Question

Find the slope of the line tangent to the following polar curves at the given points. At the points where the curve intersects the origin (when this occurs), find the equation of the tangent line in polar coordinates. $$r^{2}=4 \cos 2 \theta ;\left(0, \pm \frac{\pi}{4}\right)$$

Short answer

The equations of the tangent lines in polar coordinates are: 1. \(r(\sin{\theta} - \cos{\theta}) = 0\) and 2. \(r(\sin{\theta} + \cos{\theta}) = 0\)

Step-by-step solution

Convert the polar equation to rectangular coordinates

Recall the relation between rectangular coordinates \((x, y)\) and polar coordinates \((r,\theta)\) is \(x = r \cos \theta\) and \(y = r \sin \theta\). Squaring both equations, we get \(x^{2}=r^{2}\cos^{2}\theta\) and \(y^{2}=r^{2}\sin^{2}\theta\). Moreover, since the polar equation is \(r^{2}=4\cos2\theta\), we can substitute \(x^{2}\) and \(y^{2}\) to get $$x^{2}+y^{2}=4\cos2\theta$$

Differentiate with respect to \(\theta\)

Differentiate the equation with respect to \(\theta\) to find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):$$\frac{d}{d\theta}(x^{2}+y^{2})=\frac{d}{d\theta}(4\cos2\theta)$$Differentiate both sides (\(x\) and \(y\) in terms of \(\theta\)): $$2x \frac{dx}{d\theta}+2y\frac{dy}{d\theta}=-8\sin2\theta$$

Evaluate \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\) at given points

To find the slope of the tangent line, we need to find the values of \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\) at the given points \((0, \pm \frac{\pi}{4})\). We first find the corresponding rectangular coordinates at these points: $$(0, \pm\frac{\pi}{4}) \rightarrow (0, 0)$$(since \(r = 0\) for both points) Now, substituting the rectangular coordinates back into the differentiated equation:$$2(0) \frac{dx}{d\theta}+2(0)\frac{dy}{d\theta}=-8 \sin{2\theta}$$gives \(\frac{dx}{d\theta}=0\) and \(\frac{dy}{d\theta}=0\).

Find the slope of the tangent line

Now, find the slope of the tangent line by evaluating \(\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\):$$\frac{dy}{dx}=\frac{0}{0}$$Since there isn't a specific value for the slope of the tangent line, instead of finding the polar equation directly, we will find the equation of the tangent lines when the curve intersects the origin.

Find equation of tangent line when curve intersects origin

When the curve intersects the origin, \(r=0\). Plugging this into the polar curve equation, we get: $$0^{2}=4 \cos{2\theta}.$$Since the points given are \((0, \pm\frac{\pi}{4})\), allowing us to calculate the tangent line's equation in polar coordinates. For \(\theta = \frac{\pi}{4}\):$$r\sin{\theta} - r\cos{\theta} = 0$$which simplifies to$$r(\sin{\theta} - \cos{\theta}) = 0$$Similarly, for \(\theta = -\frac{\pi}{4}\), the tangent line's equation in polar coordinates is given by:$$r(\sin{\theta} + \cos{\theta}) = 0$$Thus, the equations of the tangent lines in polar coordinates when the curve intersects the origin are $$r(\sin{\theta} - \cos{\theta}) = 0$$and$$r(\sin{\theta} + \cos{\theta}) = 0$$

Key concepts

Slope of Tangent Line

Understanding the slope of a tangent line to a curve is a fundamental concept in calculus and geometry. The slope represents the rate of change of the function at a given point, and geometrically, it is the inclination of the tangent line to the curve.

In terms of polar coordinates, finding the slope involves a bit of a different approach than with rectangular coordinates. Given a polar curve with equation in the form of \( r = f(\theta) \), the slope of the tangent line at any point can be found using the derivatives of \( r \) with respect to \( \theta \). If we denote \( \frac{dr}{d\theta} \) as \( r' \) and use it along with the original polar equation, we can find the slope \( m \) of the tangent line using the formula: \[ m = \frac{dy}{dx} = \frac{r' \sin(\theta) + r \cos(\theta)}{r' \cos(\theta) - r \sin(\theta)} \]. However, if both \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \) are zero, as in our exercise, the slope can be undefined, which indicates a vertical or horizontal tangent line — special cases that require a separate analysis.

Remember that when a curve passes through the origin, the slope of the tangent may not be well-defined there because multiple lines can pass through the origin and be tangent to the curve. The key takeaway is that to find the slope of a tangent in polar coordinates, we need to calculate the derivatives of the polar equations with respect to \( \theta \).

Rectangular to Polar Conversion

Conversion between rectangular and polar coordinates is a crucial skill when working with different types of functions and equations. In our exercise, we started with a polar equation and converted it into rectangular form to simplify the process of finding slopes of tangents.

The transformation is grounded on the relationships \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \), representing the projections on the x-axis and y-axis, respectively. These relations stem from the trigonometric definitions in a unit circle and can be used to rewrite any polar equation in terms of \( x \) and \( y \). For instance, the square of the distance from the origin is expressed in rectangular coordinates as \( x^2 + y^2 \), which is equivalent to \( r^2 \) in polar coordinates.

By adopting these relations, we can navigate between both coordinate systems with ease, translating the rich geometrical interpretations of polar coordinates into the more commonly used rectangular coordinates. Students should practice these conversions, as they are not only significant for solving calculus problems but are also essential in many physics and engineering applications.

Differentiation of Polar Equations

Differentiation in polar coordinates might seem a bit more intricate than in the standard Cartesian system. The process involves finding the derivative of \( r \) with respect to \( \theta \), which gives us the change in the radial distance as the angle changes. In more technical terms, it helps us understand how the curve moves as we sweep around the origin.

The direct application of differentiation to polar equations is not enough to find the slope of the tangent line directly because the slope in polar coordinates is not the ratio of derivatives of \( r \) and \( \theta \), as it is in rectangular coordinates. Instead, after differentiating, we generally require a conversion to rectangular form to obtain \( \frac{dy}{dx} \), or we must use the chain rule.

In our exercise, we followed the latter by differentiating \( x^2 + y^2 = r^2 \) to get an equation in terms of \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \), which allowed us to find the changes in \( x \) and \( y \) with respect to \( \theta \), ultimately leading us to the slope of the tangent line. This process highlights the interplay between different representations of a curve and the corresponding calculus operations needed to analyze its properties.