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Chapter 12: Problem 33

Find the critical points of the following functions. Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, local minimum, or saddle point. Confirm your results using a graphing utility. \(f(x, y)=y e^{x}-e^{y}\)

Short Answer

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Question: Determine the critical points of the function \(f(x, y) = ye^x - e^y\) and classify them as local maxima, local minima, or saddle points. Solution: We followed a step-by-step procedure that involved: 1. Finding the partial derivatives of the function \(f(x, y)\) with respect to x and y. 2. Setting the partial derivatives equal to zero and solving the system of equations to find the critical points. 3. Applying the Second Derivative Test to classify the critical points as local maxima, local minima, or saddle points. 4. Confirming the result with a graphing utility. The critical point of the function is \((0,0)\), and according to the Second Derivative Test, it is a saddle point.

Step by step solution

01

1. Find the partial derivatives

To find the critical points, first we need to find the partial derivatives of the function \(f(x, y) = ye^x - e^y\) with respect to x and y. \(\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(y e^x - e^y) = ye^x\) \(\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(y e^x - e^y) = e^x - e^y\)
02

2. Find the critical points

To find the critical points, set both partial derivatives equal to zero, and solve for x and y. \(ye^x = 0\) (partial derivative respect to x) \(e^x - e^y = 0\) (partial derivative respect to y) From the first equation, we can see that \(y=0\) or \(e^x=0\). However, \(e^x\) can never be 0, because the exponential function is always positive. Thus, \(y=0\) is the only possibility, and plugging this into the second equation: \(e^x - e^0 = e^x - 1 = 0 \Rightarrow x = 0\) Thus, we have found a single critical point: \((0, 0)\).
03

3. Apply the Second Derivative Test

To apply the Second Derivative Test, we must find the second-order partial derivatives. \(\frac{\partial ^2 f}{\partial x^2} = \frac{\partial}{\partial x} \left( ye^x \right) = ye^x\) \(\frac{\partial ^2 f}{\partial y^2} = \frac{\partial}{\partial y} \left( e^x - e^y \right) = -e^y\) \(\frac{\partial ^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} \left( e^x - e^y \right) = e^x\) (equal to \(\frac{\partial^2 f}{\partial y \partial x}\) due to Schwarz's Theorem) Now, we will use these to compute the discriminant: \(D(x, y) = \left(\frac{\partial^2 f}{\partial x^2}\right) \left(\frac{\partial^2 f}{\partial y^2}\right) - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2\) At the critical point \((0,0)\): \(D(0,0) = (0) (-1) - (1)^2 = -1\) Since \(D(0,0) < 0\), the Second Derivative Test tells us that the critical point is a saddle point.
04

4. Confirm the result with a graphing utility

Use a graphing utility to plot the function \(f(x, y) = y e^x - e^y\). You should see that it confirms the results we found analytically: there is a saddle point at \((0,0)\).

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Key Concepts

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

Partial Derivatives
When analyzing multi-variable functions like the one given by the function f(x, y) = ye^x - e^y, partial derivatives are key. They measure how the function changes as each variable changes, while all other variables are held constant. To find these, we differentiate f(x, y) with respect to 'x' while treating 'y' as a constant and vice versa. The resulting partial derivatives 鈭俧/鈭倄 and 鈭俧/鈭倅 are the slopes of the tangent lines parallel to the x-axis and the y-axis, respectively.

These derivatives are crucial for locating critical points, which are points where the partial derivatives are zero or do not exist. In this example, the partial derivatives are ye^x and e^x - e^y. Solving these for zero helps us find the critical points of the function, paving the way to understanding the function's behavior more deeply.
Second Derivative Test
The Second Derivative Test is a handy tool for classifying the nature of critical points found in multivariable calculus. After identifying the critical points by setting the first partial derivatives to zero, we move onto the second partial derivatives. These second-order derivatives 鈭偮瞗/鈭倄虏, 鈭偮瞗/鈭倅虏, and mixed partial derivatives 鈭偮瞗/鈭倄鈭倅 provide information on the curvature of the surface the function represents.

We calculate the discriminant using these second derivatives, which can tell us if a critical point is a local maximum, local minimum, or a saddle point. For our given function, the discriminant at the critical point (0,0) is negative, indicating the presence of a saddle point.
Saddle Point
A saddle point is a particular type of critical point on a surface where the concavity changes direction. Unlike local maxima or minima, where the surface curves upwards or downwards respectively, a saddle point curves up in one direction and down in another. Imagine a mountain pass or the shape of a saddle on a horse, hence the name.

