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Directional derivatives and gradient vectors (Sect. 14.5).IDirectional derivative of functions of two variables.IPartial derivatives and directional derivatives.IDirectional derivative of functions of three variables.IThe gradient vector and directional derivatives.IProperties of the the gradient vector.

Directional derivative of functions of two variables.Remark: The directional derivative generalizes the partialderivatives to any direction.

Directional derivative of functions of two variables.Remark: The directional derivative generalizes the partialderivatives to any direction.DefinitionThe directional derivative of the function f : D R2 R at thepoint P0 (x0 , y0 ) D in the direction of a unit vectoru hux , uy i is given byDu f P0 1 f (x0 ux t, y0 uy t) f (x0 , y0 ) ,t 0 t limif the limit exists.

Directional derivative of functions of two variables.Remark: The directional derivative generalizes the partialderivatives to any direction.DefinitionThe directional derivative of the function f : D R2 R at thepoint P0 (x0 , y0 ) D in the direction of a unit vectoru hux , uy i is given byDu f P0 1 f (x0 ux t, y0 uy t) f (x0 , y0 ) ,t 0 t limif the limit exists.Notation: The directional derivative is also denoted as df dtu,P 0.

Directional derivatives generalize partial derivatives.ExampleThe partial derivativesfx and fy are particular cases of directional derivatives Du f P limt 0 1t f (x0 ux t, y0 uy t) f (x0 , y0 ) :0

Directional derivatives generalize partial derivatives.ExampleThe partial derivativesfx and fy are particular cases of directional derivatives Du f P limt 0 1t f (x0 ux t, y0 uy t) f (x0 , y0 ) :0 I u h1, 0i i, then Di f fx (x0 , y0 ).P0

Directional derivatives generalize partial derivatives.ExampleThe partial derivativesfx and fy are particular cases of directional derivatives Du f P limt 0 1t f (x0 ux t, y0 uy t) f (x0 , y0 ) :0 I u h1, 0i i, then Di f fx (x0 , y0 ).P 0I u h0, 1i j , then Dj f fy (x0 , y0 ).P0

Directional derivatives generalize partial derivatives.ExampleThe partial derivativesfx and fy are particular cases of directional derivatives Du f P limt 0 1t f (x0 ux t, y0 uy t) f (x0 , y0 ) :0 I u h1, 0i i, then Di f fx (x0 , y0 ).P 0I u h0, 1i j , then Dj f fy (x0 , y0 ).P0zf ( x, y0 )f ( x, y )zfx (x0, y0 )f (x,y)( Du f )Pf y (x0, y0 )0f ( x0, y 1111001i( x0, y0 1111001yuP0 (x0 , y0 ) u 1

Directional derivative of functions of two variables.Remark: The condition u 1 implies that the parameter t in theline r(t) hx0 , y0 i u t is the distance between the points(x(t), y (t)) (x0 ux t, y0 uy t) and (x0 , y0 ).

Directional derivative of functions of two variables.Remark: The condition u 1 implies that the parameter t in theline r(t) hx0 , y0 i u t is the distance between the points(x(t), y (t)) (x0 ux t, y0 uy t) and (x0 , y0 ).Proof.d hx x0 , y y0 i , hux t, uy ti , t u ,that is, d t .

Directional derivative of functions of two variables.Remark: The condition u 1 implies that the parameter t in theline r(t) hx0 , y0 i u t is the distance between the points(x(t), y (t)) (x0 ux t, y0 uy t) and (x0 , y0 ).Proof.d hx x0 , y y0 i , hux t, uy ti , t u ,that is, d t .Remark: The directional derivative of f (x, y ) at P0 (x0 , y0 )along u, denoted as Du f P , is the pointwise rate of change of f0with respect to the distance along the line parallel to u passingthrough (x0 , y0 ).

Directional derivatives and gradient vectors (Sect. 14.5).IDirectional derivative of functions of two variables.IPartial derivatives and directional derivatives.IDirectional derivative of functions of three variables.IThe gradient vector and directional derivatives.IProperties of the the gradient vector.

