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