Slope \(\frac{dy}{dx}\)

To find the slope of the tangent line to the curve, use the formula,\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-1}{-\sqrt{3}} = \frac{1}{\sqrt{3}} \]

Step 5: Conclusion

The slope of the tangent line to the polar curve \( r = 1 + 2 \cos \theta \) at \( \theta = \frac{\pi}{3} \) is \( \frac{1}{\sqrt{3}} \).

Key Concepts

Polar Coordinates

Polar coordinates are a set of two values, usually written as \((r, \theta)\), that describe a point's location in a plane. Here, \(r\) represents the radial distance from a fixed origin point, and \(\theta\) is the angle measured from a fixed direction, usually the positive x-axis. This system is particularly useful in situations involving circular patterns or when the relation of a point to a central origin is more intuitive.

• Radial Distance \(r\): The length from the origin to the point.
• Angle \(\theta\): The direction from the origin to the point, measured in radians or degrees.

Polar coordinates are often used in navigation, physics, and engineering to simplify solving problems involving angles and curves. For example, expressing circular or spiral paths using polar coordinates can make equations more manageable.

Cartesian Coordinates Conversion

Converting from polar coordinates to Cartesian coordinates involves transitioning from a radius-angle system to an x-y coordinate system. This transformation allows for easier use of traditional algebraic and calculus techniques, especially when dealing with slopes or linear equations.

• The conversion formulas are:
• \(x = r \cos\theta\): Computes the horizontal distance (x-coordinate).
• \(y = r \sin\theta\): Computes the vertical distance (y-coordinate).

Through this conversion, various mathematical operations such as finding distances, slopes, or intersections become more straightforward. In the exercise, the point \((r, \theta) = (1 + 2 \cos \theta, \pi/3)\) is converted to Cartesian as \((\frac{1}{2}, \frac{\sqrt{3}}{2})\). This form is ready for differentiation or further analysis.

Differential Calculus

Differential calculus deals with the concept of change and rates of change. In the context of our problem, we are interested in how the position of the point changes as \(\theta\) changes, which leads to the calculation of the slope of a tangent line to a curve.

• Derivative of a Function: Provides the slope of the function at any point, indicating how the function value changes with respect to change in the input.
• In this exercise, polar coordinates are differentiated:
• \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\) are computed to find how x and y change with \(\theta\).
• The slope \(\frac{dy}{dx}\) is found by dividing \(\frac{dy}{d\theta}\) by \(\frac{dx}{d\theta}\).

The result, \(\frac{1}{\sqrt{3}}\), gives a precise description of the steepness and direction of the curve at a specific point. This calculation is pivotal in various scientific and engineering fields where understanding the instantaneous rate of change is crucial.