Functions of Several Variables - UCONNstein/math210/Slides/math210-04slides.pdf · Functions of Several Variables A function of several variables is just what it sounds like. It may

Functions of Several Variables

A function of several variables is just what it sounds like.

It may beviewed in at least three different ways. We will use a function oftwo variables as an example.

I z = f (x , y) may be viewed as a function of the twoindependent variables x , y .

I It may be viewed as a function defined at different points(x , y) in the plane.

I It may be viewed as a function whose domain is the set ofvectors < x , y > or x i + y j.

Alan H. SteinUniversity of Connecticut


A function of several variables is just what it sounds like. It may beviewed in at least three different ways. We will use a function oftwo variables as an example.























Limits of Functions of Several Variables

We define a limit of a function of several variables essentially thesame way we define a limit for an ordinary function:

Definition (Limit)

limx→c f (x) = L if ∀ε > 0, ∃δ > 0 such that |f (x)− L| < εwhenever 0 < |x − c | < δ.

Definition (Limit)

limx→c f (x) = L if ∀ε > 0, ∃δ > 0 such that |f (x)− L| < εwhenever 0 < |x− c| < δ.




Definition (Limit)


Definition (Limit)





Definition (Limit)


Definition (Limit)



Properties of Limits

Rule of Thumb: If a property of limits makes sense whentranslated to refer to a limit of a function of several variables, thenit is valid for a function of several variables.

For example, the limit of a sum will be the sum of the limits, thelimit of a difference will be the difference of the limits, the limit ofa product will be the product of the limits and the limit of aquotient will be the quotient of the limits, provided the latter limitexists.




For example, the limit of a sum will be the sum of the limits,

thelimit of a difference will be the difference of the limits, the limit ofa product will be the product of the limits and the limit of aquotient will be the quotient of the limits, provided the latter limitexists.




For example, the limit of a sum will be the sum of the limits, thelimit of a difference will be the difference of the limits,

the limit ofa product will be the product of the limits and the limit of aquotient will be the quotient of the limits, provided the latter limitexists.




For example, the limit of a sum will be the sum of the limits, thelimit of a difference will be the difference of the limits, the limit ofa product will be the product of the limits

and the limit of aquotient will be the quotient of the limits, provided the latter limitexists.




For example, the limit of a sum will be the sum of the limits, thelimit of a difference will be the difference of the limits, the limit ofa product will be the product of the limits and the limit of aquotient will be the quotient of the limits, provided the latter limitexists.


Continuity

The definition of continuity for a function of several variables isessentially the same as the definition for an ordinary function.

Definition (Continuity)

A function f is continuous at c iflimx→c f (x) = f (c).

Definition (Continuity for a Function of Several Variables)

A function f is continuous at c if limx→c f (x) = f (c).

As with ordinary functions, functions of several variables willgenerally be continuous except where there’s an obvious reason forthem not to be.


Continuity








Continuity








Continuity








Partial Derivatives

For a function of several variables, we have partial derivatives withrespect to each of its variables. The definition is based on thedefinition of an ordinary derivative.

Definition (Derivative)

Let f : R → R.df

dx(x) = limh→0

f (x + h)− f (x)

h.

Definition (Partial Derivative)

Let f : R2 → R.∂f

∂x(x , y) = limh→0

f (x + h, y)− f (x , y)

h,

∂f

∂y(x , y) = limh→0

f (x , y + h)− f (x , y)

h.

The obvious generalizations hold for functions with more than twoindependent variables.


Partial Derivatives



Let f : R → R.df

dx(x) = limh→0

f (x + h)− f (x)

h.



∂x(x , y) = limh→0

f (x + h, y)− f (x , y)

h,

∂f

∂y(x , y) = limh→0

f (x , y + h)− f (x , y)

h.



Partial Derivatives



Let f : R → R.df

dx(x) = limh→0

f (x + h)− f (x)

h.



∂x(x , y) = limh→0

f (x + h, y)− f (x , y)

h,

∂f

∂y(x , y) = limh→0

f (x , y + h)− f (x , y)

h.



Partial Derivatives



Let f : R → R.df

dx(x) = limh→0

f (x + h)− f (x)

h.



∂x(x , y) = limh→0

f (x + h, y)− f (x , y)

h,

∂f

∂y(x , y) = limh→0

f (x , y + h)− f (x , y)

h.



Calculation of Partial Derivatives

Effectively, we calculate the partial derivative of a function withrespect to one of its independent variables by acting as if the otherindependent variables were actually constants.


Notation

The following notations for the partial derivatives of a functionz = f (x , y) are equivalent.

fx =∂f

∂x=

∂z

∂x= f1 = D1f = Dx f

fy =∂f

∂y=

∂z

∂y= f2 = D2f = Dy f


Notation


fx =∂f

∂x=

∂z

∂x= f1 = D1f = Dx f

fy =∂f

∂y=

∂z

∂y= f2 = D2f = Dy f


Notation


fx =∂f

∂x=

∂z

∂x= f1 = D1f = Dx f

fy =∂f

∂y=

∂z

∂y= f2 = D2f = Dy f


Higher Order Derivatives

Since a partial derivative is itself a function of several variables, ithas its own partial derivatives.

(fx)y = fxy = f12 =∂

∂y

(∂f

∂x

)=

∂2f

∂y∂x=

∂2z

∂y∂x

(fy )x = fyx = f21 =∂

∂x

(∂f

∂y

)=

∂2f

∂x∂y=

∂2z

∂x∂y




(fx)y = fxy = f12 =∂

∂y

(∂f

∂x

)=

∂2f

∂y∂x=

∂2z

∂y∂x

(fy )x = fyx = f21 =∂

∂x

(∂f

∂y

)=

∂2f

∂x∂y=

∂2z

∂x∂y




(fx)y = fxy = f12 =∂

∂y

(∂f

∂x

)=

∂2f

∂y∂x=

∂2z

∂y∂x

(fy )x = fyx = f21 =∂

∂x

(∂f

∂y

)=

∂2f

∂x∂y=

∂2z

∂x∂y


Changing the Order of Differentiation

Theorem (Clairaut’s Theorem)

If fxy and fyx are both continuous on a disk containing (a, b), thenfxy (a, b) = fyx(a, b).

