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Magazine Chapter 10: Problem 64
\(63-68\) Find the points on the given curve where the tangent line is horizontal or vertical. $$r=1-\sin \theta$$
Short Answer
Expert verified
The points are \((0,0)\) for both horizontal and vertical tangents, and \((0,-2)\) for a horizontal tangent.
Step by step solution
01
Understand Horizontal and Vertical Tangents
For a polar curve, a tangent is horizontal where \(\frac{dy}{d\theta} = 0\) and vertical where \(\frac{dx}{d\theta} = 0\). For the curve given by \(r = 1 - \sin\theta\), we need to convert it to Cartesian coordinates and then find where these conditions hold.
02
Convert Polar to Cartesian Coordinates
The relationship between polar and Cartesian coordinates is \(x = r \cos \theta\) and \(y = r \sin \theta\). Substituting \(r = 1 - \sin\theta\), we get: \(x = (1 - \sin\theta) \cos \theta\) and \(y = (1 - \sin\theta) \sin \theta\).
03
Compute Derivatives
Compute \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) using the product and chain rules. This gives us:\[\frac{dx}{d\theta} = -(1-\sin\theta) \sin\theta + \cos\theta \cos\theta\] \[\frac{dy}{d\theta} = (1-\sin\theta) \cos\theta + \sin\theta \sin\theta\].
04
Set \(\frac{dy}{d\theta} = 0\) for Horizontal Tangents
Solving \((1-\sin\theta) \cos\theta + \sin\theta \sin\theta = 0\), simplify to \(\cos\theta = 0\) or \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}\). Calculate \(r\) for these \(\theta\) values to find corresponding points.
05
Set \(\frac{dx}{d\theta} = 0\) for Vertical Tangents
Solving \(-(1-\sin\theta) \sin\theta + \cos^2\theta = 0\), simplify to \(\sin\theta = 1\), then \(\theta = \frac{\pi}{2}\). Calculate \(r\) to find the corresponding point.
06
Calculate Points from \(r = 1 - \sin \theta\)
For \(\theta = \frac{\pi}{2}\), \(r = 1 - 1 = 0\), thus the point is \((0,0)\). For \(\theta = \frac{3\pi}{2}\), \(r = 1 - (-1) = 2\), thus the point is \((0,-2)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Horizontal Tangents
In polar coordinates, a tangent line to a curve is horizontal when the rate of change of the y-coordinate with respect to the angle \( \theta \) is zero. Mathematically, this means we're looking for where \( \frac{dy}{d\theta} = 0 \).
- Identifying Horizontal Tangents: For the curve \( r = 1 - \sin \theta \), the y-coordinate in Cartesian form is \( y = (1 - \sin \theta) \sin \theta \). Using the chain and product rules of derivatives, we find \( \frac{dy}{d\theta} \).
- Solving: Set \( (1-\sin\theta) \cos\theta + \sin^2\theta = 0 \). Simplifying, we find \( \cos\theta = 0 \), leading to the values \( \theta = \frac{\pi}{2} \text{ and } \theta = \frac{3\pi}{2} \).
- Corresponding Points: Calculate \( r \) using these \( \theta \) values:
- For \( \theta = \frac{\pi}{2} \), \( r = 1 - 1 = 0 \), resulting in the point \((0,0)\).
- For \( \theta = \frac{3\pi}{2} \), \( r = 1 - (-1) = 2 \), resulting in the point \((0,-2)\).
Vertical Tangents
For a tangent line to be vertical, the sensitivity of the x-coordinate to changes in \( \theta \) should be zero. This corresponds to \( \frac{dx}{d\theta} = 0 \) for a vertical tangent.
- Identifying Vertical Tangents: The x-coordinate equation is \( x = (1 - \sin\theta) \cos\theta \). Using the derivatives, we find \( \frac{dx}{d\theta} \).
- Solving: We solve \( -(1-\sin\theta) \sin\theta + \cos^2\theta = 0 \). Simplifying, we set \( \sin\theta = 1 \), leading to \( \theta = \frac{\pi}{2} \).
- Corresponding Point: For \( \theta = \frac{\pi}{2} \), \( r = 1 - 1 = 0 \), giving the point \((0,0)\).
Cartesian Coordinates
Converting from polar coordinates to Cartesian coordinates is crucial for understanding and solving problems related to tangents. Polar coordinates \( (r, \theta) \) express a point in terms of the distance from the origin and the angle from the positive x-axis.
- Conversion Formulas: The relationships are given by:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
- Application: For the specific curve \( r = 1 - \sin\theta \):
- We substitute the value for \( r \) into the equations above to convert polar to Cartesian:
- \( x = (1 - \sin \theta) \cos \theta \)
- \( y = (1 - \sin \theta) \sin \theta \)
- We substitute the value for \( r \) into the equations above to convert polar to Cartesian:
Derivatives
Derivatives are tools that help us determine how a function changes at any given point, which in turn helps us understand the geometry of curves.
- Role of Derivatives in Tangents:
- For tangents, we determine the slopes along the x and y directions using derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \).
- Horizontal tangents occur where \( \frac{dy}{d\theta} = 0 \).
- Vertical tangents occur where \( \frac{dx}{d\theta} = 0 \).
- Calculating Derivatives:
- For \( x = (1 - \sin \theta) \cos \theta \), apply the product and chain rules to find \( \frac{dx}{d\theta} \).
- For \( y = (1 - \sin \theta) \sin \theta \), apply the same differentiation techniques to find \( \frac{dy}{d\theta} \).
