Problem 20 - Thomas Calculus

Question

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 \(r=\cos 2 \theta ; \quad \theta=0, \pm \pi / 2, \pi\)

Step-by-Step Solution

Answer

The slopes at \(\theta = 0, \pm \pi / 2, \pi\) are all zero (horizontal tangents).

Step 1: Convert Polar to Cartesian Coordinates

The equation given is in polar form, \(r = \cos 2\theta\). To find the slope of the tangent to the curve, we first convert the polar equation to Cartesian coordinates \((x, y)\). Using the formulas \(x = r\cos\theta\) and \(y = r\sin\theta\), we have:\[ x = \cos 2\theta \cdot \cos \theta \] and \[ y = \cos 2\theta \cdot \sin \theta \].

Step 2: Find Derivatives with Respect to \(\theta\)

To find the slope of the tangent, we need \(\frac{dy}{dx}\), which requires \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\). Differentiate \(x\) and \(y\) with respect to \(\theta\):\[ \frac{dx}{d\theta} = \frac{d}{d\theta}(\cos 2\theta \cdot \cos \theta) = -2\sin 2\theta \cdot \cos \theta - \cos 2\theta \cdot \sin \theta \]\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(\cos 2\theta \cdot \sin \theta) = -2\sin 2\theta \cdot \sin \theta + \cos 2\theta \cdot \cos \theta \].

Step 3: Calculate \(\frac{dy}{dx}\) at Specific \(\theta\) Values

The slope \(\frac{dy}{dx}\) is given by \(\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\). Substituting \(\theta = 0, \pm \frac{\pi}{2}, \pi\):- At \(\theta = 0\): \(\frac{dy}{dx} = \frac{\cos 0}{-\sin 0} = 0\). - At \(\theta = \frac{\pi}{2}\): \(\frac{dy}{dx} = \frac{-\cos \pi}{2\sin \pi} = 0\).- At \(\theta = -\frac{\pi}{2}\): \(\frac{dy}{dx} = \frac{-\cos \pi}{2\sin \pi} = 0\).- At \(\theta = \pi\): \(\frac{dy}{dx} = \frac{\cos 2\pi}{-\sin 2\pi} = 0\).

Step 4: Sketch the Curve and Tangents

The four-leaved rose \(r = \cos 2\theta\) forms a symmetrical figure with leaves touching the axes at \(\theta = 0, \pm \frac{\pi}{2}, \pi\). The tangents at these points are horizontal, as the slope calculations show \(\frac{dy}{dx} = 0\). Draw a neat sketch showing the curve and the horizontal tangent lines at these points.

Key Concepts

Polar to Cartesian Conversion

When dealing with curves defined by polar equations, such as a four-leaved rose, it's often helpful to convert these equations into Cartesian form. This method uses transformations that relate the polar coordinates \((r, \theta)\) to Cartesian coordinates \((x, y)\). The conversion formulas are:
\[ x = r\cos\theta \] and \[ y = r\sin\theta \].
In our specific case, for the four-leaved rose given by \( r = \cos 2\theta \), we substitute \( r \) in the conversion formulas. Thus, the Cartesian coordinates become \[ x = \cos 2\theta \cdot \cos\theta \] and \[ y = \cos 2\theta \cdot \sin\theta \].
These expressions allow us to compute derivatives, making it possible to find the slope of tangents, which is a vital step in many calculus computations.

Differentiation

Differentiation is a core concept in calculus that allows us to find how a function changes. For our curve, the four-leaved rose, we need to differentiate in terms of \( \theta \) to find the slope of the tangent.
To achieve this, we need \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \), which involve differentiating \( x \) and \( y \) as functions of \( \theta \).
The derivatives are computed as follows:

• \( \frac{dx}{d\theta} = -2\sin 2\theta \cdot \cos \theta - \cos 2\theta \cdot \sin \theta \)
• \( \frac{dy}{d\theta} = -2\sin 2\theta \cdot \sin \theta + \cos 2\theta \cdot \cos \theta \)

These derivatives are crucial for finding the tangent slope \( \frac{dy}{dx} \) at specific points on the curve by using \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \).

Four-leaved Rose

The four-leaved rose, described by the polar equation \( r = \cos 2\theta \), is a fascinating mathematical figure with symmetrical petals. It is an example of a type of curve known as a polar plot, displaying a balanced spread of its four 'leaves' from the origin.
This curve intersects the axes at \( \theta = 0, \pm \frac{\pi}{2}, \text{and } \pi \), pinpointing where its 'leaves' meet the axes.

• \( \theta = 0 \) and \( \theta = \pi \) correspond to points on the positive x-axis and negative x-axis, respectively.
• \( \theta = \pm \frac{\pi}{2} \) correspond to points on the positive and negative y-axis.

These symmetrical properties make the four-leaved rose both beautiful and illustrative for learning about polar-to-Cartesian conversions and tangent line calculations.

Tangent Slopes

Finding the slopes of tangents to curves in polar form, like the four-leaved rose, involves a mix of geometry and calculus. The slope of the tangent line is simply \( \frac{dy}{dx} \), which describes the steepness or flatness of the tangent at a given point.
For the four-leaved rose curve, compute \( \frac{dy}{dx} \) using the derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \). The specific points given, \( \theta = 0, \pm \frac{\pi}{2}, \text{and } \pi \), indicate where \( \frac{dy}{dx} = 0 \), meaning all tangents at these points are horizontal.
Horizontal tangents denote locations where the rise-over-run aspect of slopes equals zero. This characteristic shows up at the 'leaves' endpoints of the rose, making these tangents quite easy to visualize and sketch.