Problem 35 - Thomas Calculus

Question

Mass of wire with variable density Find the mass of a thin wire lying along the curve \(\mathbf{r}(t)=\sqrt{2} t \mathbf{i}+\sqrt{2} t \mathbf{j}+\left(4-t^{2}\right) \mathbf{k}\) \(0 \leq t \leq 1,\) if the density is (a) \(\delta=3 t\) and (b) \(\delta=1 .\)

Step-by-Step Solution

Answer

Mass for (a) is \(4\sqrt{2} - 2\), and for (b) is \(\frac{4\sqrt{2}}{3} - \frac{2}{3}\).

Step 1: Understand the Problem

We need to find the mass of the wire along the curve \(\mathbf{r}(t) = \sqrt{2}t \mathbf{i} + \sqrt{2}t \mathbf{j} + (4-t^2) \mathbf{k}\) for \(0 \leq t \leq 1\) with given density functions \(\delta = 3t\) and \(\delta = 1\). The mass \(m\) of a wire with variable density \(\delta(t)\) is calculated by the integral \(m = \int_{a}^{b} \delta(t) \cdot \|\mathbf{r}'(t)\| \ dt\).

Step 2: Calculate the Derivative of r(t)

First, find the derivative of \(\mathbf{r}(t)\):\[\mathbf{r}'(t) = \frac{d}{dt}(\sqrt{2}t \mathbf{i}) + \frac{d}{dt}(\sqrt{2}t \mathbf{j}) + \frac{d}{dt}((4-t^2) \mathbf{k})\]Calculation:\[\mathbf{r}'(t) = \sqrt{2} \mathbf{i} + \sqrt{2} \mathbf{j} - 2t \mathbf{k}\]

Step 3: Compute the Magnitude of r'(t)

The magnitude of \(\mathbf{r}'(t)\) is given by:\[\|\mathbf{r}'(t)\| = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2 + (-2t)^2} = \sqrt{2 + 2 + 4t^2} = \sqrt{4 + 4t^2} = 2\sqrt{1 + t^2}\]

Step 4: Set Up Integral for Each Density Function

For \(\delta(t) = 3t\), the mass is:\[m_1 = \int_{0}^{1} 3t \cdot 2\sqrt{1 + t^2} \ dt\]For \(\delta(t) = 1\), the mass is:\[m_2 = \int_{0}^{1} 2\sqrt{1 + t^2} \ dt\]

Step 5: Solve the Integral for (a) δ=3t

Evaluate the integral:\[m_1 = 2 \int_{0}^{1} 3t \sqrt{1 + t^2} \ dt\]Let \( u = 1 + t^2 \) then \( du = 2t \, dt \), changing bounds, when \( t = 0, u = 1 \) and when \( t = 1, u = 2 \):\[m_1 = 3 \int_{1}^{2} \sqrt{u} \, du = 3 \left[ \frac{2}{3} u^{3/2} \right]_{1}^{2} = 3 \left( \frac{2}{3} (2^{3/2}) - \frac{2}{3} (1^{3/2}) \right)\]Calculate:\[m_1 = 3 \left( \frac{2}{3} \times 2\sqrt{2} - \frac{2}{3} \times 1 \right) = 3 \left( \frac{4\sqrt{2} - 2}{3} \right)\]Finally simplify:\[m_1 = 4\sqrt{2} - 2\]

Step 6: Solve the Integral for (b) δ=1

Evaluate the integral:\[m_2 = 2 \int_{0}^{1} \sqrt{1 + t^2} \ dt\]Use substitution \(u = 1 + t^2\), \(du = 2t \, dt\); the substituted integral:\[m_2 = \int_{1}^{2} \sqrt{u} \, du = \left[ \frac{2}{3}u^{3/2} \right]_{1}^{2} = \frac{2}{3} \left( 2^{3/2} - 1^{3/2} \right)\]Calculate:\[m_2 = \frac{2}{3} (2\sqrt{2} - 1)\]Finally simplify:\[m_2 = \frac{4\sqrt{2}}{3} - \frac{2}{3}\]

Step 7: Conclude the Masses

For density \(\delta = 3t\), the mass \(m_1\) is \(4\sqrt{2} - 2\). For density \(\delta = 1\), the mass \(m_2\) is \(\frac{4\sqrt{2}}{3} - \frac{2}{3}\).

Key Concepts

Variable Density

When dealing with real-world physical objects, such as wires or beams, the density can often change along their length. This is what's known as **variable density**. Unlike a uniform density, which remains constant, variable density changes and impacts the weight or mass of different sections.
In our exercise, the wire followed a path along a curve defined by \( \mathbf{r}(t) = \sqrt{2}t \mathbf{i} + \sqrt{2}t \mathbf{j} + (4-t^2) \mathbf{k} \).
The density functions provided were \( \delta = 3t \) and \( \delta = 1 \).

To find out the mass of the wire, we integrate the density function over the length of the wire times the magnitude of the derivative of the curve.
This integration sums up the contributions to the total mass at each point along the curve. It's a straightforward procedure once you understand the idea of density changing with location (in this case, represented by the parameter \( t \)).

• Understanding these concepts helps in applying correct formulas and solving real physical problems.
• The concept of variable density is crucial when designing structures or materials where weight distribution can significantly affect performance.

Integral Calculus

Integral calculus is vital in solving problems involving accumulation of quantities, such as area, volume, and in this case, mass.
The general process involves setting up an integral that sums up small changes (area, volume, etc.) over an interval.
In our problem, you're calculating the mass by integrating the density function multiplied by the curve's rate of change.

Mathematics tells us that if we have a curve defined by a vector function \( \mathbf{r}(t) \), the mass \( m \) is given by:
\[ m = \int_{a}^{b} \delta(t) \cdot \| \mathbf{r}'(t) \| \ dt \]
This allows for the calculation of a continuously changing quantity over a range. The specifics involve:

• Calculating \( \mathbf{r}'(t) \), the derivative of the position vector.
• Finding \( \| \mathbf{r}'(t) \| \), the magnitude representing the rate of traversal along the wire.
• Setting up the integral with respective limits and density function.

Integral calculus thus provides the tools necessary to analyze and solve complex real-world problems related to change and accumulation.

Vector Calculus

Very often in physics and engineering, we deal with vectors rather than scalars. That's where **vector calculus** becomes essential.

In our task, the position of the wire was given as a vector \( \mathbf{r}(t) \).
This vector encapsulates the x, y, and z components of the wire's path in a 3D space, allowing us to describe movements and interactions in multiple dimensions concisely.

A key concept here is finding the derivative of this vector, \( \mathbf{r}'(t) \).

• This derivative provides us with a tangent vector to the curve, representing the wire's direction or path rate of change concerning time \( t \).
• The magnitude of this tangent vector \( \| \mathbf{r}'(t) \| \) is crucial to figuring out how much 'travel' happens over small time intervals. This helps solve physics problems by capturing both the direction and speed of movement.

Understanding vector calculus allows us to extend these ideas to more complex surfaces and flows, facilitating solutions in fields like fluid dynamics, electromagnetism, and computer graphics.