Chapter 15: Problem 25
Find the work done by \(\mathbf{F}\) in moving a particle once counterclockwise around the given curve. $$\mathbf{F}=2 x y^{3} \mathbf{i}+4 x^{2} y^{2} \mathbf{j}$$ C: The boundary of the "triangular" region in the first quadrant enclosed by the \(x\) -axis, the line \(x=1,\) and the curve \(y=x^{3}\)
Short Answer
Expert verified
The work done is \(-\frac{89}{132}\).
Step by step solution
01
Analyze the Region and Curve
The region described is in the first quadrant. It is bounded by the x-axis (where \(y = 0\)), the line \(x = 1 \), and the curve \(y = x^3\). The curve starts at the origin (0,0), moves along the curve \(y = x^3\) until \(x = 1\), then goes vertically down to (1,0), and finally returns to the origin along the x-axis.
02
Check Line Integral Convertibility
To find the work done by \( \mathbf{F} \), we need to evaluate the line integral \(\oint_C \mathbf{F} \cdot d\mathbf{r} \). Since the field \(\mathbf{F}\) is not immediately conservative, we will need to check the curl of \(\mathbf{F}\) to see if it simplifies the integral by Stoke's theorem.
03
Compute the Curl of \( \mathbf{F} \)
The curl \( abla \times \mathbf{F} \) in 2D is determined by \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \) where \(\mathbf{F}=M\mathbf{i}+N\mathbf{j}\). Here, \(M=2xy^3\) and \(N=4x^2y^2\). Thus, \(abla \times \mathbf{F} = 8xy - 6xy^2 = 2xy(4-3y)\). Since the curl is non-zero, \(\mathbf{F}\) is not conservative.
04
Parameterize the Curve Segments
Parameterize each segment of the boundary curve: 1. Along \(y = x^3\) from \((0, 0)\) to \((1, 1)\), let \(\mathbf{r}_1 = \langle t, t^3 \rangle\), \(0 \leq t \leq 1 \).2. Along \(x = 1\) from \((1, 1)\) to \((1, 0)\), let \(\mathbf{r}_2 = \langle 1, 1-t \rangle\), \(0 \leq t \leq 1 \).3. Along the x-axis from \((1, 0)\) back to \((0, 0)\), let \(\mathbf{r}_3 = \langle 1-t, 0 \rangle\), \(0 \leq t \leq 1 \).
05
Evaluate Line Integral over Each Segment
Compute \(\int_C \mathbf{F} \cdot d\mathbf{r}\) for each curve segment:1. For \( \mathbf{r}_1 \), \(dx=dt, dy=3t^2dt\), and the integral is \(\int_0^1 {\left(2t(t^3)^3 + 4t^2(t^3)^2 \right) dt}\).2. For \( \mathbf{r}_2 \), \(dx=0, dy=-dt\), and the integral is \(\int_0^1 4 \cdot 1^2 (1-t)^2 \cdot (-dt)\).3. For \( \mathbf{r}_3 \), \(dx=-dt, dy=0\), and the integral is \(\int_0^1 2(1-t)(0)^3 \cdot (-dt)\).
06
Calculate Each Integral
Compute each integral:1. \(\int_0^1 \left( 2t^7 + 4t^{10} \right) dt = \left[\frac{2}{8}t^8 + \frac{4}{11}t^{11} \right]_0^1 = \frac{1}{4} + \frac{4}{11} =\frac{15}{44}\).2. \(-4 \int_0^1 (1-t)^2 dt = -4 \left[\frac{1}{3} - \frac{2}{3} + \frac{1}{3}\right]_0^1 = -\frac{4}{3} \).3. The integral over \( \mathbf{r}_3 \) is zero since there's no \(x\) in the term.Add resulting values: \(\frac{15}{44} - \frac{4}{3} = -\frac{89}{132}\).
07
Conclude the Work Done
The total work done by the force field \(\mathbf{F}\) as the particle moves once around the curve \(C\) is the sum of the integrals computed over each segment. This gives \(-\frac{89}{132}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Work Done by Force Fields
In physics, a force field represents a vector field that describes a distribution of forces in a given space. The work done by a force field on a particle as it moves through a path is defined by the line integral of the force field along that path. It can be mathematically represented as \( W = \int_C \mathbf{F} \cdot d\mathbf{r} \), where \( \mathbf{F} \) is the force field, \( C \) is the path, and \( d\mathbf{r} \) is a differential element of the path.
When trying to find the work done by a vector field \( \mathbf{F} \), the key focus is on the path taken by the particle.
When trying to find the work done by a vector field \( \mathbf{F} \), the key focus is on the path taken by the particle.
- If the path is closed, the closed-loop line integral is evaluated.
- The integral will take into account that force may exert differently along different segments of the path.
Parameterization of Curves
Parameterization of curves is a technique used in calculus to simplify the process of integrating vector fields over curves. By representing a curve in a parametric form, the problem of finding the work done can be effectively managed in steps.
To parameterize a curve, we employ a parameter (often denoted as \( t \)) to describe both \( x \) and \( y \) components of the curve in terms of \( t \). For example:
To parameterize a curve, we employ a parameter (often denoted as \( t \)) to describe both \( x \) and \( y \) components of the curve in terms of \( t \). For example:
- The curve given by \( y = x^3 \) for \( x \) in \((0, 1)\) can be parameterized as \( \mathbf{r}(t) = \langle t, t^3 \rangle \) for \( 0 \leq t \leq 1 \).
- For the vertical segment where \(x = 1\) moving from \((1,1)\) to \((1,0)\), the parameterization is \( \mathbf{r}(t) = \langle 1, 1-t \rangle \).
- Along the x-axis from \((1,0)\) to \((0,0)\), the parameterization is \( \mathbf{r}(t) = \langle 1-t, 0 \rangle \).
Non-Conservative Vector Fields
Non-conservative vector fields are areas where the vector field does not conserve mechanical energy, meaning the work done depends on the specific path taken, unlike conservative fields.
Key characteristics include:
Key characteristics include:
- A non-zero curl (\( abla \times \mathbf{F} \)), indicating that the field is rotational.
- Path dependency of the work done, which can result in different outcomes for different paths, even if they have the same end points.
- Two-dimensional planar fields can be checked for conservativeness using the curl calculated by \( abla \times \mathbf{F} \), specifically using \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \).