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.\(r=6+3 \cos \theta ;(3, \pi)\)

Short answer

The slope of the tangent line at \(\theta = \pi\) is undefined (vertical tangent).

Step-by-step solution

Convert Polar to Parametric Equations

We are given the polar equation \(r = 6 + 3\cos\theta\) and need to find the parametric form. Using the formulas \(x = r \cos \theta\) and \(y = r \sin \theta\), we substitute \(r = 6 + 3\cos\theta\):\[\begin{align*}x &= (6 + 3\cos\theta)\cos\theta, \y &= (6 + 3\cos\theta)\sin\theta.\end{align*}\]Thus, the parametric form is obtained as \(x(\theta) = (6 + 3\cos\theta)\cos\theta\) and \(y(\theta) = (6 + 3\cos\theta)\sin\theta\).

Find Derivatives x'(θ) and y'(θ)

To find the slope of the tangent line at a given \(\theta\), we first find the derivatives of \(x(\theta)\) and \(y(\theta)\). Differentiate \(x(\theta) = (6 + 3\cos\theta)\cos\theta\) using the product and chain rules to get:\[x'(\theta) = -3\sin^2\theta - (6 + 3\cos\theta)\sin\theta.\]Similarly, differentiate \(y(\theta) = (6 + 3\cos\theta)\sin\theta\) to find:\[y'(\theta) = 3\cos\theta\sin\theta + (6 + 3\cos\theta)\cos\theta.\]

Calculate the Slope dy/dx

The slope of the tangent line is given by the derivative \(\frac{dy}{dx} = \frac{y'(\theta)}{x'(\theta)}\). Substitute \(x'(\theta)\) and \(y'(\theta)\) from the previous step to find:\[\frac{dy}{dx} = \frac{3\cos\theta\sin\theta + (6 + 3\cos\theta)\cos\theta}{-3\sin^2\theta - (6 + 3\cos\theta)\sin\theta}.\]

Evaluate at θ = π

We need to evaluate the slope at the point where \(\theta = \pi\). Substitute \(\theta = \pi\) into \(\frac{dy}{dx}\):- \(\cos\pi = -1\) and \(\sin\pi = 0\), so:\[\begin{align*}y'(\pi) &= 3(-1)(0) + (6 + 3(-1))(-1) = -3, \x'(\pi) &= -3(0)^2 - (6 + 3(-1))(0) = 0.\end{align*}\]The expression for slope is \(-3/0\), indicating a vertical tangent.

Key concepts

Tangent Line Slope in Polar Curves

Finding the tangent line slope is important when working with polar curves. In polar coordinates, we often start with an equation in the form \(r = f(\theta)\). To determine the slope of the tangent line, we must convert this polar equation into a parametric form. This process involves interpreting the curve's behavior, making it essential for analysis and graphing.

To find the slope of the tangent line at a specific point \((r, \theta)\), we use derivatives. For polar coordinates, the slope \(\frac{dy}{dx}\) is determined using the derivatives of parameter expressions \(x(\theta)\) and \(y(\theta)\). The tangent line's slope at a given point tells us the direction the curve is heading at that specific location.

• The tangent slope can be expressed as \(\frac{dy}{dx} = \frac{y'(\theta)}{x'(\theta)}\).
• Evaluate \(\theta\) at points of interest to find specific slopes.
• Understanding this slope helps identify features like vertical and horizontal tangents.

Understanding Parametric Equations in Polar Conversion

Parametric equations are a key tool in transforming polar equations into a form more manageable for calculus operations. Understanding these allows us to see how the polar curve relates in a Cartesian plane.

When converting a polar equation \(r = f(\theta)\) into parametric equations, we use transformations:- \(x = r \cos \theta = (f(\theta)) \cos \theta\)- \(y = r \sin \theta = (f(\theta)) \sin \theta\)

This conversion is vital as it provides a connection between polar and Cartesian coordinates, making it easier to visualize and compute derivatives.

• This conversion helps express the curve in terms of a parameter \(\theta\).
• Once in parametric form, you can find the corresponding \(x(\theta)\) and \(y(\theta)\) expressions easily.
• This aids in understanding how changes in \(\theta\) affect the curve's position and behavior.

Calculating Derivatives in Polar Coordinates

Taking derivatives in polar coordinates involves handling the derivatives of the parametric forms of the equations. This step is crucial for determining the slope of the tangent line.

After converting a polar equation to its parametric form, differentiate \(x(\theta)\) and \(y(\theta)\) with respect to \(\theta\). This provides the rates of change for the \(x\) and \(y\) components along the curve.

• Use the product rule and chain rule to differentiate \(x(\theta)\) and \(y(\theta)\).
• Once obtained, these derivatives \(x'(\theta)\) and \(y'(\theta)\) can be used to find \(\frac{dy}{dx}\).
• Evaluating these derivatives at a specific \(\theta\) gives insights into the curve's properties at that point, such as indicating the presence of horizontal or vertical tangents.

These derivatives are fundamental in the study and application of calculus to polar curves.