In the context of our function, after applying the Second Derivative Test, a negative discriminant at the critical point (0,0) confirms that this point is indeed a saddle point. Here, the function doesn't achieve a local extreme value but rather forms a 'valley' in one direction and a 'hill' in another, right at the critical point.
Exponential Functions
Exponential functions, like e^x in our given function f(x, y) = ye^x - e^y, have noteworthy properties. They are always positive, regardless of the input value, which indicates that e^x is never zero. This behavior of the exponential function helps us swiftly conclude that in the equation ye^x = 0, only y can be zero. Notably, exponential functions exhibit a constant rate of growth or decay and this unique quality makes them inherently important in various applications such as compound interest, population growth models, and decay processes.
Discriminant in Calculus
The discriminant in calculus plays a significant role in the Second Derivative Test. It's calculated using the formula D(x, y) = (f_xx * f_yy) - (f_xy)^2, where f_xx and f_yy are the second partial derivatives of f with respect to 'x' and 'y', respectively, and f_xy is the mixed partial derivative. The sign of the discriminant reveals whether the critical point is a local maximum (positive), local minimum (also positive), or a saddle point (negative). In our example, since the discriminant at the critical point is negative, it confirmed the existence of a saddle point.
Graphing Utility
A graphing utility can be an invaluable resource for visualizing functions and confirming analytical results. By plotting the function f(x, y) = ye^x - e^y, we can graphically confirm the presence of a saddle point at (0,0). Visualizing the surface can enhance our understanding of the function's behavior near the critical point and provide a graphical verification of the conclusions we've drawn from the Second Derivative Test. This approach is especially helpful for complex functions where the behavior is not immediately obvious from the equation alone.

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Most popular questions from this chapter

(Adapted from 1938 Putnam Exam) Consider the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1\) and the plane \(P\) given by \(A x+B y+C z+1=0\). Let \(h=\left(A^{2}+B^{2}+C^{2}\right)^{-1 / 2}\) and \(m=\left(a^{2} A^{2}+b^{2} B^{2}+c^{2} C^{2}\right)^{1 / 2}\) a. Find the equation of the plane tangent to the ellipsoid at the point \((p, q, r)\) b. Find the two points on the ellipsoid at which the tangent plane is parallel to \(P\) and find equations of the tangent planes. c. Show that the distance between the origin and the plane \(P\) is \(h\) d. Show that the distance between the origin and the tangent planes is \(h m\) e. Find a condition that guarantees that the plane \(P\) does not intersect the ellipsoid.

A function of one variable has the property that a local maximum (or minimum) occurring at the only critical point is also the absolute maximum (or minimum) (for example, \(f(x)=x^{2}\) ). Does the same result hold for a function of two variables? Show that the following functions have the property that they have a single local maximum (or minimum), occurring at the only critical point, but that the local maximum (or minimum) is not an absolute maximum (or minimum) on \(\mathbb{R}^{2}\). a. \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) b. \(f(x, y)=\left(2 y^{2}-y^{4}\right)\left(e^{x}+\frac{1}{1+x^{2}}\right)-\frac{1}{1+x^{2}}\) This property has the following interpretation. Suppose that a surface has a single local minimum that is not the absolute minimum. Then water can be poured into the basin around the local minimum and the surface never overflows, even though there are points on the surface below the local minimum. (Source: Mathematics Magazine, May 1985, and Calculus and Analytical Geometry, 2nd ed., Philip Gillett, 1984)

Determine whether the following statements are true and give an explanation or counterexample. a. The plane passing through the point (1,1,1) with a normal vector \(\mathbf{n}=\langle 1,2,-3\rangle\) is the same as the plane passing through the point (3,0,1) with a normal vector \(\mathbf{n}=\langle-2,-4,6\rangle\) b. The equations \(x+y-z=1\) and \(-x-y+z=1\) describe the same plane. c. Given a plane \(Q\), there is exactly one plane orthogonal to \(Q\). d. Given a line \(\ell\) and a point \(P_{0}\) not on \(\ell\), there is exactly one plane that contains \(\ell\) and passes through \(P_{0}\). e. Given a plane \(R\) and a point \(P_{0}\), there is exactly one plane that is orthogonal to \(R\) and passes through \(P_{0}\) f. Any two distinct lines in \(\mathbb{R}^{3}\) determine a unique plane. g. If plane \(Q\) is orthogonal to plane \(R\) and plane \(R\) is orthogonal to plane \(S\), then plane \(Q\) is orthogonal to plane \(S\).

Potential functions arise frequently in physics and engineering. A potential function has the property that \(a\) field of interest (for example, an electric field, a gravitational field, or a velocity field is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter 14 .) The electric field due to a point charge of strength \(Q\) at the origin has a potential function \(\varphi=k Q / r,\) where \(r^{2}=x^{2}+y^{2}+z^{2}\) is the square of the distance between a variable point \(P(x, y, z)\) and the charge, and \(k>0\) is a physical constant. The electric field is given by \(\mathbf{E}=-\nabla \varphi,\) where \(\nabla \varphi\) is the gradient in three dimensions. a. Show that the three-dimensional electric field due to a point charge is given by $$ \mathbf{E}(x, y, z)=k Q\left\langle\frac{x}{r^{3}}, \frac{y}{r^{3}}, \frac{z}{r^{3}}\right\rangle $$ b. Show that the electric field at a point has a magnitude \(|\mathbf{E}|=k Q / r^{2} .\) Explain why this relationship is called an inverse square law.

Traveling waves (for example, water waves or electromagnetic waves) exhibit periodic motion in both time and position. In one dimension, some types of wave motion are governed by the one-dimensional wave equation $$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u(x, t)\) is the height or displacement of the wave surface at position \(x\) and time \(t,\) and \(c\) is the constant speed of the wave. Show that the following functions are solutions of the wave equation. \(u(x, t)=A f(x+c t)+B g(x-c t),\) where \(A\) and \(B\) are constants and \(f\) and \(g\) are twice differentiable functions of one variable
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