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.For the cardioid \(r=1+\sin \theta\), find the slope of the tangent line when \(\theta=\frac{\pi}{3}\).

Short answer

Slope of the tangent line is \(\frac{1}{2}\).

Step-by-step solution

Convert Polar to Parametric Form

For the cardioid given by \(r = 1 + \sin \theta\), express the parametric equations as: \(x = (1 + \sin \theta) \cos \theta\) and \(y = (1 + \sin \theta) \sin \theta\).

Differentiate Parametric Equations

Find the derivatives with respect to \(\theta\). For \(x\), \(\frac{dx}{d\theta} = \frac{d}{d\theta}[(1+\sin \theta)\cos \theta]\). For \(y\), \(\frac{dy}{d\theta} = \frac{d}{d\theta}[(1+\sin \theta)\sin \theta]\). Use the product and chain rules to differentiate both expressions.

Apply Product Rule to \(\frac{dx}{d\theta}\)

Differentiate \((1+\sin \theta) \cos \theta\) using product rule: \(u(x) = 1 + \sin \theta\) and \(v(x) = \cos \theta\). Then, \(\frac{du}{d\theta}=\cos \theta\) and \(\frac{dv}{d\theta}=-\sin \theta\). Hence, \(\frac{dx}{d\theta} = \cos \theta \cos \theta - (1+\sin \theta)\sin \theta\).

Apply Product Rule to \(\frac{dy}{d\theta}\)

Differentiate \((1+\sin \theta) \sin \theta\) similarly: \(\frac{dy}{d\theta} = \cos \theta \sin \theta + (1+\sin \theta)\cos \theta\).

Substitute \(\theta = \frac{\pi}{3}\) to Find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\)

Calculate \(\frac{dx}{d\theta}\) by substituting \(\theta = \frac{\pi}{3}\) into the differentiated expression for \(\frac{dx}{d\theta}\) found in Step 3. Similarly, find \(\frac{dy}{d\theta}\) using Step 4.

Calculate Slope of Tangent Line

Find the slope of the tangent line by computing \(\frac{dy}{dx}\), which equals \(\frac{dy}{d\theta} \div \frac{dx}{d\theta}\). Insert the values found when \(\theta = \frac{\pi}{3}\). Simplify the expression to get the slope.

Key concepts

Understanding Parametric Equations

Parametric equations provide us a way to represent a curve with a pair of functions, typically connecting the functions of a parameter, often denoted as \( t \) or \( \theta \) in the case of polar coordinates. These equations allow the expression of coordinates in terms of a third parameter. In the context of polar coordinates, this parameter usually is \( \theta \), the angle.
Here the relationship between polar and parametric forms is revealed. Instead of directly mapping \( r \) against \( \theta \), the equations transform into Cartesian coordinates using:

• \( x = (1 + \sin \theta)\cos \theta \)
• \( y = (1 + \sin \theta)\sin \theta \)

These parametric equations define the \( x \) and \( y \) coordinates as a function of \( \theta \). Translating the polar equation \( r = f(\theta) \) into this form enables us to analyze the curve's geometry in a more structured way. This approach is especially beneficial when calculating properties like derivatives.

Derivative of Trigonometric Functions

Finding the derivative of trigonometric functions is paramount in understanding the behavior of curves in both polar and parametric forms. Calculating derivatives in this context requires using the chain rule and product rule which are vital concepts in calculus.
To understand the derivative of the function \( x = (1 + \sin \theta)\cos \theta \):

• Apply the product rule: \((uv)' = u'v + uv'\), where \( u = 1 + \sin \theta \) and \( v = \cos \theta \).
• Differentiate \( u \) to get \( u' = \cos \theta \) and \( v \) to obtain \( v' = -\sin \theta \).
• Combine these to get: \( \frac{dx}{d\theta} = \cos \theta \cos \theta - (1 + \sin \theta)\sin \theta \).

Repeat the process for \( y \),

• again using the product rule on \( y = (1 + \sin \theta)\sin \theta \).
• Following similar steps, we arrive at \( \frac{dy}{d\theta} = \cos \theta \sin \theta + (1 + \sin \theta)\cos \theta \).

Being able to find these derivatives is essential for further calculations, specifically when finding the slope of the tangent line.

Calculating the Slope of a Tangent Line

The slope of a tangent line to a curve at a given point provides insight into the curve's behavior at that specific location. To find this slope in terms of \( \theta \), we need to calculate \( \frac{dy}{dx} \). This involves dividing \( \frac{dy}{d\theta} \) by \( \frac{dx}{d\theta} \).
Here's a step-by-step process:

• First, compute \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) using their formulas derived previously.
• With \( \theta = \frac{\pi}{3} \), substitute the angle into these derivative expressions.
• For example, at \( \theta = \frac{\pi}{3} \), calculate exact values using known trigonometric values: \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \) and \( \cos \frac{\pi}{3} = \frac{1}{2} \).
• The final step is to derive \( \frac{dy}{dx} \), which represents the slope of the tangent line: use \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \).

By substituting the numerical values into this quotient, one can effectively determine the slope, which tells how steeply the curve slopes at \( \theta = \frac{\pi}{3} \). Understanding these calculations is crucial for analyzing the geometry of the curve in polar form.