Question

For the following exercises, find the slope of a tangent line to a polar curve \(r=f(\theta) .\) Let \(x=r \cos \theta=f(\theta) \cos \theta\) and \(y=r \sin \theta=f(\theta) \sin \theta\), so the polar equation \(r=f(\theta)\) is now written in parametric form.\(r=2 \sin (3 \theta) ;\) tips of the leaves

Short answer

The slope of the tangent line at the tip of a leaf is \(\frac{1}{3}\).

Step-by-step solution

Derive Parametric Equations

Given the polar curve is \(r = 2 \sin(3\theta)\), we can write the parametric equations as \(x = f(\theta) \cos \theta = 2 \sin(3\theta) \cos \theta\) and \(y = f(\theta) \sin \theta = 2 \sin(3\theta) \sin \theta\). These represent the transformation from polar coordinates \((r, \theta)\) to cartesian coordinates \((x, y)\).

Differentiate Parametric Equations

To find the slope of the tangent line, we need \(\frac{dy}{dx}\). First, differentiate \(x\) and \(y\) with respect to \(\theta:\) \(\frac{dx}{d\theta} = 2\left(\cos(3\theta)\cos(\theta) - 3\sin(3\theta)\sin(\theta)\right)\) and \(\frac{dy}{d\theta} = 2\left(\cos(3\theta)\sin(\theta) + 3\sin(3\theta)\cos(\theta)\right)\).

Find \(\frac{dy}{dx}\)

The slope of the tangent line \(\frac{dy}{dx}\) is given by the ratio of the derivatives: \(\frac{dy}{d\theta} / \frac{dx}{d\theta}\). Substitute the expressions found in Step 2: \(\frac{dy}{dx} = \frac{ \cos(3\theta)\sin(\theta) + 3\sin(3\theta)\cos(\theta)}{ \cos(3\theta)\cos(\theta) - 3\sin(3\theta)\sin(\theta)}\).

Evaluate at the Tips of the Leaves

The tips of the leaves occur where the polar curve is maximum, i.e., where \(r = 2\). From \(r = 2\sin(3\theta)\), set \(2\sin(3\theta) = 2\Rightarrow \sin(3\theta) = 1\). This gives \(3\theta = \frac{\pi}{2}, \frac{5\pi}{2}, \ldots\) or \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \ldots\). Substitute these \(\theta\) values into \(\frac{dy}{dx}\) to find the slope at each leaf's tip.

Key concepts

Polar Coordinates

Polar coordinates are a way of representing points in a plane using a distance and an angle. Unlike the traditional Cartesian coordinates that use \(x\) and \(y\) to define a location, polar coordinates are given as \( (r, \theta) \). Here:

• \( r \) is the distance from the point to the origin.
• \( \theta \) is the angle measured from the positive x-axis, usually in radians.

Polar coordinates are especially useful for expressing curves that have rotational symmetry or are circular in nature.
The curve described in the exercise, \( r = 2 \sin(3\theta) \,\) is a polar equation representing a three-leaved rose.
As you vary \( \theta \), the value of \( r \) changes, tracing out the shape on a polar grid. This visualization can help you better understand the shape of the curve and the context of the exercise.

Parametric Equations

Parametric equations are used to express the coordinates of points on a curve as functions of one or more parameters. When dealing with polar curves, we often convert them into parametric form to relate to Cartesian coordinates. The equations:

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

are examples of this conversion, transforming the polar form into something we can work with using calculus. Parametric equations give us separate expressions for \( x \) and \( y \) based on \( \theta \), allowing us to analyze each independently.
This separation simplifies the calculation of properties like distances, areas, and slopes, as it decouples otherwise interlinked aspects of the curve.

Differentiation

Differentiation is the process of calculating a derivative, which shows how a function changes as its input changes. In the context of parametric equations:

• We differentiate the equations \( x(\theta) \) and \( y(\theta) \) with respect to \( \theta \).
• This gives us the rates of change of \( x \) and \( y \), noted as \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \).

Using these derivatives, we can investigate the behavior of the curve.
In the exercise, the differentiation provides the basis for the ensuing slope calculation. Communicating how \( x \) and \( y \) change as \( \theta \) varies lays the groundwork to find instantaneous rates such as the slope of a tangent line.
This is key for further analysis and interpretations of motion or value changes along the curve.

Slope of Tangent Line

The slope of a tangent line to a curve at any point provides a snapshot of the curve's direction at that point. By finding \( \frac{dy}{dx} \,\) we compute the steepness of this line for parametric equations:

• \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \)

This ratio of derivatives expresses how much \( y \) changes relative to the change in \( x \) as \( \theta \) changes.
By evaluating \( \frac{dy}{dx} \) at specific \( \theta \) values found in the problem, you determine the slope at the tips of the curve's leaves. At these points, the curve reaches its maximum radial distance from the origin.
Understanding the slope here is essential for predicting how the curve behaves around those intervening regions and for characterizing its overall geometry in a comprehensive manner.