The graph of g(x, y) =p 9 − x2− y2 on the closed disk x2+ y2 ≤ 9 is the upper hemisphere with center (0, 0, 0) and radius 3 given by S = {(x, y, z

Functions of Several Variables

Definition Let D be a subset of R², and let f : D → R be a function defined on D. Then the graph of f on D is a subset in R³ defined by

S = {(x, y, z) ∈ R³ | z = f (x, y) and (x, y) ∈ D} ⊂ R³.

Examples

ˆ If (a, b) ̸= (0, 0), the graph of the linear function f(x, y) = ax + by + c on R² is a plane in R³ given by

S = {(x, y, z) ∈ R³ | z = ax + by + c and (x, y) ∈ R²}.

ˆ The graph of g(x, y) =p

9 − x²− y² on the closed disk x²+ y² ≤ 9 is the upper hemisphere with center (0, 0, 0) and radius 3 given by

S = {(x, y, z) ∈ R³ | z =p

9 − x²− y² ≥ 0 and x²+ y² ≤ 9}

Definition Let D be a subset of R², and let f : D → R be a function defined on D. Then alevel curve of f at the level k is a subset of D given by

L_f(k) = {(x, y) ∈ D | f (x, y) = k} ⊆ D.

A collection of level curves is called a contour map of f . ExampleLet f (x, y) = −3y

x²+ y²+ 1 for (x, y) ∈ R². The following sketches show the level curves and the graph near the origin.

Remark In general, if D is a subset of Rⁿ, f : D → R is a function defined on D. Then the graph of f on D is a subset in Rⁿ⁺¹ defined by

S = {(x₁, . . . , x_n+1) ∈ Rⁿ⁺¹ | x_n+1= f (x₁, . . . , x_n) and (x₁, . . . , x_n) ∈ D} ⊂ Rⁿ⁺¹.

and a level set (or level curve, surface when n = 2, 3 respectively) of f at the level k is a subset of D given by

L_f(k) = {(x₁, . . . , x_n) ∈ D | f (x₁, . . . , x_n) = k} ⊆ D.

Limits and Continuity

Let p ∈ Rⁿ, r > 0 and let B_r(p) denote the ball of center p and radius r defined by

B_r(p) = {x ∈ Rⁿ | |x − p|² =

n

X

i=1

(x_i− p_i)² < r²},

where |x − p| is the Euclidean distance from x to p.

Definitions Let D be a subset of Rⁿ and p ∈ D. Then

ˆ p is called an interior point of D if there exists an r > 0 such that

B_r(p) = {x ∈ Rⁿ| |x − p| < r} ⊂ D ⇐⇒ if x ∈ B_r(p) then x ∈ D.

ˆ p is called a boundary point of D if it is not an interior point of D.

ˆ D is called an open subset of Rⁿ if every point in D is an interior point of D.

Remark If p is a point in D, then p is either an interior point or a boundary point of D.

Example Let D = [0, 1] ∪ {2} ⊂ R. Then (0, 1) is the set of interior points of D while {0, 1, 2}

is the set of boundary points of D.

DefinitionLet f be a function of two variables whose domainD includes points arbitrarily close to (a, b).Then

lim

(x,y)→(a,b)f (x, y) = L ∈ R

if for every ε > 0 there is a corresponding δ > 0 such that

if 0 <|(x, y) − (a, b)|< δ then (x, y) ∈ D and |f (x, y) − L| < ε

⇐⇒ if (x, y) ∈ B_δ((a, b)) \ {(a, b)} then (x, y) ∈ D and |f (x, y) − L| < ε Note that if lim

(x,y)→(a,b)f (x, y) exists, then it is unique.

Algebraic Properties of Limits Let f, g be functions of two variables whose domain D includes points arbitrarily close to (a, b). If

lim

(x,y)→(a,b)f (x, y) = L ∈ R and lim

(x,y)→(a,b)g(x, y) = M ∈ R, then

ˆ (Sum and Difference Law) lim

(x,y)→(a,b)[f ± g](x, y) = L ± M

ˆ (Product Law) lim

(x,y)→(a,b)[f × g](x, y) = L × M

ˆ (Quotient Law) lim

(x,y)→(a,b)

f (x, y) g(x, y) = L

M provided that g(x, y) ̸= 0 for (x, y) close to (a, b) and the limit of the denominator is not 0.

Proposition (Squeeze Theorem) Let f, ℓ, r : D → R be functions of two variables whose domain D includes points arbitrarily close to (a, b). Suppose that

ℓ(x, y) ≤ f (x, y) ≤ r(x, y) for all (x, y) ∈ D \ {(a, b)}, and

lim

(x,y)→(a,b)ℓ(x, y) = L = lim

(x,y)→(a,b)r(x, y).

Then lim

(x,y)→(a,b)f (x, y) = L.

