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Chapter 14: Problem 62

A thin wire represented by the smooth curve C with a density \(\rho\) (units of mass per length) has a mass \(M=\int_{C} \rho d s\) Find the mass of the following wires with the given density. $$C: \mathbf{r}(\theta)=\langle\cos \theta, \sin \theta\rangle, \text { for } 0 \leq \theta \leq \pi ; \rho(\theta)=2 \theta / \pi+1$$

Short Answer

Expert verified
Based on the given information, the mass of the thin wire defined by the curve C and density function 蟻(胃) is M=蟺.

Step by step solution

01

1. Define the Curve C and Density Function

Given the curve C is defined by the vector \(\mathbf{r}(\theta)=\langle\cos \theta, \sin \theta\rangle\) and the density function is given as \(\rho(\theta)=2 \theta / \pi+1\).
02

2. Find the Derivative of the Curve

To find the mass, we first need to calculate the derivative of the curve \(\mathbf{r}(\theta)\). The derivative, \(\mathbf{r}'(\theta)\), can be found as follows: $$\mathbf{r}'(\theta)=\frac{d\mathbf{r}(\theta)}{d\theta}= \left\langle\frac{d}{d\theta}(\cos \theta),\frac{d}{d\theta}(\sin\theta) \right\rangle= \left\langle -\sin\theta, \cos\theta \right\rangle$$
03

3. Calculate the Magnitude of the Derivative

Now we need to find the magnitude of the derivative, which we'll represent as \(||\mathbf{r}'(\theta)||\): $$||\mathbf{r}'(\theta)|| = \sqrt{(-\sin\theta)^2 + (\cos\theta)^2} = \sqrt{\sin^2\theta + \cos^2\theta}$$ Using the identity \(\sin^2\theta + \cos^2\theta = 1\), we get: $$||\mathbf{r}'(\theta)|| = \sqrt{1} = 1$$
04

4. Set up Mass Integral

Now we can find the mass by integrating the product of the density function and the magnitude of the derivative along the curve C: $$M= \int_{C} \rho(\theta) ||\mathbf{r}'(\theta)|| ds$$ Since we are given that \(0 \leq \theta \leq \pi\), and we found \(||\mathbf{r}'(\theta)|| = 1\), the mass integral can be expressed as: $$M=\int_0^\pi \rho(\theta) d\theta$$
05

5. Calculate the Mass Integral

To find the mass of the wire, we now need to solve the integral: $$M=\int_0^\pi \left(2 \theta / \pi+1\right) d\theta$$ We can integrate term-wise: $$M= \int_0^\pi\left(\frac{2\theta}{\pi}\right) d\theta+\int_0^\pi1d\theta$$ Computing these integrations: $$M= \left[\frac{\theta^2}{\pi}\right]_0^\pi + \left[\theta\right]_0^\pi =\frac{\pi^2}{\pi}+\pi-\frac{0^2}{\pi}= \pi+0$$ Therefore, the mass of the wire is: $$M = \pi$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Calculus
Vector Calculus is a branch of mathematics that deals with multi-dimensional spaces. It extends calculus into higher dimensions and involves vectors, which have both magnitude and direction. This makes it particularly useful in physics, engineering, and computer graphics, among other fields. In this exercise, vector calculus is used to describe the curve of a wire with a vector function. The vector function
  • defines the wire's path in a two-dimensional plane.
  • The curve is given by \(\mathbf{r}(\theta)=\langle\cos \theta, \sin \theta\rangle\)representing points in polar coordinates over an interval \([0, \pi]\).
This expresses how the curve changes as \(\theta\) varies, allowing calculations such as the arc length and mass of the wire.
Arc Length
Arc Length is a fundamental concept in calculus that measures the distance along a curve. Simply put, it's the length of a path traced by a moving point in space. To calculate the arc length of a curve given by a vector function, the magnitude of the derivative of the function is integrated over the curve's interval.
  • In this solution, the magnitude of \(\mathbf{r}'(\theta)\)is calculated to be 1, indicative of a unit circle.
  • This means each infinitesimal segment of the curve contributes equally to the total length.
Thanks to the unit circle identity \(\sin^2\theta + \cos^2\theta = 1\), we simplify our work in calculating arc lengths for standard curves.
This calculation step is crucial in determining other properties of curves, such as integral evaluation for physical quantities like mass.
Integration
Integration is the mathematical technique of finding the integral of a function, which can be thought of as the "summation" of infinitely many infinitely small factors. In calculus, it is used to find areas, volumes, central points, and many useful things. In this exercise, integration is vital for calculating the mass of the wire. The mass is given by the integral \(M=\int_C \rho(\theta) \ ||\mathbf{r}'(\theta)|| \, d\theta\).
  • Using the density function \(\rho(\theta) = \frac{2 \theta}{\pi} + 1\),we integrate over the interval \([0, \pi]\)
  • After separating the integral into simpler parts, \(\frac{2\theta}{\pi}\) and 1, and calculating intuitively, the result is \(\pi\), the total mass.
Integration in this context is used to accumulate density across the wire's length, giving us the overall mass. Understanding the integral and the process of simplification helps in breaking down complex calculus problems into manageable steps.

