MATH
HW5

# HW5 - Section 13.2 C13SO2.003 If either:r or y is nonzero...

• Notes
• 5
• 100% (2) 2 out of 2 people found this document helpful

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

• Fall '07
• Staff

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern