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 approximately 1.828.

Step 1: Parameterize the Wire

The position vector of the wire is given by \( \mathbf{r}(t) = (t^2 - 1) \mathbf{j} + 2t \mathbf{k} \) where \( t \) is the parameter of the wire ranging from \( t = 0 \) to \( t = 1 \). The wire curve is therefore fully described over this parameter range.

Step 2: Find the Derivative

Compute the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \) to find the tangent vector \( \mathbf{r}'(t) \). This gives us \( \mathbf{r}'(t) = \frac{d}{dt}((t^2 - 1) \mathbf{j} + 2t \mathbf{k}) = (2t) \mathbf{j} + 2 \mathbf{k} \).

Step 3: Compute the Magnitude of the Derivative

Find the magnitude of the derivative, \( \|\mathbf{r}'(t)\| = \sqrt{(2t)^2 + 2^2} = \sqrt{4t^2 + 4} = \sqrt{4(t^2 + 1)} = 2\sqrt{t^2 + 1} \). This magnitude represents the differential length element of the wire.

Step 4: Set Up the Mass Integral

The mass of the wire can be found by integrating the product of the density function \( \delta(t) = \frac{3}{2}t \) and the differential arc length along the wire. This integral is \( m = \int_{0}^{1} \delta(t) \|\mathbf{r}'(t)\| \, dt = \int_{0}^{1} \frac{3}{2} t \cdot 2\sqrt{t^2 + 1} \, dt \).

Step 5: Simplify the Integral Expression

The integral becomes \( m = \int_{0}^{1} 3t \sqrt{t^2 + 1} \, dt \). This requires substitution for solving.

Step 6: Substitution and Integration

Let \( u = t^2 + 1 \), then \( du = 2t \, dt \) implying \( t \, dt = \frac{1}{2} du \). When \( t = 0, u = 1 \) and when \( t = 1, u = 2 \). The integral becomes \( m = 3 \int_{1}^{2} \frac{1}{2} \sqrt{u} \, du = \frac{3}{2} \int_{1}^{2} u^{1/2} \, du \).

Step 7: Evaluate the Integral

Compute the integral: \( \frac{3}{2} \int u^{1/2} \, du = \frac{3}{2} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{2} = \left[ u^{3/2} \right]_{1}^{2} \). The evaluated integral is \( (2^{3/2} - 1^{3/2}) = (2\sqrt{2} - 1) \).

Step 8: Calculate the Mass

The mass of the wire is \( m = 2\sqrt{2} - 1 \), which simplifies to approximately 1.828.

Key Concepts

Density Function

In this context, a density function helps determine how much mass is distributed along the wire at any given point. It tells us how dense or heavy the wire is as it twists along the curve. For our exercise, the density function is given as \( \delta(t) = \frac{3}{2}t \), which implies that the density increases linearly with the parameter \( t \). To find out the total mass, you need to multiply this density with the infinitesimal length of the wire at each point along the curve and then integrate over the entire length. This step links the density of the wire to its total mass.

Arc Length Integral

The arc length integral is a vital concept for finding the mass of a wire when you know its density. It's essentially the way to accumulate the density along the entire length of the wire. The arc length integral can be written as:

• Calculate the derivative of the wire's position vector, \( \mathbf{r}'(t) \).
• Find the magnitude of this derivative, \( \|\mathbf{r}'(t)\| \).
• Integrate the density function multiplied by this magnitude over the given range of \( t \).

In our exercise, the integral to find mass is \( \int_{0}^{1} 3t \sqrt{t^2 + 1} \, dt \). This represents summing up little pieces of density times length along the wire, giving the total mass.

Tangent Vector

The tangent vector is a direction vector that points along the curve at any given point. It shows where the curve is heading and helps calculate changes along the wire. You obtain it by taking the derivative of the position vector of the curve. From the exercise, the tangent vector \( \mathbf{r}'(t) = (2t) \mathbf{j} + 2 \mathbf{k} \) indicates how the wire's position changes as \( t \) changes. The tangent vector allows us to measure how much of the path is covered as we increment \( t \), an essential step in determining the differential arc length.

Parametric Equations

Parametric equations provide a way to express geometric shapes in terms of parameters like time. For a curve in space, these equations will give the coordinates as functions of a parameter \( t \). In our exercise, the wire's path is described through \( \mathbf{r}(t) = (t^2 - 1) \mathbf{j} + 2t \mathbf{k} \). Here, \( t \) defines each point along the wire. Understanding parametric equations helps you break down complex curves into manageable mathematical expressions. This is important for further calculations like finding tangent vectors or integrating density functions over a curve.