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**Unformatted text preview: **Section 13.2 C13SO2.003: If either :r or y is nonzero, then 2:2 +y2 > 0, and so f(m, y) is deﬁned—but net if :1: = y = O. . Hence the domain of f consists of all points (1:, y) in the plane other than the origin. (313302.005: The real number 2 has a unique cube root z1/3 regardless of the value of 2. Hence the domain
of f(a:, y) = (y — 2:2)1/3 consists of all points in the zy—plane. C13SO2.006: The real number z has a unique cube root zl/3 regardless of the value of 2. But V21: is real
if and only if :c 2 0. Therefore the domain of ﬂat, y) = 02:)”2 + (3y)‘/3 consists of all those points (z, y)
for which a: 2. 0. 013802.033: The level curves of f (z, y) = 4:22 +y2 are ellipses centered at the origin, with major axes on
the m-axis and minor axes on the y—axis. C13802.052: The graph of f(x, y) = 23,3 - 3y2 — 12y + 2:2 is shown in Fig. 13.2%. The fact that the
derivative of g(y) = 23,13 —- 3y2 — 12y is zero when y = —1 and when y = 2 accounts forthe “waviness” of
the ﬁgure in the y—direction. Section 13.3 (313803.006: The sum, product, and quotient of continuous functions is continuous where deﬁned, so 9 - 3:2
f($. y) — 1 +333}
is continuous where my aé —1. Consequently
. 9 — 22 5
hm = = _,
(min—42.3) ﬂx’ y) 1+ 2 ~ 3 7 (313503.051: Given: 22:23;
$4 + y? ' f(x. 1/) = Suppose that (2:, y) approaches (0, 0) along the nonvertical straight line with equation y = mx. If m 74 O,
then—on that line— 3 lim f(x y) — lim 2mg: - lim 2mm -- 0 —0
(ND-40.0) ’ =~0 33“ + 7,1212 _ 1-4) 1:2 +7712 _. m2 _ ' Clearly if m = 0 the result is the same. And if (z, y) approaches (0, 0) along the y-axis, then—on that
line—- 2 - 0 - y
l. 1 = ‘
(amigos) flx’ y) 11,135 0 + :13 Therefore as (x, y) approaches (0, 0) along any straight line, the limit of f (x, y) is zero. But on the curve
y = 12 we have 4
lim f(:z:, y) = lim 2”” = 1
(rm—40,0) we :4 + x4 ’ and therefore the limit of f (2:, y) does not exist at (0, 0) (see the Remark following Example 9). For related paradoxical results involving functions of two variables, see Problem 60 of Section 13.5 and the miscellaneous
problems of Chapter 13. 013303.053: Given: After we convert to polar coordinates, we have 2 .
f(T19)= w =cos€ sine = ésin26’.
T On the hyperbolic spiral r0 = l, we have 6 —> +oo as r approaches zero through positive Values. Hence f (r, 6) takes on all values between -% and % inﬁnitely often as r —> 0+. Therefore, as we discovered in
Example 9, f(z, y) has no limit as (:5, y) —> (O, 0). Section 13.4 C13SO4.003: If f(m, y) = e‘(cosy — siny), then 3f; 2: 21(c05y — sing) and 51; = "5 (cosy +31ny)
7.2 _ 82
C13504.017: If f(r, s) = r2 + 52. then
0f 47‘32 3f 4r23 Tea—+723? “d shew)? 013804.024: If z(:c, y) = my exp(—zy), then 2:05, y) = y (KM-my) - wy’ eXP(-zy). Mac. 1/) = z exp(-zy) - zzy expl-my),
zzy(w. y) = -xy expl-xy) - (my — 1) eXP(—wy) + xy(a:y — 1) eXP(-$y) = ($2312 -.3-'ry + 1) eXp(—=cy) = 293705, :1). . C . ‘ . .
