Problem 18 - Thomas Calculus

Question

Find the slopes of the curves in Exercises \(17-20\) at the given points. Sketch the curves along with their tangents at these points. Cardioid \(\quad r=-1+\sin \theta ; \quad \theta=0, \pi\)

Step-by-Step Solution

Answer

Find slopes by differentiating Cartesian conversion, evaluate at given angles.

Step 1: Understanding the Problem

We need to find the slopes of the tangents to the curve described by the polar equation \(r = -1 + \sin \theta\) at specific points where \(\theta = 0\) and \(\theta = \pi\).

Step 2: Convert to Cartesian Coordinates

The slope of the tangent can be found by converting the polar equation to Cartesian coordinates. Using the formulas \(x = r \cos \theta\) and \(y = r \sin \theta\), substitute \(r = -1 + \sin \theta\) to express \(x\) and \(y\) in terms of \(\theta\).

Step 3: Differentiate to Find Slopes

The slope in Cartesian coordinates is \(\frac{dy}{dx}\). We find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\), then use \(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\). This involves differentiating \(x = (-1 + \sin \theta) \cos \theta\) and \(y = (-1 + \sin \theta) \sin \theta\).

Step 4: Evaluate Slopes at \(\theta = 0\)

Substitute \(\theta = 0\) into the expressions for \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\), then calculate the slope at \(\theta = 0\).

Step 5: Evaluate Slopes at \(\theta = \pi\)

Substitute \(\theta = \pi\) into the expressions for \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\), then calculate the slope at \(\theta = \pi\).

Step 6: Sketch the Curves and Tangents

Sketch the cardioid described by \(r = -1 + \sin \theta\) on polar coordinates. Draw the tangent lines at the points corresponding to \(\theta = 0\) and \(\theta = \pi\) using the slopes found in previous steps.

Key Concepts

Polar Coordinates

Polar coordinates are a way of expressing locations in a plane using a radius and an angle. Instead of specifying a point by its horizontal and vertical distances from an origin—like in Cartesian coordinates—polar coordinates use:

• Radius \(r\), which is the distance from the origin to the point.
• Angle \(\theta\), measured from the positive x-axis.

This system is especially useful for problems with circular or rotational symmetry, as it can simplify the equations involved. To convert polar coordinates to Cartesian coordinates, we use:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)

The choice between Cartesian and polar coordinates often depends on the context or the symmetry of the problem.

Cardioid

A cardioid is a type of curve that looks like a heart. It belongs to a family of shapes known as limaçons. The particular cardioid given by the equation \(r = -1 + \sin \theta\) is traced out in the polar coordinate plane as \(\theta\) varies from 0 to \(2\pi\).
Key properties of a cardioid include:

• It has a cusp at the origin (when \(\theta = \pi/2\)).
• The shape is symmetrical with respect to the horizontal axis.
• The maximum distance from the origin is when \(\theta = \pi/2\), there \(r = 0\) gives the cusp.

To visualize it, think of drawing with a compass, fixing one point, then tracing around a circular path where the tracing point touches all sides of the circle as you move the center.

Tangent Line

A tangent line to a curve at a particular point is a line that just "touches" the curve at that point, sharing the same instantaneous direction as the curve.

• In practical terms, it's the line that depicts the instantaneous rate of change of the function at a given point.
• For polar curves like our cardioid, finding a tangent line involves converting the polar equations into Cartesian coordinates and then differentiating them.

When you find the derivative \(\frac{dy}{dx}\), it gives the slope of the tangent line at any point on the curve.
For the cardioid \(r = -1 + \sin \theta\), you would need to find the slope at specific \(\theta\) values, such as 0 and \(\pi\), to sketch the tangent lines at those points.

Derivative

The derivative is a fundamental concept in calculus, representing the rate at which a function is changing at any given point. For a function \(y = f(x)\), the derivative \(\frac{dy}{dx}\) gives the slope of the tangent to the curve of the function at a particular x-value.When dealing with polar coordinates, we first express the x and y coordinates in terms of \(\theta\), and then calculate:

• \(\frac{dy}{d\theta}\), change in \(y\) with respect to \(\theta\).
• \(\frac{dx}{d\theta}\), change in \(x\) with respect to \(\theta\).

The slope in Cartesian terms, \(\frac{dy}{dx}\), is found by:\[\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\]This approach lets us find the exact slope of the tangent line at any point specified by \(\theta\), such as when \(\theta = 0\) or \(\theta = \pi\) for our cardioid. Understanding derivatives in different coordinate systems like polar is crucial for tackling a wide range of calculus problems.