Question

For \(w=8 x^{4}+y^{4}-2 x y^{2},\) find \(\partial^{2} w / \partial x^{2}\) and \(\partial^{2} w / \partial y^{2}\) at the points where \(\partial w / \partial x=\partial w / \partial y=0\).

Short answer

At (0,0): \(\frac{\partial^2 w}{\partial x^2} = 0, \frac{\partial^2 w}{\partial y^2} = 0\). At \(\frac{1}{4}, \frac{1}{2}\): \(\frac{\partial^2 w}{\partial x^2} = 6, \frac{\partial^2 w}{\partial y^2} = 2\). At \(\frac{1}{4}, -\frac{1}{2}\): \(\frac{\partial^2 w}{\partial x^2} = 6, \frac{\partial^2 w}{\partial y^2} = 2\).

Step-by-step solution

- Compute First Partial Derivatives

First, find the first partial derivatives of the function w with respect to x and y. \[\frac{\partial w}{\partial x} = \frac{\partial}{\partial x}(8x^4 + y^4 - 2xy^2)\] Using the power rule and product rule, \[\frac{\partial w}{\partial x} = 32x^3 - 2y^2\] Now, compute the first partial derivative with respect to y: \[\frac{\partial w}{\partial y} = \frac{\partial}{\partial y}(8x^4 + y^4 - 2xy^2)\] \[\frac{\partial w}{\partial y} = 4y^3 - 4xy\]

- Find Critical Points

Set the first partial derivatives equal to zero to find the critical points. \[32x^3 - 2y^2 = 0\] \[4y^3 - 4xy = 0\] From these, solve for x and y. \[16x^3 = y^2\] \[y(y^2 - x) = 0\] From the second equation, y could be 0 or y^2 = x.

- Solve the System of Equations

Solve the system of equations obtained from Step 2. If y = 0, \[16x^3 = 0\] which implies x = 0. So, one critical point is (0, 0). If y^2 = x, substitute into the first equation: \[16(y^2)^3 = y^2\] The equation simplifies to \[16y^6 = y^2\] If y ≠ 0, divide both sides by y^2: \[16y^4 = 1\] \[y^4 = \frac{1}{16}\] \[y = \pm \frac{1}{2}\] Therefore, \[x = y^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\] The other critical points are \(\frac{1}{4}, \frac{1}{2}\) and \(\frac{1}{4}, -\frac{1}{2}\).

- Compute Second Partial Derivatives

Next, compute the second partial derivatives to find \(\frac{\partial^2 w}{\partial x^2}\) and \(\frac{\partial^2 w}{\partial y^2}\). Given: \[\frac{\partial w}{\partial x} = 32x^3 - 2y^2\] \[\frac{\partial^2 w}{\partial x^2} = \frac{\partial}{\partial x} (32x^3 - 2y^2) = 96x^2\] And, given: \[\frac{\partial w}{\partial y} = 4y^3 - 4xy\] \[\frac{\partial^2 w}{\partial y^2} = \frac{\partial}{\partial y} (4y^3 - 4xy) = 12y^2 - 4x\]

- Evaluate at Critical Points

Evaluate the second partial derivatives at the critical points (0,0), \(\frac{1}{4}, \frac{1}{2}\), and \(\frac{1}{4}, -\frac{1}{2}\). For (0,0): \[\frac{\partial^2 w}{\partial x^2} = 96(0)^2 = 0\] \[\frac{\partial^2 w}{\partial y^2} = 12(0)^2 - 4(0) = 0\] For \(\frac{1}{4}, \frac{1}{2}\): \[\frac{\partial^2 w}{\partial x^2} = 96\left(\frac{1}{4}\right)^2 = 6\] \[\frac{\partial^2 w}{\partial y^2} = 12\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{4}\right) = 2\] For \(\frac{1}{4}, -\frac{1}{2}\): \[\frac{\partial^2 w}{\partial x^2} = 96\left(\frac{1}{4}\right)^2 = 6\] \[\frac{\partial^2 w}{\partial y^2} = 12\left(-\frac{1}{2}\right)^2 - 4\left(\frac{1}{4}\right) = 2\]

Key concepts

Critical Points

Critical points are values of x and y where the first partial derivatives of a function are zero. To find these points in your function, set \(\frac{\text{d}w}{\text{d}x} = 0\) and \(\frac{\text{d}w}{\text{d}y} = 0\). This step helps identify potential maxima, minima, or saddle points of the function. For the given problem, solving the equations \(32x^3 - 2y^2 = 0\) and \(4y^3 - 4xy = 0\) leads to finding the points where these conditions are satisfied, like (0,0), (1/4, 1/2), and (1/4, -1/2). These are the critical points we test further.

Partial Derivatives

Partial derivatives show how a function changes as each variable changes independently. They are crucial for multi-variable functions. For the given problem, we need to find the partial derivatives of \(w = 8x^4 + y^4 - 2xy^2\) with respect to both x and y. This is done by treating one variable as constant while differentiating with respect to the other. The results are \(32x^3 - 2y^2\) for the partial derivative with respect to x, and \(4y^3 - 4xy\) for the partial derivative with respect to y. Calculating these derivatives helps in finding the critical points and understanding the function's behavior.

Power Rule

The power rule is a basic, yet powerful tool in calculus used for finding derivatives of power functions. In simple terms, for any function \(x^n\), the derivative is \(nx^{n-1}\). This was applied in the given problem when finding the first and second partial derivatives. For example: \(8x^4\) becomes \(32x^3\) when applying the power rule. Similarly, \(y^4\) becomes \(4y^3\). Understanding the power rule allows students to quickly and accurately find derivatives of polynomial functions.

Evaluating Derivatives

Evaluating derivatives at specific points helps determine the behavior of the original function at those points. For this problem, after finding the first and second partial derivatives, we substitute the values of the critical points into the second partial derivatives: \(96x^2\) and \(12y^2 - 4x\), respectively. For instance, at the point (0,0), the evaluation of the second partial derivative \(\frac{\text{d}^2w}{\text{d}x^2} = 0\), and \(\frac{\text{d}^2w}{\text{d}y^2} = 0\). Evaluating these derivatives helps in analyzing whether the critical points are a maximum, minimum, or saddle point.