Proof.Let φ(h) = f (x + h, y + h)− f (x , y + h)− f (x + h, y) + f (x , y).

The motivation comes from writing either fxy or fyx as a limit.

We may write φ(h) = α(y + h)− α(y), whereα(t) = f (x + h, t)− f (x , t). The Mean Value Theorem impliesα(y + h)− α(y) = α′(t)h for some t between y and y + h. Sinceα′(t) = f2(x + h, t)− f2(x , t), we haveφ(h) = [f2(x + h, t)− f2(x , t)]h.

If we write β(s) = f2(s, t), thenf2(x + h, t)− f2(x , t) = β(x + h)− β(x).





Proof.Let φ(h) = f (x + h, y + h)− f (x , y + h)− f (x + h, y) + f (x , y).The motivation comes from writing either fxy or fyx as a limit.








We may write φ(h) = α(y + h)− α(y), whereα(t) = f (x + h, t)− f (x , t).

The Mean Value Theorem impliesα(y + h)− α(y) = α′(t)h for some t between y and y + h. Sinceα′(t) = f2(x + h, t)− f2(x , t), we haveφ(h) = [f2(x + h, t)− f2(x , t)]h.

















Clairault’s Theorem

β(s) = f2(s, t), f2(x + h, t)− f2(x , t) = β(x + h)− β(x).

By the Mean Value Theorem, β(x + h)− β(x) = β′(s)h for some sbetween x and x + h. Since β′(s) = f21(s, t), we getf2(x + h, t)− f2(x , t) = f21(s, t)h, so φ(h) = f21(s, t)h2.

Thusφ(h)

h2= f21(s, t) → f21(x , y) as h → 0, since f21 is continuous

at (x , y).

A similar calculation showsφ(h)

h2= f12(s, t) → f12(x , y) as h → 0,

showing f12(x , y) = f21(x , y).




By the Mean Value Theorem, β(x + h)− β(x) = β′(s)h for some sbetween x and x + h.

Since β′(s) = f21(s, t), we getf2(x + h, t)− f2(x , t) = f21(s, t)h, so φ(h) = f21(s, t)h2.

Thusφ(h)


at (x , y).


h2= f12(s, t) → f12(x , y) as h → 0,





By the Mean Value Theorem, β(x + h)− β(x) = β′(s)h for some sbetween x and x + h. Since β′(s) = f21(s, t),

we getf2(x + h, t)− f2(x , t) = f21(s, t)h, so φ(h) = f21(s, t)h2.

Thusφ(h)


at (x , y).


h2= f12(s, t) → f12(x , y) as h → 0,





By the Mean Value Theorem, β(x + h)− β(x) = β′(s)h for some sbetween x and x + h. Since β′(s) = f21(s, t), we getf2(x + h, t)− f2(x , t) = f21(s, t)h, so

φ(h) = f21(s, t)h2.

Thusφ(h)


at (x , y).


h2= f12(s, t) → f12(x , y) as h → 0,






Thusφ(h)


at (x , y).


h2= f12(s, t) → f12(x , y) as h → 0,






Thusφ(h)


at (x , y).


h2= f12(s, t) → f12(x , y) as h → 0,






Thusφ(h)


at (x , y).


h2= f12(s, t) → f12(x , y) as h → 0,






Thusφ(h)


at (x , y).


h2= f12(s, t) → f12(x , y) as h → 0,



Tangent Planes

Consider a surface z = f (x , y) and suppose we are interested inthe plane tangent to the surface at the point (a, b, c), wherec = f (a, b).

Since∂z

∂xrepresents about how much z will change if x changes by

1 and y is fixed, here, and elsewhere as we look at tangent planes,tangent plane approximations and differentials, the partialderivative shown really means the partial derivative’s value at therelevant point, in this case (a, b), it seems reasonable to expect the

vector < 1, 0,∂z

∂x> to be tangent to the surface.

Similarly, it is reasonable to expect the vector < 0, 1,∂z

∂y> to be

tangent to the surface.


Tangent Planes


Since∂z


1 and y is fixed,

here, and elsewhere as we look at tangent planes,tangent plane approximations and differentials, the partialderivative shown really means the partial derivative’s value at therelevant point, in this case (a, b), it seems reasonable to expect the

vector < 1, 0,∂z



∂y> to be



Tangent Planes


Since∂z


1 and y is fixed, here, and elsewhere as we look at tangent planes,tangent plane approximations and differentials, the partialderivative shown really means the partial derivative’s value at therelevant point, in this case (a, b),

it seems reasonable to expect the

vector < 1, 0,∂z



∂y> to be



Tangent Planes


Since∂z



vector < 1, 0,∂z



∂y> to be



Tangent Planes


Since∂z



vector < 1, 0,∂z



∂y> to be



Tangent Planes

We thus expect n =

∣∣∣∣∣∣∣∣∣i j k

1 0∂z

∂x

0 1∂z

∂y

∣∣∣∣∣∣∣∣∣ = −∂z

∂xi− ∂z

∂yj + k to be a

normal vector to the tangent plane.

We thus take n =< −∂z

∂x,−∂z

∂y, 1 >.

