Problem 19 - 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 \(r=\sin 2 \theta ; \quad \theta=\pm \pi / 4, \pm 3 \pi / 4\)

Step-by-Step Solution

Answer

The slopes are 0 at \(\theta = \pm \frac{\pi}{4}\) and undefined at \(\theta = \pm \frac{3\pi}{4}\).

Step 1: Understand the Polar Curve

The given polar equation is a four-leaved rose described by \( r = \sin 2\theta \). This curve has petal-like structures, and our goal is to find the slope of the tangent to this curve at the specified angles \( \theta = \pm \frac{\pi}{4}, \pm \frac{3\pi}{4} \).

Step 2: Convert to Cartesian Coordinates

The polar coordinates \((r, \theta)\) can be converted to Cartesian coordinates \((x,y)\) using the formulas: \( x = r \cos \theta \) and \( y = r \sin \theta \). Substituting \( r = \sin 2\theta \), we have: \( x = \sin 2\theta \cos \theta \) and \( y = \sin 2\theta \sin \theta \).

Step 3: Derive the Parametric Equations

Substitute the double angle identities: \( \sin 2\theta = 2\sin\theta \cos\theta \) into the equations: \( x = 2\sin\theta\cos^2\theta \) and \( y = 2\sin^2\theta \cos\theta \), to achieve parametric form for \(x\) and \(y\) with respect to \(\theta\).

Step 4: Find the Derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\)

Compute the derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) of the parametric equations. Using the product rule for derivatives, derive \( \frac{dx}{d\theta} = 4\cos^3\theta - 2\sin^2\theta \cos\theta \) and \( \frac{dy}{d\theta} = 4\sin\theta \cos^2\theta - 2\sin^3\theta \).

Step 5: Calculate the Slope of the Tangent Line

The slope of the tangent line is given by \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \). Substitute each \(\theta\) value and solve for the slope at \(\theta = \pm \frac{\pi}{4}, \pm \frac{3\pi}{4}\).

Step 6: Evaluate Slopes at Given Angles

For \(\theta = \frac{\pi}{4}\) and \(\theta = -\frac{\pi}{4}\), compute \(\frac{dy}{dx} = \frac{0}{2} = 0\) indicating horizontal tangents. At \(\theta = \frac{3\pi}{4}\) and \(\theta = -\frac{3\pi}{4}\), compute \(\frac{dy}{dx} = \text{Undefined} \), indicating vertical tangents.

Step 7: Sketch the Curve and Tangents

Sketch the polar rose curve with the calculated points. Use the calculated slopes to draw tangent lines: horizontal tangents at \(\theta = \pm \frac{\pi}{4}\) and vertical tangents at \(\theta = \pm \frac{3\pi}{4}\).

Key Concepts

The Tangent Line

A tangent line is a straight line that touches a curve at just one point and has the same slope as the curve at that point. In our exercise with the four-leaved rose curve described by the polar equation \( r = \sin 2\theta \), the tangent lines help us understand how the curve behaves locally around specific angles.

At the points \( \theta = \pm \frac{\pi}{4} \), the slope of the tangent line was computed to be 0. This means the tangent lines at these points are horizontal, indicating a flat slope.

On the other hand, at the angles \( \theta = \pm \frac{3\pi}{4} \), the slope was found to be undefined. This implies that the tangent lines are vertical, extending straight up and down.

Understanding tangent lines in the context of curves allows for insights into the curve's behavior with respect to changes in direction or rapidity of slope at different points.

Understanding Derivatives

Derivatives measure how a function changes as its input changes, providing vital information about the slope or steepness at any given point on a curve. In our context, finding the derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) involved calculating how both \(x\) and \(y\), as functions of \(\theta\), change. This involves using rules like the product rule when dealing with products of functions.

The derivative \( \frac{dy}{dx} \), which represents the slope of the tangent line, is derived by dividing \( \frac{dy}{d\theta} \) by \( \frac{dx}{d\theta} \). Calculating this at different angles \( \theta \) gives us the slope of the tangent line at those specific points on the curve.

Derivatives are key to turning the complex nature of curves into understandable, quantifiable slopes. Whether the tangent is horizontal or vertical, it’s derivatives that help pinpoint the exact nature at any given point.

Parametric Equations of a Curve

In many cases, especially in calculus, it's beneficial to represent a curve parametrically. This means expressing the coordinates \( x \) and \( y \) in terms of a third variable, commonly \( \theta \) for polar equations.

For our four-leaved rose, we had to convert the polar equation into parametric form using trigonometric identities. This was done using \( x = 2\sin\theta\cos^2\theta \) and \( y = 2\sin^2\theta\cos\theta \), making the behavior of the curve easier to handle with derivatives.

Parametric equations allow us to tackle distinct aspects of the curve by focusing on how \( x \) and \( y \) components evolve separately as \( \theta \) changes. This way, computations for slopes and understanding the geometry of the curve become simpler and more direct, offering a precise grasp of the curve’s characteristics at different angles or points.