Question

Find the points at which the following polar curves have a horizontal or a vertical tangent line. $$r=1-\sin \theta$$

Short answer

The polar curve has horizontal tangent lines at (\(0, \frac{\pi}{2}\)) and (\(2, \frac{3\pi}{2}\)), and vertical tangent lines at (\(1, 0\)) and (\(1, \pi\)).

Step-by-step solution

Convert to Cartesian coordinates

To convert the polar curve \(r=1-\sin\theta\) to Cartesian coordinates, recall that \(x = r\cos\theta\) and \(y = r\sin\theta\). Substitute the polar equation \(r = 1-\sin\theta\) into these expressions to obtain the Cartesian representation: $$x = (1-\sin\theta)\cos\theta$$ $$y = (1-\sin\theta)\sin\theta$$

Compute the partial derivatives

In order to find the required derivatives, we can differentiate both sides of these Cartesian coordinate equations with respect to \(\theta\): $$\frac{dx}{d\theta} = -\sin\theta\cos\theta + (1-\sin\theta)(-\sin\theta)$$ $$\frac{dy}{d\theta} = \sin\theta\cos\theta + (1-\sin\theta)\cos\theta$$ Next, find the derivative \(\frac{dy}{dx}\) by dividing \(\frac{dy}{d\theta}\) by \(\frac{dx}{d\theta}\): $$\frac{dy}{dx}=\frac{\sin\theta\cos\theta + (1-\sin\theta)\cos\theta}{-\sin\theta\cos\theta + (1-\sin\theta)(-\sin\theta)}$$ Simplify the expression: $$\frac{dy}{dx}=\frac{\cos\theta(\sin\theta+1-\sin\theta)}{\cos\theta(-\sin\theta) +(1-\sin\theta)(-\sin\theta)}$$ Further simplify: $$\frac{dy}{dx}=\frac{\cos\theta}{1-\sin^2\theta}$$

Determine the horizontal and vertical tangent lines

The tangent line will be horizontal when \(\frac{dy}{dx} = 0\) and will be vertical when \(\frac{dy}{dx}\) is undefined. For the horizontal tangent line, we look for when \(\cos\theta = 0\). The solutions for \(\theta\) are: $$\theta = \frac{\pi}{2}, \frac{3\pi}{2}$$ For the vertical tangent line, we look for when \(1-\sin^2\theta = 0\). The solutions for \(\theta\) are: $$\theta = 0, \pi$$

Find the corresponding points on the curve

Now, we need to find the corresponding points on the polar curve for these values of \(\theta\). Recall that the polar equation is \(r = 1-\sin\theta\). For the horizontal tangent line, the points are: When \(\theta = \frac{\pi}{2}\): $$r = 1 - \sin\frac{\pi}{2} = 0$$ The point is (\(0, \frac{\pi}{2}\)). When \(\theta = \frac{3\pi}{2}\): $$r = 1 - \sin\frac{3\pi}{2} = 2$$ The point is (\(2, \frac{3\pi}{2}\)). For the vertical tangent line, the points are: When \(\theta = 0\): $$r = 1 - \sin0 = 1$$ The point is (\(1, 0\)). When \(\theta = \pi\): $$r = 1 - \sin\pi = 1$$ The point is (\(1, \pi\)). So, the points at which we have horizontal or vertical tangent lines are (\(1, 0\)), (\(1, \pi\)), (\(0, \frac{\pi}{2}\)), and (\(2, \frac{3\pi}{2}\)).

Key concepts

Polar Coordinates

Polar coordinates provide a powerful way to represent points in a plane using a distance and an angle. Instead of the usual Cartesian coordinates where points are described with an (x, y) pair, polar coordinates use (r, θ). Here, "r" is the radius or the distance from the origin, and "θ" (theta) is the angle measured counterclockwise from the positive x-axis. This system is incredibly useful for curves and shapes that have rotational or cyclical properties. For example, circles and spirals often appear simpler and more intuitive in polar form than in Cartesian form.

• Keep in mind:
  • The angle θ is usually measured in radians.
  • If r is negative, the point lies in the opposite direction of the given angle.
  • Coordinate conversion between Cartesian and polar systems is often used in calculus problems involving curves.
• Understanding polar coordinates helps in simplifying calculations of complex curves such as the one given by the equation \(r = 1 - \sin \theta\).

Tangent Lines

In calculus, a tangent line to a curve at a given point is a straight line that "just touches" the curve at that point. It has the same slope as the curve at the point of tangency. The concept of tangent lines is crucial in understanding the behavior of curves, as they give insight into the direction in which the curve is moving.

• Horizontal tangent lines occur where the slope of the curve (derivative) is zero. This means that the curve is temporarily flat, like the top of a hill or the bottom of a valley.
• Vertical tangent lines occur where the slope is undefined. These are places where the curve direction is changing instantaneously in a manner that doesn't produce a defined slope.

By finding where these tangent lines occur on a polar curve like \(r = 1 - \sin \theta\), we understand key points where the curve changes its nature in terms of increasing, decreasing, or momentarily pausing.

Cartesian Conversion

Converting between polar and Cartesian coordinates allows us to apply different mathematical tools conveniently.

• The conversion from polar to Cartesian coordinates uses the equations:\(x = r\cos\theta\) and \(y = r\sin\theta\).
• This transformation translates a polar equation like \(r = 1 - \sin \theta\) into its Cartesian form, making it easier to analyze with familiar algebraic techniques.

Using both forms, we gain multiple perspectives on the geometry of the curve, allowing us to switch between them for ease of graphing or calculation. For this particular exercise, translating the polar coordinates into Cartesian coordinates helps compute the derivatives needed to find tangent lines. It's a crucial step in examining the nature of the curve more intuitively.

Derivative

Derivatives play a fundamental role in calculus by describing how a function changes. For each point on a curve, the derivative gives the slope of the tangent line to the curve at that point, revealing information about the curve's direction and steepness at that location.

• The derivative \(\frac{dy}{dx}\) is obtained by taking the ratio of \(\frac{dy}{d\theta}\) to \(\frac{dx}{d\theta}\) when the function is parametrized by \(\theta\).
• In our context, simplification of this derivative shows where the slope is zero (horizontal tangent) or undefined (vertical tangent).
• This process helps identify key characteristics of the curve, like local maxima, minima, or cusp points.

Calculating derivatives correctly is key to understanding the structures and key points of interest on any curve, especially within polar coordinates as it offers insights into the nature of non-linear or complex curves like \(r = 1 - \sin \theta\).