From Math 2220 Class 10 ample Higher Partials in Polar Cordinates - - PowerPoint PPT Presentation

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

From Math 2220 Class 10

• Dr. Allen Back
• Sep. 19, 2014

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives

Definition: A function f : U ⊂ Rn → R is C 2 or twice continuously differentiable if f and all partial derivatives of first

• r second order exist and are continuous. Similarly f is C k for
• ther positive integers k.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives

We earlier had a theorem that f a C 1 function implies f is differentiable.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives

Theorem: If f : U ⊂ Rn → R is C 2, then ∂2f ∂xi∂xj = ∂2f ∂xj∂xi .

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives

The proof of this theorem is based on showing that the symmetric expression lim

(∆x,∆y)→(0,0) [f (x0, y0) − f (x0 + ∆x, y0)

−f (x0, y0 + ∆y) + f (x0 + ∆x, y0 + ∆y)] / ∆x∆y. exists and equals both fxy and fyx at (x0, y0).

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives

The proof of this theorem is based on showing that the symmetric expression lim

(∆x,∆y)→(0,0) [f (x0, y0) − f (x0 + ∆x, y0)

−f (x0, y0 + ∆y) + f (x0 + ∆x, y0 + ∆y)] / ∆x∆y. exists and equals both fxy and fyx at (x0, y0). The basic tool is the mean value estimate g(x0 + ∆x) − g(x0) = g′(c)∆x for some c between x0 and x0 + ∆x.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives

The proof of this theorem is based on showing that the symmetric expression lim

(∆x,∆y)→(0,0) [f (x0, y0) − f (x0 + ∆x, y0)

−f (x0, y0 + ∆y) + f (x0 + ∆x, y0 + ∆y)] / ∆x∆y. exists and equals both fxy and fyx at (x0, y0). We would apply this to the function g(x) = f (x, y0 + ∆y) − f (x, y0). since the numerator looks like g(x0 + ∆x) − g(x0).

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives

Second partials show up in max-min as well as physical

• modeling. For example the vibration of a string on a musical

instrument is governed by the 1 dimensional wave equation ∂2z ∂x2 = 1 c2 ∂2z ∂t2 where z(x, t) is the displacement of the string (up or down) at position x along the string and time t. The constant c is the speed of the vibration.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives

Show that z(x, t) = A sin (x − ct) satisfies the wave equation above.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives

Or A sin (x + ct). Or indeed for any twice differentiable one variable function g(u), z(x, t) = g(x − ct) will also satisfy the wave equation. Why?

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives

A typical application of mixed partials arises out of your homework problem 3.1 #22 where you show that the wave equation ∂2w ∂x2 = ∂2w ∂y2 becomes after a change of coordinates x = u + v, y + u − v ∂2w ∂u∂v = 0

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives

A typical application of mixed partials arises out of your homework problem 3.1 #22 where you show that the wave equation ∂2w ∂x2 = ∂2w ∂y2 becomes after a change of coordinates x = u + v, y + u − v ∂2w ∂u∂v = 0

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partial Derivatives

A typical application of mixed partials arises out of your homework problem 3.1 #22 where you show that the wave equation ∂2w ∂x2 = ∂2w ∂y2 x = u + v, y + u − v ∂2w ∂u∂v = 0 implying the general solution w = f (u) + g(v) = f (x − y 2 ) + g(x + y 2 ) for any C 2 functions f and g.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

A Counterexample

When f is not C 2, fxy need not equal fyx even when both exist.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

A Counterexample

An example illustrating this is f (x, y) =

• xy(x2−y2)

x2+y2

if (x, y) = (0, 0) if (x, y) = (0, 0) This function

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

A Counterexample

An example illustrating this is f (x, y) =

• xy(x2−y2)

x2+y2

if (x, y) = (0, 0) if (x, y) = (0, 0) This function is clearly C 2 for (x, y) = (0, 0) since it is a rational function (quotient of polynomials) with nonzero denominator.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

A Counterexample

An example illustrating this is f (x, y) =

• xy(x2−y2)

x2+y2

if (x, y) = (0, 0) if (x, y) = (0, 0) This function satisfies f (x, y) = −f (y, x) implying fxy = fyx whenever x = y. (A good chain rule exercise . . . )

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

A Counterexample

An example illustrating this is f (x, y) =

• xy(x2−y2)

x2+y2

if (x, y) = (0, 0) if (x, y) = (0, 0) This function satisfies fx(0, 0) = 0 and fy(0, 0) = 0 since f vanishes along the axes.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

