Problem 57 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the region between the curve \(x=e^{2 t}, y=e^{-t}\), and the \(x\)-axis from \(t=0\) to \(t=\ln 5\). Make a sketch.

Step-by-Step Solution

Answer

The area is 8.

Step 1: Parametric Equations

The given parametric equations are \( x = e^{2t} \) and \( y = e^{-t} \). These describe a curve in the \(xy\)-plane as \(t\) varies.

Step 2: Determine Bounds for x

The parameter \(t\) varies from \(0\) to \(\ln 5\). For these values, the range for \(x\) can be found: at \(t = 0\), \(x = e^{2 \cdot 0} = 1\), and at \(t = \ln 5\), \(x = e^{2 \ln 5} = 25\). Thus, \(x\) varies from 1 to 25.

Step 3: Set Up Integral for Area

The area under the parametric curve and above the \(x\)-axis can be found using \( A = \int_{t_1}^{t_2} y \frac{dx}{dt} \, dt \). First, find \( \frac{dx}{dt} \): since \( x = e^{2t} \), \( \frac{dx}{dt} = 2e^{2t} \). Substitute \(y = e^{-t}\) into the formula, leading to \( A = \int_{0}^{\ln 5} e^{-t} \cdot 2e^{2t} \, dt \).

Step 4: Simplify the Integral

Simplify the integrand: \( e^{-t} \cdot 2e^{2t} = 2 \cdot e^{t} \). Now the integral becomes \( \int_{0}^{\ln 5} 2 \cdot e^{t} \, dt \).

Step 5: Evaluate the Integral

Evaluate the integral: \( \int 2 \cdot e^{t} \, dt = 2 \cdot e^{t} \). Thus, \( A = 2 \cdot \left[ e^{t} \right]_{0}^{\ln 5} = 2 \cdot (e^{\ln 5} - e^{0}) = 2 \cdot (5 - 1) = 8 \).

Step 6: Make a Sketch

Draw the curve on the \(xy\)-plane. Plot a smooth curve passing through the points obtained from substituting \(t=0\) (\(x=1, y=1\)) and \(t=\ln 5\) (\(x=25, y=\frac{1}{5}\)). The curve is above the \(x\)-axis within these bounds.

Key Concepts

Parametric Equations

In calculus, parametric equations are used to define a curve in a plane or space using a parameter. In our exercise, the curve is described by the parametric equations:

• \( x = e^{2t} \)
• \( y = e^{-t} \)

These equations imply that both coordinates \(x\) and \(y\) depend on another variable \(t\), called a parameter. When \(t\) changes, it traces a path as a pair of values \((x, y)\) in the plane.
A common application of parametric equations is understanding motion along a path. Different functions for \(x\) and \(y\) allow representation of more complex curves compared to standard Cartesian equations. In this context, they provide a useful method for solving real-world problems where variables are interdependent.
To visualize, imagine drawing the curve as \(t\) goes from 0 to \(\ln 5\). Each value of \(t\) generates a point on the curve, forming a smooth path in the plane.

Area Under a Curve

Finding the area under a curve often involves integration. For parametric curves, this process can be done via a specific formula:
\[ A = \int_{t_1}^{t_2} y \frac{dx}{dt} \, dt \]
where \( \frac{dx}{dt} \) represents the rate of change of \(x\) with respect to \(t\). In our problem, this involves:

• \( \frac{dx}{dt} = 2e^{2t} \)

Substituting the parametric equation \(y = e^{-t}\) into the formula gives us the integral we solve to find the area.
The limits of integration are determined by the initial and final values of \(t\): from \(t = 0\) to \(t = \ln 5\). The result of this integration provides the area under the curve between these bounds.
This method is essential in scenarios where the curve does not meet the x-axis at standard intervals, or is defined via parameters rather than a single function.

Integration

Integration is a fundamental concept in calculus used to find areas under curves, amongst other applications. In the context of parametric equations, integration requires transforming the curve's description into an integral expression.
For our curve, the integral for the area from \(t = 0\) to \(t = \ln 5\) is:

• \( \int_{0}^{\ln 5} e^{-t} \, 2e^{2t} \, dt \)

Simplifying the integrand:
\[ e^{-t} \, 2e^{2t} = 2 \, e^{t} \]
reduces the complexity of the integration process. This leaves us with:
\[ \int_{0}^{\ln 5} 2 \, e^{t} \, dt \]
The antiderivative of \(2 \, e^{t}\) is \(2 \, e^{t}\). Evaluating this from 0 to \(\ln 5\) gives:

• \( 2 \, \left[ e^{t} \right]_{0}^{\ln 5} = 2 \, (5 - 1) = 8 \)

This calculation concludes with the area being 8, showing the beauty and efficiency of integration in solving real-world problems.