MA102: Multivariable Calculus Rupam Barman and Shreemayee Bora - PowerPoint PPT Presentation

MA102: Multivariable Calculus Rupam Barman and Shreemayee Bora Department of Mathematics IIT Guwahati R. Barman & S. Bora MA-102 (2017)

Differentiability of f : U ⊂ R n → R m Definition: Let U ⊂ R n be open. Then f : U ⊂ R n → R m is differentiable at X 0 ∈ U if there exists a linear map L : R n → R m such that � f ( X 0 + H ) − f ( X 0 ) − L ( H ) � lim = 0 . � H � H → 0 R. Barman & S. Bora MA-102 (2017)

Differentiability of f : U ⊂ R n → R m Definition: Let U ⊂ R n be open. Then f : U ⊂ R n → R m is differentiable at X 0 ∈ U if there exists a linear map L : R n → R m such that � f ( X 0 + H ) − f ( X 0 ) − L ( H ) � lim = 0 . � H � H → 0 The linear map L is called the derivative of f at X 0 and is denoted by D f ( X 0 ) , that is, L = D f ( X 0 ) . Other notations: f ′ ( X 0 ) , d f d X ( X 0 ) . R. Barman & S. Bora MA-102 (2017)

Characterization of differentiability Theorem: Consider f : R n → R m with f ( X ) = ( f 1 ( X ) , . . . , f m ( X )) , where f i : R n → R . Then f is differentiable at X 0 ∈ R n ⇐ ⇒ f i is differentiable at X 0 for i = 1 , 2 , . . . , m . Further D f ( X 0 )( H ) = ( ∇ f 1 ( X 0 ) • H , . . . , ∇ f m ( X 0 ) • H ) . R. Barman & S. Bora MA-102 (2017)

Characterization of differentiability Theorem: Consider f : R n → R m with f ( X ) = ( f 1 ( X ) , . . . , f m ( X )) , where f i : R n → R . Then f is differentiable at X 0 ∈ R n ⇐ ⇒ f i is differentiable at X 0 for i = 1 , 2 , . . . , m . Further D f ( X 0 )( H ) = ( ∇ f 1 ( X 0 ) • H , . . . , ∇ f m ( X 0 ) • H ) . The matrix of D f ( X 0 ) is called the Jacobian matrix of f at X 0 and is denoted by J f ( X 0 ) . R. Barman & S. Bora MA-102 (2017)

Jacobian matrix of f : R n → R m J f ( X 0 ) is an m × n matrix with ( i , j ) -th entry a ij := ∂ j f i ( X 0 ) . R. Barman & S. Bora MA-102 (2017)

Jacobian matrix of f : R n → R m J f ( X 0 ) is an m × n matrix with ( i , j ) -th entry a ij := ∂ j f i ( X 0 ) . ∇ f 1 ( X 0 ) � ∂ f i   � . . ( X 0 ) J f ( X 0 ) = = . .   ∂ x j m × n ∇ f m ( X 0 ) m × n R. Barman & S. Bora MA-102 (2017)

Jacobian matrix of f : R n → R m J f ( X 0 ) is an m × n matrix with ( i , j ) -th entry a ij := ∂ j f i ( X 0 ) . ∇ f 1 ( X 0 ) � ∂ f i   � . . ( X 0 ) J f ( X 0 ) = = . .   ∂ x j m × n ∇ f m ( X 0 ) m × n • f ( x , y ) = ( f 1 ( x , y ) , f 2 ( x , y ) , f 3 ( x , y )   ∂ x f 1 ( a , b ) ∂ y f 1 ( a , b ) J f ( a , b ) = ∂ x f 2 ( a , b ) ∂ y f 2 ( a , b )   ∂ x f 3 ( a , b ) ∂ y f 3 ( a , b ) R. Barman & S. Bora MA-102 (2017)

