Chapter 16: Problem 20
Evaluate the line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(C\) is given by the vector function \(\mathbf{r}(t)\) $$ \begin{array}{l}{\mathbf{F}(x, y, z)=\left(x+y^{2}\right) \mathbf{i}+x z \mathbf{j}+(y+z) \mathbf{k}} \\ {\mathbf{r}(t)=t^{2} \mathbf{i}+t^{3} \mathbf{j}-2 t \mathbf{k}, \quad 0 \leqslant t \leqslant 2}\end{array} $$
Short Answer
Step by step solution
Parametrize the Curve
Compute the Derivative \( d\mathbf{r} \)
Evaluate \( \mathbf{F}( \mathbf{r}(t) ) \)
Find \( \mathbf{F} \cdot d\mathbf{r} \)
Integrate over the Interval [0, 2]
Solve the Definite Integral
Final Calculation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Calculus
In our problem, the vector field \( \mathbf{F}(x, y, z) = \left(x+y^{2}\right) \mathbf{i}+xz \mathbf{j}+(y+z) \mathbf{k} \) is given, indicating the complexity of interactions in a three-dimensional space. Evaluating the line integral involves moving through this space according to a specific path or curve \( C \), guided by the vector function \( \mathbf{r}(t) \). This movement is represented through the dot product, separating vector calculus distinctly from other areas of calculus due to the multi-dimensional nature involved.
Developing an intuition for vector calculus allows students to solve real-world problems involving fluid dynamics, electromagnetism, and more by breaking down forces, velocities into manageable calculus problems.
Parametrization of Curves
In the given exercise, the curve \( C \) is parametrized by \( \mathbf{r}(t) = t^2 \mathbf{i} + t^3 \mathbf{j} - 2t \mathbf{k} \), which describes the trajectory of the curve using the parameter \( t \) ranging from 0 to 2. This process of parametrization translates a geometric idea of the path \( C \) into an algebraic form that we can manipulate mathematically.
- \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors along the x, y, and z axes respectively.
- The expressions \( t^2 \), \( t^3 \), and \(-2t \) represent the x, y, and z coordinates as functions of \( t \).
Dot Product
In the exercise, we calculate \( \mathbf{F}(\mathbf{r}(t)) \cdot d\mathbf{r} \), where \( d\mathbf{r} \) is a derivative of the parametrized curve \( \mathbf{r}(t) \) with respect to \( t \). The result of this operation is used in the integral for finding the work done by the force field along the curve \( C \).
- The dot product formula for vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \).
- The result is a scalar and not a vector, indicating the magnitude of the vectors' interaction.
Definite Integration
In the problem provided, we integrate the function derived from the dot product, \( [(2t^3 + 2t^7) - 6t^5 - 2t^4 + 4t] \) from \( t=0 \) to \( t=2 \). The result gives a measure of how much the field \( \mathbf{F} \) does work along the curve \( C \).
- Evaluate the integral to find the accumulation of values over a specified range.
- Apply the antiderivative to the limits of integration to compute the total effect between \( t=0 \) and \( t=2 \).