From Math V2 Partial From Math 2220 Class 4pi.math.cornell.edu/~back/m222_f14/slides/sep5_v2.pdfFrom Math 2220 Class 4 V2 Partial derivatives Derivative Def Chain Rule Tangent Planes

From Math2220 Class 4

V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

From Math 2220 Class 4

Dr. Allen Back

Sep. 5, 2014


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Partial derivatives

Def: For f (x , y), the partial derivative with respect to x atp0 = (x0, y0) is

∂f

∂x= lim

h→0

f (x0 + h, y0)− f (x0, y0)h

or∂f

∂x= lim

h→0

f (p0 + he1)− f (p0)h

where e1 =

[10

]is the first standard basis vector of R2.[]


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Partial derivatives

Interpretation: Geometrically, for z = f (x , y), fx(x0, y0) is therate of increase of z as one moves along a horizontal line in thexy -plane through (x0, y0). Similarly fy (x0, y0) would relate tochange along a vertical line in the domain thru f (x0, y0).


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Partial derivatives

Calculation: For ∂f∂x just treat all variables other than x asconstants and use 1 variable differentiation rules.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Partial derivatives

Examples: Find fx and fy for

(a) f (x , y) = x3y4 (b) f (x , y) =√

x2 + y2

(c) f (x , y) = xy

On (c) use x = e ln x so xy = ey ln x .


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Partial derivatives

A Basic Application:

∆f ∼ fx∆x + fy ∆y .


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Partial derivatives

Problem: A right triangle with legs of length 20 and 99 has ahypotenuse of length exactly 101. About how much does thehypotenuse increase in length if each leg increases by .1?


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Partial derivatives

Approximations: Using ∆x = ∆y = .1 and computing fx , fy for

f (x , y) =√

x2 + y2 at (x , y) = (20, 99):

∆z ∼ (.2)(.1) + (1)(.1) =.12∆z ∼ (.1980)(.1) + (.9802)(.1) =.1178

z ∼ 101.1178

The exact length to 4 decimal places is 101.1179.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Partial derivatives

One uses terminology like “marginal” or “partial slope” forsuch partial derivatives.Notations Include: For z = f (x , y) at the point (x0, y0), any of

fx∂f

∂x

∂z

∂x

(∂z

∂x

)y

D1f fx(x0, y0)∂f

∂x

∣∣∣∣(x0,y0)


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Partial derivatives

Important because for (x,y) near (x0, y0) for f (x , y),

∆f ∼ fx∆x + fy ∆y


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Partial derivatives



In the above, the terms more precisely are:

∆f = f (x , y)− f (x0, y0)∆x = x − x0∆y = y − y0fx = fx(x0, y0)

fy = fy (x0, y0)


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Partial derivatives



This is true because

∆f = f (x , y)− f (x0, y0)= f (x , y)− f (x0, y)+ f (x0, y)− f (x0, y0)∼ fx∆x + fy ∆y

by the one variable approximation properties of derivatives.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Partial derivatives

Example: If the sides of a rectangle increase by 2% and 3%,approximately how much will the area increase?


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Derivative Def

For f : U ⊂ Rn → Rm, we write

f (p) = (f1(p), f2(p), . . . fm(p))

where each fi : U ⊂ Rn → R is an ordinary real valued functioncalled a component function of f .


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Derivative Def

For example the polar coordinate formulas

x = r cos θ

y = r sin θ

may be expressed in terms of the function f : R2 → R2 definedby

f (r , θ) = (r cos θ, r sin θ).

(Specifically (x , y) = f (r , θ).)

Here the first component function is

f1(r , θ) = r cos θ

and the second isf2(r , θ) = r sin θ.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Derivative Def

Def: A linear transformation T is DF(p) if

limx→pf (x)− f (p)− T (x − p)

‖x − p‖= 0


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Derivative Def

Thm: If f : U ⊂ Rn → Rm is diff., then the entries of Df(p)must be the partial derivatives of the component functions off .


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Derivative Def

Thm: If f : U ⊂ Rn → Rm is diff., then the entries of Df(p)must be the partial derivatives of the component functions off .So for f : Rn → Rm, the matrix T = Df (p0) will be m × n.One row for each component function. The row i col j entry

will be∂fi∂xj

.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Derivative Def



‖x − p‖= 0

Thm: If f : U ⊂ Rn → Rm is diff., then the entries of Df(p)must be the partial derivatives of the component functions off .Problem: f (r , θ) = (r cos θ, r sin θ). Find the derivative.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Derivative Def



‖x − p‖= 0

Thm: If f : U ⊂ Rn → Rm is diff., then the entries of Df(p)must be the partial derivatives of the component functions off .Why?


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Derivative Def

First remember from linear algebra, that for an m × n matrixT , the i’th column of T is where ei is mapped by Y ; i.e. Tei .


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Derivative Def

If the limit exists, restrict x = p + tei to lie on a line through pin the direction of the i’th standard basis vector ei .


