Problem 101 - Calculus Volume 3

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 \(x=0\).

Step-by-Step Solution

Answer

The bounded area is 1.

Step 1: Understanding the Problem

We need to find the area bounded by the parametric curve \(x = \cos t\) and \(y = e^t\) for \(0 \leq t \leq \frac{\pi}{2}\), along with the lines \(y = 1\) and \(x = 0\). We will integrate with respect to the parameter \(t\).

Step 2: Find Intersection with Line y = 1

Since the curve is represented by \(y = e^t\), setting \(e^t = 1\) gives us \(t = 0\). This shows that at \(t = 0\), the point on the curve where it intersects \(y=1\) is \(x = \cos(0) = 1\).

Step 3: Setup Area Integral

The region is bounded by the curve on the right (given by \(x = \cos t\)) from \(t = 0\) to \(t'\), where \(t'\) is the point of intersection with \(x = 0\). Also, vertically from \(y = 1\) to \(y = e^t\). We compute the area as \(\int_0^{\frac{\pi}{2}} [\cos t - 0] \ dt - \int_0^{t'} [\cos t - 0] \ dt\), where \(t'\) is the \(t\) value where \(x = \cos t = 0\), i.e., \(t = \frac{\pi}{2}\).

Step 4: Integrate with Respect to t

Calculate the definite integral from \(t=0\) to \(t=\frac{\pi}{2}\) of \(\cos t\) with respect to the parameter \(t\):\[\int_0^{\frac{\pi}{2}} \cos t \ dt = [\sin t]_0^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(0) = 1 - 0 = 1.\]

Step 5: Subtract Inner Area Between Curve and y = 1

Subtract the area bounded from \(y=1\) to \(y = e^t\) over interval \(t = 0\) to \(t=0 \) (no area, \(y=1\) and \(y = e^0 = 1\)), which results in 0.

Step 6: Calculate Total Bounded Area

The calculation results in:- Area above \(y = 1\) from \(t=0\) to \(t=\frac{\pi}{2}\) is 1.- Subtracting any potential "negative plates" gives total area = 1.

Key Concepts

Area Calculation

To calculate the area bounded by a parametric curve, we use integration. The idea is to integrate the function that defines one side of the bounded region with respect to its parameter, instead of directly using standard Cartesian coordinates.
In this problem, we look at the curve defined by the equations \(x = \cos t\) and \(y = e^t\), and we seek the area enclosed by this curve, along with the lines \(y = 1\) and \(x = 0\).
The steps involve computing the integral of \(\cos t\) from \(t = 0\) to \(t = \frac{\pi}{2}\). This gives us the area under the curve \(x = \cos t\) and above \(x=0\).
Next, we realize there are no negative areas involved, as the line \(y = e^t\) never dips below \(y = 1\) between the bounds of integration on the \(t\) scale, reinforcing our integral calculation. Through these steps, we ultimately find the total area to be 1.

Definite Integrals

The definite integral is a fundamental concept in calculus. It helps find the area under a curve between specific limits. In this exercise, the definite integral is used to calculate the area under the parametric curve \(x = \cos t\) from \(t = 0\) to \(t = \frac{\pi}{2}\).
This integration results in the integral \( \int_0^{\frac{\pi}{2}} \cos t \ dt \), which yields \([\sin t]_0^{\frac{\pi}{2}}\).

• The solution of the integral comes out to be \(1\), revealing the area under the curve \(x = \cos t\) over the interval.
• Integrals can be visualized as the sum of infinitesimally small rectangles under the curve, which when added up give the total space, or area, beneath it.

Definite integrals also account for literal boundaries — as defined by these \(t\) limits — ensuring calculations stay precise and applicable only to specific sections of a graph.

Bounded Regions

In mathematics, a bounded region is a confined area formed by intersecting curves or lines. When dissecting a problem like this exercise, understanding bounded regions is essential.
Let’s break it down:

• The curve \(x = \cos t\), varying over \(0 \leq t \leq \frac{\pi}{2}\), forms the right-hand boundary of our region.
• The horizontal line \(y = 1\) provides a constant lower limit to the region we are interested in.

The problem ensures the area calculation only involves portions visually enclosed within these boundaries, simplifying the integral evaluations. Such region constraints prevent calculations from spiraling out to include unwanted sections of the curve or plane.
Understanding these boundaries is crucial because they determine the entire configuration and feasibility of calculating areas with parametric equations and integrations. By ensuring all dimensions of the enclosure are clear, you can calculate accurately and avoid simplification errors influenced by an incorrect setup.