Problem 68 - Calculus Volume 3

Question

In the following exercises, find the work done by force field \(\mathbf{F}\) on an object moving along the indicated path. Let \(\mathbf{F}\) be vector field \(\mathbf{F}(x, y)=\left(y^{2}+2 x e^{y}+1\right) \mathbf{i}+\left(2 x y+x^{2} e^{y}+2 y\right) \mathbf{j}\) Compute the work of integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}, \quad\) where \(C\) is the path \(\mathbf{r}(t)=\sin t \mathbf{i}+\cos t \mathbf{j}, 0 \leq t \leq \frac{\pi}{2}\).

Step-by-Step Solution

Answer

Calculate the definite integral of the dot product over the interval [0, π/2] to find the work done.

Step 1: Parameterizing the Path

The path \( C \) is given by the parameterization \( \mathbf{r}(t) = \sin t \mathbf{i} + \cos t \mathbf{j} \) for \( 0 \leq t \leq \frac{\pi}{2} \). This represents a quarter-circle arc in the unit circle from \((0, 1)\) to \((1, 0)\).

Step 2: Diferentiating r(t) with respect to t

The derivative \( \frac{d\mathbf{r}(t)}{dt} \) is needed to compute the line integral. Computation gives: \( \mathbf{r}'(t) = \cos t \mathbf{i} - \sin t \mathbf{j} \).

Step 3: Evaluating the Vector Field on the Path

Substitute \( x = \sin t \) and \( y = \cos t \) into \( \mathbf{F}(x, y) \): \[ \mathbf{F}(\sin t, \cos t) = (\cos^2 t + 2 \sin t e^{\cos t} + 1) \mathbf{i} + (2 \sin t \cos t + \sin^2 t e^{\cos t} + 2 \cos t) \mathbf{j}. \]

Step 4: Computing the Dot Product

Compute the dot product \( \mathbf{F}(\sin t, \cos t) \cdot \mathbf{r}'(t) \):\[ (\cos^2 t + 2 \sin t e^{\cos t} + 1) \cos t + (2 \sin t \cos t + \sin^2 t e^{\cos t} + 2 \cos t)(-\sin t). \]Use algebraic simplification to express this in a form that can be integrated.

Step 5: Evaluating the Integral

Now integrate the simplified expression from Step 4 with respect to \( t \) over the interval \( [0, \frac{\pi}{2}] \).This integral will evaluate the work done by the force field along the curve \( C \).

Step 6: Solving the Integral

During the simplification and integration, it may be helpful to break down into components, evaluate definite integrals, and use properties of trigonometric and exponential functions. Sum the results to get the final answer.

Key Concepts

Vector Field

A vector field assigns a vector to each point in space. In simpler terms, imagine that at every point in a region, there is an arrow that indicates a certain direction and magnitude. For example, consider the vector field \( \mathbf{F}(x, y) = (y^2 + 2xe^y + 1) \mathbf{i} + (2xy + x^2e^y + 2y) \mathbf{j} \).
Here, for any point \((x, y)\), the vector has:

• A component in the direction of \(\mathbf{i}\) (horizontal), which is \(y^2 + 2xe^y + 1\).
• A component in the direction of \(\mathbf{j}\) (vertical), which is \(2xy + x^2e^y + 2y\).

This concept is useful in physics and engineering, as it can represent things like the flow of fluids, or, in this problem, a force field acting on an object.

Parametric Equations

Parametric equations express the coordinates of the points that make up a curve as functions of a variable, often time \( t \). For the given problem, the path \( C \) is represented by the parametric equations \( \mathbf{r}(t) = \sin t \mathbf{i} + \cos t \mathbf{j} \), where \( 0 \leq t \leq \frac{\pi}{2} \).
This setup describes a quarter-circle arc of the unit circle. In parametric equations:

• The \(\sin t\) term describes the change in the horizontal position as \( t \) varies.
• The \(\cos t\) term describes the change in the vertical position.

Differentiating these equations helps us find the velocity vector at any point, \( \mathbf{r}'(t) = \cos t \mathbf{i} - \sin t \mathbf{j} \),which is crucial for computing other quantities like the line integral.

Dot Product

The dot product is a way to multiply two vectors, resulting in a scalar (a number). In the context of this problem, you need the dot product to find the line integral of the vector field along a path.
To find the work done by the vector field \( \mathbf{F} \) along the path \( \mathbf{r}(t) \), we calculate the dot product \[ \mathbf{F}(\sin t, \cos t) \cdot \mathbf{r}'(t) \]. This is computed by multiplying corresponding components and adding them together:

• The term \((\cos^2 t + 2 \sin t e^{\cos t} + 1) \cos t\) comes from the \(\mathbf{i}\) components.
• The term \((2 \sin t \cos t + \sin^2 t e^{\cos t} + 2 \cos t)(-\sin t)\) comes from the \(\mathbf{j}\) components.

Once determined, this scalar needs to be integrated over the given interval of \( t \).

Work in Physics

In physics, work is defined as the transfer of energy that occurs when a force makes an object move. It is calculated by integrating the dot product of the force vector and the displacement vector over a path. In the context of this problem, the work done by the force field \( \mathbf{F} \) on the object as it moves along the path \( C \) is given by the line integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \).
To compute this integral, we:

• First, parameterize the path using parametric equations.
• Evaluate the vector field on this path.
• Find the derivative of the parameterization (the velocity vector), and then the dot product of the vector field with this velocity vector.
• Finally, integrate this dot product over the given interval of \( t \).

The result of this integral tells us the total work done by the force field on the object over the specified path.