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Derivative Def

If the limit exists, restrict x = p + tei to lie on a line through pin the direction of the i’th standard basis vector ei .Then

limt→0+f (p + tei )− f (p)

t= Tei

showing that the i’th column of T consists of the i’th partialderivatives of the component functions of f .


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

The chain rule in multivariable calculus is in some ways verysimple. But it can lead to extremely intricate sorts ofrelationships (try thermodynamics in physical chemistry . . . ) aswell as counter-intuitive looking formulas like

∂y

∂x= –

∂z∂x∂z∂y

.

(The above in a context where f (x , y , z) = C .)


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

First let’s try the conceptually simple point of view, using thefact that derivatives of functions are linear transformations.(Matrices.)


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

Think about differentiable functions

g : U ⊂ Rn → Rm

andf : V ⊂ Rm → Rp

where the image of f (f (U)) is a subset of the domain V of g .The chain rule is about the derivative of the composition f ◦ g .


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

Here’s a picture:


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

For g : U ⊂ Rn → Rm and f : V ⊂ Rm → Rp, let’s use

p to denote a point of Rn

q to denote a point of Rm

r to denote a point of Rp.

So more colloquially, we might write

q = g(p)

r = f (q)

and so of course f ◦ g gives the relationship r = f (g(p)).(The latter is (f ◦ g)(p).)


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

Fix a point p0 with g(p0) = q0 and f (q0) = r0. Let thederivatives of g and f at the relevant points be

T = Dg(p0) S = Df (q0).

How are the changes in p, q, and r related?


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

Fix a point p0 with g(p0) = q0 and f (q0) = r0. Let thederivatives of g and f at the relevant points be

T = Dg(p0) S = Df (q0).

How are the changes in p, q, and r related?By the linear approximation properties of the derivative,

∆q ∼ T∆p ∆r ∼ S∆q

And so plugging the first approximate equality into the secondgives the approximation

∆r ∼ S(T∆p) = (ST )∆p.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

∆r ∼ (ST )∆p.

What is this saying?


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

∆r ∼ (ST )∆p.

What is this saying?For g : U ⊂ Rn → Rm and f : V ⊂ Rm → Rp,

T = Df (p0) is an m × n matrix

S = Dg(q0) is an p ×m matrix

So the product ST is a p × n matrix representing the derivativeat p0 of g ◦ f .


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

So the chain rule theorem says that if f is differentiable at p0with f (p0) = q0 and g is differentiable at q0, then g ◦ f is alsodifferentiable at p0 with derivative the matrix product

(Dg(q0)) (Df (p0)) .


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Chain Rule

Problem: Suppose we have the polar coordinate map

g(r , θ) = (r cos θ, r sin θ)

and (r , θ) = f (u, v) is given by f (u, v) = (uv , v). Find thederivative of g ◦ f .


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Tangent Planes

The tangent plane to the graph of z = f (x , y) at(x , y) = (x0, y0) is defined to be the plane given by

z − z0 = fx(x0, y0)(x − x0) + fy (x0, y0)(y − y0).

(The approximation ∆z ∼ fx∆x + fy ∆y is replaced by an exactequality on the tangent plane.)


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Tangent Planes

Tangent plane to z = x2 − y2 at (−1, 0, 1).

Note the tangent plane needn’t meet the surface in just onepoint.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Planes in R3


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Planes in R3

If the normal ~n =< a, b, c >, ~r =< x , y , z > is a general pointand P0 = (x0, y0, z0), then ~n · (~r − P0) = 0 becomes

a(x − x0) + b(y − y0) + c(z − z0) = 0.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Planes in R3

Find the equation of the plane through the three pointsP0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3).


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Planes in R3

Find the equation of the plane through the three pointsP0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3).Solution: First find the normal

~n =−−−→P0P1 ×

−−−→P0P2


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Planes in R3


~n =−−−→P0P1 ×

−−−→P0P2

The cross product: ∣∣∣∣∣∣î ĵ k̂−2 1 10 2 2

∣∣∣∣∣∣which is


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Planes in R3


~n =−−−→P0P1 ×

−−−→P0P2

The cross product: ∣∣∣∣∣∣î ĵ k̂−2 1 10 2 2

∣∣∣∣∣∣which is

î

∣∣∣∣ 1 12 2∣∣∣∣− ĵ ∣∣∣∣ −2 10 2

∣∣∣∣+ k̂ ∣∣∣∣ −2 10 2∣∣∣∣ =< 0, 4,−4 > .


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Planes in R3

Find the equation of the plane through the three pointsP0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3).

î

∣∣∣∣ 1 12 2∣∣∣∣− ĵ ∣∣∣∣ −2 10 2

∣∣∣∣+ k̂ ∣∣∣∣ −2 10 2∣∣∣∣ =< 0, 4,−4 > .

