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Magazine Chapter 1: Problem 253
For the following exercises, find the points at which the following polar curves have a horizontal or vertical tangent line. The cardioid $$r=1+\sin \theta$$
Short Answer
Expert verified
Horizontal: \( (2, \frac{\pi}{2}), (0, \frac{3\pi}{2}) \). Vertical: \( (\frac{3}{2}, \frac{\pi}{6}), (0, \frac{3\pi}{2}), (\frac{3}{2}, \frac{5\pi}{6}) \).
Step by step solution
01
Understand Polar Curves and Tangents
A cardioid is a heart-shaped curve represented in polar coordinates. The equation given is \( r = 1 + \sin \theta \). Our objective is to find points where the tangent line is horizontal or vertical. In polar coordinates, horizontal tangents occur when \( \frac{dy}{d\theta} = 0 \) and vertical tangents occur when \( \frac{dx}{d\theta} = 0 \).
02
Express x and y in Polar Coordinates
In polar coordinates, \( x = r\cos \theta \) and \( y = r\sin \theta \). Substituting the given \( r = 1 + \sin \theta \), we get: \[ x = (1 + \sin \theta) \cos \theta \] and \[ y = (1 + \sin \theta) \sin \theta \].
03
Compute Derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \)
Apply the product and chain rules to find the derivatives: For \( x \), \[ \frac{dx}{d\theta} = \frac{d}{d\theta}((1+\sin \theta)\cos \theta) = \cos \theta \cdot \cos \theta - (1+\sin \theta)\sin \theta = \cos^2 \theta - \sin \theta - \sin^2 \theta \]. For \( y \), \[ \frac{dy}{d\theta} = \frac{d}{d\theta}((1+\sin \theta)\sin \theta) = \cos \theta \cdot \sin \theta +(1+\sin \theta)\cos \theta = \cos \theta + \cos \theta \sin \theta \].
04
Set \( \frac{dy}{d\theta} = 0 \) for Horizontal Tangents
From the derivative \( \frac{dy}{d\theta} = \cos \theta + \cos \theta \sin \theta \), factor out \( \cos \theta \): \[ \cos \theta (1 + \sin \theta) = 0 \]. Set each factor to zero: \( \cos \theta = 0 \) which gives \( \theta = \frac{\pi}{2}, \frac{3\pi}{2} \). And \( 1 + \sin \theta = 0 \) gives \( \sin \theta = -1 \) or \( \theta = \frac{3\pi}{2} \) in the unit circle.
05
Set \( \frac{dx}{d\theta} = 0 \) for Vertical Tangents
From \( \frac{dx}{d\theta} = \cos^2 \theta - \sin \theta - \sin^2 \theta \), simplify: Use identity \( \cos^2 \theta = 1 - \sin^2 \theta \): \[ 1 - \sin^2 \theta - \sin \theta - \sin^2 \theta = 0 \]. Thus, \( 1 - 2\sin^2 \theta - \sin \theta = 0 \). Solve this quadratic for \( \sin \theta \): Letting \( x = \sin \theta \), \( 2x^2 + x - 1 = 0 \). Solve: \[ x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]. Giving \( x = \frac{1}{2} \) or \( x = -1 \): \( \sin \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6}, \frac{5\pi}{6} \); \( \sin \theta = -1 \Rightarrow \theta = \frac{3\pi}{2} \).
06
Find Points on the Cardioid
Substitute \( \theta \) back into \( r = 1 + \sin \theta \) to find \( r \) at each \( \theta \): - \( \theta = \frac{\pi}{2} \), \( r = 1 + 1 = 2 \); Point: \( (2, \frac{\pi}{2}) \). - \( \theta = \frac{3\pi}{2} \), \( r = 1 - 1 = 0 \); Point: \( (0, \frac{3\pi}{2}) \). - \( \theta = \frac{\pi}{6} \), \( r = 1 + \frac{1}{2} = \frac{3}{2} \); Point: \( (\frac{3}{2}, \frac{\pi}{6}) \). - \( \theta = \frac{5\pi}{6} \), \( r = 1 + \frac{1}{2} = \frac{3}{2} \); Point: \( (\frac{3}{2}, \frac{5\pi}{6}) \).
07
Conclusion: Summary of Tangent Points
Horizontal tangents occur at \( (2, \frac{\pi}{2}) \) and \( (0, \frac{3\pi}{2}) \); vertical tangents occur at \( (\frac{3}{2}, \frac{\pi}{6}) \), \( (0, \frac{3\pi}{2}) \), and \( (\frac{3}{2}, \frac{5\pi}{6}) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cardioid
A cardioid is a special type of polar curve that is heart-shaped. It is generally given by the equation \( r = 1 + ext{{some function of }} \theta \). In our exercise, this is \( r = 1 + \sin \theta \). This configuration creates a unique curve that often appears in mathematical and engineering problems.
Some characteristics of a cardioid include:
Some characteristics of a cardioid include:
- Symmetry around the vertical axis for this equation form.
- The curve touches the origin, which in polar coordinates is marked as \( r = 0 \).
- Useful in modeling situations where heart-shaped curves are needed.
Horizontal Tangent
In polar coordinates, a horizontal tangent line represents a point on the curve where the slope is zero. This means that the curve is flat at that point.
To find horizontal tangents, we need to calculate where \( \frac{dy}{d\theta} = 0 \). For the cardioid \( r = 1 + \sin \theta \), we use the sine and cosine trigonometric derivatives to determine the points of horizontality.
To find horizontal tangents, we need to calculate where \( \frac{dy}{d\theta} = 0 \). For the cardioid \( r = 1 + \sin \theta \), we use the sine and cosine trigonometric derivatives to determine the points of horizontality.
- First, express \( x \) and \( y \) using polar to Cartesian conversions: \( x = (1 + \sin \theta) \cos \theta \), \( y = (1 + \sin \theta) \sin \theta \).
- Then, derive \( y \) with respect to \( \theta \) to get \( \frac{dy}{d\theta} = \cos \theta + \cos \theta \sin \theta \).
- Factorizing gives us \( \cos \theta (1 + \sin \theta) = 0 \).
- Solve for \( \theta \) where \( \cos \theta = 0 \) and \( 1 + \sin \theta = 0 \), leading to horizontal points at specific angles.
Vertical Tangent
Vertical tangents occur where the derivative of \( x \), \( \frac{dx}{d\theta} \), is zero in polar coordinates. Such points indicate a vertical direction at that part of the curve. To find vertical tangents for the cardioid \( r = 1 + \sin \theta \):
- Find the derivative \( \frac{dx}{d\theta} = \cos^2 \theta - \sin \theta - \sin^2 \theta \).
- Set this equal to zero and use the trigonometric identity \( \cos^2 \theta = 1 - \sin^2 \theta \).
- Solve for \( \sin \theta \) using the equation \( 1 - 2\sin^2 \theta - \sin \theta = 0 \).
- This quadratic solution gives us angles where the vertical tangents occur: \( \sin \theta = \frac{1}{2} \) or \( \sin \theta = -1 \).
- The respective angles leading to vertical tangents are \( \theta = \frac{\pi}{6}, \frac{5\pi}{6}, \text{and } \frac{3\pi}{2} \).
Derivatives in Polar Coordinates
Derivatives in polar coordinates provide crucial information about the behavior of curves regarding tangency and other geometric features. Calculating derivatives for both \( x \) and \( y \) with respect to \( \theta \) helps find horizontal and vertical tangents.
Key steps in finding these derivatives include:
Key steps in finding these derivatives include:
- Convert polar equations using \( x = r \cos \theta \) and \( y = r \sin \theta \).
- Apply product and chain rules while differentiating, as each component of \( r \) may depend on \( \theta \).
- Analyze each derivative for critical points by setting them to zero and solving for \( \theta \).
- This process helps locate points on the curve where the tangents become horizontal or vertical.
