Problem 12 - Calculus Early Transcendentals

Question

Find all critical points of the following functions. $$f(x, y)=3 x^{2}-4 y^{2}$$

Step-by-Step Solution

Answer

Answer: The critical point of the function is at the point (0, 0).

Step 1: Calculate partial derivatives

First, determine the partial derivatives of the function $$f(x, y)$$ with respect to x and y. To do this, differentiate the function with respect to x while treating y as a constant, and vice versa. Partial derivative with respect to x: $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (3x^2 - 4y^2) = 6x$$ Partial derivative with respect to y: $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (3x^2 - 4y^2) = -8y$$

Step 2: Set partial derivatives to zero

Now, set both partial derivatives equal to zero to find the points where the function has critical points. $$6x = 0$$ $$-8y = 0$$

Step 3: Solve the equations

Solve each equation for x and y, respectively. For the first equation: Divide by 6: $$x = 0$$ For the second equation: Divide by -8: $$y = 0$$

Step 4: Determine the critical point

The critical point occurs at the point where both x and y equal zero. Therefore, the critical point of the function $$f(x, y) = 3x^2 - 4y^2$$ is at the point (0, 0).

Key Concepts

Partial Derivatives

In multivariable calculus, partial derivatives are a fundamental concept used to analyze functions with more than one variable. They help you understand how a function changes as each variable is varied independently.
For example, let's look at the function \(f(x, y) = 3x^2 - 4y^2\). To find the partial derivative with respect to \(x\), treat \(y\) as a constant and differentiate \(3x^2\), ignoring the \(-4y^2\). The result: \(\frac{\partial f}{\partial x} = 6x\).
Similarly, for the partial derivative with respect to \(y\), treat \(x\) as a constant and differentiate \(-4y^2\), momentarily ignoring \(3x^2\), yielding \(\frac{\partial f}{\partial y} = -8y\).

These derivatives provide crucial information about how the function behaves in the direction of each variable, helping you find where changes in output are very sensitive to slight changes in input.

Function Optimization

Function optimization involves finding the maximum or minimum values of a function, which often occurs at critical points. Critical points are where the gradient (or in simpler terms, the slopes given by the partial derivatives) equals zero.
In the context of our function \(f(x, y) = 3x^2 - 4y^2\), we find critical points by setting the partial derivatives equal to zero, like so: \(6x = 0\) and \(-8y = 0\). Solving these gives us the critical point \((0,0)\).
But what's next? After finding a critical point, we often perform additional tests, such as the Second Derivative Test, to classify these points. This helps in determining if they are maximum, minimum, or saddle points.

Optimization is key in various applications, from economics to engineering, wherever choosing the best option from a set of alternatives is necessary.

Multivariable Calculus

Multivariable calculus extends the principles of single-variable calculus to functions of two or more variables. It's essential for understanding systems that depend on multiple inputs.
A key technique in multivariable calculus is handling functions like \(f(x, y) = 3x^2 - 4y^2\) using tools such as partial derivatives and gradients. These methods allow us to explore how functions behave not just along linearly independent axes, but in any direction in the input space.
When you find critical points in a function of two variables, you're engaging with the core of multivariable calculus. You're addressing questions about geometry, like how the surface defined by a function rises, falls, or stays flat at various places. Multivariable calculus thus provides a deeper understanding of physical, natural, and abstract systems in multiple dimensions.