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 square units.

Step 1: Parameterize the Region

The curve is given in parametric form where \( x = e^{2t} \) and \( y = e^{-t} \). For \( t \) values running from \( 0 \) to \( \ln 5 \), we need to express the limits for \( x \). The values of \( x \) are:1. At \( t = 0 \), \( x = e^{2 \times 0} = 1 \).2. At \( t = \ln 5 \), \( x = e^{2 \ln 5} = 25 \). Thus, \( x \) varies from 1 to 25.

Step 2: Set Up the Integral for Area

In parametric equations, the area \( A \) between a parametric curve and the x-axis from \( t = a \) to \( t = b \) is given by:\[ A = \int_{a}^{b} y \frac{dx}{dt} \, dt \]For our curve:- \( y = e^{-t} \)- \( \frac{dx}{dt} = \frac{d}{dt}(e^{2t}) = 2e^{2t} \)Thus, the integral becomes:\[ A = \int_{0}^{\ln 5} e^{-t} \cdot 2e^{2t} \, dt \]

Step 3: Simplify and Calculate the Integral

Simplify the integral expression obtained:\[ A = \int_{0}^{\ln 5} 2e^{2t-t} \, dt \]This simplifies to:\[ A = \int_{0}^{\ln 5} 2e^{t} \, dt \]Now calculate the integral:\[ A = 2 \int_{0}^{\ln 5} e^{t} \, dt = 2 \left[ e^{t} \right]_{0}^{\ln 5} \]Evaluate the bounds:\[ = 2 \left( e^{\ln 5} - e^{0} \right) = 2 (5 - 1) \]\[ = 2 \times 4 = 8 \]

Step 4: Sketch the Curve

To sketch the curve in the xy-plane:1. Recognize the parametric form describes a portion of a parabola.2. As \( t \) increases from 0 to \( \ln 5 \), the x-coordinate moves from 1 to 25, and the y-coordinate moves from 1 to \( \frac{1}{5} \).3. Sketch should roughly show a curve in the first quadrant starting at (1,1) and ending at (25, 0.2).The area under this path and above the x-axis from x = 1 to x = 25 is the region whose area we calculated.

Key Concepts

Area of Parametric Curves

Finding the area under parametric curves is a fascinating and useful application of calculus concepts. A parametric curve is defined by separate equations for the x and y coordinates, each as a function of a third variable, usually denoted as \(t\). This approach allows complex curves to be described easily. To find the area between such a curve and the x-axis, you use a specialized integral formula. The area \(A\) is given by:

• \( A = \int_{a}^{b} y \frac{dx}{dt} \, dt \)

Here, \(y\) is the parametric equation for the y-coordinate, and \( \frac{dx}{dt} \) is the derivative of the parametric equation for the x-coordinate with respect to \(t\). This allows us to find the exact area of the region bounded by these curves from \(t = a\) to \(t = b\). You simply plug the appropriate equations into the integral formula and solve for \(A\).

Integral Calculus

Integral calculus is a branch of mathematics that focuses on finding the accumulated total of quantities, often called the integral. It's a fundamental concept that helps in determining areas under curves, volumes, and other related measures. At its core, it deals with two main operations: finding derivatives (differential calculus) and finding integrals (integral calculus).

• Definite integrals help calculate exact areas and other quantities between specified limits.
• Indefinite integrals find antiderivatives, which are more general solutions to differential equations.

In this exercise, we use a definite integral to find the area under a parametric curve. The process involves plugging in the limits of integration and the parametric expressions into the integral formula. This highlights integral calculus's power and versatility in solving geometrical problems and other applications.

Parametric Equations

Parametric equations allow us to express a curve by defining its coordinates using a third variable, typically \(t\). This can be particularly useful for representing complex curves that don't easily fit into the standard Cartesian format \(y = f(x)\). With parametric equations, both x and y are described separately:

• \(x = f(t)\)
• \(y = g(t)\)

These equations enable us to create more versatile and intricate curves. In the given problem, the parametric equations \(x = e^{2t}\) and \(y = e^{-t}\) describe a curve in the xy-plane. By varying \(t\) over a particular range, we trace out a path in the plane that can be analyzed further, in this case, finding the area with respect to the x-axis.

Definite Integrals

Definite integrals are a powerful tool in calculus used to compute the exact accumulation of some quantity over a particular interval. When calculated, a definite integral gives you a numeric value that represents this total accumulation, such as area under a curve.

• It is expressed as \(\int_{a}^{b} f(x) \, dx\)
• The limits \(a\) and \(b\) are boundaries over which the function \(f(x)\) is integrated.

In our parametric curve, the definite integral \(\int_{0}^{\ln 5} e^{-t} \cdot 2e^{2t} \, dt\) was used to find the area under the curve from \(t = 0\) to \(t = \ln 5\). By evaluating this integral, we effectively computed the total area between the curve, the x-axis, and the boundaries at \(x = 1\) and \(x = 25\), showcasing how definite integrals provide precise solutions to geometric problems.