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Chapter 12: Problem 13

Compute the gradient of the following functions and evaluate it at the given point \(P\). $$f(x, y)=x e^{2 x y} ; P(1,0)$$

Short Answer

Expert verified
Question: Determine the gradient of the function \(f(x, y) = x e^{2xy}\) at the point \(P(1, 0)\). Answer: The gradient of the function at the point \(P(1, 0)\) is \(\left\< 1, 0 \right\>\).

Step by step solution

01

Compute the partial derivative of f with respect to x

First, we'll find the partial derivative of \(f\) with respect to \(x\). We treat \(y\) as a constant: $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x e^{2xy})$$ Apply the product rule of differentiation: $$\frac{\partial f}{\partial x} = e^{2xy} + x\frac{\partial}{\partial x}(e^{2xy})$$ Now, differentiate the exponential term while treating \(2xy\) as a function of x: $$\frac{\partial f}{\partial x} = e^{2xy} + x(2y \cdot e^{2xy})$$ Simplify: $$\frac{\partial f}{\partial x} = e^{2xy}(1 + 2xy)$$
02

Compute the partial derivative of f with respect to y

Next, we'll find the partial derivative of \(f\) with respect to \(y\). We treat \(x\) as a constant: $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x e^{2xy})$$ Since there is only one term with respect to \(y\), apply the chain rule of differentiation: $$\frac{\partial f}{\partial y} = x(2x \cdot e^{2xy})$$ Simplify: $$\frac{\partial f}{\partial y} = 2x^2y e^{2xy}$$
03

Form the gradient

The gradient is a vector that combines the partial derivatives with respect to \(x\) and \(y\). So, the gradient of \(f(x, y)\) is given by: $$\nabla f(x, y) = \left\< \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\>$$ Substitute the partial derivatives: $$\nabla f(x, y) = \left\< e^{2xy}(1 + 2xy), 2x^2y e^{2xy} \right\>$$
04

Evaluate the gradient at the point P

Now, we need to evaluate the gradient at the given point \(P(1,0)\). Put the values of \(x=1\) and \(y=0\) into the gradient: $$\nabla f(1, 0) = \left\< e^{2(1)(0)}(1 + 2(1)(0)), 2(1)^2(0) e^{2(1)(0)} \right\>$$ Simplify: $$\nabla f(1, 0) = \left\< 1, 0 \right\>$$ So, the gradient of the function \(f(x, y) = x e^{2xy}\) at the point \(P(1, 0)\) is \(\left\< 1, 0 \right\>\).

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Key Concepts

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

Partial Derivatives
When working with functions of multiple variables, understanding the concept of partial derivatives is crucial. A partial derivative signifies the rate at which a function changes with respect to one variable, while keeping all other variables constant. Imagine you're in a hilly terrain and you only consider the slope in the east-west direction, disregarding any change in the north-south direction - that's analogous to finding a partial derivative.

Using the function from our problem, \( f(x, y) = x e^{2xy} \), we need to determine how \( f \) changes as \( x \) and \( y \) vary independently. To find the rate of change concerning \( x \), we calculate \( \frac{\partial f}{\partial x} \) by treating \( y \) as a constant value. This process requires applying various differentiation rules such as the product and chain rule, highlighting how derivatives for multivariable functions are calculated. As an advice for exercise improvement, make sure to understand the geometric interpretation of partial derivatives as slopes along the axes in the multi-dimensional space.
Product Rule of Differentiation
The product rule is a derivative rule used when a function is the product of two other functions. The rule states that the derivative of a product is the derivative of the first function times the second function plus the first function times the derivative of the second function. This rule is essential when dealing with functions like \( f(x, y) = x e^{2xy} \), where \( x \) and \( e^{2xy} \) are multiplied.

For the computation of \( \frac{\partial f}{\text{partial} x} \), we apply this rule and find that the derivative of the first function (\( x \)) is 1, and the derivative of the second function (\( e^{2xy} \)) needs the chain rule for differentiation because it's a composite function. When learning the product rule, remember that order does not matter; you can start with either function. Plotting functions and examining their product visually can also enhance understanding of how the product rule operates in calculus.
Chain Rule of Differentiation
The chain rule is indispensable in calculus, especially when dealing with composite functions - that is, functions within another function. When a function's variable is also expressed through another function, like \( e^{2xy} \) within our main function \( f \), the chain rule helps us find the derivative accurately.

