Problem 38 - Calculus Early Transcendentals

Question

Find the mass of a thin wire shaped in the form of the curve \(x=e^{t} \cos t, y=e^{t} \sin t(0 \leq t \leq 1)\) if the density function \(\delta\) is proportional to the distance from the origin.

Step-by-Step Solution

Answer

The mass of the wire is \( \frac{\sqrt{2} k}{2} (e^2 - 1) \).

Step 1: Parametrize the Curve

The curve is already given in parametric form with parameter \( t \), where \( x = e^t \cos t \) and \( y = e^t \sin t \). This describes the position of the wire in the plane for \( 0 \leq t \leq 1 \).

Step 2: Express the Density Function

The density function \( \delta(t) \) is proportional to the distance from the origin. This means \( \delta(t) = k \cdot \sqrt{x^2 + y^2} \) for some constant \( k \). Substitute the parameterizations of \( x \) and \( y \) to get \( \delta(t) = k \cdot \sqrt{(e^t \cos t)^2 + (e^t \sin t)^2} = k \cdot e^t \).

Step 3: Calculate the Differential Arc Length

The differential arc length \( ds \) is given by \( ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \). Compute \( \frac{dx}{dt} = e^t \cos t - e^t \sin t \) and \( \frac{dy}{dt} = e^t \sin t + e^t \cos t \). Substitute these into the formula for \( ds \).

Step 4: Simplify the Arc Length

Compute \( \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 = (e^t \cos t - e^t \sin t)^2 + (e^t \sin t + e^t \cos t)^2 = 2 e^{2t} \). So, \( ds = \sqrt{2 e^{2t}} \, dt = \sqrt{2} e^t \, dt \).

Step 5: Set Up the Integral for the Mass

The mass \( M \) is the integral of the density function over the length of the wire: \( M = \int_{0}^{1} \delta(t) \, ds \). Substitute \( \delta(t) = k e^t \) and \( ds = \sqrt{2} e^t \, dt \) into the integral to get \( M = \int_{0}^{1} k e^t \cdot \sqrt{2} e^t \, dt \).

Step 6: Evaluate the Integral

Combine and simplify the integrand: \( M = \sqrt{2} k \int_{0}^{1} e^{2t} \, dt \). Evaluate the integral: \( \int e^{2t} \, dt = \frac{1}{2} e^{2t} \). So, \( M = \sqrt{2} k \left[ \frac{1}{2} e^{2t} \right]_{0}^{1} = \sqrt{2} k \left( \frac{1}{2} e^2 - \frac{1}{2} \right) \).

Step 7: Final Expression for Mass

Simplify the expression: \( M = \frac{\sqrt{2} k}{2} (e^2 - 1) \). This gives the mass of the wire in terms of \( k \).

Key Concepts

Density Function

In this exercise, the density function (\(\delta(t)\)) is a crucial concept to determine the wire's mass. A density function indicates how mass is distributed along an object. Here, the density is variable and depends on the position of the wire in the plane.
The problem specifies that the density function is proportional to the distance from the origin. This means that the farther a point on the wire is from the origin, the higher its mass density will be. Mathematically, proportionality can be expressed as:

• \(\delta(t) = k \cdot \text{distance from origin} = k \cdot \sqrt{x^2 + y^2}\)

where \(k\) is a constant of proportionality. In our specific case with parametric equations given by \(x = e^t \cos t\) and \(y = e^t \sin t\), substituting gives:

• \(\delta(t) = k e^t\)

This functional form will be pivotal when calculating the mass of the wire.

Arc Length

Arc length is a fundamental concept when calculating dimensions along a curve in parametric form. It helps to measure the extent of a curve or path. The differential arc length (\(ds\)) expresses an infinitesimally small segment of the curve.
Given the parametric equations for the curve, the derivative components are:

• \(\frac{dx}{dt} = e^t \cos t - e^t \sin t\)
• \(\frac{dy}{dt} = e^t \sin t + e^t \cos t\)

The formula for the differential arc length is given by:

• \(ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\)

By computing these derivatives and combining them, we find:

• \((\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 = 2 e^{2t}\)

Thus, the differential arc length becomes:

• \(ds = \sqrt{2} e^t \, dt\)

This small segment (\(ds\)) is crucial for evaluating the integral that results in the wire's mass.

Mass of a Wire

The mass of a wire distributed over a curve requires integrating the density function along the arc length. By combining these components, the mass (\(M\)) of the wire can be calculated by evaluating the integral of the density over its length:

• \(M = \int_{0}^{1} \delta(t) \, ds\)

In our problem, the density function is \(\delta(t) = k e^t\), and the differential arc length is \(ds = \sqrt{2} e^t \, dt\). Substituting these into the integral gives:

• \(M = \sqrt{2} k \int_{0}^{1} e^{2t} \, dt\)

Now, evaluating the integral for \(e^{2t}\):

• \(\int e^{2t} \, dt = \frac{1}{2} e^{2t}\)

Calculate from \(t = 0\) to \(t = 1\):

• \(M = \sqrt{2} k \left( \frac{1}{2} e^2 - \frac{1}{2} \right)\)

Finally, we express the mass of the wire in terms of the proportionality constant \(k\):

• \(M = \frac{\sqrt{2} k}{2} (e^2 - 1)\)

The result provides how mass is distributed along the wire based on its configuration and density properties.