Chapter 5: Problem 30
In the following exercises, find the work done by force field \(\mathbf{F}\) on an object moving along the indicated path. $$ \begin{aligned} &\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}-5 z \mathbf{k} \\ &C: \mathbf{r}(t)=2 \cos t \mathbf{i}+2 \sin t \mathbf{j}+t \mathbf{k}, 0 \leq t \leq 2 \pi \end{aligned} $$
Short Answer
Step by step solution
Understand the Problem
Set Up the Integral for Work
Compute the Derivative \( \mathbf{r}'(t) \)
Evaluate \( \mathbf{F}(\mathbf{r}(t)) \)
Formulate the Dot Product \( \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \)
Integrate to Find Work
Present the Result
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Force Field
When you have a force field:
- The first part, \(x \mathbf{i}\), determines the force's influence in the x-direction, which is proportional to x.
- The \(y \mathbf{j}\) component describes how the force affects motion along the y-direction, also directly linked to the y-coordinate.
- The term \(-5z \mathbf{k}\) indicates that the force in the z-direction is opposite to the increase in z, and it scales linearly with \(-5z\).
Parameterized Path
In this exercise, the path \(C\) is parameterized by:
- \( \mathbf{r}(t) = 2 \cos t \mathbf{i} \) for the x-coordinate, shaping a circular path in the x-direction.
- \( \mathbf{r}(t) = 2 \sin t \mathbf{j} \) for the y-coordinate, maintaining circular symmetry in the y-direction.
- \( \mathbf{r}(t) = t \mathbf{k} \) for the z-coordinate, representing a linear change as \(t\) increases, creating a helical path.
- For circular paths, the radius in the xy-plane is constant, determined by the coefficients of \(\cos t\) and \(\sin t\).
- The linear term in the z-component causes the object to rise or fall as it revolves around the circle, resembling a staircase in the air.
Work Done
The mathematical expression for calculating the work done by a force as it acts along a path is given by the line integral: \[ W = \int_C \mathbf{F} \cdot d\mathbf{r} \]Here, the line integral sums up the force's influence over each small segment \(d\mathbf{r}\) of the path \(C\). Notably, this exercise involves calculating the dot product:
- The dot product \( \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \) translates the force components relative to their influence on the path's direction and magnitude.
- The negative sign in the scalar result \(-5t\) indicates a directionally opposing or energetically resisting movement in the z-direction along our parameterized helix.