We thus get < −∂z

∂x,−∂z

∂y, 1 > · < x − a, y − b, z − c >= 0 as an

equation for the tangent plane, or

−∂z

∂x(x − a)− ∂z

∂y(y − b) + (z − c) = 0, or

z − c =∂z

∂x(x − a) +

∂z

∂y(y − b).

This should be reminiscent of the Point-Slope Formula for theequation of a line.


Tangent Planes

We thus expect n =

∣∣∣∣∣∣∣∣∣i j k

1 0∂z

∂x

0 1∂z

∂y

∣∣∣∣∣∣∣∣∣ = −∂z

∂xi− ∂z

∂yj + k to be a



∂x,−∂z

∂y, 1 >.


∂x,−∂z

∂y, 1 > · < x − a, y − b, z − c >= 0 as an


−∂z

∂x(x − a)− ∂z

∂y(y − b) + (z − c) = 0, or

z − c =∂z

∂x(x − a) +

∂z

∂y(y − b).



Tangent Planes

We thus expect n =

∣∣∣∣∣∣∣∣∣i j k

1 0∂z

∂x

0 1∂z

∂y

∣∣∣∣∣∣∣∣∣ = −∂z

∂xi− ∂z

∂yj + k to be a



∂x,−∂z

∂y, 1 >.


∂x,−∂z

∂y, 1 > · < x − a, y − b, z − c >= 0 as an

equation for the tangent plane,

or

−∂z

∂x(x − a)− ∂z

∂y(y − b) + (z − c) = 0, or

z − c =∂z

∂x(x − a) +

∂z

∂y(y − b).



Tangent Planes

We thus expect n =

∣∣∣∣∣∣∣∣∣i j k

1 0∂z

∂x

0 1∂z

∂y

∣∣∣∣∣∣∣∣∣ = −∂z

∂xi− ∂z

∂yj + k to be a



∂x,−∂z

∂y, 1 >.


∂x,−∂z

∂y, 1 > · < x − a, y − b, z − c >= 0 as an


−∂z

∂x(x − a)− ∂z

∂y(y − b) + (z − c) = 0,

or

z − c =∂z

∂x(x − a) +

∂z

∂y(y − b).



Tangent Planes

We thus expect n =

∣∣∣∣∣∣∣∣∣i j k

1 0∂z

∂x

0 1∂z

∂y

∣∣∣∣∣∣∣∣∣ = −∂z

∂xi− ∂z

∂yj + k to be a



∂x,−∂z

∂y, 1 >.


∂x,−∂z

∂y, 1 > · < x − a, y − b, z − c >= 0 as an


−∂z

∂x(x − a)− ∂z

∂y(y − b) + (z − c) = 0, or

z − c =∂z

∂x(x − a) +

∂z

∂y(y − b).



Tangent Planes

We thus expect n =

∣∣∣∣∣∣∣∣∣i j k

1 0∂z

∂x

0 1∂z

∂y

∣∣∣∣∣∣∣∣∣ = −∂z

∂xi− ∂z

∂yj + k to be a



∂x,−∂z

∂y, 1 >.


∂x,−∂z

∂y, 1 > · < x − a, y − b, z − c >= 0 as an


−∂z

∂x(x − a)− ∂z

∂y(y − b) + (z − c) = 0, or

z − c =∂z

∂x(x − a) +

∂z

∂y(y − b).



Tangent Hyperplanes

It generalizes to

y − b =∑n

i=1

∂y

∂xi(xi − ai )

as an equation for the hyperplane tangent to the hypersurfacey = f (x1, x2, . . . , xn) at the point (a1, a2, . . . , an, b).


Tangent Plane Approximations and Differentials

If we take z − c =∂z

∂x(x − a) +

∂z

∂y(y − b) and solve for z , we get

z = c +∂z

∂x(x − a) +

∂z

∂y(y − b)

This should be reminiscent of the Tangent Line Approximation forordinary functions.

We may use this formula to approximate f (x , y) at a point (x , y)close to a point (a, b).

Definition (Differentials)

dx = ∆x = x − ady = ∆y = y − b

dz =∂z

∂x(x − a) +

∂z

∂y(y − b)




∂x(x − a) +

∂z


z = c +∂z

∂x(x − a) +

∂z

∂y(y − b)




dx = ∆x = x − ady = ∆y = y − b

dz =∂z

∂x(x − a) +

∂z

∂y(y − b)




∂x(x − a) +

∂z


z = c +∂z

∂x(x − a) +

∂z

∂y(y − b)




dx = ∆x = x − ady = ∆y = y − b

dz =∂z

∂x(x − a) +

∂z

∂y(y − b)




∂x(x − a) +

∂z


z = c +∂z

∂x(x − a) +

∂z

∂y(y − b)




dx = ∆x = x − ady = ∆y = y − b

dz =∂z

∂x(x − a) +

∂z

∂y(y − b)




∂x(x − a) +

∂z


z = c +∂z

∂x(x − a) +

∂z

∂y(y − b)




dx = ∆x = x − ady = ∆y = y − b

dz =∂z

∂x(x − a) +

∂z

∂y(y − b)


Differentials

We may use the differential dz to approximate the change∆z = ∆f of a function f (x , y) if the independent variables x andy change by amounts dx and dy .

This generalizes in the obvious way to functions of more than twovariables.


Differentials

We may use the differential dz to approximate the change∆z = ∆f of a function f (x , y) if the independent variables x andy change by amounts dx and dy .

This generalizes in the obvious way to functions of more than twovariables.


Differentiability

Recall that for an ordinary function y = f (x) which was

differentiable at a point, we founddy −∆y

∆x→ 0 as ∆x → 0.

We take the analogue of this as a definition of differentiability forfunctions of several variables. We state the definition for the caseof a function of two variables; the variation for more variablesshould be obvious.

