On Quadratics and 2Nd Derivative Test Wednesday, October 17, 2012

MA 2233 Lecture 19 - More on quadratics and 2nd derivative test

Wednesday, October 17, 2012.

Objectives: Continue behavior of quadratics around vertices and introduce 2nd derivative test.

The Theorem

OK. You’ve been completing the square on quadratic functions in two variables. Let’s do this in general. We’ll go through the same steps. We have (1) f(x, y) = ax2 + bxy + cy2. We group the terms with x’s and factor out the a. b (2) f(x, y) = a x2 + xy+ + cy2. a Half of the x-coefficient is b (3) y, 2a and this squared is b2 (4) y2. 4a2 This is what we’re going to add in, but this is being multiplied by a, so we will subtract 2 b 2 (5) y . 4a So to complete the square, we get b b2 b2 (6) f(x, y) = a x2 + xy + y2 + cy2 − y2. a 4a2 4a Simplifying, we get 2 2 b b 2 (7) f(x, y) = a x + y + c − y . 2a 4a Whew! To determine whether we have an elliptic paraboloid, a hyperbolic paraboloid, or a parabolic cylinder, we need to know, if the two coefficients have the same sign, different signs, or if one of them is zero. We can get this information by multiplying the coefficients together, which actually gives us a simpler expression b2 b2 4ac − b2 (8) a · c − = ac − = . 4a 4 4 If this thing is positive, then the signs on the two coefficients were the same. If it’s negative, the coefficients had different signs, and if it’s zero, then at least one of the coefficients is zero. (If the a was zero, then we would have completed the square on the y’s, and we would have gotten the same expression.) In this context, the 4 in the denominator doesn’t really matter. Theorem 1. For a quadratic function of the form f(x, y) = ax2 + bxy + cy2,

• 4ac − b2 > 0, then the graph of f is an elliptic paraboloid (bowl) • 4ac − b2 < 0, then the graph of f is a hyperbolic paraboloid (saddle) • 4ac − b2 =0, then the graph of f is a parabolic cylinder

Furthermore, an elliptic paraboloid will open upwards, if a > 0, and downwards, if a < 0. We will call 4ac − b2 the discriminant. 1 2

Quiz 19

1. Determine if each of the following quadratics is bowl-shaped up or down, saddle, or parabolic cylinder.

a. f(x, y)=3x2 − 2xy +5y2. b. f(x, y)=2x2 +10xy + y2.

c. f(x, y) = x2 +2xy + y2. d. f(x, y) = −3x2 − xy − y2.

2. This is preparation for later. For f(x, y) = ax2 + bxy + cy2, compute all ﬁrst and second partial derivatives.

3. Continuing Problem 2, compute f(0, 0), fx(0, 0), fy(0, 0), fxx(0, 0), fxy(0, 0), fyx(0, 0), and fyy(0, 0).

The Second Derivative Test

Since fxx(0, 0)=2a, fyy(0, 0)=2c, and fxy(0, 0) = b, we have that 2 2 (9) 4ac − b = fxx(0, 0) · fyy(0, 0) − ( fxy(0, 0) ) .

Remember that a critical point is point where both ﬁrst partial derivatives are zero (i.e., where there is a horizontal tangent plane). For a general quadratic function, f(x, y) = ax2 + bxy + cy2 + dx + ey + k, the second partial derivatives look the same. At a critical point, at least, the second partial derivatives completely control the shape of the graph. Basic Principle 1. Essentially, the second derivatives of a general quadratic function tell us what the coeﬃcients of the x2-, xy-, and y2-terms, if we moved the critical point to the origin.

For functions in general, we have non-zero third and higher partial derivatives. These have an eﬀect on the shape of the graph, but by far, the second derivatives have the most obvious inﬂuence. As a result, at critical points on functions of two variables, the graphs look very much like quadratics with the same second derivatives. Summing this up, we have the following theorem.

Theorem 2. Consider a function of two variables f(x, y), and suppose (x0, y0) is a critical point of f. The “best ﬁt” quadratic to f at the critical point will be determined by the discriminant D = fxx(x0, y0)fyy(x0, y0)− 2 fxy(x0, y0).

• If D > 0, then the graph of f looks like an elliptic paraboloid (bowl) • If D < 0, then the graph of f looks like a hyperbolic paraboloid (saddle) • If D =0, then the graph of f looks like a parabolic cylinder

Furthermore, the elliptic paraboloid will open upwards, if fxx(x0, y0) > 0, and downwards, if fxx(x0, y0) < 0.

Relative Maximums and Minimums

Relative maximums and minimums (generically referred to as relative extrema) are as you would expect. They are the highest and lowest points on a small piece of a graph. At a nice critical point, there are three possibilities. We have a relative maximum or minimum, and otherwise, we have some sort of saddle. I’m − 2 skipping some of the details, but the discriminant D = fxxfyy fxy being positive or negative deﬁnitely determines that we have a max, min, or saddle. 3

If D = 0, on the other hand, the subtle effects of the higher order partial derivatives come into play. Essentially, if D = 0, then the second derivative has no effect on the shape of the graph in at least one direction. Let’s go back to functions of one variable to see what’s going on. Consider the function f(x) = x4. The graph is ∪-shaped, but it isn’t a parabola. Note that at x = 0, f(0) = 0, f 0(0) = 0, f 00(0) = 0, f 000(0) = 0, but f iv(0) = 24. For comparison, consider the function, g(x) = −x2 + x4, which has a non-zero second derivative at zero, f 00(0) = −2. Look at the graphs of these two functions in Maple. What I want you to take from this is that if the second derivative is non-zero, the higher order derivatives (third, fourth, etc.) don’t matter much (near a critical point). If the second derivative is zero, then the very subtle effects of the higher order derivatives can make a difference. OK. Now it’s time for our theorem.

Theorem 3. Second Derivative Test Consider a function of two variables f(x, y), and suppose (x0, y0) − 2 is a critical point of f. The sign of the discriminant D = fxx(x0, y0)fyy(x0, y0) fxy(x0, y0) tells us the following.

• If D > 0 and fxx(x0, y0) < 0, then we have a relative maximum at (x0, y0). • If D > 0 and fxx(x0, y0) > 0, then we have a relative minimum at (x0, y0). • If D < 0, then we have a saddle at (x0, y0). • If D =0, then we don’t know.

Homework 19

1. Consider f(x, y) = x4 − y2. Find the critical point. Compute D at (0, 0). What does the Second Derivative Test tell us. Look carefully at the function. You should be able to ﬁgure this out without the Second Derivative Test. Is this a max, min, or saddle?

2. Consider f(x, y) = x4 + y2. Do the same as for Problem 13.

3. Let f(x, y) = x2y +3xy − y2. Find all critical points. What does the Second Derivative Test tell us about the nature of these critical points?

Answers: 1) Second Derivative Test tells us nothing, but in the x-direction the graph is ∪-shaped, and in the y-direction, it is ∩-shaped, so this is a saddle. 2) Similar to Problem 13, but it’s ∪-shaped in both directions, so this is a relative min. 3) Saddle at (0, −1) and saddle at (0, 1).