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Critical points Find all critical points of the following functions. $$f(x, y)=\frac{x^{3}}{3}-\frac{y^{3}}{3}+3 x y$$

Short Answer

Expert verified
Question: Determine the critical points of the function, \(f(x, y) = \frac{x^{3}}{3} - \frac{y^{3}}{3} + 3xy\). Answer: The critical points are (0, 0), (-9, 3√3), and (-9, -3√3).

Step by step solution

01

1. Find the partial derivative with respect to x.

To find the partial derivative of the given function \(f(x, y)\) with respect to \(x\), we differentiate it with respect to \(x\) while treating \(y\) as a constant. $$\frac{\partial f(x, y)}{\partial x} = \frac{\partial}{\partial x} \left(\frac{x^{3}}{3}-\frac{y^{3}}{3}+3 x y\right)$$ Keeping \(y\) as a constant, we get: $$\frac{\partial f(x, y)}{\partial x} = x^{2} + 3y$$
02

2. Find the partial derivative with respect to y.

To find the partial derivative of the given function \(f(x, y)\) with respect to \(y\), we differentiate it with respect to \(y\) while treating \(x\) as a constant. $$\frac{\partial f(x, y)}{\partial y} = \frac{\partial}{\partial y} \left(\frac{x^{3}}{3}-\frac{y^{3}}{3}+3 x y\right)$$ Keeping \(x\) as a constant, we get: $$\frac{\partial f(x, y)}{\partial y} = -y^{2} + 3x$$
03

3. Solve the system of equations for critical points.

Now that we have both partial derivatives, we will set up a system of equations and solve for the critical points. To do this, we set both partial derivatives equal to zero: $$x^{2} + 3y = 0$$ $$-y^{2} + 3x = 0$$ We can solve these equations to find the critical points. From the second equation, we get: $$y^2 = 3x$$ Substitute this expression into the first equation: $$x^2 + 3(3x) = 0 \implies x^2 + 9x = 0$$ Factor the equation: $$x(x+9) = 0$$ So, we have two possible solutions for x: \(x = 0\) and \(x = -9\) For \(x = 0\), we get \(y^2 = 0\), which implies \(y = 0\) For \(x = -9\), we get \(y^2 = 27\), which implies \(y = \pm 3\sqrt{3}\) So, we have three critical points: \((0,0)\), \((-9, 3\sqrt{3})\), and \((-9, -3\sqrt{3})\).

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Key Concepts

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

Partial Derivatives
The concept of partial derivatives is crucial when dealing with functions of multiple variables. In our given problem, the function \(f(x, y)\) involves two variables: \(x\) and \(y\). To find the critical points, we first need to compute the partial derivatives with respect to each variable.

When taking the partial derivative of \(f(x, y)\) with respect to \(x\), we treat \(y\) as a constant. The result of this differentiation gives us \(\frac{\partial f}{\partial x} = x^2 + 3y\). This expression allows us to see how the function changes as \(x\) changes, while \(y\) remains constant.

Similarly, when calculating the partial derivative with respect to \(y\), we hold \(x\) constant. This gives us \(\frac{\partial f}{\partial y} = -y^2 + 3x\). Hence, this derivative shows how the function responds to changes in \(y\), keeping \(x\) fixed.

Understanding these gradients is essential as they are used to locate points where the function can have maximum, minimum, or saddle points—collectively known as critical points.
System of Equations
Once we have the partial derivatives, the next step is to set them to zero because critical points occur where these derivatives are zero. This setup leads us to form a system of equations.

The equations are:
  • \(x^2 + 3y = 0\)
  • \(-y^2 + 3x = 0\)
These equations must be solved simultaneously to find the values of \(x\) and \(y\) that satisfy both conditions.

To solve this system, we can express one variable in terms of the other. For instance, from the second equation, \(-y^2 + 3x = 0\), we derive \(y^2 = 3x\). Substituting \(y^2\) in terms of \(x\) into the first equation helps us simplify and solve the system.

Through solving, we discover that \(x(x+9) = 0\), giving us potential solutions for \(x\) and consequently for \(y\), leading to the identification of critical points.
Multivariable Calculus
Multivariable calculus extends the concepts of calculus to functions with more than one variable, as seen in our function \(f(x, y)\). An integral part of multivariable calculus is exploring critical points, where the function exhibits special behaviors such as local maxima, minima, or saddle points.

Given the function \(f(x, y)\), critical points are determined by finding when the gradient (a vector comprising partial derivatives) is zero. For this, the role of partial derivatives becomes pivotal, as they construct components of the vector, \(abla f(x, y)\), indicating the direction of greatest increase of the function.

This technique is not just limited to finding critical points but extends to real-world applications, such as optimizing multivariable functions to find maximum profits or minimum costs. By understanding each variable's impact, multivariable calculus provides a robust toolset for tackling complex problems across various fields. In our function, the critical points \((0,0)\), \((-9, 3\sqrt{3})\), and \((-9, -3\sqrt{3})\) exemplify how we can analyze multivariable functions effectively using these techniques.

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Most popular questions from this chapter

A snapshot of a water wave moving toward shore is described by the function \(z=10 \sin (2 x-3 y),\) where \(z\) is the height of the water surface above (or below) the \(x y\) -plane, which is the level of undisturbed water. a. Graph the height function using the window $$[-5,5] \times[-5,5] \times[-15,15]$$ b. For what values of \(x\) and \(y\) is \(z\) defined? c. What are the maximum and minimum values of the water height? d. Give a vector in the \(x y\) -plane that is orthogonal to the level curves of the crests and troughs of the wave (which is parallel to the direction of wave propagation).

Find the points at which the following surfaces have horizontal tangent planes. $$z=\cos 2 x \sin y \text { in the region }-\pi \leq x \leq \pi,-\pi \leq y \leq \pi$$

Use what you learned about surfaces in Sections 13.5 and 13.6 to sketch a graph of the following functions. In each case, identify the surface and state the domain and range of the function. $$H(x, y)=\sqrt{x^{2}+y^{2}}$$

Suppose the elevation of Earth's surface over a 16 -mi by 16 -mi region is approximated by the function \(z=10 e^{-\left(x^{2}+y^{2}\right)}+5 e^{-\left((x+5)^{2}+(y-3)^{2}\right) / 10}+4 e^{-2\left((x-4)^{2}+(y+1)^{2}\right)}\) a. Graph the height function using the window \([-8,8] \times[-8,8] \times[0,15]\) b. Approximate the points \((x, y)\) where the peaks in the landscape appear. c. What are the approximate elevations of the peaks?

Traveling waves in general Generalize Exercise 79 by considering a set of waves described by the function \(z=A+\sin (a x-b y),\) where \(a, b,\) and \(A\) are real numbers. a. Find the direction in which the crests and troughs of the waves are aligned. Express your answer as a unit vector in terms of \(a\) and \(b\). b. Find the surfer's direction- that is, the direction of steepest descent from a crest to a trough. Express your answer as a unit vector in terms of \(a\) and \(b\).

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