Directional derivative and partial derivatives.zf (x,y)Remark: The directional derivative Du f P is the0derivative of f along theline r(t) hx0 , y0 i u t.( Du f 11111001yuP0 (x0 , y0 ) u 1

Directional derivative and partial derivatives.zf (x,y)Remark: The directional derivative Du f P is the0derivative of f along theline r(t) hx0 , y0 i u t.( Du f 11111001yuP0 (x0 , y0 ) u 1TheoremIf the function f : D R2 R is differentiable at P0 (x0 , y0 )and u hux , uy i is a unit vector, then Du f P fx (x0 , y0 ) ux fy (x0 , y0 ) uy .0

Directional derivative and partial derivatives.Proof.The line r(t) hx0 , y0 i hux , uy i t has parametric equations:x(t) x0 ux t and y (t) y0 uy t;

Directional derivative and partial derivatives.Proof.The line r(t) hx0 , y0 i hux , uy i t has parametric equations:x(t) x0 ux t and y (t) y0 uy t;Denote f evaluated along the line as fˆ(t) f (x(t), y (t)).

Directional derivative and partial derivatives.Proof.The line r(t) hx0 , y0 i hux , uy i t has parametric equations:x(t) x0 ux t and y (t) y0 uy t;Denote f evaluated along the line as fˆ(t) f (x(t), y (t)).Now, on the one hand, fˆ0 (0) Du f P , since0 1 fˆ0 (0) lim fˆ(t) fˆ(0) ,t 0 t 1 lim f (x0 ux t, y0 uy t) f (x0 , y0 ) ,t 0 t Du f (x0 , y0 ).

Directional derivative and partial derivatives.Proof.The line r(t) hx0 , y0 i hux , uy i t has parametric equations:x(t) x0 ux t and y (t) y0 uy t;Denote f evaluated along the line as fˆ(t) f (x(t), y (t)).Now, on the one hand, fˆ0 (0) Du f P , since0 1 fˆ0 (0) lim fˆ(t) fˆ(0) ,t 0 t 1 lim f (x0 ux t, y0 uy t) f (x0 , y0 ) ,t 0 t Du f (x0 , y0 ).On the other hand, the chain rule implies:fˆ0 (0) fx (x0 , y0 ) x 0 (0) fy (x0 , y0 ) y 0 (0).

Directional derivative and partial derivatives.Proof.The line r(t) hx0 , y0 i hux , uy i t has parametric equations:x(t) x0 ux t and y (t) y0 uy t;Denote f evaluated along the line as fˆ(t) f (x(t), y (t)).Now, on the one hand, fˆ0 (0) Du f P , since0 1 fˆ0 (0) lim fˆ(t) fˆ(0) ,t 0 t 1 lim f (x0 ux t, y0 uy t) f (x0 , y0 ) ,t 0 t Du f (x0 , y0 ).On the other hand, the chain rule implies:fˆ0 (0) fx (x0 , y0 ) x 0 (0) fy (x0 , y0 ) y 0 (0). Therefore, Du f P fx (x0 , y0 ) ux fy (x0 , y0 ) uy .0

Directional derivative and partial derivatives.ExampleCompute the directional derivative of f (x, y ) sin(x 3y ) at thepoint P0 (4, 3) in the direction of vector v h1, 2i.

Directional derivative and partial derivatives.ExampleCompute the directional derivative of f (x, y ) sin(x 3y ) at thepoint P0 (4, 3) in the direction of vector v h1, 2i.Solution: We need to find a unit vector in the direction of v.

Directional derivative and partial derivatives.ExampleCompute the directional derivative of f (x, y ) sin(x 3y ) at thepoint P0 (4, 3) in the direction of vector v h1, 2i.Solution: We need to find a unit vector in the direction of v.v1Such vector is u u h1, 2i. v 5

Directional derivative and partial derivatives.ExampleCompute the directional derivative of f (x, y ) sin(x 3y ) at thepoint P0 (4, 3) in the direction of vector v h1, 2i.Solution: We need to find a unit vector in the direction of v.v1Such vector is u u h1, 2i. v 5 We now use the formula Du f P fx (x0 , y0 ) ux fy (x0 , y0 ) uy .0