co](.).:»dSi::036if tlThe gratphfof the given equation 2 = 3x +4y is a plane, so it is its own tangent plane and the
es 0 1e pom o tangency don’t matter Answer An uation of the 13.11 t
. . n
, of z = 31: + 4y at the point P is z = 3:: + 4y. eq P e a gent to the graph ?%:SU)4f137: Given: f(1:, y) = my and the point P(1, _1, —1) on its graph. Then fab, y) = y and
; y i]?! - :r, 50 f;(1, —-1)= .—1 and fy(1, —-l) = 1. By Eq. (11) an equation of the plane tangent to the
grap ofz=f(:r, y) at the pothisz+1= “(I—1)+(y—-1); that is, z—y+z= 1' C13804.043: If f¢(:z:, y) = cos2 my and fy(a:, y) = sin2 my, then
fxy(a:, y) = -2:csin my cos 1y 94 —2y sin my coszry = fyx(a:, y). By the Note preceding and following Eq. (14), there can be no function f (x, y) having the given ﬁrst-order partial derivatives. C13804.044: Given fx(ac, y) = cosa: sing and fy(z, y) = sin: cosy, we ﬁnd that fag/(Ix, y) = (2951 cosy = fuz<xy 9) So it’s not impossible that there exists a function f (2:, y) having the given ﬁrst-order partial derivatives. Indeed, by inspection, one such function is f (3:, y) = sinz sin y. 013804.055: If u(1:, t) = exp(—n2kt)sin 711: where n and k are constants, then u¢(z, t) = ~n2kexp(-n2kt)sin nx, u1(z, t) = nexp(—n2kt)cos ms, and umm, t) = —n2 exp(—n2kt)sinn:c. Therefore in 2 kn“. for any choice of the constants k and n. 5013804057: Part (a): If y($, t) = sin(:c + at) (where a. is a constant), then y¢(z, t) = acos(a: + at), y;(::, t) 2: cos(z + at), y“(:c, t) = —a2 sin(:z: + at), yum, t) = — sin(:r + at).
Therefore y“ = azyu. Part (b): If y(:r, t) = cosh (3(z — at)), then y¢(.’n, t) = ——3asinh (3(m — at», y;(:c, t) = 35inh (3(9: — at)), yu(z, t) =' 9a? cosh (3(z — at)), ymbr, t) = 9cosh (3(a: -— at)).
Therefore :9“ = 02y“. Part (c): If y(z, t) = sin k2: cos kat (where k is a constant), then y¢(z, t) = —kasin k3: sin kat, y:(z, t) = kcos k1: cos kat,
yu(x, t) = ~k2a2 sin km cos Icat, yum, t) = —k2 sin kmcos kat. Therefore y” = a222,. C13SO4.060: Given: The constant (1 and the function q '1 ,Z =_——'-—_—_‘
d>(£ y ) ,——-——xz+y2+zz Then _ _______ir____
¢x(z, y, z) —— ($2 + y? + 22)“? and gum? — yg — z?) «ﬁnes, y. z) = W By the symmetries in 4’: among x, y, and 2, it follows that q(222 _ $2 __ y2) (1(21/2 - $2 — 22)
—— —————— and 452205, 2/, Z) = W (153111031 ya Z) " (22 +y2 + 22)5/2 It is now clear that q'; satisﬁes the three-dimensional Laplace equation 45“ + d)“ + ¢zz = 0 C13SO4.065: If f(:c, y) = 2:2 + 23y + 2y2 - 6m + 8y, then the equations f:($, y)=01 fy(1',y)=0
are
2I+2y—6=O, 22:+4y=_sY which have the unique solution x = 10, y = —-7. So the surface that is the graph of z = f(a:, y) contains
exactly one point at which the tangent plane is horizontal, and that point is (10, —7, -58) a 013804.068: If 2(2, y) = ln(cos 1:) —— lu(cosy), then 23(33, y) = - tanx! Z‘U($’ y) = tany’ 221505! y) = _ 8862 I? 2311(31 :9) E 7 and zyy(:c, y) = 8602 y.
Therefore '2 2 (1+23)zm — ZZzzyzxy +(1+ 2%)‘2‘9'5/ = ‘Sec 17 39023} — 0 +590 5‘7 536231 E 0' 013804.070: The sum of two harmonic functions is harmonic. Here is a. proof for the case of functions of
two variables. Suppose that f and g are harmonic. Then fun: + fW = 0 and g” + 9w = 0. Let Mas, y) = f(:r:, y) +g(m, y). Then [tn-FILM=fm+ym+fyy+gyy=fzx+fw+gu+gw=0+0=0- Therefore f + g is harmonic. This concludes the proof. By induction, you may extend it to show that the
sum of any ﬁnite number of harmonic functions is harmonic. This answers the question in Problem 70. ...

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