To differentiate the term \( e^{2xy} \) concerning \( y \), we observe \( 2xy \) as a separate function inside the exponential function and thus, its own derivative is taken into account, multiplying it with the derivative of the outside function. This is precisely what happens during the calculation of \( \frac{\partial f}{\partial y} \). Notably, it's important to track and multiply each derivative accurately. Tips to avoid mistakes include clearly identifying the inner and outer functions before starting the differentiation process and practicing with various composite functions to gain confidence in applying the chain rule.

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Most popular questions from this chapter

Consider the following functions \(f\). a. Is \(f\) continuous at (0,0)\(?\) b. Is \(f\) differentiable at (0,0)\(?\) c. If possible, evaluate \(f_{x}(0,0)\) and \(f_{y}(0,0)\). d. Determine whether \(f_{x}\) and \(f_{y}\) are continuous at (0,0). e. Explain why Theorems 12.5 and 12.6 are consistent with the results in parts \((a)-(d)\). $$f(x, y)=\sqrt{|x y|}$$

The angle between two planes is the angle \(\theta\) between the normal vectors of the planes, where the directions of the normal vectors are chosen so that \(0 \leq \theta<\pi\) Find the angle between the planes \(5 x+2 y-z=0\) and \(-3 x+y+2 z=0\)

When two electrical resistors with resistance \(R_{1}>0\) and \(R_{2}>0\) are wired in parallel in a circuit (see figure), the combined resistance \(R,\) measured in ohms \((\Omega),\) is given by \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}.\) a. Estimate the change in \(R\) if \(R_{1}\) increases from \(2 \Omega\) to \(2.05 \Omega\) and \(R_{2}\) decreases from \(3 \Omega\) to \(2.95 \Omega\) b. Is it true that if \(R_{1}=R_{2}\) and \(R_{1}\) increases by the same small amount as \(R_{2}\) decreases, then \(R\) is approximately unchanged? Explain. c. Is it true that if \(R_{1}\) and \(R_{2}\) increase, then \(R\) increases? Explain. d. Suppose \(R_{1}>R_{2}\) and \(R_{1}\) increases by the same small amount as \(R_{2}\) decreases. Does \(R\) increase or decrease?

Find an equation for a family of planes that are orthogonal to the planes \(2 x+3 y=4\) and \(-x-y+2 z=8\)

Let \(w=f(x, y, z)=2 x+3 y+4 z\) which is defined for all \((x, y, z)\) in \(\mathbb{R}^{3}\). Suppose that we are interested in the partial derivative \(w_{x}\) on a subset of \(\mathbb{R}^{3}\), such as the plane \(P\) given by \(z=4 x-2 y .\) The point to be made is that the result is not unique unless we specify which variables are considered independent. a. We could proceed as follows. On the plane \(P\), consider \(x\) and \(y\) as the independent variables, which means \(z\) depends on \(x\) and \(y,\) so we write \(w=f(x, y, z(x, y)) .\) Differentiate with respect to \(x\) holding \(y\) fixed to show that \(\left(\frac{\partial w}{\partial x}\right)_{y}=18,\) where the subscript \(y\) indicates that \(y\) is held fixed. b. Alternatively, on the plane \(P,\) we could consider \(x\) and \(z\) as the independent variables, which means \(y\) depends on \(x\) and \(z,\) so we write \(w=f(x, y(x, z), z)\) and differentiate with respect to \(x\) holding \(z\) fixed. Show that \(\left(\frac{\partial w}{\partial x}\right)_{z}=8,\) where the subscript \(z\) indicates that \(z\) is held fixed. c. Make a sketch of the plane \(z=4 x-2 y\) and interpret the results of parts (a) and (b) geometrically. d. Repeat the arguments of parts (a) and (b) to find \(\left(\frac{\partial w}{\partial y}\right)_{x}\) \(\left(\frac{\partial w}{\partial y}\right)_{z},\left(\frac{\partial w}{\partial z}\right)_{x},\) and \(\left(\frac{\partial w}{\partial z}\right)_{y}\)
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