Advanced Calculus (II)

Advanced Calculus (II)

W^EN-C^HINGL^IEN

Department of Mathematics National Cheng Kung University

2009

WEN-CHINGLIEN Advanced Calculus (II)

Ch11: Differentiability on R

11.3: Derivatives, Differentials, and Tangent Planes

Theorem (11.20)

Let α ∈R, a ∈ Rⁿ, and suppose that f and g are vector functions. If f and g are differentiable ata, then f + g, αf , and f · g are all differentiable ata. In fact,

(7) D(f +g)(a) = Df (a)+Dg(a),

(8) D(αf )(a) = αDf (a),

and

(9) D(f · g)(a) = g(a)Df (a) + f (a)Dg(a).

Definition (11.21 A tangent Hyperplane)

(1) Let S be a subset ofR^m andc ∈ S. A hyperplane Π with normaln is said to be tangent to S at c if and only if c ∈ Π and

(11) n · ck − c

kc_k − ck → 0

for all sequencesck ∈ S\{c} that converge to c.

(2)n · (x − c) = 0

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (11.22)

Suppose that V is open inRⁿ, thata ∈ V , and that f : V →R. If f is differentiable at a, then the surface

S := {(x, z) ∈ Rⁿ⁺¹ :z = f (x) and x ∈ V } has a tangent hyperplane at (a, f (a)) with normal (12) n = (∇f (a), −1) := (fx₁(a), fx₂(a), . . . , fxn(a), −1).

Proof.

Letck ∈ S with c_k 6= (a, f (a)) and c_k → (a, f (a)).Then c_k = (a_k,f (a_k))for some a_k ∈ V and a_k → a as k → ∞.

For smallh, set ε(h) = f (a + h) − f (a) − ∇f (h) and define n by (12).Since

kc_k − ck =pka_k − ak²+ |f (a_k) −f (a)|² ≥ ka_k − ak, it is clear by (12) that

0 ≤    

n · ck − c kc_k − ck

≤ |ε(ak − a)|

ka_k − ak .

Since ^ε(h)_khk → 0 as h → 0, it follows from the Squeeze Theorem thatn satisfies (11) for c := (a, f (a)).

WEN-CHINGLIEN Advanced Calculus (II)

Proof.

Letck ∈ S with c_k 6= (a, f (a)) and c_k → (a, f (a)). Then c_k = (a_k,f (a_k))for some a_k ∈ V and a_k → a as k → ∞.

For smallh, set ε(h) = f (a + h) − f (a) − ∇f (h) and define n by (12). Since

kc_k − ck =pka_k − ak²+ |f (a_k) −f (a)|² ≥ ka_k − ak, it is clear by (12) that

0 ≤    

n · ck − c kc_k − ck

≤ |ε(ak − a)|

ka_k − ak .

Since ^ε(h)_khk → 0 as h → 0, it follows from the Squeeze Theorem thatn satisfies (11) for c := (a, f (a)).

Proof.

Letck ∈ S with c_k 6= (a, f (a)) and c_k → (a, f (a)). Then c_k = (a_k,f (a_k))for some a_k ∈ V and a_k → a as k → ∞.

For smallh, set ε(h) = f (a + h) − f (a) − ∇f (h) and define n by (12).Since

kc_k − ck =pka_k − ak²+ |f (a_k) −f (a)|² ≥ ka_k − ak, it is clear by (12) that

0 ≤    

n · ck − c kc_k − ck

≤ |ε(ak − a)|

ka_k − ak .

Since ^ε(h)_khk → 0 as h → 0, it follows from the Squeeze Theorem thatn satisfies (11) for c := (a, f (a)).

WEN-CHINGLIEN Advanced Calculus (II)

Proof.

Letck ∈ S with c_k 6= (a, f (a)) and c_k → (a, f (a)). Then c_k = (a_k,f (a_k))for some a_k ∈ V and a_k → a as k → ∞.

For smallh, set ε(h) = f (a + h) − f (a) − ∇f (h) and define n by (12). Since

kc_k − ck =pka_k − ak²+ |f (a_k) −f (a)|² ≥ ka_k − ak, it is clear by (12) that

0 ≤    

n · ck − c kc_k − ck

≤ |ε(ak − a)|

ka_k − ak .

Since ^ε(h)_khk → 0 as h → 0,it follows from the Squeeze Theorem thatn satisfies (11) for c := (a, f (a)).

Proof.

Letck ∈ S with c_k 6= (a, f (a)) and c_k → (a, f (a)). Then c_k = (a_k,f (a_k))for some a_k ∈ V and a_k → a as k → ∞.

For smallh, set ε(h) = f (a + h) − f (a) − ∇f (h) and define n by (12). Since

kc_k − ck =pka_k − ak²+ |f (a_k) −f (a)|² ≥ ka_k − ak, it is clear by (12) that

0 ≤    

n · ck − c kc_k − ck

≤ |ε(ak − a)|

ka_k − ak .

Since ^ε(h)_khk → 0 as h → 0, it follows from the Squeeze Theorem thatn satisfies (11) for c := (a, f (a)).

WEN-CHINGLIEN Advanced Calculus (II)

Proof.

