18 Second derivative test II

Review

Big idea

The second derivative test gives us a way to classify critical values of \(f(x,y)\).

Remark

The pure second derivatives ask how much a function looks like a paraboloid at the origin. \[f(x,y) = \tfrac{1}{2} ( x^2 + y^2 ) \quad \implies \quad f_{xx}(0,0) = 1 = f_{yy}(0,0) \text{ and } f_{xy}(0,0) = 0.\]
The mixed second derivative \(f_{xy}(x_0,y_0)\) asks how much a function looks like a saddle at the origin. \[f(x,y) = xy \quad \implies \quad f_{xx}(0,0) = 0 = f_{yy}(0,0) \text{ and } f_{xy}(0,0) = 1.\]

The \(D\) of the second derivative test is a tug-of-war between the pure second derivatives and the mixed second derivative: does the function look more like a paraboloid or a saddle near the critical point?

We can visualize this tug of war by considering the function \(f(x,y) = \frac{1}{2}(x^2 + y^2) - a xy\), as the parameter \(a\) varies, using either contours or the \(z = f(x,y)\) surface plot.

Manipulate[
  ContourPlot[
    (x^2 + y^2)/2 - a x y,
    {x, -2, 2}, {y, -2, 2},
    PerformanceGoal -> "Quality"
  ],
  {{a, 0}, -2, 2}
]

Manipulate[
  Plot3D[
    (x^2 + y^2)/2 - a x y,
    {x, -2, 2}, {y, -2, 2},
    PlotRange -> 2, BoxRatios -> 1,
    ClippingStyle -> None,
    PerformanceGoal -> "Quality"
  ],
  {{a, 0}, -2, 2}
]

More examples

Procedure: Second derivative test

Given \(f(x,y)\), we want to find and classify its critical points: maybe we want to find its maximum or minimum values, or are interested in some other phenomenon.

Compute the gradient, \(\nabla f = \langle f_x, f_y \rangle\),
Set \(\nabla f = 0\) (a system of equations) and solve to find critical points,
Compute the second derivatives (\(f_{xx}, f_{yy}, f_{xy}\)) and write \(D = f_{xx} f_{yy} - f_{xy}^2\), and
Classify the critical points via the second derivative test.

Mathematica

The algebra (not the calculus!) in Step 2 is the hardest part of this Procedure. You should practice solving these problems by hand!—but you should also use Mathematica to check your work. Mathematica can:

Compute partial derivatives

f = x^4 + y^4 - 4 x y
fx = D[f, x]
fy = D[f, y]
fxx = D[fx, x]
fyy = D[fy, y]
fxy = D[fx, y]

x^4 - 4 x y + y^4
4 x^3 - 4 y
-4 x + 4 y^3
12 x^2
12 y^2
-4

Solve systems of equations

solutions = Solve[
  {fx == 0, fy == 0},
  {x, y}, Reals
]

{
  {x -> -1, y -> -1},
  {x -> 0, y -> 0},
  {x -> 1, y -> 1}
}

The inclusion of Reals tells Mathematica to only consider real solutions.

Evaluate functions at specific points.

fxx*fyy - fxy^2 /. solutions

{128, -16, 128}

The expressions of the form {x -> -1, y -> -1} are called rules, which we can use to evaluate expressions at specific values. Note that these outputs come in the order of the rules.

Example: critical points of Patrick’s pants

Let’s see how the algebra works out. We have the system \(4x^3 - 4y = 0\) and \(4y^3 - 4x = 0\). Thus

flowchart LR
  A("2x = λ 2x <br> 4y = λ 2y <br> x² + y² = 1")
  A --> B("2x (1-λ) = 0 <br> 2y (2-λ) = 0 <br> x² + y² = 1")
  B -->|"x=0"| C("2y (2-λ) = 0 <br> y² = 1")
  B -->|"λ=1"| D("2y = 0 <br> x² + y² = 1")
  C -->|"y=0"| E("0 = 1"):::contradiction
  C -->|"λ=2"| F("y² = 1")
  F -->|"y=±1"| G("(x,y,λ) = (0,1,2), (0,-1,2)"):::solution
  D -->|"y=0"| H("x² = 1")
  H -->|"x=±1"| I("(x,y,λ) = (1,0,1), (-1,0,1)"):::solution

Using the results from the previous block, we see that

\((-1,-1)\) and \((1,1)\) are extrema of \(f(x,y)\)
\((0,0)\) is a saddle point.

