Problem 24 - Calculus Early Transcendentals

Question

Use the formula $$ A=\frac{1}{2} \oint_{C}-y d x+x d y $$ to find the area of the region swept out by the line from the origin to the hyperbola \(x=a \cosh t, y=b \sinh t\) if \(t\) varies from \(t=0\) to \(t=t_{0}\left(t_{0} \geq 0\right)\)

Step-by-Step Solution

Answer

The area is \( \frac{1}{2} ab t_0 \).

Step 1: Parameterize the Curve

The hyperbola is given by the parametric equations: \[ x = a \cosh t, \quad y = b \sinh t \]We will use these equations to evaluate the line integral for the area. In this parameterization, the variable \(t\) varies from \(0\) to \(t_0\).

Step 2: Compute the Differentials

To apply the formula, we need the differentials \(dx\) and \(dy\). Compute these using the derivatives of the parametric equations with respect to \(t\):\[ dx = \frac{d}{dt}(a \cosh t) \, dt = a \sinh t \, dt \]\[ dy = \frac{d}{dt}(b \sinh t) \, dt = b \cosh t \, dt \]

Step 3: Substitute into the Area Integral

Substituting the values of \(x\), \(y\), \(dx\), and \(dy\) into the area formula:\[ A = \frac{1}{2} \int_0^{t_0} \left(-y \, dx + x \, dy\right) \]Substitute \(x = a \cosh t\), \(y = b \sinh t\), \(dx = a \sinh t \, dt\), and \(dy = b \cosh t \, dt\):\[ A = \frac{1}{2} \int_0^{t_0} (-b \sinh t \cdot a \sinh t + a \cosh t \cdot b \cosh t) \, dt \]

Step 4: Simplify the Integral

Simplify the expression inside the integral:\[ A = \frac{1}{2} \int_0^{t_0} (-ab (\sinh^2 t) + ab (\cosh^2 t)) \, dt \]Using the identity \(\cosh^2 t - \sinh^2 t = 1\), the integral simplifies to:\[ A = \frac{1}{2} \int_0^{t_0} ab \, dt \]

Step 5: Evaluate the Integral

Now, compute the integral:\[ A = \frac{1}{2} ab \int_0^{t_0} dt \]\[ A = \frac{1}{2} ab [t]_0^{t_0} = \frac{1}{2} ab (t_0 - 0) \]

Step 6: Simplified Area Formula

We have:\[ A = \frac{1}{2} ab t_0 \]This is the area of the region swept out by the line from the origin to the hyperbola as \(t\) varies from \(0\) to \(t_0\).

Key Concepts

Parametric Equations

Parametric equations allow us to express complex curves and shapes using parameters. Instead of relating variables directly, each variable is expressed in terms of a third variable, usually denoted as \( t \), time, or another suitable parameter.
For the hyperbola in this problem, the parametric equations are:

• \( x = a \cosh t \)
• \( y = b \sinh t \)

These equations define a set of coordinates, \( (x, y) \), for every value of \( t \). As \( t \) varies from \( 0 \) to \( t_0 \), it traces out the curve of the hyperbola.
By using parametric equations, we gain more control over the shape and orientation of curves, making them highly useful in calculus and geometry. They are particularly helpful in problems involving paths or motion.

Line Integral

Line integrals allow us to integrate functions over a curve. Think of it like "summing up" values along a path or a surface. These integrals can take into account not just the path of integration but also the direction along the path.
In this exercise, we're using the following line integral formula to find the area enclosed by a loop:\[ A = \frac{1}{2} \oint_{C} -y \, dx + x \, dy \]Here:

• \( -y \, dx \) and \( x \, dy \) are part of the integral that tracks the change in \( y \) and \( x \) along the curve.

This formula is specifically useful for finding areas bounded by parametric curves, making it perfect for our hyperbola problem. The use of line integrals in this context provides a link between calculus and geometry, allowing for elegant solutions to area problems like this one.

Hyperbola

A hyperbola is a type of conic section that resembles two mirrored open curves. It's defined by its rectangular hyperbola equation, but can also be expressed using parametric equations with hyperbolic functions.
For our problem, the hyperbola is defined as:

• \( x = a \cosh t \)
• \( y = b \sinh t \)

This parametric form utilizes the hyperbolic sine and cosine functions, \( \sinh \) and \( \cosh \), which are analogs to the regular trigonometric functions.
Hyperbolas have unique properties that make them important in various scientific fields, including physics and engineering. In this exercise, the hyperbola plays a central role in determining the region whose area is to be calculated.

Differentiation

Differentiation is the process of finding the rate at which a function is changing at any point. It is a fundamental concept in calculus that involves calculating derivatives.
In the context of this problem, we compute the differentials \( dx \) and \( dy \) from the parametric equations:

• \( dx = \frac{d}{dt}(a \cosh t) \cdot dt = a \sinh t \cdot dt \)
• \( dy = \frac{d}{dt}(b \sinh t) \cdot dt = b \cosh t \cdot dt \)

By employing differentiation, we determine how \( x \) and \( y \) change with respect to \( t \). This is crucial for setting up the line integral that follows. Understanding differentiation helps in grasping how changes in parameters affect the curve, ultimately leading to solving for areas like in this exercise.