Definition Let D be a subset of R², f : D → R be a function defined on D and let (a, b) be an interior point of D. Then f is said to be continuous at (a, b)if

lim

(x,y)→(a,b)f (x, y) = f (a, b), i.e. for every ε > 0 there is a corresponding δ > 0 such that

if |(x, y) − (a, b)|< δ then (x, y) ∈ D and |f (x, y) − f (a, b)| < ε

⇐⇒ if (x, y) ∈ B_δ((a, b)) then (x, y) ∈ D and |f (x, y) − f (a, b)| < ε We say that f is continuous on D if f is continuous at every point (a, b) in D.

Algebraic Properties of Continuous Functions Let f and g be functions of two variables whose domain D includes points arbitrarily close to (a, b). If f and g are continuous at (a, b), i.e.

lim

(x,y)→(a,b)f (x, y) = f (a, b) and lim

(x,y)→(a,b)g(x, y) = g(a, b), then so is the

ˆ (Sum and Difference) f ± g since lim

(x,y)→(a,b)[f ± g](x, y) = f (a, b) ± g(a, b).

ˆ (Product) f × g since lim

(x,y)→(a,b)

[f × g](x, y) = f (a, b) × g(a, b).

ˆ (Quotient) f

g provided that g(a, b) ̸= 0 since lim

(x,y)→(a,b)

f (x, y)

g(x, y) = f (a, b) g(a, b). Examples

1. Show that lim

(x,y)→(0,0)

sin(x²+ y²) x² + y² = 1.

2. Show that lim

(x,y)→(0,0)

x²− y²

x²+ y² does not exist.

3. Evaluate lim

(x,y)→(1,2)

(x²y³− x³y²+ 3x + 2y).

4. Where is the function f (x, y) = x²− y²

x²+ y² continuous?

Remark In general, if D is a subset of Rⁿ and f : D → R is a real-valued function defined on D includes points arbitrarily close to p, then we say that lim

x→pf (x) = Lif for every number ε > 0 there is a corresponding number δ > 0 such that

if 0 < |x − p| < δ then x ∈ D and |f (x) − L| < ε,

⇐⇒ if x ∈ B_δ(p) \ {p} then x ∈ D and |f (x) − L| < ε.

If p ∈ D, then we say that f is continuous at p if lim

x→pf (x) = f (p),i.e. if for every ε > 0 there is a corresponding δ > 0 such that

if |x − p| < δ then x ∈ D and |f (x) − f (p)| < ε

⇐⇒ if x ∈ Bδ(p) then x ∈ D and |f (x) − f (p)| < ε

Definition Let D be a subset of R², p = (a, b) be an interior point of D and let f : D → R be a real-valued function defined on D. Then the partial derivative of f with respect to x at (a, b), denoted by fx(a, b) or ∂f

∂x(a, b), is defined to be f_x(a, b) = lim

h→0

f (a + h, b) − f (a, b)

h provided that the limit exists, and the partial derivative of f with respect to y at (a, b), denoted by f_y(a, b) or ∂f

∂y(a, b), is defined to be

f_y(a, b) = lim

h→0

f (a, b + h) − f (a, b)

h provided that the limit exists.

Remark Recall that if f : R → R is differentiable at x = a, then lim

x→a

f (x) − f (a)

x − a exists and f is continuous at x = a since

x→alim[f (x)−f (a)] = lim

x→a

 f (x) − f (a)

x − a · (x − a)



= lim

x→a

f (x) − f (a) x − a ·lim

x→a(x−a) = 0 =⇒ lim

x→af (x) = f (a).

However, the existence of partial derivatives for a function of several variables do not always guarantee the continuity of the function.

ExampleLet f : R² → R be a function of two variables defined by

f (x, y) =





 2xy

x²+ y² if (x, y) ̸= (0, 0), 0 if (x, y) = (0, 0).

1. Show that f is not continuous at (0, 0).

2. Find f_x(0, 0) and f_y(0, 0).

RemarkLet D be an open subset of R² and let f : D → R be a real-valued function defined on D. Suppose that the partial derivatives f_x and f_y exist at every point in D, then the functions f_x, f_y : D → R are defined by

fx(x, y) = lim

h→0

f (x + h, y) − f (x, y)

h differentiate f with respect to x by treating y as a constant, f_y(x, y) = lim

h→0

f (x, y + h) − f (x, y)

h differentiate f with respect to y by treating x as a constant.

Examples

1. If f (x, y) = x³+ x²y³− 2y², find fx(2, 1) and fy(2, 1).

2. If f (x, y) = 4 − x²− 2y², find fx(1, 1) and fy(1, 1) and interpret these numbers as slopes.

3. If f (x, y, z) = e^xylnz, find f_x, f_y, and f_z.

4. If f (x, y) = x³+ x²y³ − 2y², find the second partial derivatives f_xx = (f_x)_x, f_xy = (f_x)_y, f_yx = (f_y)_x, and f_yy = (f_y)_y.

Clairaut’s Theorem Suppose f is defined on a disk D that contains the point (a, b). If the functions f_xy and f_yx are both continuous on D, then

f_xy(a, b) = f_yx(a, b).

Examples

1. Let f : R² → R be a function of two variables defined by

f (x, y) =







xy(x²− y²)

x²+ y² if (x, y) ̸= (0, 0), 0 if (x, y) = (0, 0).

