Problem 49 - Essential Calculus

Question

Use a graphing utility to (a) graph the polar equation, (b) draw the tangent line at the given value of \(\boldsymbol{\theta}\), and (c) find \(d y / d x\) at the given value of \(\theta\). (Hint: Let the increment between the values of \(\theta\) equal \(\pi / 24 .)\) $$ r=3(1-\cos \theta), \theta=\frac{\pi}{2} $$

Step-by-Step Solution

Answer

The graph we obtain is a cardioid. Drawing the tangent line and using the derivative formula for polar functions allows us to find the slope of that tangent, which is the value of \( \frac{dy}{dx} \) at \( \theta=\frac{\pi}{2} \).

Step 1: Graph the Polar Equation

To graph \( r=3(1-\cos \theta) \), one can use a graphing utility and plot points for various values of \( \theta \) within the range \( 0 \leq \theta \leq 2\pi \). The result will be a well-known polar curve called a cardioid.

Step 2: Draw the Tangent Line at the given value of \( \theta \)

Next, we need to draw the tangent line at \( \theta=\frac{\pi}{2} \). We first find the corresponding point on the graph, then visually determine the tangent line at that point. Remember, a tangent line at a given point on a curve touches the curve at that point only and no other.

Step 3: Find \( \frac{dy}{dx} \) at \( \theta=\frac{\pi}{2} \)

Finally, we need to find \( \frac{dy}{dx} \) at \( \theta=\frac{\pi}{2} \). This is the rate of change of \( y \) with respect to \( x \), or in other words, the slope of the tangent line at that point. To do this, we can employ the formula for the derivative of a polar function \( y=r(\theta) \), which is \( \frac{dy}{dx} = \frac{r'(\theta) \sin(\theta) + r(\theta) \cos(\theta)}{r'(\theta) \cos(\theta) - r(\theta) \sin(\theta)} \), where \( r'(\theta) \) is the derivative of our polar function with respect to \( \theta \). Applying this formula will provide the desired slope.

Key Concepts

Graphing Polar Curves

Polar equations, like the one given by r = 3(1 - \(cos \theta\)), sketch out curves on a polar grid, which contrasts with the familiar Cartesian grid. Polar curves can take on many fascinating shapes such as spirals, roses, and limaçons, of which the cardioid is a particular example.

Graphing these curves involves plotting points for different values of \theta and connecting them to represent the relationship between \theta and the radial distance r. For instance, with our given equation, by plugging in increments of \theta like \pi/24, one can begin to visualize the shape of the curve. Tools like graphing calculators or computer software can automate this plotting process and produce the graph of the cardioid, revealing its heart-like structure.

Tangent Lines in Polar Coordinates

Tangent lines to polar curves require a slightly different approach than in Cartesian coordinates. Imagine visualizing the curve of r = 3(1 - \(cos \theta\)) and picking a point given by a specific angle, such as \theta = \pi/2. Here, the tangent line represents the direction in which the curve is moving at that exact point.

The key lies in visualizing the tangent as brushing the curve at exactly one point, without crossing the curve at the point of contact. For polar curves, knowing the derivative of the polar function with respect to \theta is essential to draw the tangent, which can be evaluated using calculus techniques specific to polar coordinates.

Slope of Tangent Line

In the context of polar coordinates, the slope of the tangent line at a particular point on the curve is a measure of how steep that line is. For the curve given by r = 3(1 - \(cos \theta\)), finding this slope involves differentiating with respect to \theta and using trigonometric identities to express dy/dx.

This slope tells us the inclination of the tangent line and can be calculated by the ratio of the change in y to the change in x, given by the derivative dy/dx. It's a crucial step for drawing an accurate tangent line, as it gives us the angle at which the tangent line deviates from the horizontal.

Polar Functions Derivatives

Understanding derivatives in polar coordinates is slightly more complex than in Cartesian coordinates. For polar functions, we use the concept of differentiation to find the rate of change of r with respect to \theta, which helps in determining the behavior of the curve and the slope of its tangent line at any given point.

The formula \frac{dy}{dx} = \frac{r'(\theta) \(sin \theta\) + r(\theta) \(cos \theta\)}{r'(\theta) \(cos \theta\) - r(\theta) \(sin \theta\)} incorporates the derivatives of sine and cosine functions to express dy/dx. In our case, by applying the derivative rules to r = 3(1 - \(cos \theta\)), we can calculate r'(\theta) and then find the slope of the tangent line at \theta = \pi/2. This calculation is fundamental for motion analysis and curve sketching in polar coordinates.