Question

Verify that each of the following force fields is conservative. Then find, for each, a scalar potential \(\phi\) such that \(\mathbf{F}=-\mathbf{\nabla} \phi\).\(F=2 x \cos ^{2} y i-\left(x^{2}+1\right) \sin 2 y j\)

Short answer

The force field is conservative, and the scalar potential is \(\varphi (x,y) = -x^2 \text{cos}^2 y - \cos 2y + c\).

Step-by-step solution

- Check if the Force Field is Conservative

A force field \textbf{F} is conservative if the curl of \textbf{F} is zero. Calculate the curl of \textbf{F}. For \textbf{F} = 2x \text{cos}^2 y \textbf{i} - (x^2 + 1) \text{sin} 2y \textbf{j}, the components are: \( F_1 = 2x \text{cos}^2 y \)\( F_2 = -(x^2 + 1) \text{sin} 2y \). Compute the partial derivatives: \( \frac{\text{d} F_1}{\text{d} y} = \frac{\text{d}}{\text{d} y} (2x \text{cos}^2 y) = -2x \text{sin} 2y \)\( \frac{\text{d} F_2}{\text{d} x} = \frac{\text{d}}{\text{d} x} [-(x^2 + 1) \text{sin} 2y] = -2x \text{sin} 2y \).Since \( \frac{\text{d} F_1}{\text{d} y} = \frac{\text{d} F_2}{\text{d} x} \), the force field is conservative.

- Find the Scalar Potential \varphi

To find the scalar potential \varphi, we know that \mathbf{F} = -\mathbf{\abla} \varphi. Therefore, \mathbf{F} represents the gradient of \varphi:\(\frac{\text{d} \varphi}{\text{d} x} = -2x \text{cos}^2 y \)\(\frac{\text{d} \varphi}{\text{d} y} = -(-(x^2 + 1) \text{sin} 2y) = (x^2 + 1) \text{sin} 2y \). Integrate \(\frac{\text{d} \varphi}{\text{d} x} \) with respect to x:\( \varphi (x,y) = \int -2x \text{cos}^2 y \text{d} x = -x^2 \text{cos}^2 y + g(y) \), where \g(y) is a function of y.Next, differentiate \varphi with respect to y and set equal to the corresponding component of \mathbf{F}:\( \frac{\text{d} \varphi}{\text{d} y} = \frac{\text{d}}{\text{d} y} (-x^2 \text{cos}^2 y + g(y)) \)\(-x^2 (-2 \text{cos} y \text{sin} y) + g'(y) = (x^2 + 1) \text{sin} 2y \).This simplifies to:\(2x^2 \text{sin} y \text{cos} y + g'(y) = (x^2 + 1) \text{sin} 2y \)Using the identity \text{sin} 2y = 2 \text{sin} y \text{cos} y:\(2x^2 \text{sin} y \text{cos} y + g'(y) = (x^2 + 1) 2 \text{sin} y \text{cos} y \).Simplify to:\(g'(y) = 2 \text{sin} y \text{cos} y \).Integrate g'(y) to find \g(y):\( g(y) = \int 2 \text{sin} y \text{cos} y \text{d} y = \int \text{sin} 2y \text{d} y = -\cos 2y + c\).So, the scalar potential is \varphi (x,y) = -x^2 \text{cos}^2 y - \cos 2y + c.

Key concepts

scalar potential

Let's start by understanding the scalar potential, often denoted by \( \varphi \). A scalar potential is a function whose gradient equals a given vector field. In simpler terms, if you take the gradient (we'll explain what that is later) of \( \varphi \), you'll get back the vector field. This \( \varphi \) essentially helps us describe a force field more simply.
The beauty of a scalar potential is that it helps us visualize complex force fields in a more straightforward manner. Imagine you have a hilly terrain; the scalar potential tells you the height at every point. The force field, on the other hand, shows you in which direction and how strongly a marble will roll down the hill. By understanding \( \varphi \), we're essentially understanding the 'landscape' of our force field.

gradient

Now, let's dive into the gradient. The gradient of a scalar function is a vector field that points in the direction of the greatest rate of increase of that function. Mathematically, if we have a scalar function \( \varphi \), the gradient of \( \varphi \) is denoted by \( \mathbf{abla} \varphi \).
The gradient is like taking a road map and converting it into a 3D model where each point has a steepness and direction. For a 2D function \( \varphi(x, y) \), its gradient is \[\mathbf{abla} \varphi = \left( \frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y} \right)\]. These components (partial derivatives) reveal how quickly the function changes in each direction.
This concept is crucial for understanding how a force field relates to its scalar potential. Simply put, the gradient tells us how steep or flat a surface is.

partial derivatives

Partial derivatives are a fundamental part of calculus, especially when dealing with functions of multiple variables. If we have a function \( f(x, y) \), its partial derivative with respect to \( x \) is represented by \( \frac{\partial f}{\partial x} \), and it tells us how \( f \) changes as only \( x \) changes, keeping \( y \) constant.
For example, in our original exercise, we had a vector field \( \mathbf{F} = (2x \cos^2 y) \mathbf{i} - (x^2 + 1) \sin 2y \mathbf{j} \). We computed partial derivatives \( \frac{\partial F_1}{\partial y} \) and \( \frac{\partial F_2}{\partial x} \) to check if the field is conservative. These partial derivatives help us understand how each component of the field changes with respect to different variables.
Understanding partial derivatives allows us to see the local behavior of functions and is vital for computing gradients and curls.

curl of a vector field

Finally, let's explore the curl of a vector field. The curl indicates how much and in what direction a vector field 'rotates' around a point. For a 2D vector field \( \mathbf{F} = (F_1, F_2) \), the curl is given by \[\mathbf{abla} \times \mathbf{F} = \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y},\] where the result is a scalar in 2D systems.
If the curl is zero everywhere in the field, it means no rotational tendency at any point, implying that the field is conservative. In our exercise, we confirmed the field was conservative by showing \( \frac{\partial F_2}{\partial x} = \frac{\partial F_1}{\partial y} = -2x \sin 2y \).
Understanding curl helps us determine if a force field can be described using a scalar potential. It’s an essential tool in vector calculus, helping to unveil deeper properties of fields.