Show that

f_x(x, y) =







x⁴y + 4x²y³− y⁵

(x²+ y²)² if (x, y) ̸= (0, 0) by direct differentiation,

0 if (x, y) = (0, 0) by definition of partial derivative,

f_y(x, y) =







−y⁴x − 4y²x³+ x⁵

(y²+ x²)² if (x, y) ̸= (0, 0) by direct differentiation,

0 if (x, y) = (0, 0) by definition of partial derivative, and that f_xy(0, 0) = −1 ̸= 1 = f_yx(0, 0) by definition of partial derivatives.

2. Show that the function u(x, y) = e^xsin y is a solution of the Laplace equation, that is

∆u = u_xx+ u_yy = 0.

Definition Let D ⊂ R², (a, b) be an interior point of D and let f : D → R be a function defined on D. Then f is called differentiable at (a, b) if there exists (ℓ1, ℓ2) ∈ R² such that

lim

(x,y)→(a,b)

|f (x, y) − f (a, b) − ℓ₁(x − a) − ℓ₂(y − b)|

|(x, y) − (a, b)| = 0.

⇐⇒ lim

(x,y)→(a,b)

|ε(x, y)|

p(x − a)²+ (y − b)² = 0, where ε(x, y) = f (x, y) − f (a, b) − ℓ1(x − a) − ℓ2(y − b) TheoremIf f is differentiable at (a, b), then the partial derivatives f_x and f_y exist at (a, b) and (ℓ₁, ℓ₂) = (f_x(a, b), f_y(a, b)).

Proof Since 0 = lim

(x,b)→(a,b)

|f (x, b) − f (a, b) − ℓ1(x − a) − ℓ2(b − b)|

|(x, b) − (a, b)| = lim

x→a

|f (x, b) − f (a, b) − ℓ1(x − a)|

|x − a|

=⇒ 0 = lim

x→a

f (x, b) − f (a, b) − ℓ₁(x − a)

x − a = lim

x→a

f (x, b) − f (a, b)

x − a − ℓ₁ = f_x(a, b) − ℓ₁, and

0 = lim

(a,y)→(a,b)

|f (a, y) − f (a, b) − ℓ₁(a − a) − ℓ₂(y − b)|

|(a, y) − (a, b)| = lim

y→b

|f (a, y) − f (a, b) − ℓ₂(y − b)|

|y − b|

=⇒ 0 = lim

y→b

f (a, y) − f (a, b) − ℓ₂(y − b)

y − b = lim

y→b

f (a, y) − f (a, b)

y − b − ℓ₂ = f_y(a, b) − ℓ₂.

Theorem If the partial derivatives f_x and f_y exist near (a, b) and are continuous at (a, b), then f is differentiable at (a, b), that is,

lim

(x,y)→(a,b)

|f (x, y) − f (a, b) − fx(a, b)(x − a) − fy(a, b)(y − b)|

|(x, y) − (a, b)| = 0.

Proof For each ε > 0, since f_x and f_y exist near (a, b) and are continuous at p = (a, b), there exists a δ > 0 such that if (x, y) ∈ B_δ(p), then

|f_x(x, y) − f_x(a, b)| + |f_y(x, y) − f_y(a, b)| < ε.

Let g : B_δ(p) → R be defined by

g(x, y) = f (x, y) − f (a, b) − f_x(a, b)(x − a) − f_y(a, b)(y − b) for (x, y) ∈ B_δ(p).

Then g(a, b) = 0,

gx(x, y) = fx(x, y) − fx(a, b) and gy(x, y) = fx(x, y) − fy(a, b).

Let the line segment in Bδ(p) from (x, y) to p = (a, b) be given by r(t) = (a, b) + t(x − a, y − b), t ∈ [0, 1], and consider the function

g(r(t)) = g(x(t), y(t)) = g(a + t(x − a), b + t(y − b)) for t ∈ [0, 1].

By the Mean Value Theorem, there is a 0 < t0 < 1 with r(t0) = (x0, y0), such that

|g(r(1)) − g(r(0))| = |d

dtg(r(t))|_t=t0(1 − 0)| = |d

dtg(x(t), y(t))|_t=t0|

= |g_x(x₀, y₀)x^′(t₀) + g_y(x₀, y₀)y^′(t₀)|

= |[f_x(x₀, y₀) − f_x(a, b)](x − a) + [f_y(x₀, y₀) − f_y(a, b)](y − b)|

≤ (|f_x(x₀, y₀) − f_x(a, b)| + |f_y(x₀, y₀) − f_y(a, b)|)p

(x − a)²+ (y − b)²

< εp

(x − a)²+ (y − b)² = ε|(x, y) − (a, b)|

Since

g(r(1)) = g(x, y) = f (x, y) − f (a, b) − f_x(a, b)(x − a) − f_y(a, b)(y − b), g(r(0)) = g(a, b) = 0, we have

|f (x, y) − f (a, b) − f_x(a, b)(x − a) − f_y(a, b)(y − b)| = |g(r(1)) − g(r(0))| < ε|(x, y) − (a, b)|, which implies that

|f (x, y) − f (a, b) − f_x(a, b)(x − a) − f_y(a, b)(y − b)|

|(x, y) − (a, b)| < ε.

