Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the polar equation for the curve described by the following Cartesian equations. (a) \(y=45\) (b) \(x^{2}+y^{2}=36\) (c) \(x^{2}-y^{2}=1\) (d) \(4 x y=1\) (e) \(y=3 x+2\) (f) \(3 x^{2}+4 y=2\) (g) \(x^{2}+2 x+y^{2}-4 y-25=0\) Computers and graphing calculators offer a wonderful opportunity to experiment with the graphing of polar equations of the form \(r=f(\theta)\). In some cases these aids require that the equations be recast in a parametric form. Since \(x=r \cos \theta=f(\theta) \cos \theta\) and \(y=r \sin \theta=f(\theta) \sin \theta\), you can use the parametric graphing capabilities to graph \(x=f(t) \cos t\) and \(y=f(t) \sin t\) as a set of parametric equations.

Step-by-Step Solution

Answer

(a) \(r = \frac{45}{\sin \theta}\), (b) \(r = 6\), (c) \(r^2 = \frac{1}{\cos 2\theta}\), (d) \(r^2 = \frac{1}{2 \sin 2\theta}\), (e) \(r = \frac{2}{\sin \theta - 3 \cos \theta}\), (f) Solve for \(r\), (g) \((r \cos \theta + 1)^2 + (r \sin \theta - 2)^2 = 30\).

Step 1: Convert Equation (a) to Polar Form

For the equation \(y = 45\), use the relation \(y = r \sin \theta\). Set \(r \sin \theta = 45\). Hence the polar equation is \(r = \frac{45}{\sin \theta}\).

Step 2: Convert Equation (b) to Polar Form

For \(x^2 + y^2 = 36\), substitute \(x = r \cos \theta\) and \(y = r \sin \theta\). This gives \((r \cos \theta)^2 + (r \sin \theta)^2 = 36\). Simplifying gives \(r^2 = 36\), so \(r = 6\) or \(r = -6\).

Step 3: Convert Equation (c) to Polar Form

For \(x^2 - y^2 = 1\), use \(x = r \cos \theta\) and \(y = r \sin \theta\). Substitute to get \((r \cos \theta)^2 - (r \sin \theta)^2 = 1\). Simplifying, \(r^2 (\cos^2 \theta - \sin^2 \theta) = 1\), or \(r^2 \cos 2\theta = 1\). Hence, \(r = \sqrt{\frac{1}{\cos 2\theta}}\).

Step 4: Convert Equation (d) to Polar Form

For \(4xy = 1\), substitute \(x = r \cos \theta\) and \(y = r \sin \theta\) to get \(4(r \cos \theta)(r \sin \theta) = 1\). This simplifies to \(4r^2 \cos \theta \sin \theta = 1\), or \(2r^2 \sin 2\theta = 1\). So, \(r^2 = \frac{1}{2 \sin 2\theta}\).

Step 5: Convert Equation (e) to Polar Form

For \(y = 3x + 2\), use \(x = r \cos \theta\) and \(y = r \sin \theta\). Substitute and get \(r \sin \theta = 3(r \cos \theta) + 2\), simplifying gives \(r(\sin \theta - 3 \cos \theta) = 2\). Hence, \(r = \frac{2}{\sin \theta - 3 \cos \theta}\).

Step 6: Convert Equation (f) to Polar Form

For \(3x^2 + 4y = 2\), substitute \(x = r \cos \theta\) and \(y = r \sin \theta\) to get \(3(r \cos \theta)^2 + 4(r \sin \theta) = 2\). This simplifies to \(3r^2 \cos^2 \theta + 4r \sin \theta = 2\). Eventually solve for \(r\).

Step 7: Convert Equation (g) to Polar Form

The equation is \(x^2 + 2x + y^2 - 4y - 25 = 0\). Complete the squares for \(x\) and \(y\). First, \((x+1)^2 - 1 + (y-2)^2 - 4 - 25 = 0\) results in \((x+1)^2 + (y-2)^2 = 30\). Substitute \(x = r \cos \theta\), \(y = r \sin \theta\); hence \((r \cos \theta + 1)^2 + (r \sin \theta - 2)^2 = 30\).

Key Concepts

Parametric Equations

Parametric equations are a way of defining mathematical curves using parameters. Instead of expressing points in terms of x and y alone, parametric equations involve an independent parameter, often denoted by \(t\), that represents a point on the curve. Using parametric equations, we can graph equations more flexibly by expressing both x and y coordinates as functions of \(t\).
Here’s how they work:

• Each coordinate, \(x\) and \(y\), is a function of a parameter \(t\).
• This allows the expression of more complex curves than those possible in standard form.
• It is especially useful for describing paths and directions, like the circle which can be represented with \(x = \cos(t)\) and \(y = \sin(t)\).

Parametric equations offer a different perspective by adding movement along the curve over time, enriching the concept of graphing and visualization in mathematical contexts.

Cartesian to Polar Conversion

Converting Cartesian coordinates to polar coordinates involves a new way of defining location. In this conversion, every point is expressed in terms of radius \(r\) and angle \(\theta\), rather than \(x\) and \(y\).
Steps for conversion:

• The radius \(r\) is calculated as \(r = \sqrt{x^2 + y^2}\).
• The angle \(\theta\) is determined using \(\tan \theta = \frac{y}{x}\).
• Substitute these values back to find \(r\) and \(\theta\).

This approach transforms simpler equations into their geometric counterparts, simplifying integration and solving in many scenarios. Polar coordinates are particularly powerful for curves like circles and spirals.

Polar Equations

Polar equations express mathematical functions and curves in terms of polar coordinates. Instead of \(x\) and \(y\), polar equations use \(r\) and \(\theta\).
Here’s what to know about polar equations:

• Simple polar equations often involve \(r = f(\theta)\), where \(r\) directly depends on the angle \(\theta\).
• Some familiar Cartesian shapes, like circles, look cleaner in polar form, ex: \(r = a \cos \theta\) for a circle.
• Polar equations can represent more complex curves, such as spirals and conic sections.

Understanding polar equations opens the door to more dynamic graphing experiences that differ from linear Cartesian methods.

Graphing Polar Functions

Graphing polar functions allows us to visualize how \(r\) (radius) changes as \(\theta\) (angle) progresses, offering a unique insight into curves and their properties.
How to graph polar functions:

• Start by computing values of \(r\) for various angles \(\theta\).
• Plot these (\(r, \theta\)) points on a polar grid.
• Connect the points to see the shape or curve created by the function.
• Tools like graphing calculators can simplify this process, allowing dynamic manipulation of \(\theta\).

Graphing in polar coordinates highlights rotational symmetry and periodicity, revealing beautiful patterns integral to fields like engineering and physics.