/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 37\. Determine all the relative ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 37

37\. Determine all the relative extrema for \(V=x^{3}+y^{2}-\) \(3 x y+(3 / 2) x\), and for each one determine whether it is a maximum or minimum. Compare your answer with that of Euler.

Short Answer

Expert verified
Answer: The function V has one relative minimum at the point \((1, \frac{3}{2})\).

Step by step solution

01

Find the partial derivatives

To find the partial derivatives, we will differentiate V with respect to x and y, keeping the other variable constant. The partial derivatives are given by: $$\frac{\partial V}{\partial x} = 3x^2 - 3y + \frac{3}{2}$$ $$\frac{\partial V}{\partial y} = 2y - 3x$$
02

Find the critical points

To find the critical points, we need to set both partial derivatives equal to zero and solve for x and y: $$3x^2 - 3y + \frac{3}{2} = 0$$ $$2y - 3x = 0$$ Solving this system of equations, we get two critical points: \((x_1, y_1) = (0, 0)\) and \((x_2, y_2) = (1, \frac{3}{2})\).
03

Apply the second partial derivative test

To apply the second partial derivative test, we need to find the second partial derivatives and calculate the determinant of the Hessian matrix. First, let's find the second partial derivatives: $$\frac{\partial^2 V}{\partial x^2} = 6x$$ $$\frac{\partial^2 V}{\partial y^2} = 2$$ $$\frac{\partial^2 V}{\partial x \partial y} = \frac{\partial^2 V}{\partial y \partial x} = -3$$ The determinant of the Hessian matrix is given by: $$D(x,y) = \frac{\partial^2 V}{\partial x^2} \cdot \frac{\partial^2 V}{\partial y^2} - \left(\frac{\partial^2 V}{\partial x \partial y}\right)^2 = (6x)(2) - (-3)^2$$ Now let's evaluate the determinant at both critical points: - At \((x_1, y_1) = (0, 0)\): $$D(0, 0) = (6 \cdot 0)(2) - (-3)^2 = 0 - 9 < 0$$ Since the determinant is negative, \((0, 0)\) is a saddle point, and hence not a relative extrema. - At \((x_2, y_2) = (1, \frac{3}{2})\): $$D(1, \frac{3}{2}) = (6 \cdot 1)(2) - (-3)^2 = 12 - 9 = 3 > 0$$ Since the determinant is positive, we need to check the value of \(\frac{\partial^2 V}{\partial x^2}\) at this point: $$\frac{\partial^2 V}{\partial x^2}(1, \frac{3}{2}) = 6 \cdot 1 = 6 > 0$$ Since \(\frac{\partial^2 V}{\partial x^2} > 0\), the point \((1, \frac{3}{2})\) is a relative minimum of V. In conclusion, we found one relative extrema for the function V: a minimum at \((1, \frac{3}{2})\). This result is consistent with Euler's.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Partial Derivatives
Partial derivatives are essential in calculus, especially when dealing with functions of multiple variables. In this exercise, the function \(V(x, y) = x^3 + y^2 - 3xy + \frac{3}{2}x\) involves two variables, \(x\) and \(y\). When we compute the partial derivative with respect to \(x\), we treat \(y\) as a constant. Conversely, computing the partial derivative with respect to \(y\) treats \(x\) as a constant.

  • For \(\frac{\partial V}{\partial x}\), we differentiate each term that includes \(x\) while keeping \(y\) constant, yielding: \(3x^2 - 3y + \frac{3}{2}\).
  • For \(\frac{\partial V}{\partial y}\), we differentiate each term that includes \(y\), resulting in: \(2y - 3x\).
Once partial derivatives are calculated, they help us identify points where the function may attain a relative extremum or saddle point by solving \(\frac{\partial V}{\partial x} = 0\) and \(\frac{\partial V}{\partial y} = 0\). This is a crucial prelude to finding critical points.
Critical Points
Critical points occur where the gradient of the function \(V(x, y)\) is zero or undefined. In this exercise, the critical points are found by solving the system of equations obtained from setting both partial derivatives to zero.

  • \(3x^2 - 3y + \frac{3}{2} = 0\)
  • \(2y - 3x = 0\)
By solving these equations simultaneously, we obtain the critical points \((0, 0)\) and \((1, \frac{3}{2})\). At these locations, the function \(V\) potentially has local maxima, minima, or saddle points.