Since ε > 0 is an arbitrary positive number, we have lim

(x,y)→(a,b)

|f (x, y) − f (a, b) − f_x(a, b)(x − a) − f_y(a, b)(y − b)|

|(x, y) − (a, b)| = 0

and f is differentiable at (a, b).

Equation of a Tangent Plane Let D ⊂ R² and f : D → R be a function with continuous partial derivatives. Then the plane tangent to the surface

S = {(x, y, z) ∈ R³ | z = f (x, y) and (x, y) ∈ D}, at the point P (x₀, y₀, z₀) ∈ S has an equation

z − z₀ = f_x(x₀, y₀) (x − x₀) + f_y(x₀, y₀) (y − y₀).

ProofLet C1 and C2 be the curves obtained by intersecting the vertical planes y = y0 and x = x0

with the surface S. Then the point P lies on both C₁ and C₂. Let T₁ and T₂ be the tangent lines to the curves C₁ and C₂ at the point P.

Then the plane tangent to the surface S at the point P is defined to be the plane that contains both tangent lines T₁ and T₂.

Since C1 = S ∩ {(x, y0, z) | (x, z) ∈ R²} and C2 = S ∩ {(x0, y, z) | (y, z) ∈ R²} are curves in S, we may parametrize C₁ and C₂ by vector function

C₁ : r₁(x) = (x, y₀, z)^C1= (x, y^⊂S ₀, f (x, y₀)) for (x, y₀) ∈ D =⇒ r^′₁(x₀) = (1, 0, f_x(x₀, y₀)) // T₁ C₂ : r₂(y) = (x₀, y, z)^C2= (x^⊂S ₀, y, f (x₀, y)) for (x₀, y) ∈ D =⇒ r₂^′(y₀) = (0, 1, f_y(x₀, y₀)) // T₂ This implies that the tangent plane to the surface S at the point P is perpendicular the vector

(1, 0, f_x(x₀, y₀)) × (0, 1, f_y(x₀, y₀)) = (−f_x(x₀, y₀), −f_y(x₀, y₀), 1), and has an equation

(x − x₀, y − y₀, z − z₀) · (−f_x(x₀, y₀), −f_y(x₀, y₀), 1)⟩ = 0 since cosπ 2 = 0

⇐⇒ −f_x(x₀, y₀) (x − x₀) − f_y(x₀, y₀) (y − y₀) + (z − z₀) = 0

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

ExampleFind an equation for the plane tangent to the elliptic paraboloid z = 2x²+ y² at the point (1, 1, 3).

Definition Let D ⊂ R², (a, b) be an interior point of D and let f : D → R be a function with continuous partial derivatives. The linear function L : R² → R defined by

L(x, y) = f (a, b) + f_x(a, b) (x − a) + f_y(a, b) (y − b) for all (x, y) ∈ R²

is called thelinearization of f at (a, b) and the approximation

f (x, y) ≈ f (a, b) + f_x(a, b) (x − a) + f_y(a, b) (y − b)

is called the linear approximationor the tangent plane approximation of f at (a, b).

ExampleLet f : R² → R be a function of two variables defined by

f (x, y) =





 xy

x²+ y² if (x, y) ̸= (0, 0), 0 if (x, y) = (0, 0).

Note that f_x(0, 0) = 0 and f_y(0, 0) = 0, but f_x and f_y are not continuous at (0, 0), and the surface z = f (x, y) does not have a tangent plane at (0, 0).

Example Show that f (x, y) = xe^xy is differentiable at (1, 0) and find its linearization there.

Then use it to approximate f (1.1, −0.1).

DefinitionLet D be an open subset of R² and let f : D → R be a differentiable function defined on D. The differential df is defined by

df = f_x(x, y)dx + f_y(x, y)dy

Note that the differentials

ˆ dy = f^′(x)dx = the change in height of the tangent line,

ˆ dz = f_x(x, y)dx + f_y(x, y)dy = the change in height of the tangent plane, whereas the increments

ˆ ∆y = f(x + ∆x) − f(x) = the change in height of the curve y = f(x),

ˆ ∆z = f (x + ∆x, y + ∆y) − f (x, y) = the change in height of the surface z = f (x, y), and ∆z − dz = R = the gaps between surface and tangent plane satisfies that

lim

(∆x,∆y)→(0,0)

R

p(∆x)²+ (∆y)² = 0 by Taylor’s Theorem.

Examples

1. If z = f (x, y) = x²+ 3xy − y², find the differential dz = df.

2. If x changes from 2 to 2.05 and y changes from 3 to 2.96, compare the values of ∆z and dz.

3. The dimensions of a rectangular box are measured to be 75 cm, 60 cm, and 40 cm, and each measurement is correct to within ε cm.