A Counterexample

An example illustrating this is f (x, y) =

• xy(x2−y2)

x2+y2

if (x, y) = (0, 0) if (x, y) = (0, 0) This function satisfies fx = y(x2 − y2) x2 + y2 + 2x2y x2 + y2 − 2x2y(x2 − y2) (x2 + y2)2 . implying fx(0, y) = −y (the last two terms vanish) and ∂2f ∂y∂x (0, 0) = −1.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

A Counterexample

An example illustrating this is f (x, y) =

• xy(x2−y2)

x2+y2

if (x, y) = (0, 0) if (x, y) = (0, 0) This function fx(0, y) = −y (the last two terms vanish) and ∂2f ∂y∂x (0, 0) = −1. By fxy(x, x) = −fyx(x, x) (mentioned above) or directly we see ∂2f ∂x∂y (0, 0) = 1 and the mixed partials really differ!

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partials in Polar Cordinates

Let x = r cos θ, y = r sin θ. Given f (x, y), express ∂2f ∂r2 in terms of partials of f with respect to x and y.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partials in Polar Cordinates

Really there are two functions here; namely f (x, y) and g(r, θ) = f (r cos θ, r sin θ). But many applied fields routinely use the same letter f for both functions.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partials in Polar Cordinates

The key step is realizing that the task (via the chain rule) of relating ∂f ∂r to ∂f ∂x and ∂f ∂y is just like the task of doing the analagous thing with expressions like ∂ ∂r ∂f ∂x

• .

(The right hand side “operator notation” means partially differentiate the function ∂f ∂x of (x, y) with respect to r.)

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partials in Polar Cordinates

x = r cos θ y = r sin θ ∂f ∂r = ∂f ∂x ∂x ∂r + ∂f ∂y ∂y ∂r = ∂f ∂x cos θ + ∂f ∂y sin θ ∂ ∂r ∂f ∂x

• =

∂ ∂x ∂f ∂x ∂x ∂r + ∂ ∂y ∂f ∂x ∂y ∂r = ∂2f ∂x2

• cos θ +

∂2f ∂y∂x

• sin θ

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Higher Partials in Polar Cordinates

A similar computation with ∂2f ∂θ2 would show that the laplacian ∂2f ∂x2 + ∂2f ∂y2 is given in polar coordinates by ∂2f ∂r2 + 1 r ∂f ∂r + 1 r2 ∂2f ∂θ2

• .

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Can you estimate the value of the gradient?

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Method of Characteristics

Problem 33 on page 146 from your homework asks you to show using the chain rule that if z = f (y x ) for a 1-variable differentiable function f , then x ∂z ∂x + y ∂z ∂y = 0.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Method of Characteristics

Problem 33 on page 146 from your homework asks you to show using the chain rule that if z = f (y x ) for a 1-variable differentiable function f , then x ∂z ∂x + y ∂z ∂y = 0. You don’t need it to do this homework problem, but a natural question is Does x ∂z ∂x + y ∂z ∂y = 0 imply z = f (y x ) for some 1-variable differentiable function f ?

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Method of Characteristics

The answer is yes, and we can easily show it using what we learned in section 2.6! (i.e. essentially chain rule ideas applied to curves.)

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Method of Characteristics

The beautiful idea for the equation a(x, y)∂z ∂x + b(x, y)∂z ∂y = c(x, y, z) is to consider curves (x(t), y(t) called characteristics) satisfying dx dt = a(x(t), y(t)) dy dt = b(x(t), y(t)).

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Method of Characteristics

Here for x ∂z ∂x + y ∂z ∂y = 0 we have dx dt = x dy dt = y with solutions x(t) = x0et and y(t) = y0et.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Method of Characteristics

Here for x ∂z ∂x + y ∂z ∂y = 0 we have dx dt = x dy dt = y with solutions x(t) = x0et and y(t) = y0et. The key observation is that for a solution z(x, y) d dt [z(x(t), y(t))] = ∂z ∂x dx dt + ∂z ∂y dy dt = ∂z ∂x x+ ∂z ∂y y = 0 !

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Method of Characteristics

Here for x ∂z ∂x + y ∂z ∂y = 0 with solutions x(t) = x0et and y(t) = y0et. The key observation is that for a solution z(x, y) d dt [z(x(t), y(t))] = ∂z ∂x dx dt + ∂z ∂y dy dt = ∂z ∂x x+ ∂z ∂y y = 0 ! So z(x, y) is constant along lines y x = k for any k and thus z(x, y) = z(1, y x ) = f (y x ) as we wanted to show.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Critical Points

For a function f (x, y), these are points (x0, y0) where both fx(x0, y0) = fy(x0, y0) = (OR f is not differentiable at (x0, y0).)

