Question

In Exercises \(19-20\), a parametrized curve \(C\) is given that represents a thin wire with density \(\delta .\) Find the mass and center of mass of the thin wire. ). \(C: r(t)=\left\langle t-t^{2}, t^{2}-t^{3}, t^{3}-t^{4}\right\rangle\) for \(0 \leq t \leq 1\) \(\delta(x, y, z)=x+2 y+2 z .\) Use technology to approximate the value of each integral.

Short answer

Use technology to integrate and find approximate values for mass and center of mass coordinates.

Step-by-step solution

Set Up the Mass Integral

First, find the differential arc length along the curve \(C\). The vector derivative of \(r(t)\) is \(r'(t) = \left\langle 1 - 2t, 2t - 3t^2, 3t^2 - 4t^3 \right\rangle\). The magnitude of this derivative gives the element of arc length \(ds\), which is \(|r'(t)| = \sqrt{(1 - 2t)^2 + (2t - 3t^2)^2 + (3t^2 - 4t^3)^2}\). Compute \( |r'(t)| \) for integration.

Integrate for Mass

To find the mass \(M\), integrate the density function \(\delta(x, y, z)\) over the curve C. The mass integral is given by: \[ M = \int_{0}^{1} \delta(x(t), y(t), z(t)) \cdot |r'(t)| \, dt \]. Substitute \(x(t) = t - t^2\), \(y(t)=t^2 - t^3\), and \(z(t)=t^3 - t^4\) into the density function \(\delta(x, y, z) = (t - t^2) + 2(t^2 - t^3) + 2(t^3 - t^4)\) and evaluate the integral using technology.

Parametrize for Center of Mass

Calculate the center of mass, \(\overline{x}\), \(\overline{y}\), and \(\overline{z}\), using the formulas: \[ \overline{x} = \frac{1}{M} \int_0^1 x(t) \delta(x(t), y(t), z(t)) |r'(t)| \, dt \], \[ \overline{y} = \frac{1}{M} \int_0^1 y(t) \delta(x(t), y(t), z(t)) |r'(t)| \, dt \], \[ \overline{z} = \frac{1}{M} \int_0^1 z(t) \delta(x(t), y(t), z(t)) |r'(t)| \, dt \]. Substitute \(x(t), y(t), z(t)\) into these integrals and evaluate using technology.

Use Technology

Use a computer or calculator capable of symbolic computation to perform the integral evaluations from Steps 2 and 3. This will yield approximate values for the mass \(M\) and the coordinates of the center of mass \(\overline{x}, \overline{y}, \overline{z}\).

Key concepts

Understanding the Mass Integral

When discussing the mass of a wire or curve in calculus, we refer to the mass integral, which combines elements of geometry and physics.
For a parametrized curve, the mass, denoted as \( M \), can be calculated by integrating the density function along the path of the curve. The curve is described by a set of parametric equations, and its shape crucially affects the calculation.

Here's the process broken down:

• The first step is determining the differential arc length \( ds \), essential for computing the integral over the curve.
• This involves taking the derivative of the parametric vector \( r(t) \) to find \( r'(t) \). The magnitude of this derivative, \( |r'(t)| \), gives \( ds \).
• To find mass \( M \), integrate the density function \( \delta(x, y, z) \) multiplied by \( |r'(t)| \) over the interval \([0, 1]\), reflecting the curve's entire length.

In practical applications, exact values might be challenging to obtain manually. To help, technology such as symbolic computation on calculators or computers is often used.

Finding the Center of Mass

The center of mass, similar to a balance point, is where the mass of a body or object is concentrated. For a wire represented by a parametrized curve, the center of mass involves integrating each coordinate with the density function.
This tells us how mass is distributed across the geometry of the curve.

To calculate the center of mass's coordinates \( (\overline{x}, \overline{y}, \overline{z}) \), integrate using the formulas:

• \( \overline{x} = \frac{1}{M} \int_0^1 x(t) \delta(x(t), y(t), z(t)) |r'(t)| \, dt \)
• \( \overline{y} = \frac{1}{M} \int_0^1 y(t) \delta(x(t), y(t), z(t)) |r'(t)| \, dt \)
• \( \overline{z} = \frac{1}{M} \int_0^1 z(t) \delta(x(t), y(t), z(t)) |r'(t)| \, dt \)

The variables \( x(t) \), \( y(t) \), and \( z(t) \) represent the coordinates of the curve at each point \( t \). Each of these requires substitution into the integral. Using computational tools helps manage these calculations, especially for complex density functions.

The Role of the Density Function

The density function \( \delta(x, y, z) \) is critical in defining how much mass exists at each point on the curve. In our example, the density function is \( x + 2y + 2z \), indicating how the mass changes with position.
This function is not constant, meaning its value changes according to the coordinates \((x, y, z)\). This variability means the distribution of mass is not uniform over the curve.

Here’s how the density function is applied:

• Substitute the parametric curves \( x(t), y(t), z(t) \) into the density function to obtain \( \delta(x(t), y(t), z(t)) = (t - t^2) + 2(t^2 - t^3) + 2(t^3 - t^4) \).
• With substitutions complete, use this density function for the mass integral and center of mass calculations.

Understanding the density’s impact allows for accurate calculation of mass and center of mass, critical in fields like physics, engineering, and robotics, where material distribution is central to structural analysis.