Problemas de Funciones de Varias Variables y Sus Derivadas

7/30/2019 Problemas de Funciones de Varias Variables y Sus Derivadas

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Chapter 11. Multivariable Functions and their

Derivatives

11.2 Limits and Continuity in Higher Dimensions

56. f(x, y) = cos(x3y3

x2+y2)

Sol. lim(x,y)(0,0) cos(x3y3x2+y2

) = limr0 cos(r3 cos3 r3 sin3 r2 cos2 +r2 sin2

) = limr0 cos[r(cos3 sin3 )

1] =

cos0 = 1

66. Define f(0, 0) in a way that extends f(x, y) = xy x2y2

x2+y2to be continuouos

at the origin.

Sol. |xy(x2 y2)| = |xy||x2 y2| |x||y||x2 + y2| =

x2y2|x2 + y2|

x2 + y2

x2 + y2|x2 + y2| = (x2 + y2)2 |xy

(x2

y2

)x2+y2 | (x2

+Y2

)

2

x2+y2 =

x2+y2 (x2+y2) xy(x2y2)x2+y2

(x2+y2) lim(x,y)(0,0) xy x2y2

x2+y2= 0

by Sandwich Theorem, since lim(x,y)(0,0) (x2 + y2) = 0; thus, definef(0, 0) = 0

69. Does knowing that | sin(1/x)| 1 tell you any thing about

lim(x,y)(0,0)

y sin1

x?

Give reasons for your answer.

Sol. The limit is 0 since

|sin( 1

x)

| 1

1

sin 1

x

1

y

y sin 1

x

y

for y 0, and y y sin 1x y for y 0. Thus as (x, y)rightarrow(0, 0),both y and y approach 0 y sin 1

x 0, by Sandwich Theorem

11.3 Partial Derivarives

45. Find all second-order partial derivatives of the function in Exercise 45,46.r(x, y) = ln(x + y)

Sol. rx

= 1x+y ,

ry

= 1x+y ,

2rx2

= 1(x+y)2 ,2ry2

= 1(x+y)2 ,2rxy

= 1(x+y)2 ,2ryx

= 1(x+y)2

46. s(x, y) = tan1(y/x)

Sol. sx = [ 11+(yx)2 ] x ( yx) = ( yx2 )[ 11+(y

x)2 ] = yx2+y2 , sy = [ 11+(y

x)2 ] y ( yx ) =

( 1x

)[ 11+(y

x)2

] = xx2+y2 ,

2sx2

= y(2x)(x2+y2)2

= 2xy(x2+y2)2

, 2s

y2= x(2y)

(x2+y2)2= 2xy

(x2+y2)2,

2sxy

= 2s

yx= (x

2+y2)(1)+y(2y)(x2+y2)2

= y2x2

(x2+y2)2

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57. Find the value of z/x at the point (1, 1, 1) if the equation

xy + z3x 2yz = 0

define z as a function of the two independent variables x and y and thepartial derivative exists.

Sol. y + (3z2 zx

)x + z2 2y zx

= 0 (3xz2 2y) zx

= y z3 at (1, 1, 1)we have (3 2) z

x= 1 1 or z

x= 2

62. Find x/u and y/u if the equations u = x2 y2 and v = x2 y definex and y as functions of the independent variables u and v, and the partialderivatives exist. Then let s = x2 + y2 and find s/u.

Sol. Differentiating each equation implicitly gives 1 = (2x)xu(2y)yu and 0 =

(2x)xuyu or (2x)xu (2y)yu = 1(2x)xu yu = 0

xu =1 2y0 1

2x 2y2x 1

= 12x+4xy =

12x4xy and yu =

2x 12x 02x+4xy =

2x2x+4xy =

2x4x4xy =

112y ; next s = x

2 + y2

su

= 2xxu

+ 2y yu

= 2x( 12x4xy ) + 2y(

112y ) =

112y +

2y12y =

1+2y12y

11.4 The Chain Rule

29. Use these equations z/x = Fx/Fz and z/y = Fy/Fz to findthe values of z/x and z/y at the point in Exercise 29.z3 xy + yz + y3 2 = 0, (1, 1, 1)

Sol. Let F(x,y,z) = z3xy +yz +y32 = 0 Fx(x,y,z) = y, Fy(x,y,z) =x + z + 3y2, Fz(x,y,z) = 3z2 + y = FxFy =

y3x2+y

= y3z2+y

zx

(1, 1, 1) = 14 ;zy

= FyFz

= x+z+3y23z2+y = xz3y2

3z2+y zy (1, 1, 1) = 3442. Suppose that we substitute polar coordinates x = r cos and y = r sin in

a differentiable function w = f(x, y).

(a) Show thatw/r = fx cos + fy sin

and1

r w/ = fx sin + fy cos .(b) Solve the equaitons in part (a) to express fx and fy in terms ofw/r

and w/.

(c) Show that

(fx)2 + (fy)

2 = (w

r)2 +

1

r2(

w

)2.

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Sol.

(a) wr = fx xr + fy yr = fx cos + fy sin and w = fx(r sin ) +fy(r cos ) 1r w = fx sin + fy cos

(b) wr

sin = fx sin cos + fy sin2 and ( cos

r)w

= fx sin cos +fy cos

2 fy = (sin )wr +( cosr )w ; then wr = fx cos +[(sin )wr +( cos

r)w

](sin ) fx cos = wr (sin2 )wr ( sin cosr )w =(1 sin2 )w

r ( sin cos

r)w

fx = (cos )wr ( sin r )w(c) (fx)2 = (cos2 )(

wr

)2(2 sin cosr

)(wr

w

)+( sin2

r2)(w

)2 and (fy)2 =

(sin2 )(wr

)2 + ( 2sin cosr

)(wr

w

) + ( cos2

r2)(w

)2 (fx)2 + (fy)2 =

(wr

)2 + 1r2

(w

)2

47. Let T = f(x, y) be the temperature at the point (x, y) on the circle

x = cos t, y = sin t, 0 t 2 and suppose thatT/x = 8x 4y, T /y = 8y 4x.