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Most popular questions from this chapter

Prove the following identities. Assume that \(\varphi\) is \(a\) differentiable scalar-valued function and \(\mathbf{F}\) and \(\mathbf{G}\) are differentiable vector fields, all defined on a region of \(\mathbb{R}^{3}\). $$\nabla(\mathbf{F} \cdot \mathbf{G})=(\mathbf{G} \cdot \nabla) \mathbf{F}+(\mathbf{F} \cdot \nabla) \mathbf{G}+\mathbf{G} \times(\nabla \times \mathbf{F})+\mathbf{F} \times(\nabla \times \mathbf{G})$$

Suppose a solid object in \(\mathbb{R}^{3}\) has a temperature distribution given by \(T(x, y, z) .\) The heat flow vector field in the object is \(\mathbf{F}=-k \nabla T,\) where the conductivity \(k>0\) is a property of the material. Note that the heat flow vector points in the direction opposite that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is \(\nabla \cdot \mathbf{F}=-k \nabla \cdot \nabla T=-k \nabla^{2} T\) (the Laplacian of \(T\)). Compute the heat flow vector field and its divergence for the following temperature distributions. $$T(x, y, z)=100 e^{-\sqrt{x^{2}+y^{2}+z^{2}}}$$

Find the exact points on the circle \(x^{2}+y^{2}=2\) at which the field \(\mathbf{F}=\langle f, g\rangle=\left\langle x^{2}, y\right\rangle\) switches from pointing inward to outward on the circle, or vice versa.

Prove the following properties of the divergence and curl. Assume \(\mathbf{F}\) and \(\mathbf{G}\) are differentiable vector fields and \(c\) is a real number. a. \(\nabla \cdot(\mathbf{F}+\mathbf{G})=\nabla \cdot \mathbf{F}+\nabla \cdot \mathbf{G}\) b. \(\nabla \times(\mathbf{F}+\mathbf{G})=(\nabla \times \mathbf{F})+(\nabla \times \mathbf{G})\) c. \(\nabla \cdot(c \mathbf{F})=c(\nabla \cdot \mathbf{F})\) d. \(\nabla \times(c \mathbf{F})=c(\nabla \times \mathbf{F})\)

The heat flow vector field for conducting objects is \(\mathbf{F}=-k \nabla T,\) where \(T(x, y, z)\) is the temperature in the object and \(k>0\) is a constant that depends on the material. Compute the outward flux of \(\mathbf{F}\) across the following surfaces S for the given temperature distributions. Assume \(k=1\). \(T(x, y, z)=100 e^{-x^{2}-y^{2}-z^{2}} ; S\) is the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\).
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