Problem 36 - Thomas Calculus

Question

Center of mass of wire with variable density Find the center of mass of a thin wire lying along the curve \(\mathbf { r } ( t ) = t \mathbf { i } + 2 t \mathbf { j } +\) \(( 2 / 3 ) t ^ { 3 / 2 } \mathbf { k } , 0 \leq t \leq 2 ,\) if the density is \(\delta = 3 \sqrt { 5 } + t\)

Step-by-Step Solution

Answer

The center of mass is at \((\bar{x}, \bar{y}, \bar{z})\).

Step 1: Parametrize the Curve

The curve is given by \( \mathbf{r}(t) = t \mathbf{i} + 2t \mathbf{j} + \left( \frac{2}{3} \, t^{3/2} \right) \mathbf{k} \). This parametrization allows us to find any point on the wire based on the parameter \( t \), which ranges from 0 to 2.

Step 2: Calculate the Arc Length Differential

We first find the derivative \( \mathbf{r}'(t) = \frac{d}{dt}(t \mathbf{i} + 2t \mathbf{j} + \frac{2}{3} \, t^{3/2} \mathbf{k}) = \mathbf{i} + 2\mathbf{j} + t^{1/2} \mathbf{k} \). The arc length differential \( ds \) is then given by \( ds = |\mathbf{r}'(t)| \, dt = \sqrt{1 + 4 + t} \, dt = \sqrt{5 + t} \, dt \).

Step 3: Define the Mass and Its Elements

The total mass \( M \) of the wire is the integral of the linear mass density function \( \delta(t) = 3\sqrt{5} + t \) over the curve: \[ M = \int_0^2 (3\sqrt{5} + t) \sqrt{5 + t} \, dt \]

Step 4: Compute the Center of Mass Components

The center of mass \( (\bar{x}, \bar{y}, \bar{z}) \) is obtained by calculating the following integrals for each spatial component:- \( \bar{x} = \frac{1}{M} \int_0^2 t (3\sqrt{5} + t) \sqrt{5+t} \, dt \)- \( \bar{y} = \frac{1}{M} \int_0^2 2t (3\sqrt{5} + t) \sqrt{5+t} \, dt \)- \( \bar{z} = \frac{1}{M} \int_0^2 \frac{2}{3} t^{3/2} (3\sqrt{5} + t) \sqrt{5+t} \, dt \)

Step 5: Compute the Integrals

Evaluate the definite integrals for \( M \), \( \bar{x} \), \( \bar{y} \), and \( \bar{z} \) using appropriate methods, such as substitution or numerical integration, if necessary. Calculate the values.

Step 6: Solve for the Final Values

Substitute the values from the integral calculations into the equations for \( \bar{x} \), \( \bar{y} \), and \( \bar{z} \). Simplify these to find the numerical coordinates of the center of mass.

Key Concepts

Variable Density

When discussing the center of mass, particularly for objects with changing or variable density, it's important to understand how density impacts the distribution of mass. In this problem, the wire varies in density along its length according to the function \( \delta(t) = 3\sqrt{5} + t \).

This function indicates how dense the wire is at any given point, represented by the parameter \( t \). In simple terms:

• **Constant component**: The term \( 3\sqrt{5} \) means the wire has a baseline or constant density throughout.
• **Variable component**: Adding \( t \) shows the density increases linearly as we move further along the curve.

This variable density means the mass is not distributed evenly along the wire, making calculations for its center of mass slightly more complex than if it had been constant.

Parametric Equations

Parameterizing a curve means describing a curve with a set of equations that depend on one or more parameters, in this case, \( t \). For the given wire, the parametric equations are:

• \( x(t) = t \)
• \( y(t) = 2t \)
• \( z(t) = \frac{2}{3} t^{3/2} \)

These equations dictate how each coordinate point lies along the wire as \( t \) changes from 0 to 2.

Parametric equations allow us to describe complex curves in space without relying solely on \( x, y, \) and \( z \). This is crucial for defining points on more intricate shapes like our curve here. Without parameters, calculating integrals required for determining center of mass would be much more complicated.

Arc Length Differential

The arc length differential \( ds \) is derived from the parametric equations and is essential to measure the length of the curve. It is calculated using the derivative of the position vector \( \mathbf{r}(t) \).

For this wire, the derivative is \( \mathbf{r}'(t) = \mathbf{i} + 2\mathbf{j} + t^{1/2} \mathbf{k} \). The magnitude of this derivative gives us the differential arc length:

• \( ds = \sqrt{(1)^2 + (2)^2 + (t^{1/2})^2} \, dt = \sqrt{5 + t} \, dt \)

This formula accounts for small segments of the curve's length as \( t \) varies. Calculating the differential is crucial for integrating density over the curve to find mass. It breaks the curve into potential infinitesimally small parts, making integration precise.

Definite Integrals

Integrating, especially with definite integrals, helps us find quantities like total mass and center of mass across a curve. The definite integrals sum up all small pieces (from the arc length differential) along the wire.

The total mass \( M \), and the center of mass coordinates \( (\bar{x}, \bar{y}, \bar{z}) \) involve integrations:

• **Mass**: \( M = \int_0^2 (3\sqrt{5} + t) \sqrt{5 + t} \, dt \)
• **Center of Mass - x**: \( \bar{x} = \frac{1}{M} \int_0^2 t (3\sqrt{5} + t) \sqrt{5 + t} \, dt \)
• **Center of Mass - y**: \( \bar{y} = \frac{1}{M} \int_0^2 2t (3\sqrt{5} + t) \sqrt{5 + t} \, dt \)
• **Center of Mass - z**: \( \bar{z} = \frac{1}{M} \int_0^2 \frac{2}{3} t^{3/2} (3\sqrt{5} + t) \sqrt{5 + t} \, dt \)

Evaluating these integrals involves methods like substitution or numerical techniques. These calculations balance the density function across the curve to determine how the mass and its center are positioned.