quadric surfaces, partial derivatives. sections 12.6 and ...nirobles/files241/lecture07.pdfpartial...

Math 241: Multivariable calculus, Lecture 7Quadric Surfaces, Partial derivatives.

Sections 12.6 and 14.2, 14.3

go.illinois.edu/math241fa17

Wednesday, September 13th, 2017

go.illinois.edu/math241fa17.

Quadric Surfaces.

Definition

A Quadric Surface is the collection of points (x , y , z) that satisfy asecond degree equation:

Ax2 + By2 + Cz2 + Dxy + Eyz + Fxz + Gx + Hy + Iz + J = 0

Looks complicated!

However we can shift and rotate our surfaceand then it becomes simpler: it can always be written as

Ax2 + By2 + Cz2 + J = 0, or as Ax2 + By2 + Iz = 0

(Different A,B,C , I , J!). The type of surface you get depends onthe signs of A,B,C , . . . .

Quadric Surfaces.

Definition

Looks complicated!

However we can shift and rotate our surfaceand then it becomes simpler: it can always be written as

Quadric Surfaces.

Definition

Looks complicated! However we can shift and rotate our surfaceand then it becomes simpler:

it can always be written as

Quadric Surfaces.

Definition

Looks complicated! However we can shift and rotate our surfaceand then it becomes simpler: it can always be written as

(Different A,B,C , I , J!).

The type of surface you get depends onthe signs of A,B,C , . . . .

Quadric Surfaces.

Definition

Looks complicated! However we can shift and rotate our surfaceand then it becomes simpler: it can always be written as

Coordinate Transform

Example 0.1

Consider the equation

x2 + y2 + z2 − 2x = 3

Simplify equation by shifting coordinates

Note that we can complete the square:x2 − 2x = x2 − 2x + 1− 1 = (x − 1)2 − 1. Using this in theequation we see that it is equivalent to

(x − 1)2 + y2 + z2 − 1 = 3, or (x − 1)2 + y2 + z2 = 4

so we have a sphere with radius 2. After shifting the x-axis byx 7→ x + 1 we get a standard equation

x2 + y2 + z2 = 4.

Example 0.1

x2 + y2 + z2 − 2x = 3

(x − 1)2 + y2 + z2 − 1 = 3, or (x − 1)2 + y2 + z2 = 4

x2 + y2 + z2 = 4.

Example 0.1

x2 + y2 + z2 − 2x = 3

Note that we can complete the square:x2 − 2x = x2 − 2x + 1− 1 = (x − 1)2 − 1.

Using this in theequation we see that it is equivalent to

(x − 1)2 + y2 + z2 − 1 = 3, or (x − 1)2 + y2 + z2 = 4

x2 + y2 + z2 = 4.

Example 0.1

x2 + y2 + z2 − 2x = 3

(x − 1)2 + y2 + z2 − 1 = 3,

or (x − 1)2 + y2 + z2 = 4

x2 + y2 + z2 = 4.

Example 0.1

x2 + y2 + z2 − 2x = 3

(x − 1)2 + y2 + z2 − 1 = 3, or (x − 1)2 + y2 + z2 = 4

x2 + y2 + z2 = 4.

Example 0.1

x2 + y2 + z2 − 2x = 3

(x − 1)2 + y2 + z2 − 1 = 3, or (x − 1)2 + y2 + z2 = 4

x2 + y2 + z2 = 4.

Example 0.1

x2 + y2 + z2 − 2x = 3

(x − 1)2 + y2 + z2 − 1 = 3, or (x − 1)2 + y2 + z2 = 4

x2 + y2 + z2 = 4.

Example 0.1

x2 + y2 + z2 − 2x = 3

(x − 1)2 + y2 + z2 − 1 = 3, or (x − 1)2 + y2 + z2 = 4

x2 + y2 + z2 = 4.

Example 0.2

Sketch the graph of x2

9 + y2 + z2

9 = 1.

��

Example 0.2

9 + y2 + z2

9 = 1.

��

Example 0.2

9 + y2 + z2

9 = 1.

��

Example 0.3

Sketch the graph of x2 − y2 + z2 = 1.

��

Example 0.3

��

Example 0.3

��

Example 0.4

Sketch the graph of x2 + y2 − z = 0.

��

Example 0.4

��

Example 0.4

��

partial derivatives

• fx(a, b) = ∂f (x ,y)∂x |(x ,y)=(a,b) = limh→0

f (a+h,b)−f (a,b)h . Same for

fy (a, b).

• Easy to calculate: the partial fx is just the ordinary derivativeof f (x , y), where y is considered a constant.

• In the same way we can define second partialsfxx , fyy , fxy = fyx , for nice functions.

• In the same way we can define partials of functions of 3 ormore variables.

partial derivatives

fy (a, b).

partial derivatives

fy (a, b).

partial derivatives

fy (a, b).

partial derivatives

fy (a, b).

Math 241: Problem of the day

Level curves for a function f : R2 → R are shown.