Critical points are the gateway to deeper analysis with the Hessian matrix and Second Partial Derivative Test, which help determine their nature.
Hessian Matrix
The Hessian matrix is a square matrix of second-order partial derivatives of a function of several variables. It gives insight into the local curvature of the function. For the given function \(V(x, y)\), the Hessian matrix, \(H\), is constructed as follows:

\[H = \begin{pmatrix}\frac{\partial^2 V}{\partial x^2} & \frac{\partial^2 V}{\partial x \partial y} \\frac{\partial^2 V}{\partial y \partial x} & \frac{\partial^2 V}{\partial y^2}\end{pmatrix} = \begin{pmatrix}6x & -3 \-3 & 2\end{pmatrix} \]
The Hessian matrix is primarily used in the Second Partial Derivative Test to determine the nature (maximum, minimum, or saddle point) of each critical point.
Second Partial Derivative Test
The Second Partial Derivative Test uses the Hessian matrix to classify critical points. For each critical point, the determinant \(D(x, y)\) of the Hessian matrix is calculated. This determinant helps decide the nature of the critical point:

  • If \(D(x, y) > 0\) and \(\frac{\partial^2 V}{\partial x^2} > 0\), the function has a local minimum.
  • If \(D(x, y) > 0\) and \(\frac{\partial^2 V}{\partial x^2} < 0\), the function has a local maximum.
  • If \(D(x, y) < 0\), the point is a saddle point, indicating neither a local maximum nor minimum.
Evaluating the determinant at the critical points of \(V(x, y)\) found that \((0, 0)\) is a saddle point, as \(D(0,0) < 0\). At \((1, \frac{3}{2})\), the positive determinant combined with \(\frac{\partial^2 V}{\partial x^2} > 0\) confirms a local minimum, aligning with Euler's original findings.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The modern way to derive Kepler's area law is to break the force into its radial and transverse components, rather than the tangential and normal components used by Hermann, and use polar coordinates whose origin is the center of force. Assume then that the center of force is at the origin of a polar coordinate system. Using vector notation, \(\operatorname{set} \mathbf{u}_{r}=\mathbf{i} \cos \theta+\mathbf{j} \sin \theta\) and \(\mathbf{u}_{\theta}=-\mathbf{i} \sin \theta+j \cos \theta .\) Show that \(d \mathbf{u}_{r} / d \theta=\mathbf{u}_{\theta}\) and \(d \mathbf{u}_{\theta} / d \theta=-\mathbf{u}_{r}\). Then show that if \(\mathbf{r}=r \mathbf{u}_{r}\), then the velocity \(\mathbf{v}\) is given by \(r(d \theta / d t) \mathbf{u}_{\theta}+\) \((d r / d t) \mathbf{u}_{r}\). Show next that the radial component \(a_{r}\) and the transverse component \(a_{\theta}\) of the acceleration are given by $$ a_{r}=\frac{d^{2} r}{d t^{2}}-r\left(\frac{d \theta}{d t}\right)^{2} \quad \text { and } \quad a_{\theta}=r \frac{d^{2} \theta}{d t^{2}}+2 \frac{d r}{d t} \frac{d \theta}{d t} $$ Since the force is central, \(a_{\theta}=0 .\) Multiply the differential equation expressing that fact by \(r\) and integrate to get \(r^{2} \frac{d \theta}{d t}=k\), where \(k\) is a constant. Show finally that \(r^{2} \frac{d \theta}{d t}=\) \(\frac{d A}{d t}\), where \(A\) is the area swept out by the radius vector. This proves Kepler's law of areas.

24\. Suppose that the solution to the wave equation \(\frac{2^{2} y}{\partial t^{2}}=\frac{\partial^{2} y}{\partial x^{2}}\) is given by \(y=\Psi(t+x)-\Psi(t-x)\). Show that the initial conditions \(y(0, x)=f(x), y^{\prime}(0, x)=g(x)\) and the condition \(y(t, 0)=y(t, l)=0\) for all \(t\) lead to the requirements that \(f(x)\) and \(g(x)\) are odd functions of period \(2 l\) (d'Alembert).

Show that the equation \(a \pm c x / b=\sqrt{x^{2}+y^{2}}\) is a parabola if \(b=c\), is an ellipse if \(b>c\), and is a hyperbola if \(b

Determine a procedure for finding the differential equation of a family of orthogonal trajectories to a given family \(f(x, y, \alpha)=0\). (Use the fact that orthogonal lines have negative reciprocal slopes.) Use your procedure to find the family orthogonal to the family of hyperbolas \(x^{2}-y^{2}=a^{2}\). 20\. Determine and solve the differential equation for the family of synchrones, the family orthogonal to the family of brachistochrones.

Did eighteenth-century mathematicians prove or use the fundamental theorem of calculus in the sense it is used today? What concepts must be defined before one can even consider this theorem? How are these concepts dealt with by eighteenth-century mathematicians? Did these mathematicians consider the fundamental theorem as "fundamental"?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.