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Magazine Chapter 6: Problem 55
For the following exercises, evaluate the line integrals. Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(\mathbf{F}(x, y)=-1 \mathbf{j},\) and \(C\) is the part of the graph of \(y=\frac{1}{2} x^{3}-x\) from \((2,2)\) to \((-2,-2)\)
Short Answer
Expert verified
The line integral is 4.
Step by step solution
01
Parameterize the Curve C
The function given is \( y = \frac{1}{2} x^3 - x \), so we can let \( x = t \) and parameterize the curve as \( \mathbf{r}(t) = (t, \frac{1}{2} t^3 - t) \). Here, the parameter \( t \) varies from \( 2 \) to \( -2 \), as these correspond to the given endpoints of the curve \( C \), namely \( (2, 2) \) to \( (-2, -2) \).
02
Compute the Derivative of the Parameterization
Compute the derivative \( \mathbf{r}'(t) = \frac{d}{dt}(t, \frac{1}{2} t^3 - t) = (1, \frac{3}{2}t^2 - 1) \).
03
Express d\( \mathbf{r} \) in Terms of the Parameter t
Since \( \mathbf{r}'(t) = (1, \frac{3}{2}t^2 - 1) \), we have \( d \mathbf{r} = (1, \frac{3}{2}t^2 - 1) dt \).
04
Evaluate the Dot Product \( \mathbf{F}(\mathbf{r}(t)) \cdot d \mathbf{r} \)
Given \( \mathbf{F}(x, y) = -\mathbf{j} \), at any point on the curve, \( \mathbf{F}(t, \frac{1}{2} t^3 - t) = (0, -1) \). The dot product with \( d \mathbf{r} = (1, \frac{3}{2}t^2 - 1) dt \) is \( (0, -1) \cdot (1, \frac{3}{2}t^2 - 1) = -\left(\frac{3}{2}t^2 - 1\right) dt \).
05
Integrate Over the Bounds of the Parameterization
We need to evaluate the integral \( \int_{2}^{-2} -\left(\frac{3}{2}t^2 - 1\right) dt \). Simplifying, this is \( \int_{2}^{-2} 1 - \frac{3}{2}t^2 \, dt \).
06
Integrate the Expression
Calculate \( \int (1 - \frac{3}{2}t^2) \, dt \) which is \( t - \frac{1}{2}t^3 \). Evaluate it from \( t=2 \) to \( t=-2 \): \(\left[-2 - \frac{1}{2}(-2)^3\right] - \left[2 - \frac{1}{2}(2)^3\right] \).
07
Calculate the Definite Integral
Simplify and calculate: For \( t = -2 \), it becomes \( -2 + 4 = 2 \). For \( t = 2 \), it becomes \( 2 - 4 = -2 \). \(\text{So, the integral is } 2 - (-2) = 4 \).
08
Conclusion
The value of the line integral \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \) is \( 4 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parameterization of Curves
When dealing with line integrals, parameterization of the curve is a helpful technique. It involves expressing the curve in terms of a parameter, typically denoted as \( t \). This approach simplifies the evaluation process by transforming a problem in two or three dimensions into a single variable problem.
- **Given Function**: For this exercise, the curve \( C \) is represented by the equation \( y = \frac{1}{2}x^3 - x \).
- **Parameterization Approach**: We can choose \( x = t \) as our parameter. Thus, the vector function representing the curve becomes \( \mathbf{r}(t) = (t, \frac{1}{2}t^3 - t) \).
- **Parameter Range**: To cover the entire segment of the curve from \((2, 2)\) to \((-2, -2)\), \( t \) varies from 2 to -2.
Dot Product
The dot product is an algebraic operation that combines two vectors to produce a scalar (a single number). This scalar gives a sense of how much two vectors align with each other.
- **Formula**: For two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), the dot product is given by \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \).
- **Exercise Application**: The vector field \( \mathbf{F}(x, y) = (0, -1) \), and the derivative of parameterization is \( d\mathbf{r} = (1, \frac{3}{2}t^2 - 1) dt \).
- **Result**: Computing \( (0, -1) \cdot (1, \frac{3}{2}t^2 - 1) \) yields \( -\left(\frac{3}{2}t^2 - 1\right) \). This expression is then used in the evaluation of the integral.
Definite Integral
In mathematics, a definite integral represents the accumulation of quantities, such as areas under a curve, over a specific interval. For this exercise, evaluating a definite integral will provide the value of the line integral along the parameterized path.
- **Definite Integral Expression**: Derived from the dot product calculation above, the integral becomes \( \int_{2}^{-2} -\left(\frac{3}{2}t^2 - 1\right) dt \).
- **Simplification**: Simplifying this integral, we get \( \int_{2}^{-2} \left(1 - \frac{3}{2}t^2\right) \, dt \).
- **Evaluation**: Solving this integral from \( t = 2 \) to \( t = -2 \) reveals the value of the line integral, giving importance to correct limits and simplification.
Vector Field
A vector field assigns a vector to every point in space. They are used extensively in physics to represent various quantitative properties that have both magnitude and direction, such as force.
- **Definition**: A vector field \( \mathbf{F}(x, y) \) in two dimensions can be expressed as \( (P(x, y), Q(x, y)) \).
- **Exercise Specific**: In our problem, \( \mathbf{F}(x, y) = (0, -1) \). This indicates a constant vector field with only a downward direction across all points.
- **Effect on Line Integral**: The evaluation of the line integral involves this vector field interacting along the curve \( C \), showcasing how vector fields influence line integrals.
