Find The Slope Of The Tangent Line To The Polar Curve For The Given Value Of . R=2sin ; =(\pi )/(6) R=1+cos
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Introduction
Understanding the slope of the tangent line to a polar curve at a specific point is a fundamental concept in calculus and analytical geometry. This knowledge not only helps visualize the behavior of curves in the polar coordinate system but also has practical applications in physics, engineering, and computer graphics.
In this article, we will explore how to find the slope of the tangent line to given polar curves at specified points. Specifically, we will analyze the curves defined by the equations:
- \( R = 2 \sin \theta \), at \( \theta = \frac{\pi}{6} \)
- \( R = 1 + \cos \theta \), at a given value of \( \theta \) (though the exact angle isn't specified, we will demonstrate how to compute the slope at any point)
By understanding the derivation process and applying the appropriate calculus techniques, you will be able to determine the slopes for these types of curves with confidence.
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Understanding Polar Coordinates and Curves
Before diving into the calculations, it's essential to grasp the basics of polar coordinates and how curves are represented in this system.
What Are Polar Coordinates?
- Polar coordinates represent a point in the plane using a radius \( R \) and an angle \( \theta \).
- A point \( P \) in Cartesian coordinates \((x, y)\) can be expressed as:
- Conversely, given \( R \) and \( \theta \), the Cartesian coordinates can be obtained as above.
Polar Curves and Their Equations
- Polar curves are defined by equations involving \( R \) as a function of \( \theta \), such as \( R = f(\theta) \).
- Examples include circles, cardioids, rose curves, and limacons.
- Understanding how to compute the slope of the tangent line involves differentiating these equations with respect to \( \theta \).
Calculating the Slope of the Tangent Line in Polar Coordinates
The key to finding the slope of the tangent line \( \frac{dy}{dx} \) at a particular point on a polar curve involves differentiating the parametric equations derived from the polar form.
Parametric Equations from Polar Coordinates
Given:
\[
x = R(\theta) \cos \theta
\]
\[
y = R(\theta) \sin \theta
\]
The derivatives with respect to \( \theta \) are:
\[
\frac{dx}{d\theta} = R'(\theta) \cos \theta - R(\theta) \sin \theta
\]
\[
\frac{dy}{d\theta} = R'(\theta) \sin \theta + R(\theta) \cos \theta
\]
where \( R'(\theta) = \frac{dR}{d\theta} \).
Formula for the Slope \( \frac{dy}{dx} \)
Using the chain rule:
\[
\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{R'(\theta) \sin \theta + R(\theta) \cos \theta}{R'(\theta) \cos \theta - R(\theta) \sin \theta}
\]
This formula allows us to compute the slope at any point \( \theta \) for a given \( R(\theta) \).
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Calculating the Slope for \( R = 2 \sin \theta \) at \( \theta = \frac{\pi}{6} \)
Let's analyze the first curve:
\[
R = 2 \sin \theta
\]
and find the slope of the tangent line at \( \theta = \frac{\pi}{6} \).
Step 1: Compute \( R'(\theta) \)
\[
R'(\theta) = \frac{d}{d\theta} (2 \sin \theta) = 2 \cos \theta
\]
Step 2: Evaluate \( R(\theta) \) and \( R'(\theta) \) at \( \theta = \frac{\pi}{6} \)
\[
R\left(\frac{\pi}{6}\right) = 2 \sin \left(\frac{\pi}{6}\right) = 2 \times \frac{1}{2} = 1
\]
\[
R'\left(\frac{\pi}{6}\right) = 2 \cos \left(\frac{\pi}{6}\right) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}
\]
Step 3: Compute the numerator and denominator for \( \frac{dy}{dx} \)
\[
\frac{dy}{dx} = \frac{R'(\theta) \sin \theta + R(\theta) \cos \theta}{R'(\theta) \cos \theta - R(\theta) \sin \theta}
\]
Substituting the values:
Numerator:
\[
\sqrt{3} \times \frac{1}{2} + 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}
\]
Denominator:
\[
\sqrt{3} \times \frac{\sqrt{3}}{2} - 1 \times \frac{1}{2} = \frac{3}{2} - \frac{1}{2} = 1
\]
Step 4: Final slope calculation
\[
\boxed{
\frac{dy}{dx} = \frac{\sqrt{3}}{1} = \sqrt{3}
}
\]
Conclusion: The slope of the tangent line to the curve \( R = 2 \sin \theta \) at \( \theta = \frac{\pi}{6} \) is \( \sqrt{3} \).
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Calculating the Slope for \( R = 1 + \cos \theta \) at a Specific \( \theta \)
Next, consider the second curve:
\[
R = 1 + \cos \theta
\]
Suppose we wish to find the slope at \( \theta = \frac{\pi}{3} \) (or any other specified \( \theta \)). The process is similar.
Step 1: Compute \( R'(\theta) \)
\[
R'(\theta) = - \sin \theta
\]
Step 2: Evaluate \( R(\theta) \) and \( R'(\theta) \) at \( \theta = \frac{\pi}{3} \)
\[
R\left(\frac{\pi}{3}\right) = 1 + \cos \left(\frac{\pi}{3}\right) = 1 + \frac{1}{2} = \frac{3}{2}
\]
\[
R'\left(\frac{\pi}{3}\right) = - \sin \left(\frac{\pi}{3}\right) = - \frac{\sqrt{3}}{2}
\]
Step 3: Compute numerator and denominator for \( \frac{dy}{dx} \)
\[
\frac{dy}{dx} = \frac{R'(\theta) \sin \theta + R(\theta) \cos \theta}{R'(\theta) \cos \theta - R(\theta) \sin \theta}
\]
Substituting:
Numerator:
\[
- \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{3}{2} \times \frac{1}{2} = - \frac{3}{4} + \frac{3}{4} = 0
Denominator:
\[
- \frac{\sqrt{3}}{2} \times \frac{1}{2} - \frac{3}{2} \times \frac{\sqrt{3}}{2} = - \frac{\sqrt{3}}{4} - \frac{3 \sqrt{3}}{4} = - \frac{\sqrt{3}}{4} - \frac{3 \sqrt{3}}{4} = - \frac{4 \sqrt{3}}{4} = - \sqrt{3}
Step 4: Final slope calculation
\[
\frac{dy}{dx} = \frac{0}{- \sqrt{3}} = 0
\]
Conclusion: The slope of the tangent line to \( R = 1 + \cos \theta \) at \( \theta = \frac{\pi}{3} \) is 0, indicating a horizontal tangent.
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Additional Considerations and Applications
Understanding
