/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Find the work done by \(\mathbf{... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 19

Find the work done by \(\mathbf{F}\) over the curve in the direction of increasing \(t\) $$\begin{aligned} &\mathbf{F}=x y \mathbf{i}+y \mathbf{j}-y z \mathbf{k}\\\ &\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 1 \end{aligned}$$

Short Answer

Expert verified
The work done is \(\frac{1}{2}\).

Step by step solution

01

Express the Curve in Parameterized Form

The problem provides the parametric curve \(\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k}\). The curve is parameterized by \(t\) within the interval \(0 \leq t \leq 1\).\ To compute the work done by the force \(\mathbf{F}\) over this path, start by expressing the curve as a path integral in terms of \(t\).
02

Calculate the Derivative of the Parameterized Curve

Find \(\frac{d\mathbf{r}}{dt}\), the derivative of the curve with respect to \(t\). This gives the tangent vector along the curve:\[\frac{d\mathbf{r}}{dt} = \frac{d}{dt}(t \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k}) = \mathbf{i} + 2t \mathbf{j} + \mathbf{k}.\]
03

Substitute the Parameterized Curve into the Force Field

Substitute \(x = t\), \(y = t^2\), and \(z = t\) from \(\mathbf{r}(t)\) into the components of \(\mathbf{F}\). This gives the force field \(\mathbf{F}\) along the path:\[\mathbf{F} = (t \cdot t^2) \mathbf{i} + t^2 \mathbf{j} - t^2 \cdot t \mathbf{k} = t^3 \mathbf{i} + t^2 \mathbf{j} - t^3 \mathbf{k}.\]
04

Compute the Dot Product

Find the dot product \(\mathbf{F} \cdot \frac{d\mathbf{r}}{dt}\):\[\mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = (t^3 \mathbf{i} + t^2 \mathbf{j} - t^3 \mathbf{k}) \cdot (\mathbf{i} + 2t \mathbf{j} + \mathbf{k}) = t^3 \cdot 1 + t^2 \cdot 2t - t^3 \cdot 1.\]Simplifying, we get:\[t^3 + 2t^3 - t^3 = 2t^3.\]
05

Integrate the Dot Product over the Interval

Compute the integral of \(2t^3\) with respect to \(t\) over the interval \([0, 1]\) to find the work done:\[\text{Work} = \int_{0}^{1} 2t^3 \, dt.\]This integral simplifies to:\[= 2 \times \left[ \frac{t^4}{4} \right]_{0}^{1} = 2 \left( \frac{1^4}{4} - \frac{0^4}{4} \right) = \frac{1}{2}.\]
06

Final Result

The total work done by the force \(\mathbf{F}\) along the curve from \(t=0\) to \(t=1\) is \(\frac{1}{2}.\)

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Key Concepts

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

Parametric Equations
When dealing with curves in vector calculus, parametric equations are a powerful tool. They allow us to express a curve in terms of a parameter鈥攊n this case, 't'. Instead of directly describing the relationship between x, y, and z, we specify how these coordinates vary along the curve over a certain interval. For instance, the curve given by \(\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k}\) tells us that as 't' changes from 0 to 1:
  • The x-coordinate grows linearly with 't' (i.e., x = t).
  • The y-coordinate grows quadratically (i.e., y = t^2).
  • The z-coordinate again grows linearly with 't' (i.e., z = t).
By using parameters, we can easily manage paths that follow complex patterns, and they're particularly useful for calculating integrals along such paths.
Vector Fields
Vector fields add depth to our understanding of space, providing a vector quantity for each point within a region. These could depict forces, like gravity or electromagnetism, or fluid flow and air currents. In our example, \( \mathbf{F} = x y \mathbf{i} + y \mathbf{j} - y z \mathbf{k} \), the vector field specifies a force at each point.To calculate work done over a path in this field:
  • Translate the path coordinates using the parameterization (i.e., applying \( x = t \), \( y = t^2 \), \( z = t \) ).
  • Substitute these into the vector field expressions.
This gives a snapshot of the force as it acts along the curve's journey. This vector field theory is pivotal for comprehending how fields influence objects and energy distributions in physical spaces.
Work Calculation
The concept of 'work' in physics and mathematics can be understood as a way of quantifying the energy transferred by a force along a path. Mathematically, work is determined through the line integral of the vector field along the curve.Here's how it works:
  • First, calculate the derivative of the parametric equation to get the tangent vector: \( \frac{d\mathbf{r}}{dt} = \mathbf{i} + 2t \mathbf{j} + \mathbf{k} \).
  • Next, find the dot product of the transformed force vector \( \mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = 2t^3 \).
  • Finally, compute the integral of this product over the curve's parameter range; here, from 0 to 1.
The integration \[ \int_{0}^{1} 2t^3 \, dt = \frac{1}{2} \] indicates the total work done. Through step-by-step calculations, you can directly see how each component interacts to convey work, making this abstract concept tangible and revealing how energy transfer occurs along a path.

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Most popular questions from this chapter

Find the area of the surfaces. The surface cut from the bottom of the paraboloid \(z=x^{2}+y^{2}\) by the plane \(z=3\)

Zero circulation Use Equation (8) and Stokes" Theorem to show that the circulations of the following fields around the boundary of any smooth orientable surface in space are zero. a. \(\mathbf{F}=2 x \mathbf{i}+2 y \mathbf{j}+2 z \mathbf{k}\) b. \(\mathbf{F}=\nabla\left(x y^{2} z^{3}\right)\) d. \(\mathbf{F}=\nabla f\) c. \(\mathbf{F}=\nabla \times(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})\)

Unit vectors pointing toward the origin Find a field \(\mathbf{F}=\) \(M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) in the \(x y\) -plane with the property that at each point \((x, y) \neq(0,0), \quad \mathbf{F}\) is a unit vector pointing toward the origin. (The field is undefined at \((0,0) .)\)

You have been asked to find the path along which a force field \(\mathbf{F}\) will perform the least work in moving a particle between two locations. A quick calculation on your part shows \(\mathbf{F}\) to be conservative. How should you respond? Give reasons for your answer.

Zero curl, yet the field is not conservative Show that the curl of $$ \mathbf{F}=\frac{-y}{x^{2}+y^{2}} \mathbf{i}+\frac{x}{x^{2}+y^{2}} \mathbf{j}+z \mathbf{k} $$ is zero but that $$ \oint_{c} \mathbf{F} \cdot d \mathbf{r} $$ is not zero if \(C\) is the circle \(x^{2}+y^{2}=1\) in the \(x y\) -plane. (Theorem 7 does not apply here because the domain of \(\mathbf{F}\) is not simply connected. The field \(\mathbf{F}\) is not defined along the \(z\) -axis so there is no way to contract \(C\) to a point without leaving the domain of \(\mathbf{F}\).)
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