Directional derivative and partial derivatives.ExampleCompute the directional derivative of f (x, y ) sin(x 3y ) at thepoint P0 (4, 3) in the direction of vector v h1, 2i.Solution: We need to find a unit vector in the direction of v.v1Such vector is u u h1, 2i. v 5 We now use the formula Du f P fx (x0 , y0 ) ux fy (x0 , y0 ) uy .0 That is, Du f P cos(x0 3y0 )(1/ 5) 3 cos(x0 3y0 )(2/ 5).0

Directional derivative and partial derivatives.ExampleCompute the directional derivative of f (x, y ) sin(x 3y ) at thepoint P0 (4, 3) in the direction of vector v h1, 2i.Solution: We need to find a unit vector in the direction of v.v1Such vector is u u h1, 2i. v 5 We now use the formula Du f P fx (x0 , y0 ) ux fy (x0 , y0 ) uy .0 That is, Du f P cos(x0 3y0 )(1/ 5) 3 cos(x0 3y0 )(2/ 5).0 Equivalently, Du f P (7/ 5) cos(x0 3y0 ).0

Directional derivative and partial derivatives.ExampleCompute the directional derivative of f (x, y ) sin(x 3y ) at thepoint P0 (4, 3) in the direction of vector v h1, 2i.Solution: We need to find a unit vector in the direction of v.v1Such vector is u u h1, 2i. v 5 We now use the formula Du f P fx (x0 , y0 ) ux fy (x0 , y0 ) uy .0 That is, Du f P cos(x0 3y0 )(1/ 5) 3 cos(x0 3y0 )(2/ 5).0 Equivalently, Du f P (7/ 5) cos(x0 3y0 ).0 Then , Du f P (7/ 5) cos(10).C0

Directional derivatives and gradient vectors (Sect. 14.5).IDirectional derivative of functions of two variables.IPartial derivatives and directional derivatives.IDirectional derivative of functions of three variables.IThe gradient vector and directional derivatives.IProperties of the the gradient vector.

Directional derivative of functions of three variables.DefinitionThe directional derivative of the function f : D R3 R at thepoint P0 (x0 , y0 , z0 ) D in the direction of a unit vectoru hux , uy , uz i is given byDu f P0 1 f (x0 ux t, y0 uy t, z0 uz t) f (x0 , y0 , z0 ) ,t 0 t limif the limit exists.

Directional derivative of functions of three variables.DefinitionThe directional derivative of the function f : D R3 R at thepoint P0 (x0 , y0 , z0 ) D in the direction of a unit vectoru hux , uy , uz i is given byDu f P0 1 f (x0 ux t, y0 uy t, z0 uz t) f (x0 , y0 , z0 ) ,t 0 t limif the limit exists.TheoremIf the function f : D R3 R is differentiable at P0 (x0 , y0 , z0 )and u hux , uy , uz i is a unit vector, then Du f P fx (x0 , y0 , z0 ) ux fy (x0 , y0 , z0 ) uy fz (x0 , y0 , z0 ) uz .0

Directional derivative of functions of three variables.Example Find Du f P for f (x, y , z) x 2 2y 2 3z 2 at the point0P0 (3, 2, 1) along the direction given by v h2, 1, 1i.

Directional derivative of functions of three variables.Example Find Du f P for f (x, y , z) x 2 2y 2 3z 2 at the point0P0 (3, 2, 1) along the direction given by v h2, 1, 1i.Solution: We first find a unit vector along v,u v v 1u h2, 1, 1i.6

Directional derivative of functions of three variables.Example Find Du f P for f (x, y , z) x 2 2y 2 3z 2 at the point0P0 (3, 2, 1) along the direction given by v h2, 1, 1i.Solution: We first find a unit vector along v,u Then, Du f v v 1u h2, 1, 1i.6 is given by Du f (2x)ux (4y )uy (6z)uz .