Definition (Differentiable)

We say a function f (x , y) is differentiable at a point ifdz −∆z√

(∆x)2 + (∆y)2→ 0 as

√(∆x)2 + (∆y)2 → 0.


Differentiability



∆x→ 0 as ∆x → 0.




(∆x)2 + (∆y)2→ 0 as

√(∆x)2 + (∆y)2 → 0.


Differentiability



∆x→ 0 as ∆x → 0.




(∆x)2 + (∆y)2→ 0 as

√(∆x)2 + (∆y)2 → 0.


Differentiability

Recall√

(∆x)2 + (∆y)2 is the distance between (x , y) and thepoint in question.

Effectively, we are defining a function of several variables to bedifferentialbe when an approximation using differentials isreasonable.

We still need a reasonable way of determining whether a functionis differentiable. This is given by the following theorem.


Differentiability

Recall√





Differentiability

Recall√



We still need a reasonable way of determining whether a functionis differentiable.

This is given by the following theorem.


Differentiability

Recall√





Differentiability

TheoremIf both partial derivatives of a function z = f (x , y) are continuousin some open disc {(x , y) : (x − a)2 + (y − b)2 < r} centered at(a, b), then f (x , y) is differentiable at (a, b).

Proof. We need to showdz −∆z√

(∆x)2 + (∆y)2→ 0 as√

(∆x)2 + (∆y)2 → 0.

We may write

∆z − dz = f (x , y)− f (a, b)−(

∂z

∂x(x − a) +

∂z

∂y(y − b)

)=

f (x , y)− f (a, y)− ∂z

∂x(x − a) + f (a, y)− f (a, b)− ∂z

∂y(y − b).


Differentiability



(∆x)2 + (∆y)2→ 0 as√

(∆x)2 + (∆y)2 → 0.

We may write

∆z − dz = f (x , y)− f (a, b)−(

∂z

∂x(x − a) +

∂z

∂y(y − b)

)=

f (x , y)− f (a, y)− ∂z

∂x(x − a) + f (a, y)− f (a, b)− ∂z

∂y(y − b).


Differentiability



(∆x)2 + (∆y)2→ 0 as√

(∆x)2 + (∆y)2 → 0.

We may write

∆z − dz = f (x , y)− f (a, b)−(

∂z

∂x(x − a) +

∂z

∂y(y − b)

)=

f (x , y)− f (a, y)− ∂z

∂x(x − a) + f (a, y)− f (a, b)− ∂z

∂y(y − b).


Proof

By the Mean Value Theorem, f (x , y)− f (a, y) =∂z

∂x(x∗, y)(x − a)

for some x∗ between a and x if x is close enough to a.

Similarly, f (a, y)− f (a, b) =∂z

∂y(a, y∗)(y − b) for some y∗

between b and y if y is close enough to b.

We thus get

∆z−dz =∂z

∂x(x∗, y)(x−a)− ∂z

∂x(x−a)+

∂z

∂y(a, y∗)(y−b)− ∂z

∂y(y−

b) =

(∂z

∂x(x∗, y)− ∂z

∂x

)(x − a) +

(∂z

∂x(a, y∗)− ∂z

∂y

)(y − b).


Proof


∂x(x∗, y)(x − a)





We thus get

∆z−dz =∂z

∂x(x∗, y)(x−a)− ∂z

∂x(x−a)+

∂z

∂y(a, y∗)(y−b)− ∂z

∂y(y−

b) =

(∂z

∂x(x∗, y)− ∂z

∂x

)(x − a) +

(∂z

∂x(a, y∗)− ∂z

∂y

)(y − b).


Proof


∂x(x∗, y)(x − a)





We thus get

∆z−dz =∂z

∂x(x∗, y)(x−a)− ∂z

∂x(x−a)+

∂z

∂y(a, y∗)(y−b)− ∂z

∂y(y−

b) =

(∂z

∂x(x∗, y)− ∂z

∂x

)(x − a) +

(∂z

∂x(a, y∗)− ∂z

∂y

)(y − b).


Proof

Since both|x − a|√

(x − a)2 + (y − b)2≤ 1 and

|y − b|√(x − a)2 + (y − b)2

≤ 1, we have

|(

∂z

∂x(x∗, y)− ∂z

∂x

)(x − a)|√

(x − a)2 + (y − b)2≤

∣∣∣∣∂z

∂x(x∗, y)− ∂z

∂x

∣∣∣∣ → 0

and

|(

∂z

∂y(a, y∗)− ∂z

∂y

)(y − b)|√

(x − a)2 + (y − b)2≤

∣∣∣∣∂z

∂y(a, y∗)− ∂z

∂y

∣∣∣∣ → 0

since both partial derivatives are continuous near (a, b).


Proof


(x − a)2 + (y − b)2≤ 1 and

|y − b|√(x − a)2 + (y − b)2

≤ 1, we have

|(

∂z

∂x(x∗, y)− ∂z

∂x

)(x − a)|√

(x − a)2 + (y − b)2≤

∣∣∣∣∂z

∂x(x∗, y)− ∂z

∂x

∣∣∣∣ → 0

and

|(

∂z

∂y(a, y∗)− ∂z

∂y

)(y − b)|√

(x − a)2 + (y − b)2≤

∣∣∣∣∂z

∂y(a, y∗)− ∂z

∂y

∣∣∣∣ → 0



Proof


(x − a)2 + (y − b)2≤ 1 and

|y − b|√(x − a)2 + (y − b)2

≤ 1, we have

|(

∂z

∂x(x∗, y)− ∂z

∂x

)(x − a)|√

(x − a)2 + (y − b)2≤

∣∣∣∣∂z

∂x(x∗, y)− ∂z

∂x

∣∣∣∣ → 0

and

|(

∂z

∂y(a, y∗)− ∂z

∂y

)(y − b)|√

(x − a)2 + (y − b)2≤

∣∣∣∣∂z

∂y(a, y∗)− ∂z

∂y

∣∣∣∣ → 0



The Chain Rule

For an ordinary function, if y = f (u) and u = g(x), makingy = f ◦ g(x) a composite function, we can differentiate with

respect to x using the Chain Rule:dy

dx=

dy

du

du

dx.

