Problem 249 - Calculus Volume 3

Question

Find the divergence of \(\mathbf{F}\) at the given point. $$ \mathbf{F}(x, y, z)=e^{x} \sin y \mathbf{i}-e^{x} \cos y \mathbf{j} \text { at }(0,0,3) $$

Step-by-Step Solution

Answer

The divergence of \(\mathbf{F}\) at \((0,0,3)\) is 0.

Step 1: Recall the Divergence Formula

The divergence of a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \) is calculated using the formula \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \). Our given vector field is \( \mathbf{F}(x, y, z) = e^{x} \sin y \mathbf{i} - e^{x} \cos y \mathbf{j} \). Note that there's no \( \mathbf{k} \) component, so \( R = 0 \).

Step 2: Identify Components of \( \mathbf{F} \)

From the vector field \( \mathbf{F}(x, y, z) = e^{x} \sin y \mathbf{i} - e^{x} \cos y \mathbf{j} \), identify the components: \( P = e^x \sin y \), \( Q = -e^x \cos y \), and \( R = 0 \).

Step 3: Compute Partial Derivative \( \frac{\partial P}{\partial x} \)

Differentiating \( P = e^x \sin y \) with respect to \( x \), we obtain \( \frac{\partial P}{\partial x} = e^x \sin y \).

Step 4: Compute Partial Derivative \( \frac{\partial Q}{\partial y} \)

Differentiating \( Q = -e^x \cos y \) with respect to \( y \), we find that \( \frac{\partial Q}{\partial y} = e^x \sin y \).

Step 5: Compute Partial Derivative \( \frac{\partial R}{\partial z} \)

Since \( R = 0 \), the partial derivative \( \frac{\partial R}{\partial z} = 0 \).

Step 6: Calculate the Divergence

Substitute the partial derivatives into the divergence formula: \( abla \cdot \mathbf{F} = e^x \sin y + e^x \sin y + 0 = 2e^x \sin y \).

Step 7: Evaluate at the Point (0, 0, 3)

Substitute \( x = 0 \), \( y = 0 \) into the expression for divergence: \( 2e^0 \sin 0 = 2 \times 1 \times 0 = 0 \). The divergence at the point \((0,0,3)\) is 0.

Key Concepts

Vector Field

Imagine a vector field as a space filled with arrows. Each arrow has a direction and a magnitude (length). Think of the arrows as showing the direction in which something flows and how strong or fast that flow is at each point. In our specific case, the arrow directions and strengths are described by mathematical expressions involving variables like \(x\), \(y\), and \(z\).
In the vector field \(\mathbf{F}(x, y, z) = e^x \sin y \mathbf{i} - e^x \cos y \mathbf{j}\), each point has an associated vector (or arrow). This vector doesn't change with the \(z\) coordinate since the \(\mathbf{k}\) component is zero. We only consider the \(\mathbf{i}\) and \(\mathbf{j}\) components to understand how the vector field behaves.

• The \(\mathbf{i}\) component (\(e^x \sin y\)) shows how the field behaves along the x-axis.
• The \(\mathbf{j}\) component (\(-e^x \cos y\)) shows how the field changes along the y-axis.
• There's no change along the z-axis since \(R = 0\).

Partial Derivative

Partial derivatives help us explore how a function changes when we tweak one of its variables, while keeping others constant. They're like stepping stones into understanding complex surfaces and their behavior in a multivariable space.
For our vector field, calculating the partial derivatives involves focusing on one part of the vector at a time:

• \(\frac{\partial P}{\partial x}\) focuses on how the \(\mathbf{i}\) component changes when \(x\) varies, independent of \(y\) and \(z\).
• \(\frac{\partial Q}{\partial y}\) captures how the \(\mathbf{j}\) component responds when \(y\) varies, keeping \(x\) and \(z\) constant.
• The \(\mathbf{k}\) component is zero, so \(\frac{\partial R}{\partial z} = 0\), indicating no change in the z direction.

By assessing each of these derivatives, we understand how the vector field's behavior shifts based on its input variables.

Calculus

Calculus provides us with the tools to analyze changing systems. With concepts like differentiation, we explore how quantities erratically grow or shrink. In the context of vector fields, calculus lets us measure how vectors diverge or converge at any given point.
The critical aspect here is to grasp how differentiation helps define the intrinsic behavior of fields. By using derivatives:

• We identify the rate at which one component affects another.
• We explore variations of the vector field using these rates.
• Applying calculus, particularly through partial derivatives, helps with finding the divergence.

The divergence essentially tells us if a particular point in the field acts as a source or a sink for vectors. If vectors are drawing away from a point, we see positive divergence. Conversely, if they're converging, it's negative; when neither happens, it's zero, as in our example.

Multivariable Calculus

Multivariable calculus expands the horizon by considering scenarios where functions depend on multiple inputs. Here, we aren't merely changing one direction, but several, making it crucial to see how interdependencies work between dimensions.
This topic unifies all these concepts, as it's the discipline governing how we understand forces and energies in systems with numerous variables:

• We use multivariable calculus to untangle real-world situations into understandable models.
• By involving more than one variable, we uncover deeper insights that linear functions can't reveal.
• Divergence in a multivariable context tells us about volume flows and field sources.

Ultimately, through multivariable calculus, we bridge the abstract with tangible scenarios, allowing us to calculate and predict behaviors in advanced fields like physics and engineering.