Directional derivative of functions of three variables.Example Find Du f P for f (x, y , z) x 2 2y 2 3z 2 at the point0P0 (3, 2, 1) along the direction given by v h2, 1, 1i.Solution: We first find a unit vector along v,u 1u h2, 1, 1i.6 is given by Du f (2x)ux (4y )uy (6z)uz . 211We conclude, Du f P (6) (8) (6) ,0666Then, Du f v v

Directional derivative of functions of three variables.Example Find Du f P for f (x, y , z) x 2 2y 2 3z 2 at the point0P0 (3, 2, 1) along the direction given by v h2, 1, 1i.Solution: We first find a unit vector along v,u v v 1u h2, 1, 1i.6 is given by Du f (2x)ux (4y )uy (6z)uz . 211We conclude, Du f P (6) (8) (6) ,0666 26that is, Du f P .06Then, Du f C

The gradient vector and directional derivatives.Remark: The directional derivative of a function can be written interms of a dot product.

The gradient vector and directional derivatives.Remark: The directional derivative of a function can be written interms of a dot product.IIn the case of 2 variable functions: Du f fx ux fy uyDu f ( f ) · u,with f hfx , fy i.

The gradient vector and directional derivatives.Remark: The directional derivative of a function can be written interms of a dot product.IIn the case of 2 variable functions: Du f fx ux fy uyDu f ( f ) · u,Iwith f hfx , fy i.In the case of 3 variable functions: Du f fx ux fy uy fz uz ,Du f ( f ) · u,with f hfx , fy , fz i.

The gradient vector and directional derivatives.DefinitionThe gradient vector of a differentiable function f : D R2 R atany point (x, y ) D is the vector f hfx , fy i.The gradient vector of a differentiable function f : D R3 R atany point (x, y , z) D is the vector f hfx , fy , fz i.

The gradient vector and directional derivatives.DefinitionThe gradient vector of a differentiable function f : D R2 R atany point (x, y ) D is the vector f hfx , fy i.The gradient vector of a differentiable function f : D R3 R atany point (x, y , z) D is the vector f hfx , fy , fz i.Notation:IIFor two variable functions: f fx i fy j .For two variable functions: f fx i fy j fz k.

The gradient vector and directional derivatives.DefinitionThe gradient vector of a differentiable function f : D R2 R atany point (x, y ) D is the vector f hfx , fy i.The gradient vector of a differentiable function f : D R3 R atany point (x, y , z) D is the vector f hfx , fy , fz i.Notation:IIFor two variable functions: f fx i fy j .For two variable functions: f fx i fy j fz k.TheoremIf f : D Rn R, with n 2, 3, is a differentiable function and uis a unit vector, then,Du f ( f ) · u.

The gradient vector and directional derivatives.ExampleFind the gradient vector at any point in the domain of the functionf (x, y ) x 2 y 2 .

The gradient vector and directional derivatives.ExampleFind the gradient vector at any point in the domain of the functionf (x, y ) x 2 y 2 .Solution: The gradient is f hfx , fy i,

The gradient vector and directional derivatives.ExampleFind the gradient vector at any point in the domain of the functionf (x, y ) x 2 y 2 .Solution: The gradient is f hfx , fy i, that is, f h2x, 2y i. C

The gradient vector and directional derivatives.ExampleFind the gradient vector at any point in the domain of the functionf (x, y ) x 2 y 2 .Solution: The gradient is f hfx , fy i, that is, f h2x, 2y i. Czf(x,y) x 2 y 2yRemark:f f 2r,withr hx, y i.xyxDuf

Properties of the the gradient vector.Remark: If θ is the angle between f and u, then holdsDu f f · u f cos(θ).

Properties of the the gradient vector.Remark: If θ is the angle between f and u, then holdsDu f f · u f cos(θ).The formula above implies:IThe function f increases the most rapidly when u is in thedirection of f , that is, θ 0. The maximum increase rate off is f .

Properties of the the gradient vector.Remark: If θ is the angle between f and u, then holdsDu f f · u f cos(θ).The formula above implies:IThe function f increases the most rapidly when u is in thedirection of f , that is, θ 0. The maximum increase rate off is f .IThe function f decreases the most rapidly when u is in thedirection of f , that is, θ π. The maximum decrease rateof f is f .

