Chapter 11: Problem 20
Find the slopes of the curves in Exercises \(17 - 20\) at the given points. Sketch the curves along with their tangents at these points. Four-leaved rose \(\quad r = \cos 2 \theta ; \quad \theta = 0 , \pm \pi / 2 , \pi\)
Short Answer
Expert verified
Slopes at \( \theta = 0, \pi \) are 0; vertical at \( \theta = \pm \frac{\pi}{2} \).
Step by step solution
01
Convert Polar Equation to Cartesian Coordinates
The polar equation given is \( r = \cos 2 \theta \). To convert this to Cartesian coordinates, use the formulas \( x = r \cos \theta \) and \( y = r \sin \theta \). Substituting \( r \), we have \( x = \cos 2\theta \cos \theta \) and \( y = \cos 2\theta \sin \theta \).
02
Find Derivatives wrt \( \theta \)
To find the slope of the tangent, we need \( \frac{dy}{dx} \). First, find \( \frac{dx}{d\theta} = -2\sin 2\theta \cos \theta - \cos 2\theta \sin \theta \) and \( \frac{dy}{d\theta} = -2\sin 2\theta \sin \theta + \cos 2\theta \cos \theta \).
03
Calculate \( \frac{dy}{dx} \)
Now, use the chain rule \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \). Substitute the derivatives into the formula, simplifying to find the slope in terms of \( \theta \): \[ \frac{dy}{dx} = \frac{-2\sin 2\theta \sin \theta + \cos 2\theta \cos \theta}{-2\sin 2\theta \cos \theta - \cos 2\theta \sin \theta} \].
04
Evaluate Slopes at Given Points
Evaluate \( \frac{dy}{dx} \) at \( \theta = 0, \pm \frac{\pi}{2}, \pi \). At \( \theta = 0 \), \( r = 1 \) and slope \( \frac{dy}{dx} = 0 \). At \( \theta = \pm \frac{\pi}{2} \), \( r = 0 \) and slope is undefined due to vertical tangents. At \( \theta = \pi \), \( r = -1 \) and slope \( \frac{dy}{dx} = 0 \).
05
Sketch the Curve and Tangents
Draw the four-leaved rose which resembles a four-petaled flower. Plot points at \( \theta = 0, \pm \frac{\pi}{2}, \pi \) with tangents; horizontal at \( \theta = 0, \pi \) and vertical at \( \theta = \pm \frac{\pi}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cartesian Coordinates
Polar coordinates, such as those given by the equation \( r = \cos 2\theta \), describe points in a 2D plane using a radius and an angle. To work with these in a more familiar form, we often convert them into Cartesian coordinates, which use an \( x \) and \( y \) positioning system similar to most standard graphs.
To convert from polar to Cartesian coordinates, the following conversion formulas are utilized:
To convert from polar to Cartesian coordinates, the following conversion formulas are utilized:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
Derivative
The derivative measures how a function changes as its input changes. In other words, it provides the "rate of change." For our purposes with curves, we need to find the rate at which the \( y \) value changes regarding the \( x \) value, or \( \frac{dy}{dx} \). This is fundamental when we want to determine the slope of the tangent line at a given point on the curve.
In this exercise, we calculate the derivatives of \( x \) and \( y \) with respect to \( \theta \), given by:
This derivative \( \frac{dy}{dx} \) will help us calculate the slope of the tangent, essential for sketching the curve correctly with its tangents.
In this exercise, we calculate the derivatives of \( x \) and \( y \) with respect to \( \theta \), given by:
- \( \frac{dx}{d\theta} \)
- \( \frac{dy}{d\theta} \)
This derivative \( \frac{dy}{dx} \) will help us calculate the slope of the tangent, essential for sketching the curve correctly with its tangents.
Tangent Line
A tangent line is a straight line that touches a curve at a single point. This line represents the instantaneous direction of the curve at that point and its slope is calculated using the derivative we found, \( \frac{dy}{dx} \).
In this exercise, we calculate the tangent lines at crucial points such as \( \theta = 0, \pm \frac{\pi}{2}, \pi \). The nature of these tangent lines can describe behavior:
In this exercise, we calculate the tangent lines at crucial points such as \( \theta = 0, \pm \frac{\pi}{2}, \pi \). The nature of these tangent lines can describe behavior:
- Horizontal tangent lines at points where the slope, \( \frac{dy}{dx} = 0 \).
- Vertical tangent lines where the slope is undefined, indicating points with a dramatic direction change.
Slope Calculation
The slope of a line describes its steepness and is an essential concept in calculus. Calculating the slope of a tangent line involves finding the derivative \( \frac{dy}{dx} \).
In this task, we calculate:
In this task, we calculate:
- The slope at \( \theta = 0 \) and \( \theta = \pi \) is \( 0 \), meaning we have horizontal tangents.
- The slope at \( \theta = \pm \frac{\pi}{2} \) is undefined, indicating vertical tangents.