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Chapter 14: Problem 90

Show that each function in Exercises \(83-90\) satisfies a Laplace equation. $$f(x, y, z)=e^{3 x+4 y} \cos 5 z$$

Short Answer

Expert verified
The function satisfies Laplace's equation because the sum of its second partial derivatives is zero.

Step by step solution

01

Understand the Laplace Equation

A function satisfies Laplace's equation if the sum of its second partial derivatives with respect to each variable equals zero:\[\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} = 0\] Identify the given function: \(f(x, y, z) = e^{3x + 4y} \cos 5z\).
02

Compute the First Partial Derivatives

Find the first partial derivatives with respect to \(x\), \(y\), and \(z\):\[\frac{\partial f}{\partial x} = 3e^{3x + 4y} \cos 5z \\frac{\partial f}{\partial y} = 4e^{3x + 4y} \cos 5z \\frac{\partial f}{\partial z} = -5e^{3x + 4y} \sin 5z\] These derivatives are needed to find the second partial derivatives.
03

Compute the Second Partial Derivatives

Calculate the second partial derivatives using the first derivatives:\[\frac{\partial^2 f}{\partial x^2} = 9e^{3x + 4y} \cos 5z,\frac{\partial^2 f}{\partial y^2} = 16e^{3x + 4y} \cos 5z,\frac{\partial^2 f}{\partial z^2} = -25e^{3x + 4y} \cos 5z\] These are the terms needed to verify the Laplace equation.
04

Summation of Second Partial Derivatives

Add the second partial derivatives:\[9e^{3x + 4y} \cos 5z + 16e^{3x + 4y} \cos 5z - 25e^{3x + 4y} \cos 5z = 0\] Simplify the expression:\[(9 + 16 - 25)e^{3x + 4y} \cos 5z = 0\] It simplifies to 0, which verifies the function satisfies Laplace's equation.

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental concept in multivariable calculus. They involve taking the derivative of a function with respect to one variable while holding others constant. This is essential when dealing with functions of more than one variable, such as the function given in the exercise: \(f(x, y, z) = e^{3x + 4y} \cos 5z\). When calculating partial derivatives:
  • The partial derivative with respect to \(x\) is found by differentiating while treating \(y\) and \(z\) as constants. For our function, \(\frac{\partial f}{\partial x} = 3e^{3x + 4y} \cos 5z\).
  • Similarly, for \(y\), differentiating with \(x\) and \(z\) held constant gives \(\frac{\partial f}{\partial y} = 4e^{3x + 4y} \cos 5z\).
  • Finally, the derivative with respect to \(z\) results in \(\frac{\partial f}{\partial z} = -5e^{3x + 4y} \sin 5z\), highlighting how the trigonometric component reacts to differentiation.
Partial derivatives help describe how a function changes as one of its variables changes, providing insights into the function's behavior across dimensions.
Second Partial Derivatives
Calculating second partial derivatives involves taking the derivative of a first partial derivative again, with respect to the same variable. These are necessary for checking conditions like Laplace's equation.In our scenario, we first derived:
  • \(\frac{\partial f}{\partial x} = 3e^{3x + 4y} \cos 5z\)
  • \(\frac{\partial f}{\partial y} = 4e^{3x + 4y} \cos 5z\)
  • \(\frac{\partial f}{\partial z} = -5e^{3x + 4y} \sin 5z\)
The next step is differentiating each again:
  • \(\frac{\partial^2 f}{\partial x^2} = 9e^{3x + 4y} \cos 5z\)
  • \(\frac{\partial^2 f}{\partial y^2} = 16e^{3x + 4y} \cos 5z\)
  • \(\frac{\partial^2 f}{\partial z^2} = -25e^{3x + 4y} \cos 5z\)
Second partial derivatives are vital because they reveal the curvature or concavity of the function with respect to each variable. In optimization and physics, these derivatives help determine behavior at critical points.
Function Verification
Function verification, in the context of our exercise, focuses on validating that the given function satisfies a specific condition鈥攊n this case, Laplace's equation. This equation requires the sum of the second partial derivatives to equal zero.For the function \(f(x, y, z) = e^{3x + 4y} \cos 5z\), we computed:
  • \(\frac{\partial^2 f}{\partial x^2} = 9e^{3x + 4y} \cos 5z\)
  • \(\frac{\partial^2 f}{\partial y^2} = 16e^{3x + 4y} \cos 5z\)
  • \(\frac{\partial^2 f}{\partial z^2} = -25e^{3x + 4y} \cos 5z\)
Combining these, we get:\[ 9e^{3x + 4y} \cos 5z + 16e^{3x + 4y} \cos 5z - 25e^{3x + 4y} \cos 5z = 0 \]This simplification to zero confirms that the function indeed satisfies Laplace's equation, showcasing how mathematical verification ensures accuracy and conformity with theoretical conditions.

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Most popular questions from this chapter

The condition \(\nabla f=\lambda \nabla g\) is not sufficient Although \(\nabla f=\lambda \nabla g\) is a necessary condition for the occurrence of an extreme value of \(f(x, y)\) subject to the conditions \(g(x, y)=0\) and \(\nabla g \neq \mathbf{0},\) it does not in itself guarantee that one exists. As a case in point, try using the method of Lagrange multipliers to find a maximum value of \(f(x, y)=x+y\) subject to the constraint that \(x y=16 .\) The method will identify the two points \((4,4)\) and \((-4,-4)\) as candidates for the location of extreme values. Yet the sum \(x+y\) has no maximum value on the hyperbola \(x y=16\) The farther you go from the origin on this hyperbola in the first quadrant, the larger the sum \(f(x, y)=x+y\) becomes.

Find the critical point of $$f(x, y)=x y+2 x-\ln x^{2} y$$ in the open first quadrant \((x>0, y>0)\) and show that \(f\) takes on a minimum there.

Establish the fact, widely used in hydrodynamics, that if \(f(x, y, z)=0,\) then $$ \left(\frac{\partial x}{\partial y}\right)_{z}\left(\frac{\partial y}{\partial z}\right)_{x}\left(\frac{\partial z}{\partial x}\right)_{y}=-1 $$ (Hint: Express all the derivatives in terms of the formal partial derivatives \(\partial f / \partial x, \partial f / \partial y,\) and \(\partial f / \partial z . )\)

Consider the function \(f(x, y)=x^{2}+y^{2}+2 x y-x-y+1\) over the square \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\) a. Show that \(f\) has an absolute minimum along the line segment \(2 x+2 y=1\) in this square. What is the absolute minimum value? b. Find the absolute maximum value of \(f\) over the square.

Let \(f(x, y)=\left\\{\begin{array}{ll}{0,} & {x^{2} < y < 2 x^{2}} \\ {1,} & {\text { otherwise. }}\end{array}\right.\) Show that \(f_{x}(0,0)\) and \(f_{y}(0,0)\) exist, but \(f\) is not differentiable at \((0,0) .\)
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