Properties of the the gradient vector.Remark: If θ is the angle between f and u, then holdsDu f f · u f cos(θ).The formula above implies:IThe function f increases the most rapidly when u is in thedirection of f , that is, θ 0. The maximum increase rate off is f .IThe function f decreases the most rapidly when u is in thedirection of f , that is, θ π. The maximum decrease rateof f is f .IThe function f does not change along level curve or surfaces,that is, Du f 0. Therefore, f is perpendicular to the levelcurves or level surfaces.

Properties of the the gradient vector.ExampleFind the direction of maximum increase of the functionf (x, y ) x 2 /4 y 2 /9 at an arbitrary point (x, y ), and also at thepoints (1, 0) and (0, 1).

Properties of the the gradient vector.ExampleFind the direction of maximum increase of the functionf (x, y ) x 2 /4 y 2 /9 at an arbitrary point (x, y ), and also at thepoints (1, 0) and (0, 1).Solution: The direction of maximum increase of f is the directionof its gradient vector: f D x 2y E,.2 9

Properties of the the gradient vector.ExampleFind the direction of maximum increase of the functionf (x, y ) x 2 /4 y 2 /9 at an arbitrary point (x, y ), and also at thepoints (1, 0) and (0, 1).Solution: The direction of maximum increase of f is the directionof its gradient vector: f D x 2y E,.2 9At the points (1, 0) and (0, 1) we obtain, respectively, f D12E,0 .D 2E f 0, .9C

Properties of the the gradient vector.ExampleGiven the function f (x, y ) x 2 /4 y 2 /9, find the equation of aline tangent to a level curve f (x, y ) 1 at the pointP0 (1, 3 3/2).

Properties of the the gradient vector.ExampleGiven the function f (x, y ) x 2 /4 y 2 /9, find the equation of aline tangent to a level curve f (x, y ) 1 at the pointP0 (1, 3 3/2).Solution: We first verify that P0 belongs to the level curvef (x, y ) 1.

Properties of the the gradient vector.ExampleGiven the function f (x, y ) x 2 /4 y 2 /9, find the equation of aline tangent to a level curve f (x, y ) 1 at the pointP0 (1, 3 3/2).Solution: We first verify that P0 belongs to the level curvef (x, y ) 1. This is the case, since1 (9)(3) 1 1.44 9

Properties of the the gradient vector.ExampleGiven the function f (x, y ) x 2 /4 y 2 /9, find the equation of aline tangent to a level curve f (x, y ) 1 at the pointP0 (1, 3 3/2).Solution: We first verify that P0 belongs to the level curvef (x, y ) 1. This is the case, since1 (9)(3) 1 1.44 9The equation of the line we look for is D3 3Er(t) 1, t hvx , vy i,2where v hvx , vy i is tangent to the level curve f (x, y ) 1 at P0 .

Properties of the the gradient vector.ExampleGiven the function f (x, y ) x 2 /4 y 2 /9, find the equation of aline tangent to a level curve f (x, y ) 1 at the pointP0 (1, 3 3/2).Solution: Therefore, v f at P0 .

Properties of the the gradient vector.ExampleGiven the function f (x, y ) x 2 /4 y 2 /9, find the equation of aline tangent to a level curve f (x, y ) 1 at the pointP0 (1, 3 3/2).Solution: Therefore, v f at P0 . Since,D x 2y E, f 2 9 f 2 3 3E D11 E , , .2 9 223D1 P0

Properties of the the gradient vector.ExampleGiven the function f (x, y ) x 2 /4 y 2 /9, find the equation of aline tangent to a level curve f (x, y ) 1 at the pointP0 (1, 3 3/2).Solution: Therefore, v f at P0 . Since,D x 2y E, f 2 9 Therefore,0 v · f P0 f 2 3 3E D11 E , , .2 9 223D1 P0

Properties of the the gradient vector.ExampleGiven the function f (x, y ) x 2 /4 y 2 /9, find the equation of aline tangent to a level curve f (x, y ) 1 at the pointP0 (1, 3 3/2).Solution: Therefore, v f at P0 . Since,D x 2y E, f 2 9 f 2 3 3E D11 E , , .2 9 223D1 P0Therefore,0 v · f P0 11vx vy23

Properties of the the gradient vector.ExampleGiven the function f (x, y ) x 2 /4 y 2 /9, find the equation of aline tangent to a level curve f (x, y ) 1 at the pointP0 (1, 3 3/2).Solution: Therefore, v f at P0 . Since,D x 2y E, f 2 9 f 2 3 3E D11 E , , .2 9 223D1 P0Therefore,0 v · f P0 11vx vy23 v h2, 3i.

