Question

Find the slope of the tangent line to the given polar curve at the point given by the value of \(\theta\).\(r=\ln \theta, \theta=e\)

Short answer

The slope of the tangent line is \( \frac{\frac{1}{e} \sin e + \cos e}{\frac{1}{e} \cos e - \sin e} \).

Step-by-step solution

Understand the Polar Curve

The curve given is in polar coordinates: \( r = \ln \theta \). This describes a curve where \( r \), the radius from the origin, depends on \( \theta \), the angle from the positive x-axis.

Convert to Parametric Form

In order to find the slope of the tangent line, we first convert the polar equation \( r = \ln \theta \) into parametric form, using the relationships \( x = r \cos \theta \) and \( y = r \sin \theta \).\[ x = (\ln \theta) \cos \theta \]\[ y = (\ln \theta) \sin \theta \]

Differentiate Parametric Equations

Find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \):\[ \frac{dx}{d\theta} = \frac{d}{d\theta}((\ln \theta) \cos \theta)\]Using the product rule: \[ \frac{dx}{d\theta} = \frac{1}{\theta} \cos \theta - (\ln \theta) \sin \theta \]\[ \frac{dy}{d\theta} = \frac{d}{d\theta}((\ln \theta) \sin \theta)\]\[ \frac{dy}{d\theta} = \frac{1}{\theta} \sin \theta + (\ln \theta) \cos \theta \]

Find the Slope of the Tangent Line

The slope of the tangent line in parametric form is given by:\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \]

Calculate Derivatives at \( \theta = e \)

Substitute \( \theta = e \) into the derivatives:\[ \frac{dx}{d\theta} \bigg|_{\theta = e} = \frac{1}{e} \cos e - (\ln e) \sin e \]Since \( \ln e = 1 \), this simplifies to:\[ \frac{dx}{d\theta} \bigg|_{\theta = e} = \frac{1}{e} \cos e - \sin e \]\[ \frac{dy}{d\theta} \bigg|_{\theta = e} = \frac{1}{e} \sin e + \cos e \]

Plug into Slope Formula

Calculate the slope of the tangent line:\[ \frac{dy}{dx} \bigg|_{\theta = e} = \frac{\frac{1}{e} \sin e + \cos e}{\frac{1}{e} \cos e - \sin e} \]

Simplify if Possible

Evaluate the expression to find the exact slope. This is a numeric calculation:Substitute numerical values for \( \sin e \) and \( \cos e \) using a calculator, simplifying where possible.The slope is a real number based on the trigonometric values at \( \theta = e \).

Key concepts

Parametric Equations

Parametric equations are a way to express a set of related quantities as functions of an independent variable, often denoted as \( t \) or in this case \( \theta \).
The classic representation of parametric equations is through equations for \( x \) and \( y \) in terms of a third parameter. Here, from the polar equation \( r = \ln \theta \), we derive:

• \( x = (\ln \theta) \cos \theta \)
• \( y = (\ln \theta) \sin \theta \)

These transformations are based on fundamental polar-to-cartesian conversion formulas for curves that extend the concept of a graph, providing more flexibility.
With parametric equations, every value of the parameter (\( \theta \)) gives a different point \((x, y)\) on the curve. Therefore, it is easier to handle more complex shapes and calculate other geometric properties like arcs, slopes, or lengths.

Differentiation

Differentiation is a mathematical process to find the rate at which something changes.
Here, it helps us determine the slope of the tangent at a specific point on the curve. For parametric equations, we find derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \), representing how \( x \) and \( y \) change with \( \theta \).
We apply the product rule to differentiate both equations:

• \( \frac{dx}{d\theta} = \frac{1}{\theta} \cos \theta - (\ln \theta) \sin \theta \)
• \( \frac{dy}{d\theta} = \frac{1}{\theta} \sin \theta + (\ln \theta) \cos \theta \)

The product rule follows a simple principle: differentiate each part separately while keeping the other part constant and then add the results.
Hence, through differentiation, we can explore how these changes in parameters affect the curve geometrically at any point, especially for finding tangents.

Tangent Lines

A tangent line is a straight line that touches a curve at a single point, without crossing it at that point.
To find this line's slope using parametric equations, we use the formula:\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \] This gives us the rate of change of \( y \) with respect to \( x \) at a particular \( \theta \).
In this exercise, we calculate the slope at \( \theta = e \):

• \( \frac{dx}{d\theta} \bigg|_{\theta = e} = \frac{1}{e} \cos e - \sin e \)
• \( \frac{dy}{d\theta} \bigg|_{\theta = e} = \frac{1}{e} \sin e + \cos e \)

Substituting these values into the slope formula gives the steepness of the tangent.
This practical approach of finding tangents proves useful in analyzing curves and predicting how they behave near certain points.