Problem 33 - Thomas Calculus

Question

Mass of a wire Find the mass of a wire that lies along the curve \(\mathbf{r}(t)=\left(t^{2}-1\right) \mathbf{j}+2 t \mathbf{k}, 0 \leq t \leq 1,\) if the density is \(\delta=(3 / 2) t\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Parameters

To find the mass of a wire along a curve, we need to understand the parameters involved. The curve is given by the vector function \( \textbf{r}(t) = (t^2 - 1) \mathbf{j} + 2t \mathbf{k} \), and the density function is \( \delta(t) = \frac{3}{2}t \). We need to find the integral of the density along the curve from \( t = 0 \) to \( t = 1 \).

Step 2: Calculate the Derivative of the Curve

First, calculate the derivative of the vector function \( \mathbf{r}(t) \):\[ \frac{d\mathbf{r}}{dt} = \frac{d}{dt}((t^2 - 1) \mathbf{j} + 2t \mathbf{k}) = 2t \mathbf{j} + 2 \mathbf{k} \].

Step 3: Find the Magnitude of the Derivative

The magnitude \( \left\| \frac{d\mathbf{r}}{dt} \right\| \) gives the differential arc length:\[ \left\| \frac{d\mathbf{r}}{dt} \right\| = \sqrt{(2t)^2 + 2^2} = \sqrt{4t^2 + 4} = \sqrt{4(t^2 + 1)} = 2\sqrt{t^2 + 1} \].

Step 4: Set Up the Integral for Mass

The mass \( m \) of the wire is the integral of the density function multiplied by the differential arc length:\[ m = \int_{0}^{1} \delta(t) \cdot \left\| \frac{d\mathbf{r}}{dt} \right\| \, dt = \int_{0}^{1} \left(\frac{3}{2}t\right) \cdot 2\sqrt{t^2 + 1} \, dt = 3 \int_{0}^{1} t\sqrt{t^2 + 1} \, dt \].

Step 5: Integrate the Function

To solve \( \int_{0}^{1} t\sqrt{t^2 + 1} \, dt \), use a substitution method. Let \( u = t^2 + 1 \), then \( du = 2t \, dt \), or \( t \, dt = \frac{1}{2} du \). When \( t = 0 \), \( u = 1 \). When \( t = 1 \), \( u = 2 \). The integral becomes:\[ \int_{1}^{2} \frac{1}{2} \sqrt{u} \, du \].Calculate it:\[ = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} \Bigg|_{1}^{2} = \frac{1}{3} \left[ 2^{3/2} - 1^{3/2} \right] = \frac{1}{3} (2\sqrt{2} - 1) \].

Step 6: Calculate the Final Mass

Multiply the result by 3 to find the mass:\[ m = 3 \cdot \frac{1}{3} (2\sqrt{2} - 1) = 2\sqrt{2} - 1 \].

Key Concepts

Vector Calculus

Vector calculus plays a critical role in understanding physical phenomena in multiple dimensions. It's an extension of ordinary calculus that deals with vector fields, which are functions that assign a vector to each point in space. This branch of mathematics simplifies the processes involved in physics and engineering, making it essential for many applications.

In the context of finding the mass of a wire, vector calculus helps describe the wire's geometry through a vector function, like the curve \[ \mathbf{r}(t) = (t^2 - 1) \mathbf{j} + 2t \mathbf{k} \]. This vector function assigns a point in space for each value of \( t \), effectively tracing out the path of the wire.

To compute the mass, we must consider not only the density but also the length of the wire along the curve. This is where the derivative and magnitude of the vector function come into play, as they help determine the differential arc length needed for integration.

Density Function

The density function represents how mass is distributed along the wire. In our exercise, the density is given by \[ \delta(t) = \frac{3}{2}t \]. This means the density varies linearly with \( t \), indicating that the mass per unit length of the wire increases as \( t \) increases.

To find the total mass of the wire, you need to integrate the density function over the curve defined by \( t \) from 0 to 1. The role of the density function here is central because it determines how much each segment of the wire contributes to the total mass.

The calculation involves multiplying the density by the differential arc length obtained from the curve's derivative, ensuring that the total mass is correctly accounted based on varying density.

Curve Parametrization

Curve parametrization involves expressing a curve as a vector function of a single variable, like \( t \). This process is essential in vector calculus because it simplifies the manipulation and integration of curves in space.

In our exercise, the curve is parametrized as \[ \mathbf{r}(t) = (t^2 - 1) \mathbf{j} + 2t \mathbf{k} \] over the interval \( 0 \leq t \leq 1 \). This representation turns a potentially complex spatial problem into a one-dimensional integral over \( t \).

Parametrization also provides flexibility in analysis, allowing one to focus on the functional relationships inherent in the curve. By working with these parametric equations, you can easily compute derivatives, find arc lengths, and solve physical problems such as finding the mass of a wire by integrating along its path.