Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

A line segment \(L\) of length \(2 a\) has its two end points on the \(x\) - and \(y\)-axes, respectively. The point \(P\) is on \(L\) and is such that \(O P\) is perpendicular to \(L\). Show that the set of points \(P\) satisfying this condition is a four-leaved rose by finding its polar equation.

Step-by-Step Solution

Answer

The polar equation is \(r = a \cos(2\theta)\), a four-leaved rose.

Step 1: Define the line segment equation

Consider line segment \(L\) with endpoints \((x_1, 0)\) and \((0, y_1)\) on the \(x\)- and \(y\)-axes. Since the length of \(L\) is \(2a\), we have the relation \(x_1^2 + y_1^2 = (2a)^2\).

Step 2: Determine equation of line segment

The equation of the line segment joining \((x_1, 0)\) and \((0, y_1)\) can be parameterized as \(x = x_1(1-t)\) and \(y = y_1 t\) where \(t\) is a parameter from 0 to 1.

Step 3: Find condition for perpendicularity

Point \(P\) with coordinates \((x(t), y(t))\) divides segment \(L\) such that \(OP\) is perpendicular to \(L\). The vectors \(\vec{OP} = (x(t), y(t))\) and direction vector of \(L\) \((-x_1, y_1)\) need to satisfy the dot product condition: \(x(t)(-x_1) + y(t)y_1 = 0\).

Step 4: Substitute parameterization into dot product condition

Substitute the parameterization from Step 2 into the dot product condition to get \(-x_1^2 (1-t) + y_1^2 t = 0\). Solving for \(t\), we find that \(t = \frac{x_1^2}{x_1^2 + y_1^2}\).

Step 5: Use the condition of length to express x1 and y1

From \(x_1^2 + y_1^2 = 4a^2\), substitute back \(x_1 = 2a \cos \theta\), \(y_1 = 2a \sin \theta\), where \(\theta\) is the angle the line makes with x-axis.

Step 6: Convert to polar coordinates

Using \(x = r \cos \theta\) and \(y = r \sin \theta\), the point \((x,y)\) becomes \((2a \cos \theta (1-t), 2a \sin \theta t)\). The polar coordinates relation gives \(r(\cos^2 \theta + \sin^2 \theta) = 2a \sin 2\theta\).

Step 7: Arrive at the polar equation

Given \(r \cos(2\theta) = a\), hence \(r = a \sec(2\theta)\). Rewrite using the standard format of the \(4\)-leaved rose: \(r = a \cos(2\theta)\), which is equivalent to \(r = a \cos(2(\theta - \frac{\pi}{4}))\).

Step 8: Conclusion on the polar equation

Thus, the set of points forms the polar equation of a four-leaved rose, represented by \(r = a \cos(2\theta)\).

Key Concepts

Line Segment

A line segment is a straight path connecting two points, which are its endpoints. In this exercise, the line segment connects across the axes, specifically from the x-axis to the y-axis. Here's how it was described:

• The endpoints are denoted as \((x_1, 0)\) on the x-axis and \((0, y_1)\) on the y-axis.
• The total length of the segment is given as \(2a\), following the equation \(x_1^2 + y_1^2 = (2a)^2\).

Parameterizing the line segment, we introduce a parameter \(t\), which varies from 0 to 1, indicating any point along the segment. The coordinates of any point on the line can be expressed as \(x = x_1(1-t)\) and \(y = y_1 t\). This gives a directional and precise description of the line segment on the coordinate plane. Such a representation is crucial when analyzing conditions at any point along the line in subsequent steps, allowing the exploration of the complete set of possibilities within the given constraints.

Perpendicularity Condition

The condition of perpendicularity is fundamental in this exercise. A set of two lines or segments is considered perpendicular if they intersect at a right angle (90 degrees). To find the point \(P\) where \(OP\) is perpendicular to the line segment \(L\), we use the concept of the dot product of vectors which becomes zero for perpendicular vectors.

• The vectors involved are \(\vec{OP} = (x(t), y(t))\) and the direction vector of line segment \((-x_1, y_1)\).
• The dot product is calculated as \(x(t)(-x_1) + y(t)y_1 = 0\), allowing us to solve for the parameter \(t\).

By solving this condition, we identify specific points on the line segment where the required perpendicularity condition holds. Concretely, this means pinpointing the position of \(P\) that satisfies this orthogonal relationship, integral to further transforming and understanding the geometric arrangement and its relation to polar coordinates.

Polar Coordinates

Polar coordinates offer a different perspective from Cartesian coordinates by using a distance from the origin and an angle from a reference direction to define a point. In this transformation:

• The typical Cartesian coordinates \((x, y)\) are translated into polar form as \(x = r \cos \theta\) and \(y = r \sin \theta\).
• For our specific problem, after substituting these into the perpendicularity condition and using the length constraint \((x_1 = 2a \cos \theta, y_1 = 2a \sin \theta)\), we express the relationship in polar terms: \(r = 2a \sin 2\theta\).

Polar coordinates help simplify problems that involve circular or angular symmetry. Here, they allow us to move from requirements of perpendicularity and length into an understandable visual form: a four-leaved rose, which is iconic in polar coordinate systems.

Four-Leaved Rose

The four-leaved rose is a type of curve that can be defined in polar coordinates. It is named for its petals' floral resemblance. The rose-like pattern appears in its simplest form with equations such as \(r = a \cos(2\theta)\) or \(r = a \sin(2\theta)\).

• For our exercise, starting with the expression \(r = a \cos(2\theta)\), it was shown to satisfy the conditions set by the line segment and perpendicularity.
• This equation represents the Cartesian and polar translation, expressing a symmetric, repeated pattern that creates the petal effect. Here, 2\(\theta\) denotes that there are four petals around the origin.

Recognizing and converting this form is significant as it relates geometric configurations in a coordinate system to visually pleasing and analytically elegant curves. The rose curve encapsulates how algebraic expressions describe complex shapes in a simplified manner.