Problem 27
Question
Determine whether or not the vector field is conservative. If it is, find a potential function. $$(x-2 x y) \mathbf{i}+\left(y^{2}-x^{2}\right) \mathbf{j}$$
Step-by-Step Solution
Verified Answer
The vector field is conservative as its curl is zero and divergence is not equal to zero. Its potential function is \(f=\frac{x^{2}}{2} - x^{2} y + \frac{y^{3}}{3} + C\).
1Step 1: Compute Curl
Start by calculating the curl of the vector field. The curl of a vector field \(V = P\mathbf{i} + Q\mathbf{j}\) is defined as \(\nabla \times V=(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\mathbf{k}\). By computing \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\) for the given vector field, it is found that \(-2x\) and \(-2x\) respectively, resulting in the curl to be zero.
2Step 2: Compute Divergence
Next, calculate the divergence of the vector field. The divergence of a vector field \(V = P\mathbf{i} + Q\mathbf{j}\) is defined as \(\nabla \cdot V=(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y})\). By computing \(\frac{\partial P}{\partial x}\) and \(\frac{\partial Q}{\partial y}\) for the given vector field, it is found that \(1-2y\) and \(2y\) respectively. The divergence is \(1-2y+2y = 1\), which is not equal to zero. So the vector field is conservative.
3Step 3: Find Potential Function
To find a potential function \(f\) for the vector field \(V = P\mathbf{i} + Q\mathbf{j}\), solve the equations \(\frac{\partial f}{\partial x}=P\) and \(\frac{\partial f}{\partial y}=Q\). Integrate \(P\) with respect to \(x\) and \(Q\) with respect to \(y\). Then combine these results to derive the potential function. The equations become \(\frac{\partial f}{\partial x}=x-2 x y\) and \(\frac{\partial f}{\partial y}=y^{2}-x^{2}\). Thus, \(f=\int (x-2 x y)dx=\frac{x^{2}}{2}-x^{2} y+C_{1}(y)\) and \(f=\int (y^{2}-x^{2})dy=\frac{y^{3}}{3}-x^{2} y+C_{2}(x)\). Finally, identify the common portions of both integrations and combine constants \(C_{1}(y)\) and \(C_{2}(x)\) into one constant \(C\). So, the potential function is \(f=\frac{x^{2}}{2} - x^{2} y + \frac{y^{3}}{3} + C\).
Key Concepts
Curl of a Vector FieldDivergence of a Vector FieldPotential FunctionPartial Derivatives
Curl of a Vector Field
The curl of a vector field is a measure of the rotation or 'twist' of the field at a point. Imagine a paddle wheel placed in a flow of water; the manner in which the water rotates the wheel indicates the 'curl' at that point. Mathematically, for a vector field \(V = P\mathbf{i} + Q\mathbf{j}\), the curl is given by \(abla \times V=(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\mathbf{k}\).
When the curl of a vector field in two dimensions is zero, as in the given exercise where the calculations yield \(\frac{\partial Q}{\partial x} = -2x\) and \(\frac{\partial P}{\partial y} = -2x\), resulting in zero, the field has no net rotation around any point. This is one of the properties that suggests the field could be conservative, which has implications for the existence of a potential function.
When the curl of a vector field in two dimensions is zero, as in the given exercise where the calculations yield \(\frac{\partial Q}{\partial x} = -2x\) and \(\frac{\partial P}{\partial y} = -2x\), resulting in zero, the field has no net rotation around any point. This is one of the properties that suggests the field could be conservative, which has implications for the existence of a potential function.
Divergence of a Vector Field
The divergence of a vector field indicates the tendency of the field to 'diverge' from or 'converge' into a point. In simpler terms, it measures whether a vector field is a source or a sink at a point. The divergence is defined for \(V = P\mathbf{i} + Q\mathbf{j}\) as \(abla \cdot V=(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y})\).
In our exercise, even though the curl was zero, the divergence \(1-2y+2y = 1\) did not vanish. However, the divergence being zero is not a requirement for a vector field to be conservative. A common misconception is that a zero divergence is needed for conservativeness, but it is actually the curl that needs to vanish.
In our exercise, even though the curl was zero, the divergence \(1-2y+2y = 1\) did not vanish. However, the divergence being zero is not a requirement for a vector field to be conservative. A common misconception is that a zero divergence is needed for conservativeness, but it is actually the curl that needs to vanish.
Potential Function
A potential function is a scalar field whose gradient equals the original vector field. In the context of our exercise, a conservative vector field with a zero curl suggests that this function exists. A potential function offers a wealth of information, as it allows the computation of work done by a force field without path integration.
To find it, one would solve for \(f\) such that \(\frac{\partial f}{\partial x}=P\) and \(\frac{\partial f}{\partial y}=Q\). By integrating \(P\) with respect to \(x\) and \(Q\) with respect to \(y\), and then combining these, one can find the potential function for the vector field. In our exercise, after integrating both components separately and finding the overlapping terms, the potential function was determined as \(f=\frac{x^{2}}{2} - x^{2} y + \frac{y^{3}}{3} + C\). This function is crucial in fields like physics and engineering, enabling calculations of energy for example.
To find it, one would solve for \(f\) such that \(\frac{\partial f}{\partial x}=P\) and \(\frac{\partial f}{\partial y}=Q\). By integrating \(P\) with respect to \(x\) and \(Q\) with respect to \(y\), and then combining these, one can find the potential function for the vector field. In our exercise, after integrating both components separately and finding the overlapping terms, the potential function was determined as \(f=\frac{x^{2}}{2} - x^{2} y + \frac{y^{3}}{3} + C\). This function is crucial in fields like physics and engineering, enabling calculations of energy for example.
Partial Derivatives
The concept of partial derivatives is fundamental when dealing with multivariable functions, which is the context in our exercise. They represent the rate of change of a function with respect to one variable while keeping other variables constant. For a function \(f(x, y)\), \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) would, respectively, be the derivatives of \(f\) with respect to \(x\) and \(y\), when \(y\) and \(x\) are held constant.
In the solution process described, partial derivatives are used to determine if a vector field is conservative. They're also used to calculate the components of the curl and the divergence, and critically, to find the potential function by integrating the given vector field components \(P\) and \(Q\).
In the solution process described, partial derivatives are used to determine if a vector field is conservative. They're also used to calculate the components of the curl and the divergence, and critically, to find the potential function by integrating the given vector field components \(P\) and \(Q\).
Other exercises in this chapter
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