Question

For $w=x^3-y^3-2xy+6,$ 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

Second partial derivatives at $(0,0)$ are both 0. At $(-\frac{2}{3},\frac{2}{3})$, both are -4.

Step-by-step solution

Identify the function

Given the function: $$w=x^3-y^3-2xy+6$$.

Compute first partial derivatives

Find the first partial derivatives of $w$ with respect to $x$ and $y$: $$\frac{\partial w}{\partial x}=\frac{\partial }{\partial x}(x^3-y^3-2xy+6)=3x^2-2y$$ $$\frac{\partial w}{\partial y}=\frac{\partial }{\partial y}(x^3-y^3-2xy+6)=-3y^2-2x$$.

Set first partial derivatives to zero

Set the first partial derivatives to zero and solve for $x$ and $y$: $$3x^2-2y=0$$ $$-3y^2-2x=0$$.

Solve the system of equations

Solve the system of equations: From $3x^2-2y=0$, we get $y=\frac{3x^2}{2}$. Substitute $y$ in the second equation: $$-3{\left(\frac{3x^2}{2}\right)}^2-2x=0$$ $$-3\cdot \frac{9x^4}{4}-2x=0$$ $$-\frac{27x^4}{4}-2x=0$$ $$-27x^4-8x=0$$ $$-x(27x^3+8)=0$$ So, $x=0$ or $27x^3+8=0$. For $x=0$, $y=0$. Solving $27x^3+8=0$ gives: $$x^3=-\frac{8}{27}$$ $$x=-\frac{2}{3}$$ Then, $y=\frac{3{(-\frac{2}{3})}^2}{2}=\frac{2}{3}$. So the points are $(0,0)$ and $(-\frac{2}{3},\frac{2}{3})$.

Compute second partial derivatives

Find the second partial derivatives of $w$ with respect to $x$ and $y$: $$\frac{{\partial }^{2}w}{\partial x^2}=\frac{\partial }{\partial x}(3x^2-2y)=6x$$ $$\frac{{\partial }^{2}w}{\partial y^2}=\frac{\partial }{\partial y}(-3y^2-2x)=-6y$$.

Evaluate second partial derivatives at critical points

Evaluate the second partial derivatives at $(0,0)$: $$\frac{{\partial }^{2}w}{\partial x^2}{|}_{(0,0)}=6\times 0=0$$ $$\frac{{\partial }^{2}w}{\partial y^2}{|}_{(0,0)}=-6\times 0=0$$. Evaluate at $(-\frac{2}{3},\frac{2}{3})$: $$\frac{{\partial }^{2}w}{\partial x^2}{|}_{(-\frac{2}{3},\frac{2}{3})}=6\times (-\frac{2}{3})=-4$$ $$\frac{{\partial }^{2}w}{\partial y^2}{|}_{(-\frac{2}{3},\frac{2}{3})}=-6\times \frac{2}{3}=-4$$.

Key concepts

First Partial Derivatives

In calculus, partial derivatives measure how a function changes as its variables change.
The first partial derivatives of a function with respect to its variables indicate the rates of change along those variables.
Let's illustrate this concept using our function: $w=x^3-y^3-2xy+6$.
We need to find the first partial derivatives with respect to $x$ and $y$.

Starting with $x$:
$$\frac{\partial w}{\partial x}=\frac{\partial }{\partial x}(x^3-y^3-2xy+6)=3x^2-2y$$
Here, $3x^2$ is the part influenced by $x^3$, and $-2y$ comes from the interaction term $-2xy$.
Notice that constant terms and terms involving only $y$ disappear when we differentiate with respect to $x$.

Next, we find the partial derivative with respect to $y$:
$$\frac{\partial w}{\partial y}=\frac{\partial }{\partial y}(x^3-y^3-2xy+6)=-3y^2-2x$$
Similarly, $-3y^2$ is the part affected by $y^3$, and $-2x$ arises from the $-2xy$ term.
This method helps us understand how the function changes when one variable changes while the others are held constant.

Second Partial Derivatives

The second partial derivatives give us insight into the curvature or concavity of the function.
They are the derivatives of the first partial derivatives.

Continuing with our function, we need to find $\frac{{\partial }^{2}w}{\partial x^2}$ and $\frac{{\partial }^{2}w}{\partial y^2}$.
Starting with $\frac{{\partial }^{2}w}{\partial x^2}$, we differentiate $\frac{\partial w}{\partial x}=3x^2-2y$ with respect to $x$:
$$\frac{{\partial }^{2}w}{\partial x^2}=\frac{\partial }{\partial x}(3x^2-2y)=6x$$
Here, 3 times 2 from the $3x^2$ term gives $6x$. The $-2y$ term vanishes as $y$ is treated as a constant.

Next, for $\frac{{\partial }^{2}w}{\partial y^2}$, we differentiate $\frac{\partial w}{\partial y}=-3y^2-2x$ with respect to $y$:
$$\frac{{\partial }^{2}w}{\partial y^2}=\frac{\partial }{\partial y}(-3y^2-2x)=-6y$$
This gives $-3\times 2=-6y$ from the $-3y^2$ term, with $-2x$ dropping out as it involves $x$.
These second partial derivatives provide the curvature information needed for deeper analysis, such as finding concave or convex behavior of the function.

Solving Equations

Solving equations involves finding values for variables that make the equation true.
In our case, we set the first partial derivatives to zero to find points where the function changes direction.

The first partial derivatives are:
$$3x^2-2y=0$$ $$-3y^2-2x=0$$ To solve this system of equations, we use substitution or elimination.
First, solve $3x^2-2y=0$ for $y$:
$$y=\frac{3x^2}{2}$$
Substitute this into $-3y^2-2x=0$:
$$-3{\left(\frac{3x^2}{2}\right)}^2-2x=0$$ Simplify to get:
$$-\frac{27x^4}{4}-2x=0$$ $$-27x^4-8x=0$$ Separate the variables:
$$-x(27x^3+8)=0$$ So, $x=0$ or $27x^3+8=0$.
For $x=0$, $y=0$.
For $27x^3+8=0$, $x=-\frac{2}{3}$.
Then $y=\frac{2}{3}$.
This solving process finds critical points where the function's slope is zero in both directions.

Critical Points Analysis

Critical points are locations where the first partial derivatives of a function are zero.
They help us identify potential maxima, minima, or saddle points.
In our function's context, critical points were found at $(0,0)$ and $(-\frac{2}{3},\frac{2}{3})$.

To better understand these points, we use the second partial derivatives:
At $(0,0)$:$$\frac{{\partial }^{2}w}{\partial x^2}=6\times 0=0$$ $$\frac{{\partial }^{2}w}{\partial y^2}=-6\times 0=0$$
These second partial derivatives are zero, indicating neither concave nor convex behavior at the origin.

At $(-\frac{2}{3},\frac{2}{3})$:$$\frac{{\partial }^{2}w}{\partial x^2}=6\times (-\frac{2}{3})=-4$$
$$\frac{{\partial }^{2}w}{\partial y^2}=-6\times \frac{2}{3}=-4$$
Both second partial derivatives are negative, indicating a local maximum.
This analysis confirms our critical points and provides insight into the function's behavior around these points.