Problem 14 - Calculus

Question

Find the work in moving from (1,0) to \\{0,1) . When \(F\) is conservative, construct \(f\). Choose yout own path when \(\mathbf{F}\) is not eonservative. $$\mathbf{F}=e^{y} \mathbf{i}+x e^{y} \mathbf{j}$$

Step-by-Step Solution

Answer

The work done is -1.

Step 1: Determine if the Field is Conservative

To determine if the vector field \(\mathbf{F} = e^y \mathbf{i} + x e^y \mathbf{j}\) is conservative, check if its curl is zero. The field \(\mathbf{F}\) is conservative if \(abla \times \mathbf{F} = 0\).

Step 2: Calculate the Partial Derivatives

Compute the partial derivatives needed for the curl:- \(\frac{\partial P}{\partial y}\), where \(P = e^y\).- \(\frac{\partial Q}{\partial x}\), where \(Q = xe^y\).Thus, \(\frac{\partial P}{\partial y} = e^y\) and \(\frac{\partial Q}{\partial x} = e^y\).

Step 3: Evaluate Curl of the Vector Field

The curl in two dimensions is given by \(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\). Substituting in the partial derivatives gives:\[e^y - e^y = 0\].Since the curl is zero, \(\mathbf{F}\) is a conservative field.

Step 4: Construct the Potential Function \(f\)

Since \(\mathbf{F}\) is conservative, find a potential function \(f\) such that \(abla f = \mathbf{F}\).Integrate both components of the vector field over their respective variables:1. Integrate \(e^y\) with respect to \(x\) yielding \(f(x, y) = xe^y + C(y)\).2. Differentiate \(f(x, y) = xe^y + C(y)\) with respect to \(y\) and set it to \(xe^y\). The derivative gives \(xe^y + C'(y)\), so \(C'(y) = 0\) and \(C(y) = k\).

Step 5: Evaluate the Potential Function at Movement Points

With \(f(x,y) = xe^y\), calculate \(f(0,1) - f(1,0)\) to find work done:- \(f(0,1) = 0 \cdot e^1 = 0\).- \(f(1,0) = 1 \cdot e^0 = 1\).

Step 6: Calculate Work Done Using the Potential Function

The work done \(W\) when moving from \((1,0)\) to \((0,1)\) is given by:\[ W = f(0,1) - f(1,0) = 0 - 1 = -1 \].

Key Concepts

Curl of a Vector Field

Understanding the curl of a vector field is crucial in determining whether a vector field is conservative. A vector field is deemed conservative if its curl is zero. The curl is a measure of how much the vector field rotates around a point.
For a two-dimensional field like \( \mathbf{F}=e^{y} \mathbf{i}+x e^{y} \mathbf{j} \), the curl is simplified to the expression \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \). Calculating the partial derivatives \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \) helps establish the field's behavior.
When both derivatives equal each other, as in this case, \( e^y - e^y = 0 \), it confirms the vector field is conservative, indicating no rotational tendencies.

Potential Function

In a conservative vector field, a potential function \( f \) exists such that its gradient is equal to the vector field. This function represents the potential energy within the field and simplifies the calculation of work done.
Finding this potential involves integrating the components of the vector field. Start by integrating \( e^y \) with respect to \( x \), resulting in \( f(x, y) = xe^y + C(y) \).
To ensure this is correct, differentiate the result with respect to \( y \), and find that \( C'(y) = 0 \), leading to a constant \( C(y) = k \). This shows how crucial detailed integration and differentiation are for constructing the potential function.

Partial Derivatives

Partial derivatives break down how a function changes as one variable alters, holding others constant. They are a key tool in calculus, especially when dealing with vector fields.
Calculating partial derivatives like \( \frac{\partial P}{\partial y} = e^y \) and \( \frac{\partial Q}{\partial x} = e^y \) provides insights into the field's structure.
These derivatives are used to assess the curl of the field. They also play a pivotal role in finding and verifying the potential function. Understanding partial derivatives is essential for exploring more complex calculus concepts.

Work Calculation

In the context of vector fields, work done is a measure of energy transferred along a path. For conservative fields, calculating work simplifies using potential functions.
The work, \( W \), is evaluated as the change in potential when moving between two points, expressed as \( f(0,1) - f(1,0) \). Using the potential function \( f(x, y) = xe^y \), we get

• \( f(0,1) = 0 \cdot e^1 = 0 \)
• \( f(1,0) = 1 \cdot e^0 = 1 \)

Subtracting these gives \( W = 0 - 1 = -1 \).
This tells us that 1 unit of work is extracted from the field moving from \( (1,0) \) to \( (0,1) \). Understanding this concept is crucial for solving physics problems involving forces and movement in fields.