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.\(r=4 \cos (2 \theta) ;\) tips of the leaves

Short answer

The slope of the tangent line is evaluated by \( \frac{dy/d\theta}{dx/d\theta} \).

Step-by-step solution

Understand the polar curve

The given polar curve equation is \( r = 4 \cos(2\theta) \). This type of equation typically represents a rose curve with leaves or petals.

Convert to parametric form

Convert the polar equation \( r = f(\theta) = 4 \cos(2\theta) \) into parametric form where \( x = r \cos \theta = 4 \cos(2\theta) \cos \theta \) and \( y = r \sin \theta = 4 \cos(2\theta) \sin \theta \).

Derive the parametric equations

Now, derive the parametric equations: \( \frac{dx}{d\theta} = \frac{d}{d\theta} [4 \cos(2\theta) \cos \theta] \) and \( \frac{dy}{d\theta} = \frac{d}{d\theta} [4 \cos(2\theta) \sin \theta] \). Use product rule for differentiation and simplification.

Apply the product rule for \(dx/d\theta\)

Calculate \( \frac{dx}{d\theta} = \frac{d}{d\theta} [4 \cos(2\theta) \cos \theta] = 4 \left[ -2\sin(2\theta)\cos\theta + \cos(2\theta)(-\sin\theta) \right] \). Simplify further as needed.

Apply the product rule for \(dy/d\theta\)

Calculate \( \frac{dy}{d\theta} = \frac{d}{d\theta}[4 \cos(2\theta) \sin \theta] = 4 \left[ -2\sin(2\theta)\sin\theta + \cos(2\theta)\cos\theta \right] \). Simplify further as needed.

Find the slope of the tangent line

The slope of the tangent line to the curve is given by \( \frac{dy/d\theta}{dx/d\theta} \). Substitute the derivatives obtained in previous steps and simplify to find the value of the slope.

Key concepts

Parametric Form

In mathematics, converting equations from one form to another can make complex problems easier to solve. This is particularly true when working with polar curves, which are initially expressed in terms of the radius, \( r \), and the angle, \( \theta \). The beauty of parametric forms lies in their ability to represent complex curves through simple parameter changes.
For a polar curve given by \( r = f(\theta) \), we can write it in parametric form using:

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

This transformation involves two separate equations that describe the \( x \) and \( y \) coordinates generated by \( f(\theta) \). For our exercise, given \( r = 4 \cos(2\theta) \), the parametric forms become \( x = 4 \cos(2\theta) \cos \theta \) and \( y = 4 \cos(2\theta) \sin \theta \). This allows us to work within the Cartesian plane, making it easier to use calculus tools like derivatives to find properties like slopes.

Derivative

Derivatives are the cornerstone of calculus, representing the rate of change or the slope of a function at any point. To find the slope of the tangent line to the polar curve in parametric form, we need to determine the derivatives of \( x \) and \( y \) with respect to \( \theta \).
Given the complex nature of these parametric equations, we utilize the product rule of differentiation. The product rule states that for two functions, \( u(\theta) \) and \( v(\theta) \), their derivative can be found by:

• \( \frac{d}{d\theta} [u(\theta) v(\theta)] = u'(\theta) v(\theta) + u(\theta) v'(\theta) \)

For our exercise:

• The derivative for \( x = 4 \cos(2\theta) \cos \theta \) becomes \( \frac{dx}{d\theta} = 4 [-2\sin(2\theta)\cos\theta + \cos(2\theta)(-\sin\theta)] \).
• Similarly, for \( y = 4 \cos(2\theta) \sin \theta \), the derivative is \( \frac{dy}{d\theta} = 4 [-2\sin(2\theta)\sin\theta + \cos(2\theta)\cos\theta] \).

These derivatives define the rate of change of each coordinate as the parameter \( \theta \) changes, laying the groundwork for calculating the slope of the tangent line.

Slope of Tangent Line

The slope of a tangent line is crucial in understanding how a curve behaves at a specific point. For curves expressed in parametric form, the slope of the tangent line is determined by the ratio of the derivatives of the \( y \) and \( x \) components with respect to the parameter. Simply put, it is given by:
\[ \text{Slope} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \]This formula provides insight into how steep the tangent is at a particular point on the curve. For the specific problem where:

• \( \frac{dx}{d\theta} = 4 [-2\sin(2\theta)\cos\theta + \cos(2\theta)(-\sin\theta)] \)
• \( \frac{dy}{d\theta} = 4 [-2\sin(2\theta)\sin\theta + \cos(2\theta)\cos\theta] \)

Plugging these into our slope formula will yield the specific slope of the tangent line for each value of \( \theta \).
The slopes can help us visualize and understand interesting behavior of the polar curve, such as the orientation of leaves or petals in rose curve formations.