Problem 18 - Calculus Early Transcendentals

Question

Find all critical points of the following functions. $$f(x, y)=e^{x^{2} y^{2}-2 x y^{2}+y^{2}}$$

Step-by-Step Solution

Answer

Question: Find the critical points of the function $$f(x, y) = e^{x^{2} y^{2}-2 x y^{2}+y^{2}}.$$ Answer: The critical points of the function are $(0, 0)$, $(1, 0)$, $(0, 1)$, and $(1, 1)$.

Step 1: Compute the partial derivatives

To compute the partial derivatives, use the following rules: $$\frac{\partial}{\partial x}(uv) = u'v + uv' $$ and $$\frac{\partial}{\partial x}(e^u) = u'e^u.$$ In our case, we first compute the partial derivative with respect to x: $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(e^{x^{2} y^{2}-2 x y^{2}+y^{2}}\right).$$Using the chain rule, we obtain: $$\frac{\partial f}{\partial x} = \left(2xy^{2} - 2y^{2}\right)e^{x^{2} y^{2}-2 x y^{2}+y^{2}}.$$Then, compute the partial derivative with respect to y: $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\left(e^{x^{2} y^{2}-2 x y^{2}+y^{2}}\right).$$ Again, using the chain rule, we obtain: $$\frac{\partial f}{\partial y} = \left(2x^{2} y - 4xy + 2y\right)e^{x^{2} y^{2}-2 x y^{2}+y^{2}}.$$

Step 2: Solve the system of equations

Now, set both partial derivatives equal to zero and solve the resulting system of equations: $$ \begin{cases} (2xy^{2} - 2y^{2})e^{x^{2} y^{2}-2 x y^{2}+y^{2}} = 0 \\ (2x^{2}y - 4xy + 2y)e^{x^{2} y^{2}-2 x y^{2}+y^{2}} = 0 \end{cases} $$ The exponential factor is never zero, so we can divide both equations by it and obtain: $$ \begin{cases} 2xy^{2} - 2y^{2} = 0 \\ 2x^{2}y - 4xy + 2y = 0 \end{cases} $$ Divide both equations by 2: $$ \begin{cases} xy^{2} - y^{2} = 0 \\ x^{2}y - 2xy + y = 0 \end{cases} $$ Factor y out from the first equation: $$ y(x - 1)(x - 1)(y - 1) = 0 $$ This gives us the possible solutions: \((x, y) = (0, 0)\), \((x, y) = (1, 1)\), \((x, y) = (1, 0)\), and \((x, y) = (0, 1)\). So there are four critical points: \((0, 0)\), \((1, 0)\), \((0, 1)\), and \((1, 1)\).

Key Concepts

Partial Derivatives

Partial derivatives are like regular derivatives, but for functions of multiple variables. Think of them as examining how a function changes as one variable changes while others are held constant.
In this exercise, we're dealing with a function of two variables, \(x\) and \(y\). To find the critical points, we need to find where the function's slope is zero in every direction. This means computing the partial derivatives of the function with respect to \(x\) and \(y\).
To compute the partial derivative of \(f(x, y) = e^{x^{2} y^{2}-2 x y^{2}+y^{2}}\) with respect to \(x\), we hold \(y\) constant and differentiate with respect to \(x\). When computing the partial derivative with respect to \(y\), we're doing the opposite: holding \(x\) constant and differentiating with respect to \(y\).
Here's why this matters:

• It allows us to understand the sensitivity of the function to changes in each variable.
• Partial derivatives are used to find the critical points, which are where the gradients (or slopes) become zero.

Understanding partial derivatives is crucial for solving problems involving functions with more than one variable, such as finding critical points.

Chain Rule

The chain rule is a powerful tool in calculus used for finding the derivative of composite functions. In simpler terms, it helps us differentiate functions that are composed of other functions. In our problem, we deal with an exponential function with a complex exponent, \(x^2y^2 - 2xy^2 + y^2\).
When we compute the partial derivatives, we're not directly differentiating the exponent; instead, we're applying the chain rule to differentiate step-by-step:
First, differentiate the exponent function itself to find the 'inner' derivative. Then, multiply this 'inner' derivative by the derivative of the outer function (in this case, the exponential \(e\)).
For example, the chain rule helps transform the derivative of \(e^{u}\) into \(u'e^{u}\), where \(u\) is the inner function.

• In our exercise, applying the chain rule allows us to find how each term within the exponential impacts the entire expression. This makes calculating the partial derivatives possible.
• The chain rule is essential when faced with nested functions involving exponentials, powers, or trigonometric components.

Mastery of the chain rule is vital for tackling complex problems involving multiple computational layers.

System of Equations

A system of equations is a collection of equations that one seeks to solve simultaneously. In our example, we derived two equations from our partial derivatives:
* \(2xy^{2} - 2y^{2} = 0\)* \(2x^{2}y - 4xy + 2y = 0\)By setting each partial derivative to zero, we effectively create a system because we're looking for \(x\) and \(y\) values that satisfy both equations at the same time.

• To solve the system, we further simplify by dividing by common factors and separating out shared variables. Essentially, this is like conducting algebraic surgery to uncover all possible solutions.
• When solved, these equations give us the critical points where the slope is zero for both directions.

The solutions from our system of equations, in this case, are the points \((0, 0)\), \((1, 0)\), \((0, 1)\), and \((1, 1)\). These points describe conditions under which the directional derivatives are zero, meaning they are points of flatness or transition for the function. Understanding systems of equations is essential for solving multi-variable calculus problems efficiently and is applicable in many fields such as physics, engineering, and economics.