Question

Find the area bounded by the curve \(x=\cos t, y=e^{t}, 0 \leq t \leq \frac{\pi}{2}\) and the lines \(y=1\) and \(x=0\).

Short answer

The integral \( \int_{0}^{1}(e^{\arccos x} - 1) \, dx \) computes the area. Numerical methods or deeper function analysis are needed for exact evaluation.

Step-by-step solution

Understand the Problem

We want to find the area bounded by the parametric curve given by \( x = \cos t \) and \( y = e^{t} \) for \( 0 \leq t \leq \frac{\pi}{2} \), and the lines \( y = 1 \) and \( x = 0 \). The region of interest lies between these lines and the curve.

Set Up the Integral

The curve is defined parametrically by \( x = \cos t \) and \( y = e^t \). First, we need to find the bounds for \( x \). As \( t \) goes from \( 0 \) to \( \frac{\pi}{2} \), \( x = \cos t \) goes from \( 1 \) to \( 0 \). However, since we want the area where \( x \geq 0 \), we consider only this range directly.

Define y in Terms of x

For the curve, \( y = e^t \). Since we have \( x = \cos t \), we can rewrite \( t = \arccos x \). Substitute this to express \( y = e^{\arccos x} \). Therefore, the area under the curve is defined in terms of \( x \) from \( x = 1 \) to \( x = 0 \).

Set up the Area Integral

The area we want, bounded by \( y = e^{\arccos x} \) and the line \( y = 1 \), is \( \int_{0}^{1}(e^{\arccos x} - 1) \, dx \), because we subtract the constant line \( y = 1 \) from the curve \( y = e^{\arccos x} \).

Evaluate the Integral

Solving the integral \( \int_{0}^{1}(e^{\arccos x} - 1) \, dx \) is not trivial directly, it might require numerical methods or a deeper functional approach, but often the inverse trigonometric function and the exponential can complicate exact calculations without tables or numeric methods. Let's focus on setting it up correctly first.

Analyze Bounds and Simplify

Upon careful analysis, notice the curve wraps from (1, e^0) to (0, e^{\pi/2}), and due to \( x = 0 \) boundary, ensure that our area is calculated correctly between limits, utilizing numerical approximations if necessary for precise values.

Key concepts

Parametric Equations

Parametric equations allow us to describe a curve by using parameters, often denoted as \( t \). Instead of expressing \( y \) directly in terms of \( x \), as we might typically do with functions, parametric equations separately define both \( x \) and \( y \) in terms of \( t \).
In this exercise, our curve is defined by \( x = \cos t \) and \( y = e^t \) for \( 0 \leq t \leq \frac{\pi}{2} \). This effectively traces a specific path in the \( xy \)-plane as \( t \) changes.

• \( x = \cos t \) determines the horizontal position based on \( t \).
• \( y = e^t \) determines the vertical position based on \( t \).

By examining the behavior of the function over these \( t \) values, we gather information about the curve's shape and the region to compute.

Definite Integrals

A definite integral calculates the area under a curve between two bounds. For our problem, the integral we're interested in gives the area between the curve and the line \( y = 1 \) from \( x = 0 \) to \( x = 1 \).
The fundamental theorem of calculus answers how we can compute area. In this setting, our definite integral setup is \( \int_{0}^{1} (e^{\arccos x} - 1) \ dx \), which represents:

• The total area under the \( y = e^{\arccos x} \) curve.
• Minus the rectangle formed by \( y = 1 \) over the same interval.

This integral measures the area beneath the parametric curve and above the horizontal line \( y = 1 \). Evaluating this integral can be complex due to the nature of the curve defined by inverse trigonometric functions.

Numerical Integration

Sometimes, solving a definite integral symbolically can be very challenging, especially with complicated functions involving exponentials and inverse trigonometric expressions.
In such instances, numerical integration methods, like Simpson's Rule or the Trapezoidal Rule, come to the rescue. These methods approximate the area under the curve by summing up small segments or "trapezoids":

• Simpson's Rule uses parabolic sections for better accuracy.
• The Trapezoidal Rule uses straight line segments to estimate area.

These techniques are particularly useful when dealing with awkward integrals like \( \int_{0}^{1} e^{\arccos x} \, dx \) which do not have a straightforward antiderivative. Numerical integration helps us arrive at approximate solutions that are sufficient for practical purposes.

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \arccos \), reverse the effect of trigonometric functions: they take a ratio and provide the angle that would produce it.
In this problem, \( \arccos x \) is used to express the parameter \( t \) as a function of \( x \). Since we have \( x = \cos t \), solving for \( t \) leads to \( t = \arccos x \).
This transformation lets us rewrite \( y = e^t \) as \( y = e^{\arccos x} \), making it possible to integrate with respect to \( x \).
Understanding how to manipulate equations using inverse trigonometric functions is key to handling the heterogeneous elements of parametric and polar equations. Remaining comfortable with these functions enhances your ability to tackle complex integrals.