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Critical Points

For a function f (x, y), these are points (x0, y0) where both fx(x0, y0) = fy(x0, y0) = (OR f is not differentiable at (x0, y0).) The term critical point is just used for interior points in the domain of f .

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Critical Points

More generally we have Theorem: Suppose U is an open set and f : U ⊂ Rn → R has a local max or min at p0 ∈ U. Then if the derivative exists at p0, we have Df (p0) = 0.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Critical Points

A point (x0, y0) is a local maximum if there is some (perhaps small) open disk D about (x0, y0) so that f (x, y) ≤ f (x0, y0) for all (domain) points (x, y) which are also in the disk D.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Critical Points

Example: Find the critical points of f (x, y) = x3 − xy + y2.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Critical Points

Example: Find the critical points of f (x, y) = x3 − xy + y2. fx = 3x2 − y = fy = −x + 2y =

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Critical Points

Example: Find the critical points of f (x, y) = x3 − xy + y2. fx = 3x2 − y = fy = −x + 2y = The second equation says x = 2y. Plugging into the first: 3(2y)2 − y = 0

• r

y(12y − 1) = 0.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Critical Points

Example: Find the critical points of f (x, y) = x3 − xy + y2. fx = 3x2 − y = fy = −x + 2y = The second equation says x = 2y. Plugging into the first: 3(2y)2 − y = 0

• r

y(12y − 1) = 0. Our critical points are (0, 0) and (1 6, 1 12).

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Critical Points

Example: Find the critical points of f (x, y) = x3 − xy + y2. Our critical points are (0, 0) and (1 6, 1 12). Not obvious whether these are local minima or maxima. But they are the only possible points which can be locally extrema.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the First Derivative Test?

A point (x0, y0) is a local maximum if there is some (perhaps small) disk D about (x0, y0) so that f (x, y) ≤ f (x0, y0) for all domain points (x, y) which are also in the disk D.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the First Derivative Test?

A point (x0, y0) is a local maximum if there is some (perhaps small) disk D about (x0, y0) so that f (x, y) ≤ f (x0, y0) for all domain points (x, y) which are also in the disk D. If (x0, y0) is a local maximum, then consider fx(x0, y0) = lim

∆x→0

f (x0 + ∆x, y0) − f (x0, y0) ∆x .

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the First Derivative Test?

A point (x0, y0) is a local maximum if there is some (perhaps small) disk D about (x0, y0) so that f (x, y) ≤ f (x0, y0) for all domain points (x, y) which are also in the disk D. If (x0, y0) is a local maximum, then consider fx(x0, y0) = lim

∆x→0

f (x0 + ∆x, y0) − f (x0, y0) ∆x . Thinking about points a little bit to the right (∆x > 0) we see fx(x0, y0) = lim

∆x→0

≤ 0 > 0 = ⇒ fx(x0, y0) ≤ 0.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the First Derivative Test?

A point (x0, y0) is a local maximum if there is some (perhaps small) disk D about (x0, y0) so that f (x, y) ≤ f (x0, y0) for all domain points (x, y) which are also in the disk D. If (x0, y0) is a local maximum, then consider fx(x0, y0) = lim

∆x→0

f (x0 + ∆x, y0) − f (x0, y0) ∆x . Thinking about points a little bit to the left (∆x < 0) we see fx(x0, y0) = lim

∆x→0

≤ 0 < 0 = ⇒ fx(x0, y0) ≥ 0.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the First Derivative Test?

A point (x0, y0) is a local maximum if there is some (perhaps small) disk D about (x0, y0) so that f (x, y) ≤ f (x0, y0) for all domain points (x, y) which are also in the disk D. If (x0, y0) is a local maximum, then consider fx(x0, y0) = lim

∆x→0

f (x0 + ∆x, y0) − f (x0, y0) ∆x . Since fx(x0, y0) ≥ 0 and fx(x0, y0) ≤ 0 we must have fx(x0, y0) = 0.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the First Derivative Test?

A point (x0, y0) is a local maximum if there is some (perhaps small) disk D about (x0, y0) so that f (x, y) ≤ f (x0, y0) for all domain points (x, y) which are also in the disk D. If (x0, y0) is a local maximum, then consider fx(x0, y0) = lim

∆x→0

f (x0 + ∆x, y0) − f (x0, y0) ∆x . Note this argument required some points nearby in the domain a little to the left and a little to the right. That’s why the test

• nly works at interior points.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the First Derivative Test?

A point (x0, y0) is a local maximum if there is some (perhaps small) disk D about (x0, y0) so that f (x, y) ≤ f (x0, y0) for all domain points (x, y) which are also in the disk D. If (x0, y0) is a local maximum, then consider fx(x0, y0) = lim

∆x→0

f (x0 + ∆x, y0) − f (x0, y0) ∆x . fy(x0, y0) = 0 is similar.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the First Derivative Test?

A point (x0, y0) is a local maximum if there is some (perhaps small) disk D about (x0, y0) so that f (x, y) ≤ f (x0, y0) for all domain points (x, y) which are also in the disk D. If (x0, y0) is a local maximum, then consider fx(x0, y0) = lim

∆x→0

f (x0 + ∆x, y0) − f (x0, y0) ∆x . The same idea could be expressed by saying any directional derivative D

vf (p0) = 0 if p0 is a local extremum.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Saddle Points

A fundamental example is the saddle z = x2 − y2. Note (0, 0) is a critical point.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Saddle Points

A fundamental example is the saddle z = x2 − y2. Note (0, 0) is a critical point.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Saddle Points

A fundamental example is the saddle z = x2 − y2. Note (0, 0) is a critical point. z = x2 − y2.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Saddle Points

z = x2 − y2. The saddle’s basic property is that along some directions (e.g. a y = 0 section) it is an upward parabola (z = x2) while in

• ther directions (e.g. an (x = 0) section) it is a downward

parabola z = −y2.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Saddle Points

z = x2 − y2. The saddle’s basic property is that along some directions (e.g. a y = 0 section) it is an upward parabola (z = x2) while in

• ther directions (e.g. an (x = 0) section) it is a downward

parabola z = −y2.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Saddle Points

z = x2 − y2. In fact there are also vertical sections (e.g. along y = x) where z = 0.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Saddle Points

In fact there are also vertical sections (e.g. along y = x) where z = 0. The general pattern is that along a line y = kx we generally get a parabola, but how wide or narrow as well as which way it points depends on the value of k.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Saddle Points

The point (0, 0) here fits the definition of a saddle point. No matter how close to (0, 0) you look, there are places where the graph is above f (0, 0) = 0 as well as below.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Saddle Points

By contrast, for f (x, y) = x2 + y2, all vertical sections are “upward parabolas” and (0, 0) is a local (actually absolute) minimum.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Saddle Points

And for f (x, y) = −x2 − y2, all vertical sections are “downward parabolas” and (0, 0) is a local maximum.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Saddle Points

Note that z = xy has the same saddle shape as z = x2 − y2; upward along y = x and downward along y = −x.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Saddle Points

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Saddle Points

The algebra 2 x + y √ 2 x − y √ 2

• = x2 − y2

may be interpreted as saying z = 2xy is a 45 degree rotated z = x2 − y2.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Second Derivative Test

This is a quick and easy test that usually lets us recognize when behavior near a critical point resembles Local Minimum: z = x2 + y2

• r more generally z = z0 + a(x − x0)2 + c(y − y0)2

with a > 0, c > 0 Local Maximum: z = −x2 − y2 or more generally . . . Saddle: z = x2 − y2

• r more generally z = z0 + a(x − x0)2 + c(y − y0)2

with ac < 0; i.e. a > 0, c < 0

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Second Derivative Test

At a critical point (of a twice differentiable function), the 2nd deriv test is most naturally expressed in terms of the determinant of the matrix of second partials: ∆ =

• fxx(x0, y0)

fxy(x0, y0) fxy(x0, y0) fyy(x0, y0)

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Second Derivative Test

At a critical point (of a twice differentiable function), the 2nd deriv test is most naturally expressed in terms of the determinant of the matrix of second partials: ∆ =

• fxx(x0, y0)

fxy(x0, y0) fxy(x0, y0) fyy(x0, y0)

• The test says at a critical point:

∆ > 0 and fxx > 0 = ⇒ local minimum. ∆ > 0 and fxx < 0 = ⇒ local maximum. ∆ < 0 = ⇒ saddle point. ∆ = 0 = ⇒ inconclusive.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Second Derivative Test

The test says at a critical point: ∆ > 0 and fxx > 0 = ⇒ local minimum. ∆ > 0 and fxx < 0 = ⇒ local maximum. ∆ < 0 = ⇒ saddle point. ∆ = 0 = ⇒ inconclusive. One reason the above determinant comes up naturally is that for the function f (x, y) = ax2 + 2bxy + cy2 the matrix of second partials is just 2a 2b 2b 2c