So our plane is

< 0, 4,−4 > ·(~r− < 1, 0, 1 >) = 0


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Planes in R3

Find the equation of the plane through the three pointsP0 = (1, 0, 1), P1 = (−1, 1, 2) and P2 = (1, 2, 3).So our plane is

< 0, 4,−4 > ·(~r− < 1, 0, 1 >) = 0

or

0(x − 1) + 4(y − 0)− 4(z − 1) = 0 or 4y − 4z + 4 = 0.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Lines in R3


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Lines in R3

Find the equation of the line through the points P0 = (1, 1, 0)and P1 = (2, 2, 2).


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Lines in R3

Find the equation of the line through the points P0 = (1, 1, 0)and P1 = (2, 2, 2).

Solution:−−−→P0P1 = (2, 2, 2)− (1, 1, 0) =< 1, 1, 2 > .

So our line is

~r =< 1, 1, 0 > +t < 1, 1, 2 >=< 1 + t, 1 + t, 2t > .

where t is any real number.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Lines in R3

Solution:−−−→P0P1 = (2, 2, 2)− (1, 1, 0) =< 1, 1, 2 > .

So our line is

~r =< 1, 1, 0 > +t < 1, 1, 2 >=< 1 + t, 1 + t, 2t > .

where t is any real number.This is called the vector form of the equation of a line.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Lines in R3

Solution:−−−→P0P1 = (2, 2, 2)− (1, 1, 0) =< 1, 1, 2 > .

So our line is

~r =< 1, 1, 0 > +t < 1, 1, 2 >=< 1 + t, 1 + t, 2t > .

where t is any real number.This is called the vector form of the equation of a line.Thinking our general position vector ~r =< x , y , z >, we canexpress this as the parametric form:

x = 1 + t

y = 1 + t

z = 2t


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Lines in R3

Thinking our general position vector ~r =< x , y , z >, we canexpress this as the parametric form:

x = 1 + t

y = 1 + t

z = 2t

Solving for t shows

t = x − 1 = y − 1 = z2

which realizes this line as the intersection of the planes x = yand z = 2(y − 1) but there are many other pairs of planescontaining this line.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Cross product

The cross product of vectors in R3 is another vector.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Cross product

The cross product of vectors in R3 is another vector.It is good because:

it is geometrically meaningful

it is straightforward to calculate

it is useful (e.g. torque, angular momentum)


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Cross product

The cross product of vectors in R3 is another vector.~u = ~v × ~w is geometrically determined by the properties:

~u is perpendicular to both ~v and ~w .

~u| is the |~v ||~w | sin θ, the area of the parallelogram spannedby ~v and ~w .

Choice from the remaining two possibilities is now madebased on the “right hand rule.”


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Cross product

~u = ~v × ~w is geometrically determined by the properties:~u is perpendicular to both ~v and ~w .

~u| is the |~v ||~w | sin θ, the area of the parallelogram spannedby ~v and ~w .

Choice from the remaining two possibilities is now madebased on the “right hand rule.”


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Cross product

The algebraic definition of the cross product is based on the“determinant”

~v × ~w =

∣∣∣∣∣∣î ĵ k̂v1 v2 v3w1 w2 w3

∣∣∣∣∣∣which means

~v × ~w = î∣∣∣∣ v2 v3w2 w3

∣∣∣∣− ĵ ∣∣∣∣ v1 v3w1 w3∣∣∣∣+ k̂ ∣∣∣∣ v1 v2w1 w2

∣∣∣∣ .where ∣∣∣∣ a bc d

∣∣∣∣ = ad − bcand î =< 1, 0, 0 >, ĵ =< 0, 1, 0 >, and k̂ =< 0, 0, 1 >,


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Cross product

~v × ~w = −~w × ~v


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Cross product

Sinceî × ĵ = k̂

and cyclic (so ĵ × k̂ = î and k̂ × î = ĵ) it is sometimes easiestto use that algebra or comparison with the picture below todetermine cross products or use the right hand rule.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Cross product

For example

< 1, 1, 0 > × < 0, 0, 1 >= (̂i + ĵ)× k̂ = −ĵ + î =< 1,−1, 0 >

is easier than writing out the 3× 3 determinant.


V2

Partialderivatives

Derivative Def

Chain Rule

TangentPlanes

Planes in R3

Lines in R3

Cross product

Cross product

Cross products can be used to

find the area of a parallelogram or triangle spanned by twovectors in R3.

find the volume of a parallelopiped using the scalar tripleproduct

~u · (~v × ~w) = (~u × ~v) · ~w .

Partial derivativesDerivative DefChain RuleTangent PlanesPlanes in R3Lines in R3Cross product

Documents

From Math V2 Partial From Math 2220 Class 4pi.math.cornell.edu/~back/m222_f14/slides/sep5_v2.pdfFrom Math 2220 Class 4 V2 Partial derivatives Derivative Def Chain Rule Tangent Planes