Properties of the the gradient vector.ExampleGiven the function f (x, y ) x 2 /4 y 2 /9, find the equation of aline tangent to a level curve f (x, y ) 1 at the pointP0 (1, 3 3/2).Solution: Therefore, v f at P0 . Since,D x 2y E, f 2 9 f 2 3 3E D11 E , , .2 9 223D1 P0Therefore,0 v · f P0 11vx vy23 3 3EThe line is r(t) 1, t h2, 3i.2D v h2, 3i.C

Properties of the the gradient vector.yyf331122xxff f D12E,0 ,2E f 0, ,9D 3 3Er(t) 1, t h2, 3i.2D

Further properties of the the gradient vector.TheoremIf f , g are differentiable scalar valued vector functions, g 6 0, andk R any constant, then holds,1. (kf ) k ( f );2. (f g ) f g ;3. (fg ) ( f ) g f ( g ); f ( f ) g f ( g )4. .gg2

Tangent planes and linear approximations (Sect. 14.6).IReview: Differentiable functions of two variables.IThe tangent plane to the graph of a function.IThe linear approximation of a differentiable function.IBounds for the error of a linear approximation.The differential of a function.IIIReview: Scalar functions of one variable.Scalar functions of more than one variable.

Review: Differentiable functions of two variables.DefinitionGiven a function f : D R2 R and an interior point(x0 , y0 ) D, let L be the linear functionL(x, y ) fx (x0 , y0 ) (x x0 ) fy (x0 , y0 ) (y y0 ) f (x0 , y0 ).The function f is called differentiable at (x0 , y0 ) iff the function fis approximated by the linear function L near (x0 , y0 ), that is,f (x, y ) L(x, y ) 1 (x x0 ) 2 (y y0 )where the functions 1 and 2 0 as (x, y ) (x0 , y0 ).

Review: Differentiable functions of two variables.DefinitionGiven a function f : D R2 R and an interior point(x0 , y0 ) D, let L be the linear functionL(x, y ) fx (x0 , y0 ) (x x0 ) fy (x0 , y0 ) (y y0 ) f (x0 , y0 ).The function f is called differentiable at (x0 , y0 ) iff the function fis approximated by the linear function L near (x0 , y0 ), that is,f (x, y ) L(x, y ) 1 (x x0 ) 2 (y y0 )where the functions 1 and 2 0 as (x, y ) (x0 , y0 ).TheoremIf the partial derivatives fx and fy of a function f : D R2 R arecontinuous in an open region R D, then f is differentiable in R.

Review: Differentiable functions of two variables.ExampleShow that the function f (x, y ) x 2 y 2 is differentiable for all(x, y ) R2 . Furthermore, find the linear function L, mentioned inthe definition of a differentiable function, at the point (1, 2).

Review: Differentiable functions of two variables.ExampleShow that the function f (x, y ) x 2 y 2 is differentiable for all(x, y ) R2 . Furthermore, find the linear function L, mentioned inthe definition of a differentiable function, at the point (1, 2).Solution: The partial derivatives of f are given by fx (x, y ) 2xand fy (x, y ) 2y , which are continuous functions. Therefore, thefunction f is differentiable.

Review: Differentiable functions of two variables.ExampleShow that the function f (x, y ) x 2 y 2 is differentiable for all(x, y ) R2 . Furthermore, find the linear function L, mentioned inthe definition of a differentiable function, at the point (1, 2).Solution: The partial derivatives of f are given by fx (x, y ) 2xand fy (x, y ) 2y , which are continuous functions. Therefore, thefunction f is differentiable. The linear function L at (1, 2) isL(x, y ) fx (1, 2) (x 1) fy (1, 2) (y 2) f (1, 2).