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Second Derivative Test

One reason the above determinant comes up naturally is that for the function f (x, y) = ax2 + 2bxy + cy2 the matrix of second partials is just 2a 2b 2b 2c

• So for ∆ > 0 and fxx > 0 =

⇒ local minimum, think f (x, y) = x2 + y2 ∆ =

• 2

2

• = 4 > 0

2 > 0

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Second Derivative Test

One reason the above determinant comes up naturally is that for the function f (x, y) = ax2 + 2bxy + cy2 the matrix of second partials is just 2a 2b 2b 2c

• For ∆ > 0 and fxx < 0 =

⇒ local maximum, think f (x, y) = −x2 − y2 ∆ =

• −2

−2

• = 4 > 0

− 2 < 0

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Second Derivative Test

One reason the above determinant comes up naturally is that for the function f (x, y) = ax2 + 2bxy + cy2 the matrix of second partials is just 2a 2b 2b 2c

• For ∆ < 0 =

⇒ saddle, think f (x, y) = x2 − y2 ∆ =

• 2

−2

• = −4 < 0

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Second Derivative Test

One reason the above determinant comes up naturally is that for the function f (x, y) = ax2 + 2bxy + cy2 the matrix of second partials is just 2a 2b 2b 2c

• To see that ∆ = 0 is inconclusive, note that the following all

have (0, 0) as a critical point with ∆ = 0 : Minimum: f (x, y) = x4 + y4. Maximum: f (x, y) = −x4 − y4. Saddle: f (x, y) = x4 − y4 or f (x, y) = x3.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Second Derivative Test

One reason the above determinant comes up naturally is that for the function f (x, y) = ax2 + 2bxy + cy2 the matrix of second partials is just 2a 2b 2b 2c

• Example: Classify the critical points of f (x, y) = x3 −xy +y2.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Second Derivative Test

One reason the above determinant comes up naturally is that for the function f (x, y) = ax2 + 2bxy + cy2 the matrix of second partials is just 2a 2b 2b 2c

• Example: Classify the critical points of f (x, y) = x3 −xy +y2.

From before: fx = 3x2 − y = fy = −x + 2y = Our critical points are (0, 0) and (1 6, 1 12).

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Second Derivative Test

One reason the above determinant comes up naturally is that for the function f (x, y) = ax2 + 2bxy + cy2 the matrix of second partials is just 2a 2b 2b 2c

• Example: Find and classify the critical points of

f (x, y) = 2x4 − xy + y2.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the Second Derivative Test?

At a critical point (x0, y0), the linearization of f will be just a constant: L(x, y) = f (x0, y0) + 0(x − x0) + 0(y − y0) = f (x0, y0). This is the (only) degree 1 function whose values and first partials agree with f at (x0, y0).

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the Second Derivative Test?

When we go back to study Taylor approximation next week, we’ll see the second order Taylor polynomial P2(x, y) is defined to be P2 = f (x0, y0)+1 2

• a(x − x0)2 + 2b(x − x0)(y − y0) + c(y − y0)2

where the matrix of second partials is a b b c

• This is the (unique) degree 2 function whose values, first, and

second partials agree with f at (x0, y0). We might expect it to usually be decisive in classifying local extrema.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the Second Derivative Test?

Lets specialize to x0 = y0 = 0. The second derivative test is deciding whether z = ax2 + 2bx + cy2 looks like Local Minimum or

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the Second Derivative Test?

Lets specialize to x0 = y0 = 0. The second derivative test is deciding whether z = ax2 + 2bx + cy2 looks like Local Maximum or

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the Second Derivative Test?

Lets specialize to x0 = y0 = 0. The second derivative test is deciding whether z = ax2 + 2bx + cy2 looks like Saddle

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the Second Derivative Test?

It is shown in linear algebra that you can always rotate coordinates in z = ax2 + 2bx + cy2 to get rid of the the 2bxy term; i.e. make it look like z = λ1x2 + λ2y2 where only the signs of the λi matter.

From Math 2220 Class 10 V1cc Higher Partial Derivatives A Counterex- ample Higher Partials in Polar Cordinates Gradient Estimate from Graph Method of Characteristics Critical Points Why the First Derivative Test? Saddle Points Second Derivative Test

Why the Second Derivative Test?

When you do this

• a

b b c

• =
• λ1

λ2

• = λ1λ2

and a + c = λ1 + λ2. clrm So we can use the signs of ∆ and the signs of a or c (which wll be the same when ∆ > 0) to determine the signs of the λi.