ˆ Use differentials to estimate the largest possible error when the volume of the box is cal- culated from these measurements. [Let x, y and z be the dimensions of the box. Since its volume V = xyz and the error ∆V ≈ dV = yzdx + xzdy + xydz = 9900ε, the maximum er- ror in the calculated volume is about 9900 times larger than the error in each measurement taken.]

ˆ What is the estimated maximum error in the calculated volume if the measured dimensions are correct to within 0.2 cm. [If the largest error in each measurement is ε = 0.2 cm, then dV = 9900(0.2) = 1980, so an error of only 0.2 cm in measuring each dimension could lead to an error of approximately 1980 cm³ (1.1%) in the calculated volume 180, 000 cm³.]

Chain Rule

(a) Suppose that z = f (x, y) is a differentiable function of x and y, where x = g(t) and y = h(t) are both differentiable functions of t. Then z is a differentiable function of t and

dz dt = ∂f

∂x dx

dt +∂f

∂y dy dt.

(b) Suppose that z = f (x, y) is a differentiable function of x and y, where x = g(s, t) and y = h(s, t) are both differentiable functions of s and t. Then

∂z

∂s = ∂f

∂x

∂x

∂s +∂f

∂y

∂y

∂s and ∂z

∂t = ∂f

∂x

∂x

∂t +∂f

∂y

∂y

∂t.

(c) In general, if u is a differentiable function of n variables x₁, x₂, . . . , x_n and each x_j is a differentiable function of the m variables t₁, t₂, . . . , t_m. The u is a function of t₁, t₂, . . . , t_m and for each i = 1, 2, . . . , m, we have

∂u

∂t_i = ∂u

∂x₁

∂x₁

∂t_i + ∂u

∂x₂

∂x₂

∂t_i + · · · + ∂u

∂x_n

∂x_n

∂t_i =

n

X

j=1

∂u

∂x_j

∂x_j

∂t_i

Examples

1. If z = x²y + 3xy⁴, where x = sin 2t and y = cos t, find dz

dt when t = 0.

2. If z = e^xsin y, where x = st² and y = s²t, find ∂z

∂s and ∂z

∂t.

3. If F is differentiable on a disk containing (a, b), then the equation F (x, y) = 0 defines y implicitly as a differentiable function of x near the point (a, b) and we can apply the Chain Rule to differentiate both sides of F (x, y) = 0 with respect to x, and obtain

∂F

∂x dx dx +∂F

∂y dy

dx = 0 =⇒ dy

dx = −∂F/∂x

∂F/∂y if ∂F

∂y ̸= 0, e.g. find y^′ if x³+ y³ = 6xy.

4. Suppose that z is given implicitly as a function z = f (x, y) by an equation of the form F (x, y, z) = 0, i.e. F (x, y, f (x, y)) = 0 for all (x, y) in the domain of f.

If F and f are differentiable, then we can use the Chain Rule to differentiate the equation F (x, y, z) = 0 with respect to x and y, and obtain

∂F

∂x

∂x

∂x + ∂F

∂y

∂y

∂x + ∂F

∂z

∂z

∂x = 0 ^∂y/∂x=0=⇒

∂x/∂x=1

∂z

∂x = −∂F/∂x

∂F/∂z if ∂F

∂z ̸= 0,

∂F

∂x

∂x

∂y +∂F

∂y

∂y

∂y + ∂F

∂z

∂z

∂y = 0 ^∂x/∂y=0=⇒

∂y/∂y=1

∂z

∂y = −∂F/∂y

∂F/∂z if ∂F

∂z ̸= 0.

Directional Derivatives and the Gradient Vector

Definition Let D be a subset of R², (x₀, y₀) be an interior point of D, and let f : D → R be a function on D. Then the directional derivative of f at (x₀, y₀) in the direction of a unit vector u = (a, b) is

D_uf (x₀, y₀) = lim

h→0

f (x₀+ ha, y₀+ hb) − f (x₀, y₀)

h if this limit exists.

Remarks

(a) If u = i = (1, 0), then Dif = fx and if u = j = (0, 1), then Djf = fy.

(b) If f is a differentiable function of x and y, then f has a directional derivative in the direction of any unit vector u = (a, b) and

D_uf (x, y)= f_x(x, y)a + f_y(x, y)b= (f_x(x, y), f_y(x, y)) · (a, b) = ∇f (x, y) · (a, b), where ∇f (x, y) = (fx(x, y), fy(x, y))is called the gradient (vector) of f, or grad f,at (x, y).

Furthermore, since u = (a, b) is a unit vector, there exists an angle θ measured from the positive x-axis to u in the counterclockwise direction such that u = (cos θ, sin θ). Then

D_uf (x, y) = ∇f (x, y) · (cos θ, sin θ) is a function of x, y and θ.

Theorem Suppose f is a differentiable function of two or three variables. The maximum value of the directional derivative D_uf (p) is |∇f (p)| and it occurs when u has the same direction as the gradient vector ∇f (p).

Examples

1. Let f (x, y) = sin x + e^xy, (x, y) ∈ R². Find ∇f (1, 0) and find points (x, y) such that

∇f (x, y) = (0, 0).

