Problem 36 - Thomas Calculus

Question

Replace the polar equations in Exercises \(23-48\) by equivalent Cartesian equations. Then describe or identify the graph. $$ r=4 \tan \theta \sec \theta $$

Step-by-Step Solution

Answer

The Cartesian equation is \(x^2 = 4y\) and it represents a parabola opening upwards.

Step 1: Write the equation in terms of trigonometric identities

Given the polar equation: \[ r = 4 \tan \theta \sec \theta \]Recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \sec \theta = \frac{1}{\cos \theta} \). Substitute these identities into the equation:\[ r = 4 \left( \frac{\sin \theta}{\cos \theta} \right) \left( \frac{1}{\cos \theta} \right) \]This simplifies to:\[ r = 4 \frac{\sin \theta}{\cos^2 \theta} \]

Step 2: Convert polar to Cartesian coordinates

Recall the polar to Cartesian coordinate conversions: \( x = r \cos \theta \) and \( y = r \sin \theta \).We also know that \( x = r \cos \theta \) and \( y = r \sin \theta \), so \( r = \sqrt{x^2 + y^2} \), \( \cos \theta = \frac{x}{r} \), \( \sin \theta = \frac{y}{r} \).Using these, we have:\[ \sin \theta = \frac{y}{r} \quad \text{and} \quad \cos \theta = \frac{x}{r} \]Substituting these into the equation gives:\[ r = 4 \frac{\frac{y}{r}}{\left(\frac{x}{r}\right)^2} = 4 \frac{y r}{x^2} \]

Step 3: Solve for Cartesian equation

Multiply both sides by \(x^2\) and \(r\) to clear the fractions:\[ r x^2 = 4 y r \]We know \( r = \sqrt{x^2 + y^2} \), so substitute this in:\[ \sqrt{x^2 + y^2} x^2 = 4 y \sqrt{x^2 + y^2} \]Cancel \( \sqrt{x^2 + y^2} \) from both sides:\[ x^2 = 4y \]

Step 4: Identify the graph

The equation \( x^2 = 4y \) is in the standard form of a parabola that opens upwards along the y-axis. It is centered at the origin with a vertex at (0,0) and has a vertical axis of symmetry.

Key Concepts

Trigonometric identities

Understanding trigonometric identities is essential for converting polar equations to Cartesian form. Trigonometric identities are mathematical equations that express relationships between the trigonometric functions. These include functions like sine (\( \sin \theta \)), cosine (\( \cos \theta \)), and tangent (\( \tan \theta \)). In some problems, such as the one given in the exercise, we also use the secant function (\( \sec \theta = \frac{1}{\cos \theta} \)).
The steps to convert involve understanding how to express tangent and secant in terms of sine and cosine. \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \sec \theta = \frac{1}{\cos \theta} \]These identities help simplify the polar equation by substituting the known expressions.
In the exercise, we simplified \( r = 4 \tan \theta \sec \theta \) to \( r = 4 \frac{\sin \theta}{\cos^2 \theta} \) by using these identities. This simplification eases our transition from polar to Cartesian coordinates. Recognizing and using trigonometric identities effectively is crucial in mathematical transformations.

Cartesian coordinates

The Cartesian coordinate system uses two perpendicular axes - typically labeled x and y - to define a plane where each point is identified by a pair of numerical coordinates. In the context of converting polar to Cartesian coordinates, it's vital to understand the basic conversions:

• \( x = r \cos \theta \)


• \( y = r \sin \theta \)


• \( r = \sqrt{x^2 + y^2} \)

These conversions allow us to express polar measurements, which are based on a radius and an angle, in terms of x and y coordinates on the plane.
In the exercise, we use these conversions to transform the polar equation into a Cartesian one. By replacing \( \sin \theta \) and \( \cos \theta \) with their Cartesian equivalents, and multiplying through to eliminate fractions, we achieve the Cartesian form of the equation. Practicing these conversions helps solidify understanding and facilitates tackling more complex problems involving different coordinates.

Graph descriptions

Once a polar equation is converted to a Cartesian form, understanding the resulting graph is the next step. The Cartesian equation derived from the exercise is \( x^2 = 4y \).
Understanding the structure of this equation is key to identifying its graph. Here, the equation represents a parabola that opens upward. This form is standard for parabolas of the format \( x^2 = 4py \), where p represents the distance from the vertex to the focus.
Knowing the properties of parabolas can help in visualizing the graph:

• The vertex is at the origin (0,0).


• The parabola is symmetric about the y-axis.


• It opens upwards because the equation involves \( x^2 \).

Identifying these characteristics is crucial for sketching the graph and understanding its relationship with the original polar equation. Mastery of graph descriptions allows you to translate mathematical equations into visual understanding, enhancing problem-solving skills.