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. $$r=1-\sin \theta ;\left(\frac{1}{2}, \frac{\pi}{6}\right)$$

Short answer

The slope of the tangent line at the given point is \(-\frac{\sqrt{3}}{2}\).

Step-by-step solution

Find the derivative of the polar curve with respect to θ

We are given the polar curve \(r = 1 - \sin \theta\). To find \(\frac{dr}{d\theta}\), we can simply differentiate the equation with respect to \(\theta\). We get: $$\frac{dr}{d\theta} = -\cos \theta$$

Evaluate the derivative at the given point

We are given the point \(\left(\frac{1}{2}, \frac{\pi}{6}\right)\). To find the slope of the tangent line at this point, we need to evaluate \(\frac{dr}{d\theta}\) at \(\theta = \frac{\pi}{6}\). Plugging this into the derivative expression, we get: $$\frac{dr}{d\theta}\Bigr|_{\theta=\frac{\pi}{6}} = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2}$$ So, the slope of the tangent line at the point \(\left(\frac{1}{2}, \frac{\pi}{6}\right)\) is \(-\frac{\sqrt{3}}{2}\).

Check if the given point intersects the origin and find the equation of the tangent line in polar coordinates if necessary

The polar curve \(r = 1 - \sin \theta\) intersects the origin when \(r=0\). Solving for \(\theta\) when \(r=0\), we get: $$0 = 1 - \sin \theta \Rightarrow \sin \theta = 1$$ This holds when \(\theta = \frac{\pi}{2}\), and the point on the polar curve that intersects the origin is \(\left(0, \frac{\pi}{2}\right)\). Since the given point is not this point, it does not intersect the origin. Thus, we do not need to find the equation of the tangent line in polar coordinates.

Key concepts

Tangent Line

Understanding the concept of a tangent line is essential when dealing with curves. In polar coordinates, a tangent line at a given point on a curve provides the direction and steepness, or slope, of the curve at that specific point. Always remember that the tangent line just skims the curve at one precise spot, neither cutting through nor pulling away from it.

When we talk about the slope of the tangent line for a polar curve like the one given, where the curve is expressed as \( r = 1 - \sin \theta \), the slope indicates how much the radius \( r \) changes with a slight change in the angle \( \theta \). In other words, slope gives us a measure of how tilted the curve is around that point.

For students working with polar coordinates, solving for the slope of the tangent line involves calculating the derivative of the given polar equation with respect to \( \theta \), and then evaluating that derivative at the specific angle of interest.

Derivatives

Derivatives are a key concept here, as they help us find the slope of a tangent line. When you differentiate a polar curve, you're essentially looking for the rate of change of the radius \( r \) with respect to the angle \( \theta \).

To compute the derivative \( \frac{dr}{d\theta} \), you apply standard rules of differentiation to the given polar equation. In our exercise, the function \( r = 1 - \sin \theta \) yields a derivative of \( \frac{dr}{d\theta} = -\cos \theta \).

Once you have the derivative, plug in the angle given in the point \( \left( \frac{1}{2}, \frac{\pi}{6} \right) \) to calculate the exact slope at that point. For instance, evaluating \( -\cos \frac{\pi}{6} \) gives us \( -\frac{\sqrt{3}}{2} \), which is the slope of the tangent at the specified angle \( \theta = \frac{\pi}{6} \). This shows how steep the curve is at that point, making derivatives invaluable in analyzing curves.

Trigonometric Functions

Trigonometric functions like \( \sin \theta \) and \( \cos \theta \) play a crucial role in both expressing polar curves and deriving important values. These functions help relate the angle \( \theta \) with the radius \( r \) in a polar coordinate system.

When working with derivatives of trigonometric functions, it's important to remember the basic derivatives:

• The derivative of \( \sin \theta \) is \( \cos \theta \).
• The derivative of \( \cos \theta \) is \( -\sin \theta \).

These derivatives are highly useful in finding the change in position on polar curves, as seen in the given exercise. Moreover, with functions like \( r = 1 - \sin \theta \), you're essentially examining how a trigonometric fluctuation impacts the curve's radius.

By understanding and applying these trigonometric functions, one can easily navigate through the complexities of polar plots and their respective tangent calculations.