Letck ∈ S with c_k 6= (a, f (a)) and c_k → (a, f (a)). Then c_k = (a_k,f (a_k))for some a_k ∈ V and a_k → a as k → ∞.

For smallh, set ε(h) = f (a + h) − f (a) − ∇f (h) and define n by (12). Since

kc_k − ck =pka_k − ak²+ |f (a_k) −f (a)|² ≥ ka_k − ak, it is clear by (12) that

0 ≤    

n · ck − c kc_k − ck

≤ |ε(ak − a)|

ka_k − ak .

Since ^ε(h)_khk → 0 as h → 0,it follows from the Squeeze Theorem thatn satisfies (11) for c := (a, f (a)).

Proof.

Letck ∈ S with c_k 6= (a, f (a)) and c_k → (a, f (a)). Then c_k = (a_k,f (a_k))for some a_k ∈ V and a_k → a as k → ∞.

For smallh, set ε(h) = f (a + h) − f (a) − ∇f (h) and define n by (12). Since

kc_k − ck =pka_k − ak²+ |f (a_k) −f (a)|² ≥ ka_k − ak, it is clear by (12) that

0 ≤    

n · ck − c kc_k − ck

≤ |ε(ak − a)|

ka_k − ak .

Since ^ε(h)_khk → 0 as h → 0, it follows from the Squeeze Theorem thatn satisfies (11) for c := (a, f (a)).

WEN-CHINGLIEN Advanced Calculus (II)

Note: If f is a near-valued function of two variables that is differentiable at (a, b), then the surface z = f (x , y ) has a tangent plane at (a, b, f (a, b)) with normal

n =: (∇f (a, b), −1).

Notation (the first total differential):

z = f (x), ∆z := f (a + ∆x) − f (a)

∆x := (∆x1, . . . ∆xn) dz := ∇f (a) · ∆x :=

n

X

j=1

∂f

∂xj

(a)dxj.

Remark (11.23)

Let f :Rⁿ→ R be differentiable at a and

∆x = (∆x1, . . . , ∆xn). Then

∆z − dz

k∆xk → 0 as ∆x → 0.

In particular, the differential dz approximates ∆z.

Proof.

By definition, if f is differentiable ata, then

ε(h) := f (a + h) − f (a) − ∇f (a) · h satisfies ^ε(h)_khk → 0 as h → 0.Since ∆z = f (a + h) − f (a) for h := ∆x and dz = ∇f (a) · h, it follows that ^∆z−dz_k∆xk → 0 as ∆x → 0.

WEN-CHINGLIEN Advanced Calculus (II)

Remark (11.23)

Let f :Rⁿ→ R be differentiable at a and

∆x = (∆x1, . . . , ∆xn). Then

∆z − dz

k∆xk → 0 as ∆x → 0.

In particular, the differential dz approximates ∆z.

Proof.

By definition, if f is differentiable ata,then

ε(h) := f (a + h) − f (a) − ∇f (a) · h satisfies ^ε(h)_khk → 0 as h → 0. Since ∆z = f (a + h) − f (a) for h := ∆x and dz = ∇f (a) · h,it follows that ^∆z−dz_k∆xk → 0 as ∆x → 0.

Remark (11.23)

Let f :Rⁿ→ R be differentiable at a and

∆x = (∆x1, . . . , ∆xn). Then

∆z − dz

k∆xk → 0 as ∆x → 0.

In particular, the differential dz approximates ∆z.

Proof.

By definition, if f is differentiable ata, then

ε(h) := f (a + h) − f (a) − ∇f (a) · h satisfies ^ε(h)_khk → 0 as h → 0.Since ∆z = f (a + h) − f (a) for h := ∆x and dz = ∇f (a) · h, it follows that ^∆z−dz_k∆xk → 0 as ∆x → 0.

WEN-CHINGLIEN Advanced Calculus (II)

Remark (11.23)

Let f :Rⁿ→ R be differentiable at a and

∆x = (∆x1, . . . , ∆xn). Then

∆z − dz

k∆xk → 0 as ∆x → 0.

In particular, the differential dz approximates ∆z.

Proof.

By definition, if f is differentiable ata, then

ε(h) := f (a + h) − f (a) − ∇f (a) · h satisfies ^ε(h)_khk → 0 as h → 0. Since ∆z = f (a + h) − f (a) for h := ∆x and dz = ∇f (a) · h,it follows that ^∆z−dz_k∆xk → 0 as ∆x → 0.

Remark (11.23)

Let f :Rⁿ→ R be differentiable at a and

∆x = (∆x1, . . . , ∆xn). Then

∆z − dz

k∆xk → 0 as ∆x → 0.

In particular, the differential dz approximates ∆z.

Proof.

By definition, if f is differentiable ata, then

ε(h) := f (a + h) − f (a) − ∇f (a) · h satisfies ^ε(h)_khk → 0 as h → 0. Since ∆z = f (a + h) − f (a) for h := ∆x and dz = ∇f (a) · h, it follows that ^∆z−dz_k∆xk → 0 as ∆x → 0.

WEN-CHINGLIEN Advanced Calculus (II)

Thank you.