Review: Differentiable functions of two variables.ExampleShow that the function f (x, y ) x 2 y 2 is differentiable for all(x, y ) R2 . Furthermore, find the linear function L, mentioned inthe definition of a differentiable function, at the point (1, 2).Solution: The partial derivatives of f are given by fx (x, y ) 2xand fy (x, y ) 2y , which are continuous functions. Therefore, thefunction f is differentiable. The linear function L at (1, 2) isL(x, y ) fx (1, 2) (x 1) fy (1, 2) (y 2) f (1, 2).That is, we need three numbers to find the linear function L:fx (1, 2), fy (1, 2), and f (1, 2).

Review: Differentiable functions of two variables.ExampleShow that the function f (x, y ) x 2 y 2 is differentiable for all(x, y ) R2 . Furthermore, find the linear function L, mentioned inthe definition of a differentiable function, at the point (1, 2).Solution: The partial derivatives of f are given by fx (x, y ) 2xand fy (x, y ) 2y , which are continuous functions. Therefore, thefunction f is differentiable. The linear function L at (1, 2) isL(x, y ) fx (1, 2) (x 1) fy (1, 2) (y 2) f (1, 2).That is, we need three numbers to find the linear function L:fx (1, 2), fy (1, 2), and f (1, 2). These numbers are:fx (1, 2) 2,fy (1, 2) 4,f (1, 2) 5.

Review: Differentiable functions of two variables.ExampleShow that the function f (x, y ) x 2 y 2 is differentiable for all(x, y ) R2 . Furthermore, find the linear function L, mentioned inthe definition of a differentiable function, at the point (1, 2).Solution: The partial derivatives of f are given by fx (x, y ) 2xand fy (x, y ) 2y , which are continuous functions. Therefore, thefunction f is differentiable. The linear function L at (1, 2) isL(x, y ) fx (1, 2) (x 1) fy (1, 2) (y 2) f (1, 2).That is, we need three numbers to find the linear function L:fx (1, 2), fy (1, 2), and f (1, 2). These numbers are:fx (1, 2) 2,fy (1, 2) 4,f (1, 2) 5.Therefore, L(x, y ) 2(x 1) 4(y 2) 5.C

Tangent planes and linear approximations (Sect. 14.6).IReview: Differentiable functions of two variables.IThe tangent plane to the graph of a function.IThe linear approximation of a differentiable function.IBounds for the error of a linear approximation.The differential of a function.IIIReview: Scalar functions of one variable.Scalar functions of more than one variable.

The tangent plane to the graph of a function.Remark:The function L(x, y ) 2(x 1) 4(y 2) 5 is a plane in R3 .

The tangent plane to the graph of a function.Remark:The function L(x, y ) 2(x 1) 4(y 2) 5 is a plane in R3 .We usually write down the equation of a plane using the notationz L(x, y ),

The tangent plane to the graph of a function.Remark:The function L(x, y ) 2(x 1) 4(y 2) 5 is a plane in R3 .We usually write down the equation of a plane using the notationz L(x, y ), that is, z 2(x 1) 4(y 2) 5,

The tangent plane to the graph of a function.Remark:The function L(x, y ) 2(x 1) 4(y 2) 5 is a plane in R3 .We usually write down the equation of a plane using the notationz L(x, y ), that is, z 2(x 1) 4(y 2) 5, or equivalently2(x 1) 4(y 2) (z 5) 0.

The tangent plane to the graph of a function.Remark:The function L(x, y ) 2(x 1) 4(y 2) 5 is a plane in R3 .We usually write down the equation of a plane using the notationz L(x, y ), that is, z 2(x 1) 4(y 2) 5, or equivalently2(x 1) 4(y 2) (z 5) 0.This is a plane passing through P̃0 (1, 2, 5) with normal vectorn h2, 4, 1i.

The tangent plane to the graph of a function.Remark:The function L(x, y ) 2(x 1) 4(y 2) 5 is a plane in R3 .We usually write down the equat

Directional derivatives and gradient vectors (Sect. 14.5). I Directional derivative of functions of two variables. I Partial derivatives and directional derivatives. I Directional derivative of functions of three variables. I The gradient vector and directional derivatives. I Properties of the the gradient vector.

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