Suppose we have a function z = f (x , y), but x = g(t) andy = h(t), making z = f (g(t), h(t)) a composite function of t. Wecan come up with a variation of the Chain Rule, which holds underappropriate conditions. The conditions we will assume are that allthe relevant derivatives exist and are continuous near t and all therelevant partial derivatives exist and are continuous near(f (t), g(t)).

By the definition of a derivative,

dz

dt= limk→0

f (g(t + k), h(t + k))− f (g(t), h(t))

k.


The Chain Rule



dx=

dy

du

du

dx.

Suppose we have a function z = f (x , y), but x = g(t) andy = h(t), making z = f (g(t), h(t)) a composite function of t.

Wecan come up with a variation of the Chain Rule, which holds underappropriate conditions. The conditions we will assume are that allthe relevant derivatives exist and are continuous near t and all therelevant partial derivatives exist and are continuous near(f (t), g(t)).


dz

dt= limk→0

f (g(t + k), h(t + k))− f (g(t), h(t))

k.


The Chain Rule



dx=

dy

du

du

dx.

Suppose we have a function z = f (x , y), but x = g(t) andy = h(t), making z = f (g(t), h(t)) a composite function of t. Wecan come up with a variation of the Chain Rule, which holds underappropriate conditions.

The conditions we will assume are that allthe relevant derivatives exist and are continuous near t and all therelevant partial derivatives exist and are continuous near(f (t), g(t)).


dz

dt= limk→0

f (g(t + k), h(t + k))− f (g(t), h(t))

k.


The Chain Rule



dx=

dy

du

du

dx.



dz

dt= limk→0

f (g(t + k), h(t + k))− f (g(t), h(t))

k.


The Chain Rule



dx=

dy

du

du

dx.



dz

dt= limk→0

f (g(t + k), h(t + k))− f (g(t), h(t))

k.


The Chain Rule



dx=

dy

du

du

dx.



dz

dt= limk→0

f (g(t + k), h(t + k))− f (g(t), h(t))

k.


The Chain Rule

We can rewrite the numerator asf (g(t + k), h(t + k))− f (g(t), h(t)) = [f (g(t + k), h(t + k))−f (g(t), h(t + k))] + [f (g(t), h(t + k))− f (g(t), h(t))].

Using the Mean Value Theorem, the first difference may be written:

f (g(t + k), h(t + k))− f (g(t), h(t + k)) =f1(u, h(t + k))[g(t + k)− g(t)], where u is between g(t + k) andg(t).

But, also by the Mean Value Theorem, g(t + k)− g(t) = g ′(t∗)k,where t∗ is between t and t + k.

We thus havef (g(t + k), h(t + k))− f (g(t), h(t + k)) = f1(u, h(t + k))g ′(t∗)k

Similarly, f (g(t), h(t + k))− f (g(t), h(t)) = f2(g(t), v)h′(t∗∗)k,where v is between h(t) and h(t + k) and t∗∗ is between t andt + k.


The Chain Rule








The Chain Rule








The Chain Rule








The Chain Rule








The Chain Rule








The Chain Rule

We thus getdz

dt= limk→0

f1(u, h(t + k))g ′(t∗)k + f2(g(t), v)h′(t∗∗)k

k=

limk→0 f1(u, h(t + k))g ′(t∗) + f2(g(t), v)h′(t∗∗) =f1(g(t), h(t))g ′(t) + f2(g(t), h(t))h′(t).

Using Leibniz’ Notation, this may be written as:

dz

dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt.

This is one variation of the Chain Rule.


The Chain Rule

We thus getdz

dt= limk→0


k=



dz

dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt.



The Chain Rule

We thus getdz

dt= limk→0


k=



dz

dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt.



The Chain Rule

We thus getdz

dt= limk→0


k=



dz

dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt.



Partial Derivatives Via the Chain Rule

Suppose z = f (x , y), while x = g(s, t) and y = h(s, t).

Thenz = f (g(s, t), h(s, t)) can be thought of as a function of s and t.

We might then want to calculate the partial derivatives∂z

∂sand

∂z

∂t.

By the nature of partial differentiation, the Chain Rule we justderived can be adjusted to give formulas for these partialderivatives.

∂z

∂s=

∂z

∂x

∂x

∂s+

∂z

∂y

∂y

∂s

∂z

∂t=

∂z

∂x

∂x

∂t+

∂z

∂y

∂y

∂t

If we have functions involving more than two variables, this may beadjusted in the hopefully obvious way.



Suppose z = f (x , y), while x = g(s, t) and y = h(s, t). Thenz = f (g(s, t), h(s, t)) can be thought of as a function of s and t.


∂sand

∂z

∂t.


∂z

∂s=

∂z

∂x

∂x

∂s+

∂z

∂y

∂y

∂s

∂z

∂t=

∂z

∂x

∂x

∂t+

∂z

∂y

∂y

∂t






∂sand

∂z

∂t.


∂z

∂s=

∂z

∂x

∂x

∂s+

∂z

∂y

∂y

∂s

∂z

∂t=

∂z

∂x

∂x

∂t+

∂z

∂y

∂y

∂t






∂sand

∂z

∂t.


∂z

∂s=

∂z

∂x

∂x

∂s+

∂z

∂y

∂y

∂s

∂z

∂t=

∂z

∂x

∂x

∂t+

∂z

∂y

∂y

∂t






∂sand

∂z

∂t.