2. Let f (x, y, z) = x sin yz, (x, y, z) ∈ R³. (a) Find the gradient of f and (b) find the directional derivative of f at (1, 3, 0) in the direction of v = i + 2j − k = (1, 2, −1).

3. Let f (x, y) = xe^y, (x, y) ∈ R². (a) Find the rate of change of f at the point p = (2, 0) in the direction from p to q = (1/2, 2), and (b) determine the direction in which f has the maximum rate of change and (c) find the maximum rate of change of f at p.

Tangent Planes to Level Surfaces Suppose that

ˆ S = {(x, y, z) ∈ R³ | F (x, y, z) = k} is a level surface of F in R3,

ˆ p = (x0, y₀, z₀) is a point in S,

ˆ C ⊂ S is any differentiable curve in S passing through p, and parametrized by r(t) = (x(t), y(t), z(t)), t ∈ I = (a, b), with r(t₀) = p for some t₀ ∈ I.

Since C = {r(t) | t ∈ I} ⊂ S, and by the Chain Rule, we have F (x(t), y(t), z(t)) = k =⇒ d

dtF (x(t), y(t), z(t)) = 0 for all t ∈ I

=⇒ ∂F

∂x dx

dt +∂F

∂y dy dt +∂F

∂z dz

dt = (∂F

∂x,∂F

∂y,∂F

∂x) · (dx dt,dy

dt,dz

dt) = ∇F (r(t)) · r^′(t) = 0 for all t ∈ I.

In particular, we have

∇F (x₀, y₀, z₀) · r^′(t₀) = 0,

which implies that if ∇F (x₀, y₀, z₀) ̸= (0, 0, 0), ∇F (x₀, y₀, z₀) is perpendicular to the tangent vector r^′(t₀) to any curve C in S passing through p = r(t₀) = (x₀, y₀, z₀), and the tangent plane to the level surface F (x, y, z) = k at p = (x₀, y₀, z₀)has an equation

F_x(x₀, y₀, z₀)(x − x₀) + F_y(x₀, y₀, z₀)(y − y₀) + F_z(x₀, y₀, z₀)(z − z₀) = 0.

ExampleFind the equations of the tangent plane and normal line to ellipsoid x²

4 + y²+z²

9 = 3 at the point (−2, 1, −3).

Properties of the Gradient VectorLet f be a differentiable function of two or three variables and suppose that ∇f (p) ̸= 0 (zero vector in R² or R³). Then

ˆ The directional derivative of f at p in the direction of a unit vector u is given by Duf (p) =

∇f (p) · u.

ˆ ∇f(p) points in the direction of maximum rate of increase of f at p, and that maximum rate of change is |∇f (p)|.

ˆ ∇f(p) is perpendicular to the level curve or level surface of f through p.

Examples The figure on the right shows level sets of a height function or f (x, y) = x²− y² with a gradient vector fields.

Definition Let f : D ⊂ Rⁿ → R be a real-valued function defined on D. Then

ˆ f is said to have a local maximumvalue at p if there exists r > 0 such that B_r(p) ⊂ D and f (x) ≤ f (p) for all x ∈ B_r(p).

ˆ f is said to have a local minimumvalue at p if there exists r > 0 such that B_r(p) ⊂ D and f (x) ≥ f (p) for all x ∈ B_r(p).

Therem (First derivatives Test) If f has a local maximum or minimum at p and if the first partial derivatives of f exist at p, then ∇f (p)= (f_x1, f_x2, . . . , f_xn)(p) = (0, 0, . . . , 0)= 0 ∈ Rⁿ. Definition A point p ∈ D is called a critical point (or stationary point) of f if either ∇f (p) = 0 ∈ Rⁿ or if ∇f (p) does not exist.

ExampleLet f (x, y) = x²+ y²− 2x − 6y + 14, (x, y) ∈ R². Find the critical points and extreme values (or critical values) of f if exist.

Classification of Extreme Values Theorem (Second Derivatives Test) Let f : D ⊂ R² → R be a function defined on D, p be an interior point of D and let Br(p) ⊂ D be an open disk in D. Suppose that the second partial derivatives of f are continuous on Br(p), ∇f (p) = (f_x(p), f_y(p)) = (0, 0) and that

D = f_xx(p)f_yy(p) − [f_xy(p)]² =    

f_xx(p) f_xy(p) fyx(p) fyy(p)     . (a) If D > 0 and f_xx(p) > 0,then f (p) is a local minimum.

(b) If D > 0 and f_xx(p) < 0,then f (p) is a local maximum.

(c) If D < 0, then (p, f (p)) is a saddle point of the graph of f.

(d) If D = 0, the test is inconclusive: f could have a local maximum or local minimum at p or (p, f (p)) could be a saddle point of the graph of f.