(a) Locate the maximum and minimum temparatures on the ellipse byexamining dT/dt and d2T/dt2.

(b) Suppose that T = xy 2. Find the maximum and minimum valuesof T on the ellipse.

Sol.

(a) Tx

= 8x 4y and Ty

= 8y 4x dTdt

= Tx

dxdt

+ Ty

dydt

= (8x 4y)(

sin t) + (8y

4x)(cos t) = (8cos t

4sin t)(

sin t) + (8 sin t

4cos t)(cos t) = 4sin2 t 4cos2 t d2

Tdt2 = 16sin t cos t; dTdt = 0

4sin2 t4cos2 t = 0 sin2 t = cos2 t sin t = cos t or sin t = cos t t = 4 , 54 , 34 , 74 on the interval 0 t 2;d2Tdt2

|t=4

= 16sin 4

cos 4

> 0 T has a minimum at (x, y) =(22

,22

);d2Tdt2

|t= 34

= 16sin 34

cos 34

< 0 T has a maximum at (x, y) =(

22 ,

22 );

d2Tdt2

|t= 54

= 16sin 54

cos 54

> 0 T has a minimum at (x, y) =(

22

, 22

);d2Tdt2

|t= 74

= 16sin 74 cos74 < 0 T has a maximum at (x, y) =

(22 , 22 );

(b) T = 4x2 4xy + 4y2 Tx

= 8x 4y, and Ty

= 8y 4x so the ex-treme values occur at the four points found in part(a): T(

22

,22

) =

T(22

, 22

) = 4( 12

)4(12

)+4(12

) = 6, the maximum and T(22

,22

) =

T(22

, 22

) = 4( 12

) 4(12

) + 4( 12

) = 2, the minimum

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49. Find the derivatives of the function

F(x) =

x20

t4 + x3dt

Sol. G(u, x) =ua

g(t, x)dt where u = f(x) dGdx

= Gu

dudx

+ Gx

dxdx

=

g(u, x)g(x)+ua

gx(t, x)dt; thus F(x) =x20

t4 + x3dt F(x) =

(x2)4 + x3(2x)+x2

0x

t4 + x3dt = 2x

x8 + x3 +

x20

3x2

2t4+x3

dt

11.5 Directional Derivatives, Gradient Vectors, and Tan-

gent Planes

23. By about how much will

f(x,y,z) = ln

x2 + y2 + z2

change if the point P(x,y,z) moves from P0(3, 4, 12) a distance ofds = 0.1units in the direction of 3i + 6j 2k?

Sol. f = ( xx2+y2+z2

)i + ( yx2+y2+z2

)j + ( zx2+y2+z2

)k f(3, 4, 12) = 3169

i +4169

j + 12169

k; u = v|v| =3i+6j2k32+62+(2)2 =

37

i + 67j 2

7k f u = 9

1183and

df = (f u)ds = ( 91183

)(0.1) 0.000830. Find equations for the

(a) Tangent plane and(b) Normal line at the point P0 on the given surface.

Sol.

(a) f = (2x + 2y)i + (2x 2y)j + 2zk f(1, 1, 3) = 4j + 6k Tangent plane: 4(y + 1) + 6(z 3) = 0 2y + 3z = 7;

(b) Normal line: x = 1, y = 1 + 4t, z = 3 + 6t47. Find parametric equation for the line tangent to the curve of intersection

of the surfaces at the given point.Surfaces: x3 + 3x2y2 + y3 + 4xy z2 = 0, x2 + y2 + z2 = 11Point: (1, 1, 3)

Sol. f = (3x2 + 6xy2 + 4y)i + (6x2y + 3y2 + 4x)j 2zk f(1, 1, 3) = 13i +13j 6k; g = 2xi + 2yj + 2zk g(1, 1, 3) = 2i + 2j + 6k; v = fg v =

i j k13 13 62 2 6

= 90i 90j Tangent line: x = 2 22t,

y =

2 + 2

2t, z = 4

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57. Find the derivative of f(x,y,z) = x2 + y2 + z2 in the direction of the unit

tangent vector of the helix

r(t) = (cos t)i + (sin t)j + tk

at the points where t = /4, 0, and /4. The function f gives thesquare of the distance from a point P(x,y,z) on the helix to the origin.The derivatives calculated here give the rates at which the square of thedistance is changing with respect to t as P moves through the points wheret = /4, 0, and /4.

Sol. f = 2xi + 2yj + 2zk = (2cos t)i + (2 sin t)j + 2tk and v = ( sin t)i +(cos t)j + k u = v|v| = ( sin t)i+(cos t)j+k(sin t)2+(cos t)2+12 = (

sin t2

)i + ( cos t2

)j + ( 12

)k

(Duf)P0 =

f

u = (2cos t)( sin t

2) + (2 sin t)(cos t

2) + (2t)( 1

2) = 2t

2

(Duf)(4 ) = 22 , (Duf)(0) = 0 and (Duf)(4 ) = 22

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Problemas de Funciones de Varias Variables y Sus Derivadas

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