∂z

∂s=

∂z

∂x

∂x

∂s+

∂z

∂y

∂y

∂s

∂z

∂t=

∂z

∂x

∂x

∂t+

∂z

∂y

∂y

∂t






∂sand

∂z

∂t.


∂z

∂s=

∂z

∂x

∂x

∂s+

∂z

∂y

∂y

∂s

∂z

∂t=

∂z

∂x

∂x

∂t+

∂z

∂y

∂y

∂t






∂sand

∂z

∂t.


∂z

∂s=

∂z

∂x

∂x

∂s+

∂z

∂y

∂y

∂s

∂z

∂t=

∂z

∂x

∂x

∂t+

∂z

∂y

∂y

∂t



Implicit Differentiation

The Chain Rule may be used to derive a formula for implicitdifferentiation.

Theorem (Implicit Differentiation)

If a differentiable function y = f (x) is defined implicitly by an

equation F (x , y) = 0, thendy

dx= −Fx

Fy= −

∂F

∂x∂F

∂y

.

Note: We have assumed y = f (x) is differentiable. We are nothere dealing with how one knows whether such a function isdifferentiable. In general, if such a function is not differentiable, itwill be relatively obvious.







dx= −Fx

Fy= −

∂F

∂x∂F

∂y

.








dx= −Fx

Fy= −

∂F

∂x∂F

∂y

.

Note: We have assumed y = f (x) is differentiable.

We are nothere dealing with how one knows whether such a function isdifferentiable. In general, if such a function is not differentiable, itwill be relatively obvious.







dx= −Fx

Fy= −

∂F

∂x∂F

∂y

.

Note: We have assumed y = f (x) is differentiable. We are nothere dealing with how one knows whether such a function isdifferentiable.

In general, if such a function is not differentiable, itwill be relatively obvious.







dx= −Fx

Fy= −

∂F

∂x∂F

∂y

.




Proof.

Using the Chain Rule,dF

dx=

∂F

∂x

dx

dx+

∂F

∂y

dy

dx=

∂F

∂x+

∂F

∂y

dy

dx.

Since F (x , y) = 0, it follows thatdF

dx= 0, so

∂F

∂x+

∂F

∂y

dy

dx= 0.

Solving fordy

dx, we get

∂F

∂y

dy

dx= −∂F

∂x, so

dy

dx= −

∂F

∂x∂F

∂y

.



Proof.


dx=

∂F

∂x

dx

dx+

∂F

∂y

dy

dx=

∂F

∂x+

∂F

∂y

dy

dx.

Since F (x , y) = 0,

it follows thatdF

dx= 0, so

∂F

∂x+

∂F

∂y

dy

dx= 0.

Solving fordy

dx, we get

∂F

∂y

dy

dx= −∂F

∂x, so

dy

dx= −

∂F

∂x∂F

∂y

.



Proof.


dx=

∂F

∂x

dx

dx+

∂F

∂y

dy

dx=

∂F

∂x+

∂F

∂y

dy

dx.


dx= 0,

so∂F

∂x+

∂F

∂y

dy

dx= 0.

Solving fordy

dx, we get

∂F

∂y

dy

dx= −∂F

∂x, so

dy

dx= −

∂F

∂x∂F

∂y

.



Proof.


dx=

∂F

∂x

dx

dx+

∂F

∂y

dy

dx=

∂F

∂x+

∂F

∂y

dy

dx.


dx= 0, so

∂F

∂x+

∂F

∂y

dy

dx= 0.

Solving fordy

dx, we get

∂F

∂y

dy

dx= −∂F

∂x, so

dy

dx= −

∂F

∂x∂F

∂y

.



Proof.


dx=

∂F

∂x

dx

dx+

∂F

∂y

dy

dx=

∂F

∂x+

∂F

∂y

dy

dx.


dx= 0, so

∂F

∂x+

∂F

∂y

dy

dx= 0.

Solving fordy

dx, we get

∂F

∂y

dy

dx= −∂F

∂x,

sody

dx= −

∂F

∂x∂F

∂y

.



Proof.


dx=

∂F

∂x

dx

dx+

∂F

∂y

dy

dx=

∂F

∂x+

∂F

∂y

dy

dx.


dx= 0, so

∂F

∂x+

∂F

∂y

dy

dx= 0.

Solving fordy

dx, we get

∂F

∂y

dy

dx= −∂F

∂x, so

dy

dx= −

∂F

∂x∂F

∂y

.


Directional Derivatives

Consider a function z = f (x , y) and its graph, which will be asurface.

The partial derivative∂z

∂xmay be thought of as representing how

fast the surface is rising above one’s head if one is walking on thexy -plane in the direction of the x-axis.

Similarly, the partial derivative∂z

∂ymay be thought of as

representing how fast the surface is rising above one’s head if oneis walking on the xy -plane in the direction of the y -axis.




















Directional Derivative

For a given unit vector u, we define the directional derivative Duzto represent how fast the surface is rising above one’s head if oneis walking on the xy -plane in the direction of u.

Definition (Directional Derivative)

Let f : Rn → R and let u ∈ Rn be a unit vector. Letg(t) = f (x + ut). Duf (x) = g ′(0) is called the directionalderivative of f at x in the direction of u.

Note that if n = 1, then the directional derivative is the same asthe ordinary derivative, while the directional derivatives in thedirections of the coordinate axes are the same as the partialderivatives.














The Del Operator and the Gradient

Definition (Del Operator)

5 =

(∂

∂x,

∂

∂y

)

Note this is really just a symbolic entity. By itself, it ismeaningless, but we use it as a mneumonic device.

Definition (Gradient)

grad f = 5f =

(∂f

∂x,∂f

∂y

)The gradient turns out to be convenient when calculatingdirectional derivatives. It also generalizes to higher dimensions.