To check what type of extrema we have, we evaluate either \(f_{xx}\) or \(f_{yy}\) at the point—their signs must be the same! For good measure, we will also evaluate the function at the points, just to see what we can see.

fxx /. solutions
f /. solutions

{12, 0, 12}
{-2, 0, -2}

In particular, \(f_{xx} > 0\) at \((-1,-1)\) and \((1,1)\), so these are local minima!

Example: critical points of \(f(x,y) = (x^2 + y^2) e^{-x}\)

Computing the gradient is a bit more work, since we need to use the product rule:

\[\nabla f = \langle 2x e^{-x} - (x^2 + y^2) e^{-x}, 2y e^{-x} \rangle = e^{-x} \langle -x^2 + 2x - y^2, 2y \rangle.\]

Since \(e^{-x} > 0\) for all \(x\), the critical points are determined by the equation \(\langle -x^2 + 2x - y^2, 2y \rangle = \vec{0}\).

flowchart LR
  A("-x²+2x-y² = 0 <br> 2y = 0")
  A --> B("x(2-x) = 0 <br> y = 0")
  B -->|"x=0"| C("(x,y) = (0,0)"):::solution
  B -->|"x=2"| D("(x,y) = (2,0)"):::solution

We compute \(f_{xx} = (x^2+y^2-4x+2)e^{-x}\), \(f_{xy} = -2 e^{-x} y\), and \(f_{yy} = 2 e^{-x}\), so \[D = 2 (x^2-y^2-4x+2) e^{-2x}.\] Ultimately, we see that \((0,0)\) is a local minimum (\(D=4, f_{xx}=2\)), and \((2,0)\) is a saddle (\(D=-\frac{4}{e^4}\)).

Example: critical points of \(f(x,y) = x y - x^2 y - x y^2\)

We have \(\nabla f = \langle y(1-2x-y), x(1-x-2y) \rangle\). Thus we have:

flowchart LR
  A("y(1-2x-y) = 0 <br> x(1-x-2y) = 0")
  A -->|"y=0"| B("x(1-x)=0")
  A -->|"y=1-2x"| C("x(3x-1)=0")
  B -->|"x=0"| D("(x,y) = (0,0)"):::solution
  B -->|"x=1"| E("(x,y) = (1,0)"):::solution
  C -->|"x=0"| F("(x,y) = (0,1)"):::solution
  C -->|"x=1/3"| G("(x,y) = (1/3,1/3)"):::solution

So there are four critical points: \((0,0)\), \((1,0)\), \((0,1)\), and \((\tfrac{1}{3},\tfrac{1}{3})\). Since \(f_{xx} = -2y\), \(f_{yy} = -2x\), and \(f_{xy} = 1-2x-2y\), we have \[ D(0,0) = D(1,0) = D(0,1) = -1 \quad \text{and} \quad D(\tfrac{1}{3},\tfrac{1}{3}) = \tfrac{1}{3}.\] Since \(f_{xx}(\tfrac{1}{3},\tfrac{1}{3}) = -\tfrac{2}{3}\), the first three of these are saddles and the last is a local maximum.

Homework

(Stewart, Clegg, and Watson 2020, sec. 14.7) #9, 18, 26

Stewart, James, Daniel K. Clegg, and Saleem Watson. 2020. Calculus: Early Transcendentals. 9th ed. Cengage Learning.