Jacobian matrix of f : R n → R m J f ( X 0 ) is an m × n matrix with ( i , j ) -th entry a ij := ∂ j f i ( X 0 ) . ∇ f 1 ( X 0 ) � ∂ f i   � . . ( X 0 ) J f ( X 0 ) = = . .   ∂ x j m × n ∇ f m ( X 0 ) m × n • f ( x , y ) = ( f 1 ( x , y ) , f 2 ( x , y ) , f 3 ( x , y )   ∂ x f 1 ( a , b ) ∂ y f 1 ( a , b ) J f ( a , b ) = ∂ x f 2 ( a , b ) ∂ y f 2 ( a , b )   ∂ x f 3 ( a , b ) ∂ y f 3 ( a , b ) • f ( x , y , z ) = ( f 1 ( x , y , z ) , f 2 ( x , y , z )) � � ∂ x f 1 ( a , b , c ) ∂ y f 1 ( a , b , c ) ∂ z f 1 ( a , b , c ) J f ( a , b , c ) = ∂ x f 2 ( a , b , c ) ∂ y f 2 ( a , b , c ) ∂ z f 2 ( a , b , c ) R. Barman & S. Bora MA-102 (2017)

Examples • If f ( x , y ) = ( xy , e x y , sin y ) then   y x e x y e x J f ( x , y ) =   0 cos y R. Barman & S. Bora MA-102 (2017)

Examples • If f ( x , y ) = ( xy , e x y , sin y ) then   y x e x y e x J f ( x , y ) =   0 cos y • If f ( x , y , z ) = ( x + y + z , xyz ) then � � 1 1 1 J f ( x , y , z ) = yz xz xy R. Barman & S. Bora MA-102 (2017)

Chain rule Theorem-A: Let X : R → R n be differentiable at t 0 and f : R n → R be differentiable at X 0 := X ( t 0 ) . Then f ◦ X is differentiable at t 0 and n ∂ i f ( X 0 ) d x i ( t 0 ) d � d t f ( X ( t )) | t = t 0 = ∇ f ( X 0 ) • X ′ ( t 0 ) = . d t i = 1 R. Barman & S. Bora MA-102 (2017)

Chain rule Theorem-A: Let X : R → R n be differentiable at t 0 and f : R n → R be differentiable at X 0 := X ( t 0 ) . Then f ◦ X is differentiable at t 0 and n ∂ i f ( X 0 ) d x i ( t 0 ) d � d t f ( X ( t )) | t = t 0 = ∇ f ( X 0 ) • X ′ ( t 0 ) = . d t i = 1 Proof: Since f is differentiable at X 0 , therefore f ( X 0 + H ) = f ( X 0 ) + ∇ f ( X 0 ) • H + E ( H ) � H � · · · · · · ( ∗ ) and E ( H ) → 0 as H → 0 . Put H := X ( t ) − X ( t 0 ) in ( ∗ ) to complete the proof. R. Barman & S. Bora MA-102 (2017)

Chain rule Theorem-A: Let X : R → R n be differentiable at t 0 and f : R n → R be differentiable at X 0 := X ( t 0 ) . Then f ◦ X is differentiable at t 0 and n ∂ i f ( X 0 ) d x i ( t 0 ) d � d t f ( X ( t )) | t = t 0 = ∇ f ( X 0 ) • X ′ ( t 0 ) = . d t i = 1 Proof: Since f is differentiable at X 0 , therefore f ( X 0 + H ) = f ( X 0 ) + ∇ f ( X 0 ) • H + E ( H ) � H � · · · · · · ( ∗ ) and E ( H ) → 0 as H → 0 . Put H := X ( t ) − X ( t 0 ) in ( ∗ ) to complete the proof. n ∂ i f ( X 0 ) d x i ( t 0 ) Remark: d � d t f ( X ( t 0 )) = ∇ f ( X 0 ) • X ′ ( t 0 ) = d t i = 1 sometimes referred to as total derivative. R. Barman & S. Bora MA-102 (2017)

Chain rule for partial derivatives Theorem-B: If X : R 2 → R n , ( u , v ) �→ ( x 1 ( u , v ) , . . . , x n ( u , v )) has partial derivatives at ( a , b ) and f : R n → R is differentiable at Y := X ( a , b ) then F ( u , v ) := f ( X ( u , v )) has partial derivatives at ( a , b ) and n ∂ f ( Y ) ∂ x j ( a , b ) � ∂ u F ( a , b ) = ∇ f ( Y ) • ∂ u X ( a , b ) = , ∂ x j ∂ u j = 1 n ∂ f ( Y ) ∂ x j ( a , b ) � ∂ v F ( a , b ) = ∇ f ( Y ) • ∂ v X ( a , b ) = . ∂ x j ∂ v j = 1 R. Barman & S. Bora MA-102 (2017)

