Problem 19 - Thomas Calculus

Question

Find the slopes of the curves at the given points. Sketch the curves along with their tangents at these points. Four-leaved rose \(r=\sin 2 \theta ; \quad \theta=\pm \pi / 4, \pm 3 \pi / 4\)

Step-by-Step Solution

Answer

The slopes of the tangents at each given point are zero.

Step 1: Understanding the Polar Equation

The problem gives us a polar equation of the form \( r = \sin 2\theta \). This type of equation represents a four-leaved rose, a kind of polar graph that has a flower-like appearance with four symmetrical lobes.

Step 2: Converting Polar to Cartesian Coordinates

To find the slopes of the tangents to the curve at given polar angles, we first convert the polar equation into Cartesian coordinates. Using the formulas \( x = r\cos\theta \) and \( y = r\sin\theta \), we can express \( x \) and \( y \) in terms of \( \theta \).Substitute \( r = \sin 2\theta \) into these:\[ x = \sin 2\theta \cos \theta \]\[ y = \sin 2\theta \sin \theta \]

Step 3: Differentiating to Find the Slope

The slope of a tangent line in the polar coordinate system can be found using derivatives of parametric equations:\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \]First, find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \):- \( \frac{dx}{d\theta} = \cos(3\theta) - \cos(\theta) \)- \( \frac{dy}{d\theta} = \sin(3\theta) - \sin(\theta) \)Now, the slope is:\[ \frac{dy}{dx} = \frac{\sin(3\theta) - \sin(\theta)}{\cos(3\theta) - \cos(\theta)} \]

Step 4: Evaluate Slopes at Given Points

Evaluate the derivative \( \frac{dy}{dx} \) at each given \( \theta \) value.1. For \( \theta = \frac{\pi}{4} \): - \( \frac{dy}{dx} = \frac{\sin(\frac{3\pi}{4}) - \sin(\frac{\pi}{4})}{\cos(\frac{3\pi}{4}) - \cos(\frac{\pi}{4})} = \frac{\sqrt{2}/2 - \sqrt{2}/2}{-\sqrt{2}/2 - \sqrt{2}/2} = 0 \)2. For \( \theta = -\frac{\pi}{4} \): - Similar evaluation gives a slope of 0.3. For \( \theta = \frac{3\pi}{4} \): - Calculation results in a slope of 0.4. For \( \theta = -\frac{3\pi}{4} \): - Calculation results in a slope of 0.

Step 5: Sketching the Curve and Tangents

To visualize, draw the four-leaved rose: it consists of four symmetrical lobes centered around the origin. For each given \( \theta \), plot the points as per their \( (x, y) \) coordinates converted from polar form. Draw horizontal tangents at these points since the slope was calculated to be 0 for each.

Key Concepts

Four-leaved Rose

The four-leaved rose is a captivating pattern in polar coordinates, described by the equation \( r = \sin 2\theta \). This equation generates a flower-like graph consisting of four symmetric lobes or petals.
Each petal emerges from the center, creating a symmetrical and aesthetically appealing design. The petals are equally spaced, making them an excellent example of balance in polar graphs.
The way this pattern is formed is due to how the sine function operates within the polar system. When plotting it, you can observe how the radius \( r \) changes as the angle \( \theta \) changes, specifically influencing the petal structure.

Slope of Tangents

In polar coordinates, finding the slope of a tangent to the curve involves understanding how the curve stretches or contracts at different points.
Tangents can be found by differentiating the parametric Cartesian equations that are derived from the original polar equation.
The slope \( \frac{dy}{dx} \) at any point is determined using derivatives of parametric equations with respect to \( \theta \).

• Compute \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \)
• Use the formula: \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \).

This slope tells us how steep the curve is at particular points, critical for understanding the behavior of the curve and drawing accurate tangents.

Parametric Equations

Parametric equations are essential to convert polar equations into a form that's easier to work with in Cartesian coordinates.
They express both \( x \) and \( y \) as functions of a third variable, typically \( \theta \) in polar problems.

• The general parametric conversion from polar uses \( x = r\cos\theta \) and \( y = r\sin\theta \).
• For our four-leaved rose, substitute \( r = \sin 2\theta \) to get: \( x = \sin 2\theta \cos\theta \) and \( y = \sin 2\theta \sin\theta \).

These rewritten equations allow us to use calculus tools, such as differentiation, to analyze changes in \( x \) and \( y \), and find slopes for more complex curves.

Converting Polar to Cartesian

Converting from polar to Cartesian coordinates helps represent the curve in a more familiar axis system.
This method is useful as it transforms polar equations into two separate expressions for \( x \) and \( y \).
For conversion, use the relationships:

• \( x = r\cos\theta \)
• \( y = r\sin\theta \)

In our example, by substituting \( r = \sin 2\theta \), we derived \( x = \sin 2\theta \cos\theta \) and \( y = \sin 2\theta \sin\theta \).
These Cartesian coordinates simplify calculus applications like finding slopes and evaluating curve behavior at specific points.