Outline of the Proof Let u = (h, k) be a unit vector. For any |t| < r, since f (p + tu) has continuous second derivative for each t ∈ (−r, r) and since

d

dtf (p + tu)|_t=0=f_x(p + tu)h + f_y(p + tu)k|_t=0 = f_x(p)h + f_y(p)k = ∇f (p) · (h, k) = 0, and

d²

dt²f (p + tu)|_t=0 = d

dtfx(p + tu)h + f_y(p + tu)k|t=0

= f_xx(p + tu)h²+ f_xy(p + tu)hk + f_yx(p + tu)kh + f_yy(p + tu)k²|_t=0

= f_xx(p)h²+ 2f_xy(p)hk + f_yy(p)k² =f_xx(p)



h + f_xy(p) f_xx(p)k

2

+ k²

f_xx(p) f_xx(p)f_yy(p) − f_xy² (p)
,

so by the Taylor’s Theorem and for each |t| < r and for any unit vector u = (h, k) ∈ R², we have

f (p + tu) − f (p) =  d

dtf (p + tu)|_t=0



t + 1 2

d²

dt²f (p + tu)|_t=0



t²+ R(t)

=

"

f_xx(p)



h + f_xy(p) f_xx(p)k

2

+ k²

f_xx(p) f_xx(p)f_yy(p) − f_xy² (p)

# t²

2 + R(t),

where lim

t→0

R(t)

t² = 0. Hence, the theorem follows by using the second derivative test for functions of one variable.

Remark Setting a = f_xx(p), b = f_xy(p), c = f_yy(p), note that

ˆ if a ̸= 0 and ac − b² > 0, then

ax²+ 2bxy + cy² = a



x²+2b

a xy + b² a²y²

 +

 c − b²

a

 y²

= a

 x + b

ay

2

+ac − b² a y²

(≥ 0 if a > 0

≤ 0 if a < 0

ˆ if a ̸= 0 and ac − b² < 0, then

ax²+ 2bxy + cy² = a

 x + b

ay

2

− b²− ac a y²

= a

 x + b

ay +

√b²− ac

a y

  x + b

ay −

√b²− ac

a y



and (0, 0, 0) is a saddle point of the graph of z = ax²+2bxy+cy²since x+b ±√

b²− ac

a y = 0

are distinct lines dividing xy-plane into 4 regions around (0, 0).

ˆ if a = 0 and ac − b² < 0 =⇒ b ̸= 0, then ax²+ 2bxy + cy² = by (2x + cy) and (0, 0, 0) is a saddle point of the graph of z = ax²+ 2bxy + cy² since y = 0 and 2x + cy = 0 are distinct.

Definitions Let D be a subset of Rⁿ and let f : D ⊂ Rⁿ→ R be a function defined on D. Then

ˆ D is called a boundedsubset of Rⁿ if there exists a rectangular box R = [a₁, b₁] × [a₂, b₂] ×

· · · × [a_n, b_n] = {(x₁, x₂, . . . , x_n) ∈ Rⁿ | a_i ≤ x_i ≤ b_i, 1 ≤ i ≤ n} such that D ⊂ R, or if there exists r > 0 such that D ⊂ Br(0), where 0 ∈ Rⁿ.

ˆ D is called anopen subset of Rⁿ if for each p ∈ D there exists r > 0 such that Br(p) ⊂ D, i.e. every point in D is an interior point of D.

ˆ D is called a closed subset of Rⁿ if its complement D^c = {x ∈ Rⁿ | x /∈ D} is an open subset of Rⁿ.

ˆ f(p) is called the absolute maximum (value) of f on D if f (x) ≤ f (p) for all x ∈ D;

f (p) is called theabsolute minimum (value) of f on D if f (x) ≥ f (p) for all x ∈ D.

Theorem (Extreme Value Theorem)If f : D ⊂ Rⁿ → R is continuous on a closed, bounded set D in Rⁿ, then there exist p, q ∈ D such that

f (p) ≥ f (x) ≥ f (q) for all x ∈ D ⇐⇒ max

x∈D f (x) = f (p) and min

x∈Df (x) = f (q).

Remark If f has extreme values at p, q ∈ D ⊂ R², since p (or q) is either a critical point of f or a boundary point D, we shall find the absolute maximum and minimum of f on D as follows.

1. Find the values of f at the critical points of f in D.

2. Find the extreme values of f on the boundary of D.

3. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.

ExampleFind the absolute maximum and minimum of f (x, y) = x²− 2xy + 2y on the rectangle D = {(x, y) | 0 ≤ x ≤ 3, 0 ≤ y ≤ 2}.

Method of Lagrange Multipliers

Let f : R³ → R be a differentiable function defined on R³ and let S = {(x, y, z) | g(x, y, z) = k}

be a (level) surface defined by g(x, y, z) = k. Suppose that

ˆ ∇g ̸= 0(vector) on the surface g(x, y, z) = k,

ˆ there is a point p = (x0, y₀, z₀) ∈ S such that

either f (p) = max{f (x, y, z) | g(x, y, z) = k} or f (p) = min{f (x, y, z) | g(x, y, z) = k}.

Let C be a smooth curve passing through p on S given by the vector equation C : r(t) = (x(t), y(t), z(t)), t ∈ I = (a, b) and r(t₀) = p for some t₀ ∈ I.