Chain rule for partial derivatives Case n=2: If x = x ( u , v ) and y = y ( u , v ) have first order partial derivatives at the point ( u , v ) , and if z = f ( x , y ) is differentiable at the point ( x ( u , v ) , y ( u , v )) , then z = f ( x ( u , v ) , y ( u , v )) has first order partial derivatives at ( u , v ) given by ∂ u = ∂ z ∂ z ∂ x ∂ u + ∂ z ∂ u and ∂ z ∂ y ∂ v = ∂ z ∂ x ∂ v + ∂ z ∂ y ∂ v . ∂ x ∂ y ∂ x ∂ y Example: Find ∂ w /∂ u and ∂ w /∂ v when w = x 2 + xy and x = u 2 v , y = uv 2 . R. Barman & S. Bora MA-102 (2017)

Graph and level set Let f : R n → R . Then G ( f ) := { ( X , f ( X )) : X ∈ R n } ⊂ R n + 1 is the graph of f . G ( f ) represents a hyper-surface in R n + 1 . R. Barman & S. Bora MA-102 (2017)

Graph and level set Let f : R n → R . Then G ( f ) := { ( X , f ( X )) : X ∈ R n } ⊂ R n + 1 is the graph of f . G ( f ) represents a hyper-surface in R n + 1 . The set S ( f , α ) := { X ∈ R n : f ( X ) = α } is called a level set of f and represents a hyper-surface in R n . R. Barman & S. Bora MA-102 (2017)

Graph and level set Let f : R n → R . Then G ( f ) := { ( X , f ( X )) : X ∈ R n } ⊂ R n + 1 is the graph of f . G ( f ) represents a hyper-surface in R n + 1 . The set S ( f , α ) := { X ∈ R n : f ( X ) = α } is called a level set of f and represents a hyper-surface in R n . (e) Graph of (f) Level curve � x 2 + y 2 � x 2 + y 2 = k f ( x , y ) := R. Barman & S. Bora MA-102 (2017)

Level sets and gradients Let f : R n → R be differentiable at X 0 ∈ R n . Suppose that X 0 is point on the hyper-surface f ( X ) = α. R. Barman & S. Bora MA-102 (2017)

Level sets and gradients Let f : R n → R be differentiable at X 0 ∈ R n . Suppose that X 0 is point on the hyper-surface f ( X ) = α. Let X : ( − ε, ε ) → R n be a curve on the hyper-surface f ( X ) = α passing through X 0 , i.e, X ( 0 ) = X 0 and f ( X ( t )) = α for t ∈ ( − ε, ε ) . R. Barman & S. Bora MA-102 (2017)

Level sets and gradients Let f : R n → R be differentiable at X 0 ∈ R n . Suppose that X 0 is point on the hyper-surface f ( X ) = α. Let X : ( − ε, ε ) → R n be a curve on the hyper-surface f ( X ) = α passing through X 0 , i.e, X ( 0 ) = X 0 and f ( X ( t )) = α for t ∈ ( − ε, ε ) . Suppose that X ( t ) is differentiable at 0 . Then 0 = d f ( X ( t )) | t = 0 = ∇ f ( X 0 ) • X ′ ( 0 ) ⇒ ∇ f ( X 0 ) ⊥ X ′ ( 0 ) d t R. Barman & S. Bora MA-102 (2017)

Level sets and gradients Let f : R n → R be differentiable at X 0 ∈ R n . Suppose that X 0 is point on the hyper-surface f ( X ) = α. Let X : ( − ε, ε ) → R n be a curve on the hyper-surface f ( X ) = α passing through X 0 , i.e, X ( 0 ) = X 0 and f ( X ( t )) = α for t ∈ ( − ε, ε ) . Suppose that X ( t ) is differentiable at 0 . Then 0 = d f ( X ( t )) | t = 0 = ∇ f ( X 0 ) • X ′ ( 0 ) ⇒ ∇ f ( X 0 ) ⊥ X ′ ( 0 ) d t Since the line X 0 + t X ′ ( 0 ) is tangent to the curve X ( t ) at X 0 , ∇ f ( X 0 ) is normal to the hyper-surface f ( X ) = α at X 0 . R. Barman & S. Bora MA-102 (2017)