Problem 397 - Calculus Volume 3

Question

Determine whether the statement is true or false. Justify your answer with a proof . \(\quad\left(\frac{3}{4}, \frac{9}{16}\right) \quad\) is a critical point of \(g(x, y)=4 x^{3}-2 x^{2} y+y^{2}-2\)

Step-by-Step Solution

Answer

False. \( \left(\frac{3}{4}, \frac{9}{16}\right) \) is not a critical point.

Step 1: Determine the Partial Derivatives

First, find the partial derivatives of the function \( g(x, y) = 4x^3 - 2x^2y + y^2 - 2 \). - The partial derivative with respect to \( x \) is: \( g_x(x, y) = \frac{\partial}{\partial x}(4x^3 - 2x^2y + y^2 - 2) = 12x^2 - 4xy \).- The partial derivative with respect to \( y \) is: \( g_y(x, y) = \frac{\partial}{\partial y}(4x^3 - 2x^2y + y^2 - 2) = -2x^2 + 2y \).

Step 2: Set Partial Derivatives to Zero

To find the critical points, set each partial derivative to zero and solve the equations.- Solve \( 12x^2 - 4xy = 0 \): \[ 4x(3x - y) = 0 \] This gives two possibilities: \( x = 0 \) or \( y = 3x \).- Solve \( -2x^2 + 2y = 0 \): \[ y = x^2 \].

Step 3: Combine Conditions

Combine the conditions from setting the derivatives to zero:1. If \( x = 0 \), then \( y = x^2 = 0^2 = 0 \). Thus, one critical point is \((0, 0)\).2. If \( y = 3x \) and \( y = x^2 \), then set \( 3x = x^2 \): \[ x^2 - 3x = 0 \] \[ x(x - 3) = 0 \] Thus, \( x = 0 \) or \( x = 3 \). - For \( x = 0 \), \( y = 0 \), so we get the point \((0, 0)\). - For \( x = 3 \), \( y = x^2 = 9 \), so we get the point \((3, 9)\).

Step 4: Verify Provided Point

The provided point \( \left(\frac{3}{4}, \frac{9}{16}\right) \) should be checked against the conditions:- Check \( y = 3x \): does \( \frac{9}{16} = 3 \times \frac{3}{4} \)? - We compute: \( 3 \times \frac{3}{4} = \frac{9}{4} eq \frac{9}{16} \), so this does not hold.- As a result, \( \left(\frac{3}{4}, \frac{9}{16}\right) \) does not satisfy the necessary conditions for being a critical point of the function \( g(x, y) \).

Key Concepts

Partial Derivatives

Partial derivatives allow us to examine how a multivariable function changes with respect to one variable while keeping the other variables constant. This concept is fundamental in multivariable calculus and is akin to finding the derivative of a single-variable function. In this exercise, we are dealing with a function of two variables, so we differentiate with respect to each variable separately. For the given function, the partial derivative with respect to \( x \) is found by treating \( y \) as a constant. Similarly, the partial derivative with respect to \( y \) is obtained by treating \( x \) as a constant.

• The partial derivative with respect to \( x \) is: \( g_x(x, y) = 12x^2 - 4xy \).
• The partial derivative with respect to \( y \) is: \( g_y(x, y) = -2x^2 + 2y \).

By computing these, you can analyze the local behavior of the function near a point, which is necessary for finding critical points.

Calculus Problem Solving

Solving calculus problems often requires a structured approach, especially when dealing with functions of multiple variables. In this problem, we aim to identify critical points, which occur where the partial derivatives of a function are both zero or undefined. The steps typically involve:

• Determine Partial Derivatives: Find the derivatives of the function with respect to each variable. This step helps identify where changes in the function tend to flatten out.
• Set Derivatives to Zero: Solve for when each partial derivative equals zero. This is needed to find potential critical points.
• Combine Conditions: Use the solutions from setting the derivatives equal to zero to find consistent answers that meet all conditions. This might involve algebraic manipulations or systems of equations.
• Verify Results: Check if any proposed critical points, such as the given point \( \left(\frac{3}{4}, \frac{9}{16}\right) \), meet the established conditions for criticality.

Each of these steps helps in systematically approaching the problem and ensuring that no potential critical points are overlooked.

Multivariable Calculus

In multivariable calculus, we extend the concepts from single-variable calculus to functions of multiple variables like \( g(x, y) \). This means considering how changes in more than one input can affect the output. Critical points in this context are where the function might change from increasing to decreasing, have a saddle point, or a local maximum or minimum.Multivariable functions have a surface as opposed to a curve, which makes visualizing critical points a bit more complex. To find these points, setting the partial derivatives to zero allows us to find where the function's rate of change is zero in multiple directions. This exercise shows how:- We compute partial derivatives for functions with more than one variable.- Setting these derivatives to zero identifies critical regions on the function's surface.- Understanding how combining conditions relates to simple algebra can solve for points where changes in variables coincide causing critical points.Approaching multivariable calculus problems involves understanding how different variables interact to influence the output and recognizing the points of equilibrium or transition.