5 =

(∂

∂x,

∂

∂y

)Note this is really just a symbolic entity. By itself, it ismeaningless, but we use it as a mneumonic device.


grad f = 5f =

(∂f

∂x,∂f

∂y





5 =

(∂

∂x,

∂

∂y



grad f = 5f =

(∂f

∂x,∂f

∂y

)

The gradient turns out to be convenient when calculatingdirectional derivatives. It also generalizes to higher dimensions.




5 =

(∂

∂x,

∂

∂y



grad f = 5f =

(∂f

∂x,∂f

∂y

)The gradient turns out to be convenient when calculatingdirectional derivatives.

It also generalizes to higher dimensions.




5 =

(∂

∂x,

∂

∂y



grad f = 5f =

(∂f

∂x,∂f

∂y



Calculating Directional Derivatives

TheoremIf all the partial derivatives of z = f (x) are continuous is someopen ball centered at x, then Duf (x) = (5f ) · u.

This theorem gives us a convenient way to calculate any directionalderivative of a function and also shows that it is sufficient to beable to calculate all the partial derivatives.


Calculating Directional Derivatives

TheoremIf all the partial derivatives of z = f (x) are continuous is someopen ball centered at x, then Duf (x) = (5f ) · u.

This theorem gives us a convenient way to calculate any directionalderivative of a function and also shows that it is sufficient to beable to calculate all the partial derivatives.


Proof

We will prove the theorem for R2, but a similar proof will work forhigher dimensions; only the notation would get messier.

Proof.Consider a function f (x , y) and a unit vector u =< a, b >. Letz = g(t) be defined by letting z = f (x , y), where x = x0 + at,y = y0 + bt.

By definition, Duf (x0, y0) = g ′(0).

By the Chain Rule, g ′(t) =dz

dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt= (5z) · u.

Evaluating this at 0 gives the result.


Proof





dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt= (5z) · u.



Proof





dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt= (5z) · u.



Proof





dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt= (5z) · u.



Proof





dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt= (5z) · u.



Maximum Value of the Directional Derivative

Duf = (5f ) · u = |5f | |u| cos θ, where θ is the angle between 5fand u.

Since −1 ≤ cos θ ≤ 1, the maximal value obviously occurs whenθ = 0 and cos θ = 1, in other words, when u is in the samedirection as 5f .

There’s a catch: This depends on the property u · v = |u||v| cos θ,which we’ve seen for R2 and R3, but whose very meaning isunclear for higher dimensions.




Since −1 ≤ cos θ ≤ 1, the maximal value obviously occurs whenθ = 0 and cos θ = 1,

in other words, when u is in the samedirection as 5f .











There’s a catch:

This depends on the property u · v = |u||v| cos θ,which we’ve seen for R2 and R3, but whose very meaning isunclear for higher dimensions.





There’s a catch: This depends on the property u · v = |u||v| cos θ,

which we’ve seen for R2 and R3, but whose very meaning isunclear for higher dimensions.







Cauchy-Schwarz Inequality

We can give u · v = |u||v| cos θ meaning through theCauchy-Schwarz Inequality u · v ≤ |u||v|.

We will show theCauchy-Schwarz Inequality holds in any dimension, with equalityholding if and only if one vector is a multiple of the other.

Consider a vector u− tv. Certainly (u− tv) · (u− tv) ≥ 0, withequality holding if and only if u is a multiple t of v or v = 0.

Since(u− tv) · (u− tv) = u · u− 2tu · v + t2vv = |v|2t2 − 2u · vt + |u|2,we get |v|2t2 − 2u · vt + |u|2 ≥ 0.



We can give u · v = |u||v| cos θ meaning through theCauchy-Schwarz Inequality u · v ≤ |u||v|. We will show theCauchy-Schwarz Inequality holds in any dimension, with equalityholding if and only if one vector is a multiple of the other.












Since(u− tv) · (u− tv) = u · u− 2tu · v + t2vv = |v|2t2 − 2u · vt + |u|2,

we get |v|2t2 − 2u · vt + |u|2 ≥ 0.








It follows that the quadratic equation |v|2t2 − 2u · vt + |u|2 = 0 int can’t have more than one solution,

so the discriminant(−2u · v)2 − 4|v|2|u|2 can’t be positive.

In other words, (−2u · v)2 − 4|v|2|u|2 ≤ 0, so4(u · v)2 − 4|v|2|u|2 ≤ 0, so (u · v)2 − |v|2|u|2 ≤ 0, so(u · v)2 ≤ |v|2|u|2, so u · v ≤ |u||v|.

Equality clearly holds if and only if either u− tv = 0 or if v = 0, inother words, if and only if either u is a scalar multiple of v or ifv = 0.



It follows that the quadratic equation |v|2t2 − 2u · vt + |u|2 = 0 int can’t have more than one solution, so the discriminant(−2u · v)2 − 4|v|2|u|2 can’t be positive.






In other words, (−2u · v)2 − 4|v|2|u|2 ≤ 0,

so4(u · v)2 − 4|v|2|u|2 ≤ 0, so (u · v)2 − |v|2|u|2 ≤ 0, so(u · v)2 ≤ |v|2|u|2, so u · v ≤ |u||v|.





In other words, (−2u · v)2 − 4|v|2|u|2 ≤ 0, so4(u · v)2 − 4|v|2|u|2 ≤ 0,

so (u · v)2 − |v|2|u|2 ≤ 0, so(u · v)2 ≤ |v|2|u|2, so u · v ≤ |u||v|.





In other words, (−2u · v)2 − 4|v|2|u|2 ≤ 0, so4(u · v)2 − 4|v|2|u|2 ≤ 0, so (u · v)2 − |v|2|u|2 ≤ 0,

so(u · v)2 ≤ |v|2|u|2, so u · v ≤ |u||v|.





