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Chapter 13: Problem 21

\(21-24=\) Find the gradient vector field of \(f\) $$f(x, y)=x e^{x y}$$

Short Answer

Expert verified
The gradient vector field is \( \nabla f = \left( e^{xy}(1 + xy), x^2 e^{xy} \right) \).

Step by step solution

01

Understand the Gradient Vector Field

The gradient vector field of a scalar function \( f(x, y) \) is denoted by \( abla f(x, y) \) and represents a vector field where each vector points in the direction of the greatest rate of increase of the function. It is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
02

Differentiate with Respect to x

Compute the partial derivative of \( f(x, y) = x e^{xy} \) with respect to \( x \). Using the product rule and chain rule, we get: \[ \frac{\partial f}{\partial x} = e^{xy} + x \cdot y \cdot e^{xy} = e^{xy}(1 + xy) \].
03

Differentiate with Respect to y

Find the partial derivative of \( f(x, y) = x e^{xy} \) with respect to \( y \). Applying the chain rule, we obtain: \[ \frac{\partial f}{\partial y} = x \cdot x \cdot e^{xy} = x^2 e^{xy} \].
04

Form the Gradient Vector Field

Combine the partial derivatives from Steps 2 and 3 to form the gradient vector field: \[ abla f = \left( e^{xy}(1 + xy), x^2 e^{xy} \right) \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivative
When studying gradient vector fields, it's essential to understand partial derivatives as they form the building blocks for gradients. A partial derivative measures how a function changes as one of its variables changes while keeping the other variables constant. For a function like \( f(x, y) = x e^{xy} \), you compute the partial derivatives with respect to \( x \) and \( y \) separately.
  • To find \( \frac{\partial f}{\partial x} \), differentiate as if \( y \) is a constant.
  • Similarly, for \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant.
These derivatives help describe the behavior of the function in multiple dimensions, capturing the slope along each variable direction. This process directly feeds into forming the gradient vector, which combines these directional changes.
Chain Rule
The chain rule is a crucial mathematical tool when dealing with functions that are composed of other functions. It allows us to differentiate composite functions by breaking them down into simpler parts. For our function \( f(x, y) = x e^{xy} \), the exponential component \( e^{xy} \) is a composite function, where \( xy \) acts as the inner function.
  • The chain rule helps differentiate \( e^{xy} \) by first applying the derivative of the exponential function.
  • It then multiplies by the derivative of the inner function \( xy \), which involves differentiating \( x \) with respect to \( x \) or \( y \), depending on the variable you're considering for the partial derivative.
Through the chain rule, we efficiently handle complex differentiations within the context of multivariable calculus.
Product Rule
Sometimes, even differentiating a seemingly simple function like \( x e^{xy} \) involves interactions between its components. This is where the product rule comes in handy. The product rule is used when you need to differentiate expressions that are the multiplication of two or more functions.
  • For a function \( f(x, y) = x e^{xy} \), treat \( x \) and \( e^{xy} \) as two separate functions.
  • According to the product rule, differentiate each component separately and then add their cross-products.
In our case, when calculating \( \frac{\partial f}{\partial x} \), you differentiate \( x \) and \( e^{xy} \) separately and combine them using the product rule: \( x' \times e^{xy} + x \times (e^{xy})' \). This ensures all interactions between the parts of the function are accounted for, leading to accurate differentiation results.

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Most popular questions from this chapter

Evaluate the surface integral \(\iint_{s} \mathbf{F} \cdot d \mathbf{S}\) for the given vector field \(\mathbf{F}\) and the oriented surface \(S .\) In other words, find the flux of \(\mathbf{F}\) across \(S .\) For closed surfaces, use the positive (outward) orientation. $$\begin{array}{l}{\mathbf{F}(x, y, z)=y \mathbf{j}-z \mathbf{k}} \\ {S \text { consists of the paraboloid } y=x^{2}+z^{2}, 0 \leq \leqslant y \leqslant 1} \\\ {\text { and the disk } x^{2}+z^{2} \leqslant 1, y=1}\end{array}$$

Evaluate the surface integral. $$\begin{array}{l}{\iint_{S} y^{2} d S} \\ {S \text { is the part of the sphere } x^{2}+y^{2}+z^{2}=4 \text { that lies }} \\ {\text { inside the cylinder } x^{2}+y^{2}=1 \text { and above the } x y \text { -plane }}\end{array}$$

Maxwell's equations relating the electric field \(\mathbf{E}\) and mag- netic field \(\mathbf{H}\) as they vary vith time in a region containing no charge and no current can be stated as follows: $$\begin{array}{ll}{\operatorname{div} \mathbf{E}=0} & {\operatorname{div} \mathbf{H}=0} \\ {\operatorname{curl} \mathbf{E}=-\frac{1}{c} \frac{\partial \mathbf{H}}{\partial t}} & {\operatorname{curl} \mathbf{H}=\frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}}\end{array}$$ where \(c\) is the speed of light. Use these equations to prove the following: $$\begin{array}{l}{\quad\left(\text { a) } \nabla \times(\nabla \times \mathbf{E})=-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}\right.} \\ {\text { (b) } \nabla \times(\nabla \times \mathbf{H})=-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{H}}{\partial t^{2}}} \\\ {\text { (c) } \nabla^{2} \mathbf{E}=\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}} \quad[\text {Hint} \text { : Use Exercise } 27 .]} \\\ {\text { (d) } \nabla^{2} \mathbf{H}=\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{H}}{\partial t^{2}}}\end{array}$$

Find parametric equations for the surface obtained by rotating the curve \(x=4 y^{2}-y^{4},-2 \leqslant y \leqslant 2,\) about the \(y\) -axis and use them to graph the surface.

The surface with parametric equations $$x=2 \cos \theta+r \cos (\theta / 2)$$ $$y=2 \sin \theta+r \cos (\theta / 2)$$ $$z=r \sin (\theta / 2)$$ where \(-\frac{1}{2} \leqslant r \leqslant \frac{1}{2}\) and \(0 \leqslant \theta \leqslant 2 \pi,\) is called a Mobius strip. Graph this surface with several viewpoints. What is unusual about it?
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