Problem 237 - Calculus Volume 3

Question

Find the slope of a tangent line to a polar curve \(r=f(\theta)\). Let \(x=r \cos \theta=f(\theta) \cos \theta\) and \(y=r \sin \theta=f(\theta) \sin \theta, \quad\) so the polar equation \(r=f(\theta)\) is now written in parametric form. $$ r=4 \cos \theta ;\left(2, \frac{\pi}{3}\right) $$

Step-by-Step Solution

Answer

The slope of the tangent line is \( \frac{\sqrt{3}}{3} \).

Step 1: Convert Polar to Parametric Form

Given the polar equation \( r = 4 \cos \theta \), we can express this in parametric form as \( x = r \cos \theta = 4 \cos \theta \cdot \cos \theta \) and \( y = r \sin \theta = 4 \cos \theta \cdot \sin \theta \). This simplifies to \( x = 4 \cos^2 \theta \) and \( y = 4 \cos \theta \sin \theta \).

Step 2: Differentiate Parametric Equations

To find the slope of the tangent line, we need \( \frac{dy}{dx} \), which requires us to compute \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \). Differentiating \( x \) with respect to \( \theta \), we get:\[ \frac{dx}{d\theta} = \frac{d}{d\theta} (4 \cos^2 \theta) = 4 \cdot 2 \cos \theta (-\sin \theta) = -8 \cos \theta \sin \theta. \]Next, differentiate \( y \):\[ \frac{dy}{d\theta} = \frac{d}{d\theta} (4 \cos \theta \sin \theta) = 4(\cos^2 \theta - \sin^2 \theta). \]

Step 3: Find the Derivative \( \frac{dy}{dx} \)

The slope of the tangent in the parametric form \( \frac{dy}{dx} \) is calculated using the formula:\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}. \]Substitute the expressions:\[ \frac{dy}{dx} = \frac{4(\cos^2 \theta - \sin^2 \theta)}{-8 \cos \theta \sin \theta} = \frac{-(\cos^2 \theta - \sin^2 \theta)}{2 \cos \theta \sin \theta}. \]

Step 4: Evaluate the Derivative at Given Point

The problem states that the point on the curve is \( \left(2, \frac{\pi}{3}\right) \), corresponding to \( \theta = \frac{\pi}{3} \). Calculate \( \cos \theta = \frac{1}{2} \) and \( \sin \theta = \frac{\sqrt{3}}{2} \) at \( \theta = \frac{\pi}{3} \).Substitute these into the derivative expression:\[\frac{dy}{dx} = \frac{-(\left(\frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2)}{2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2}} = \frac{-\left(\frac{1}{4} - \frac{3}{4}\right)}{\frac{\sqrt{3}}{2}} = \frac{-(-\frac{1}{2})}{\frac{\sqrt{3}}{2}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}.\]

Step 5: Rationalize the Slope if Needed

Rationalize the denominator of the slope expression:\[ \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}. \]Thus, the slope of the tangent line at the point \( \left(2, \frac{\pi}{3}\right) \) is \( \frac{\sqrt{3}}{3}. \)

Key Concepts

Polar Coordinates

Polar coordinates offer a distinctive way to represent points on a plane. Unlike the usual Cartesian coordinates (think: the classic x, y grid), polar coordinates define a point based on its distance from the origin (r) and its angle (θ) relative to the positive x-axis. This can be particularly useful in situations involving curves that sweep around a circle or rotational symmetry. Here's how it works:

• The "r" value tells you how far away the point is from the origin. It's similar to knowing the radius of a circle.
• The "θ" value tells you the direction or angle of the point from the positive x-axis.

Converting polar equations to other forms can sometimes make them easier to work with, especially if you need to perform calculus operations like differentiation.

Parametric Equations

Parametric equations are a powerful way to express mathematical curves without directly relating y to x. Instead, they use one or more parameters to define both coordinates independently. In our scenario, the parameter is the angle θ.

• These equations allow curves to be traced out in a much more flexible way.
• They usually involve a separate equation for each coordinate (x and y), both expressed in terms of the same parameter.

By expressing a polar equation in a parametric form, you can now more easily compute derivatives. This is crucial when you're trying to find the slope of the tangent line.

Differentiation of Parametric Equations

Finding the slope of a tangent line to a curve given by parametric equations is a bit different than the regular differentiation with respect to x and y. You must calculate \( \frac{dy}{dx} \), but first you need \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \).

• Once you have these derivatives, you apply the formula \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \).
• This computation gives you the slope of the tangent line at a specific point on the curve by considering the changes in y and x with respect to θ.

The beauty of this approach is that it adapts to a variety of curves, even those without convenient x-y relationships,

Rationalization of Denominators

Rationalizing the denominator in a fraction means transforming it so that the denominator is a rational number. This process can simplify the expression and make it easier to use in future calculations. Here's how it's done:

• You multiply both the numerator and the denominator by a conjugate or suitable form of one that eliminates radicals or imaginary numbers in the denominator.
• In the slope calculation, \( \frac{1}{\sqrt{3}} \) is rationalized to \( \frac{\sqrt{3}}{3} \).

This step is often one of the last in a calculus problem to ensure the answer is in its simplest and most standard form.