• What is f (2, 0)?• What is fx(2, 0)?• What is fx(2, ε), ε > 0?• Decide whether fxy (2, 0) is positive, negative or zero.

f=−2

f=−1

Level curves for a function f : R2 → R are shown.• What is f (2, 0)?

• What is fx(2, 0)?• What is fx(2, ε), ε > 0?• Decide whether fxy (2, 0) is positive, negative or zero.

f=−2

f=−1

Level curves for a function f : R2 → R are shown.• What is f (2, 0)?• What is fx(2, 0)?

• What is fx(2, ε), ε > 0?• Decide whether fxy (2, 0) is positive, negative or zero.

f=−2

f=−1

Level curves for a function f : R2 → R are shown.• What is f (2, 0)?• What is fx(2, 0)?• What is fx(2, ε), ε > 0?

• Decide whether fxy (2, 0) is positive, negative or zero.

f=−2

f=−1

Level curves for a function f : R2 → R are shown.• What is f (2, 0)?• What is fx(2, 0)?• What is fx(2, ε), ε > 0?• Decide whether fxy (2, 0) is positive, negative or zero.

f=−2

f=−1

Partial derivatives: applications.

Recall that an ODE (ordinary differential equation) is an equationinvolving a function of one variable and its derivatives: e.g. y ′ = yis exponential growth/decay equation.

A PDE (partial differential equation) is an equation involving afunction of several variables and its derivatives:e.g. ut = uxx + uyy + uzz is the heat equation.

Recall that an ODE (ordinary differential equation) is an equationinvolving a function of one variable and its derivatives

: e.g. y ′ = yis exponential growth/decay equation.

A PDE (partial differential equation) is an equation involving afunction of several variables and its derivatives

:e.g. ut = uxx + uyy + uzz is the heat equation.

Tangent planes and differentiability.

Tangent plane to graph z = f (x , y) at (a, b) is graph of the linearapproximation:

L(a,b)(x , y) = f (a, b) + fx(a, b)(x − a) + fy (a, b)(y − b).

Compare with g : R→ R, where linear approximation at a is

La(x) = g(a) + g ′(a)(x − a).

Tangent plane: z = L(a,b)(x , y).

When is this really tangent?

La(x) = g(a) + g ′(a)(x − a).

Differentiability

We say that f (x , y) is differentiable at (a, b) if fx(a, b) and fy (a, b)exist and

0 = lim(h,k)→(0,0)

f (a + h, b + k)− L(a,b)(a + h, b + k)√h2 + k2

= lim(h,k)→(0,0)

f (a + h, b + k)− f (a, b)− fx(a, b)h − fy (a, b)k√h2 + k2

fx(a, b) and fy (a, b) can exist without f being differentiable at(a, b).

Example:

f (x , y) =

{ xyx2+y2 for (x , y) 6= (0, 0)

0 for (x , y) = (0, 0)

Differentiability

0 = lim(h,k)→(0,0)

f (a + h, b + k)− L(a,b)(a + h, b + k)√h2 + k2

= lim(h,k)→(0,0)

Example:

f (x , y) =

{ xyx2+y2 for (x , y) 6= (0, 0)

0 for (x , y) = (0, 0)

Differentiability

0 = lim(h,k)→(0,0)

f (a + h, b + k)− L(a,b)(a + h, b + k)√h2 + k2

= lim(h,k)→(0,0)

Example:

f (x , y) =

{ xyx2+y2 for (x , y) 6= (0, 0)

0 for (x , y) = (0, 0)

Differentiability

0 = lim(h,k)→(0,0)

f (a + h, b + k)− L(a,b)(a + h, b + k)√h2 + k2

= lim(h,k)→(0,0)

Example:

f (x , y) =

{ xyx2+y2 for (x , y) 6= (0, 0)

0 for (x , y) = (0, 0)

Differentiability

0 = lim(h,k)→(0,0)

f (a + h, b + k)− L(a,b)(a + h, b + k)√h2 + k2

= lim(h,k)→(0,0)

Example:

f (x , y) =

{ xyx2+y2 for (x , y) 6= (0, 0)

0 for (x , y) = (0, 0)

Differentiability

0 = lim(h,k)→(0,0)

f (a + h, b + k)− L(a,b)(a + h, b + k)√h2 + k2

= lim(h,k)→(0,0)

Example:

f (x , y) =

{ xyx2+y2 for (x , y) 6= (0, 0)

0 for (x , y) = (0, 0)

Two theorems on differentiability

Theorem. If f is differentiable at (a, b) then f is continuous at(a, b).

Theorem. If fx and fy exist and are continuous near (a, b) then fis differentiable at (a, b).

This means there is some δ > 0 so that fx and fy exist and arecontinuous on a disk of radius δ > 0 about (a, b):

(x − a)2 + (y − b)2 < δ.

Clairaut’s Theorem. If f is defined near (a, b) and fxy and fyx arecontinuous near (a, b), then fxy (a, b) = fyx(a, b).

(x − a)2 + (y − b)2 < δ.

quadric surfaces, partial derivatives. sections 12.6 and ...nirobles/files241/lecture07.pdfpartial...

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