Since f (r(t)) has an extreme value at an interior point t = t0 and g(x(t), y(t), z(t)) = k for all t ∈ I, we have

0 =d

dtf (x(t), y(t), z(t))|_t=t0 =∇f (p) · r^′(t₀), 0 =d

dtk = d

dtg(x(t), y(t), z(t))|t=t0 =∇g(p) · r^′(t0),

=⇒ ∇f (p) ⊥ r^′(t₀), ∇g(p) ⊥ r^′(t₀) for each r^′(t₀) ̸= 0 ∈ T_r(t0)S

=⇒ ∇f (p), ∇g(p) ⊥ T_pS ⊂ R³ and ∇f (p) // ∇g(p).

This suggests that we can use the following procedures(Method of Lagrange Multipliers)to find the extreme values of f (x, y, z) subject to the constraint g(x, y, z) = k.

Step 1. Find all values of x, y, z and λ such that

(∇f (x, y, z) = λ∇g(x, y, z) (3 equations of x, y, z, λ), g(x, y, z) = k (an equation of x, y, z).

Step 2. Evaluate f at all the points (x, y, z) that result from Step 1. The largest of these values is the maximum value of f and the smallest is the minimum value of f.

Example A rectangular box without a lid is to be made from 12m² of cardboard. Find the maximum volume of such a box (with x, y, z being the length, width and height, respectively).

Solution To maximize V = xyz subject to A = xy + 2xz + 2yz = 12, we find all possible x, y, z, λ such that

(V_x, V_y, V_z) = λ(A_x, A_y, A_z), A = 12

⇐⇒ yz ⁽¹⁾= λ(y + 2z), xz ⁽²⁾= λ(x + 2z), xy ⁽³⁾= λ(2x + 2y), xy + 2xz + 2yz ⁽⁴⁾= 12

x(1)−y(2)

⇐⇒

x(1)−z(3)

2λ(x − y)z = 0, xz⁽²⁾= λ(x + 2z), λ(y − 2z)x = 0, xy + 2xz + 2yz = 12

=⇒ x = y = 2z, xz⁽²⁾= λ(x + 2z), xy + 2xz + 2yz = 12

=⇒ 2z^{2 (2)}= 4λz, xy + 2xz + 2yz = 12

=⇒ z⁽²⁾= 2λ, x = y = 2z = 4λ, xy + 2xz + 2yz = 48λ² = 12

Thus we have λ = 1

2, x = y = 2, z = 1, and the maximum volume V = V (2, 2, 1) = 4.

Suppose now that we want to find the maximum and minimum values of a function f (x, y, z) subject to two constraints (side conditions) of the form g(x, y, z) = k and h(x, y, z) = c.

Following the method of Lagrange multiplier, we need to find all values of x, y, z, λ and µ such

that 





∇f (x, y, z)=λ∇g(x, y, z) + µ∇h(x, y, z), g(x, y, z) = k,

h(x, y, z) = c.

Geometrically, this means that we are looking for the extreme values of f when (x, y, z) is restricted to lie on the curve of intersection C of the level surfaces g(x, y, z) = k and h(x, y, z) = c.

Example Find the maximum value of the function f (x, y, z) = x + 2y + 3z on the curve of intersection of the plane x − y + z = 1 and the cylinder x²+ y² = 1.

Solution To maximize f (x, y, z) = x + 2y + 3z subject to g(x, y, z) = x − y + z = 1 and h(x, y, z) = x²+ y² = 1, we find all possible x, y, z, λ, µ such that







(f_x, f_y, f_z) = λ(g_x, g_y, g_z) + µ(h_x, h_y, z_z), g = 1,

h = 1

⇐⇒







(fx, fy, fz) = λ(gx, gy, gz) + µ(hx, hy, zz), g = 1,

h = 1

⇐⇒







(1, 2, 3)= λ(1, −1, 1) + µ(2x, 2y, 0)= (λ + 2µx, −λ + 2µy, λ), x − y + z = 1,

x²+ y² = 1

⇐⇒







λ = 3, 1⁽¹⁾= 3 + 2µx, 2 ⁽²⁾= −3 + 2µy, x − y + z = 1,

x²+ y² = 1

x(2)−y(1)

=⇒ λ = 3, x ⁽³⁾= −2

5y, x − y + z ⁽⁴⁾= 1, x²+ y^{2 (5)}= 1

(3)→(5)

=⇒ λ = 3, y = ± 5

√29, x⁽³⁾= ∓ 2

√29, x − y + z⁽⁴⁾= 1

=⇒(4) λ = 3, y = ± 5

√29, x⁽³⁾= ∓ 2

√29, z ⁽⁴⁾= 1 ± 7

√29

Hence f (− 2

√29, 5

√29, 1 + 7

√29, ) = 3 + √

29 and f ( 2

√29, − 5

√29, 1 − 7

√29, ) = 3 −√ 29 are respectively the maximum and minimum values of f subject to g(x, y, z) = x − y + z = 1 and h(x, y, z) = x²+ y² = 1.