In other words, (−2u · v)2 − 4|v|2|u|2 ≤ 0, so4(u · v)2 − 4|v|2|u|2 ≤ 0, so (u · v)2 − |v|2|u|2 ≤ 0, so(u · v)2 ≤ |v|2|u|2,

so u · v ≤ |u||v|.











Equality clearly holds if and only if either u− tv = 0 or if v = 0,

inother words, if and only if either u is a scalar multiple of v or ifv = 0.







Cauchy-Schwarz and Directional Derivatives

Since |u · v| ≤ |u||v|,

it follows that −1 ≤ u · v|u||v|

≤ 1.

We may thus define the angle θ between u and v by

θ = arccos

(u · v|u||v|

).

It follows that u · v = |u||v| cos θ, so the argument we used beforeabout the directional derivative being maximal in the direction ofthe gradient can legitimately be used.



Since |u · v| ≤ |u||v|, it follows that −1 ≤ u · v|u||v|

≤ 1.


θ = arccos

(u · v|u||v|

).





≤ 1.


θ = arccos

(u · v|u||v|

).





≤ 1.


θ = arccos

(u · v|u||v|

).

It follows that u · v = |u||v| cos θ,

so the argument we used beforeabout the directional derivative being maximal in the direction ofthe gradient can legitimately be used.




≤ 1.


θ = arccos

(u · v|u||v|

).



Tangent Planes and Gradients

Recall the formula for the plane tangent to the surface z = f (x , y)at a point (a, b):

z − c =∂z

∂x(x − a) +

∂z

∂y(y − b).

Using the language of gradients, this could be written in the formz − c = (5f )· < x − a, y − b > or z − c = (5f ) · (x− x0),where x =< x , y > and x0 =< a, b >.

Since one standard form for the equation of a plane isz − z0 = n · (x− x0), with n being a normal to the plane, it followsthat 5f is normal to the tangent plane.




z − c =∂z

∂x(x − a) +

∂z

∂y(y − b).






z − c =∂z

∂x(x − a) +

∂z

∂y(y − b).






z − c =∂z

∂x(x − a) +

∂z

∂y(y − b).


Since one standard form for the equation of a plane isz − z0 = n · (x− x0), with n being a normal to the plane,

it followsthat 5f is normal to the tangent plane.




z − c =∂z

∂x(x − a) +

∂z

∂y(y − b).




Tangent Planes for Surfaces Defined Implicitly

Suppose a surface is the graph of an equation φ(x , y , z) = 0.

Atmost points (where there is a tangent plane and the tangent planeisn’t vertical), a portion of the surface near the point can beconsidered the graph of a function z = f (x , y) defined implicitly bythe equation φ(x , y , z) = 0 along with some side conditions.


Tangent Planes for Surfaces Defined Implicitly

Suppose a surface is the graph of an equation φ(x , y , z) = 0. Atmost points (where there is a tangent plane and the tangent planeisn’t vertical), a portion of the surface near the point can beconsidered the graph of a function z = f (x , y) defined implicitly bythe equation φ(x , y , z) = 0 along with some side conditions.


Tangent Planes

By the formula for implicit differentiation,∂z

∂x= −

∂φ

∂x∂φ

∂z

and

∂z

∂y= −

∂φ

∂y∂φ

∂z

,

so the equation of the tangent plane may be written

z − c = −

∂φ

∂x∂φ

∂z

(x − a)−

∂φ

∂y∂φ

∂z

(y − b).


Tangent Planes

By the formula for implicit differentiation,∂z

∂x= −

∂φ

∂x∂φ

∂z

and

∂z

∂y= −

∂φ

∂y∂φ

∂z

, so the equation of the tangent plane may be written

z − c = −

∂φ

∂x∂φ

∂z

(x − a)−

∂φ

∂y∂φ

∂z

(y − b).


Tangent Planes

z − c = −

∂φ

∂x∂φ

∂z

(x − a)−

∂φ

∂y∂φ

∂z

(y − b).

Simplifying:∂φ

∂z(z − c) = −∂φ

∂x(x − a)− ∂φ

∂y(y − b),

∂φ

∂x(x − a) +

∂φ

∂y(y − b) +

∂φ

∂z(z − c) = 0.

This can also be written in the form(5φ)· < x − a, y − b, z − c >= 0.


Tangent Planes

z − c = −

∂φ

∂x∂φ

∂z

(x − a)−

∂φ

∂y∂φ

∂z

(y − b).

Simplifying:∂φ

∂z(z − c) = −∂φ

∂x(x − a)− ∂φ

∂y(y − b),

∂φ

∂x(x − a) +

∂φ

∂y(y − b) +

∂φ

∂z(z − c) = 0.



Tangent Planes

z − c = −

∂φ

∂x∂φ

∂z

(x − a)−

∂φ

∂y∂φ

∂z

(y − b).

Simplifying:∂φ

∂z(z − c) = −∂φ

∂x(x − a)− ∂φ

∂y(y − b),

∂φ

∂x(x − a) +

∂φ

∂y(y − b) +

∂φ

∂z(z − c) = 0.



Tangent Planes

z − c = −

∂φ

∂x∂φ

∂z

(x − a)−

∂φ

∂y∂φ

∂z

(y − b).

Simplifying:∂φ

∂z(z − c) = −∂φ

∂x(x − a)− ∂φ

∂y(y − b),

∂φ

∂x(x − a) +

∂φ

∂y(y − b) +

∂φ

∂z(z − c) = 0.



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Functions of Several Variables - UCONNstein/math210/Slides/math210-04slides.pdf · Functions of Several Variables A function of several variables is just what it sounds like. It may