# HW6 - C13SOG.017 Ifw = f:c y «\$2 2 then:zrdx ydy zz-0212...

• Notes
• 9
• 100% (1) 1 out of 1 people found this document helpful

This preview shows pages 1–9. 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 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: C13SOG.017: Ifw = f(:c, y) :: «\$2 + 2 , then :zrdx +ydy ﬁzz-0212 . Choose :17 = 3, y = 4, dz; = —0.03, and dy = 004. Them (110: - r 3 - (—0.03) +4 (004) = 2007 = 5.014. ‘, . 2 3,4 . 4 f(297,404) f( )+ W 500 Compare with the true value of v 10057 20 f(2.07, 4.04) = z 5014229751417. (313806.023: U11} 2 f(:z:, y, z) = 6‘33”, then dw = —e‘xyz(yz dx + 2:2 (13/ + my dz). Take :r. = l, y = 0, z = —2, (1:5 = 0.02, dy = 0.03, and dz = -0.02. Then f(1.02, 0.03, —2.02) z f(1, 0, —2) — e0(0 — 2 - (0.03) + 0) = 1+ 336 = 1.06. Compare with the exact value, which is 15453 250000 f(1.02, 0.03, —‘2.02) = exp ( > 2: 1.0637623386083891. 1/3 C13806.028: Ifw = f(x, 1/) = 31/5, then 31 dw * 31.2/3y1/5 dz _ 5316/S dy = 152:2/33/5/5 ‘éke x = 27, y = 32, (hr = —2, and 033/ = —2. Then “25,30 z 27 32 160-(—2)—:31.{.—2)=§~ 79 :64 z ) f( , )+ 15.9“ 2 7320 — 1.4817129629629630. {Purposes of comparison, the true value is 57/15 f(25, 30) = m z 1.4810023646720941. C13506.038: We begin with the equation 8 .00T . 4103(p (1T — T dp) WT, p) = 2p , for which av = __SFPT__. When p = 5, dp = —().1, T = 300, and (IT = —20, we ﬁnd that . _ 0 — 300. -—).1 28721 W24103(5 ( 2) (( ))__ ._ —__. = —229.768. 50-25 125 The actual change in the volume is 8206 V(280, 4.0) _ V(300, 5) = “Eg— z -234.4571428571428571. C13806.039: The period T of a pendulum of length L is given (approximately) by “2 1/2 dL — L d T = 271' (5) , for which dT =.(i) . W 2” g. g L 9 If L = 2, (IL = 1/12, 9 = 32,and dg = 0.2, then 177r :—z .0 78161849536596. (IT 1920 0 2 The true value of the increase in the period is T(2 + 1/12, 32.2) - T(2, 32) z 0.0274043631738259. \7, R ‘\ C13S06.043: Part (a): If (r, y) ——> (0, 0) along the line y = 2:, then (qumm‘o) f(z, y) :13}, f(:c, 1r) 1135 1 1 But if (z, y) —> (0, 0) along the line y = 0, then lim :0, 1 = lim 3, O = lim 0 = 0. (awﬂwlo) f( J) he f( l W0 Therefore f is not continuous at (0, 0). Part (b): We compute the partial derivatives of f at (0, 0) by the deﬁnition: a, U, (l 2: : = __ = . f ( ) ill—Eli h Ill—IE) h Ill—.0 h 0' . f(0,0+k)-f(0,0) . f(0,k) . 0 {l (l = : = _ z ‘ M) > be. i £122, 1; 1:0 k 0 Therefore both f;E and fy exist at (0, 0) but f is not continuous at (0, O). C13506.044: The function f(:z:, y) = (ml/3 + 141/3) 15 continuous everywhere because it is the comp051t10n of the sum of continuous functions. At the origin w e compute its partial derivative with respect to a: as follows: fz(0 (0:11”, f(0+h,0)-f(0, 0) (h1/3)3 IHO Rh =53?) h =1; similarly, fy(0, O) = 1. So only the plane z = a: + y can approximate the graph of f at and near (0, 0). But the line L1 in the vertical plane 3; = 1, through (0, 0, 0), and tangent to the graph of f has slope . flr, 18) -f(0, 0) _ Pill) xﬁ “ Ni whereas the line L2 in the vertical.plane y = 2:, through (0, 0, O), and tangent to the graph of z = z + y has slope ﬂ. Because no plane through (0, O, 0) approximates the graph of f accurately near (0, 0), the function f is not differentiable at (0, 0). C13SOG.O46: Suppose that f is a function of a single variable. We are to Show that f’(a) exists if and only if f is differentiable at :1? = u in the sense of Eq. (19), meaning that there exxsts a constant c such that “m = 0 (1) h—vO and that, if Eq. (1) holds, then c = f’(u)4 So let us suppose first that f’(a) exists. Let c 2: f’(a). Then f(u+hl—f(u) lim 2 f'(a) )L—eo h , by deﬁnition, and , ch , hm —- = c : f (a). }.-»O h Consequently lim f(a+h)‘f(a)—Ch = hm _ (11m = 0 _ 0 = 0, h—.0 h ;._.o h h—+0 h and therefore Eq. (1) holds as well. Next suppose that “m f(a + h) — f(a) — ch h—40 =0 for some constant c. Then hm f(a + h) — ﬂu) - ch h—.0 lhl = [Ill-El) h ‘ and thus hm ffu+hl -f(dl - ch h—tO h It now follows that -Ch> + 2 0+0 = c, }A—90 h h—.0 h and thus f(a + h) - f(al lim ————~—— = c. ).—.o h That is, f’(u) exists and c = f’(a). Section 13.7 (313507.001: If m = exp(—172 — 1J2), :r z: t, and y = t1”, then (1111 0111 (LE (3'11) (13/ — — - — + ,— - — 71?“am dt dy (it = -—2;1:exp(—z2 — y?) -— yt‘1/2exp(—x2 — y2) = —2t exp(—t2 —— t) — exp(—t2 — t). Alternatively, w = exp(—-t2 -— t), and hence dw dt = —(2t+1)exp(-—1i2 -—t). C13807.012: Because r(3:, y, z) = exp(1¢z — any) + exp(my - yz) + exp(yz —:rz), we ﬁnd that 8r 8—1” = yeXP(wy - yz) — zexpmz - \$2) + (z — y) eXP(9»‘Z — 1y), (31' By ~ zexp(yz — x2) — zexp(mz - my) + (z — z)exp(zy — yz), and (97‘ 3—2 = xexp(\$z — 17y) - yexp(my — yz) + (y —— x) exp(yz — 2:2). C13507.013: If p = f(\$, y), :r, = \$(u, 'v, w), and y = y(u, 'u, w), then 013807.015: Up = f(u, v, w), u = u(x, y, 2), v = 11(x, y, z), and w = w(r, y, 2), then @_0f an (9f ('31; 3f (9m ('31: ("M (")Jr 31) (9m 3w . E, @_8f 3'11. 5f ('37) 3f 8w By an (33/ (3y and @_0f‘c9u+6_f‘0v af 6111 (92 3a (32 (91) 32+E'E' (313807.033: Suppose that the square base of the box measures :1: (inches) on each side and that its height is 2. Suppose also that time t is measured in hours. Then the volume of the box is V = 3:22, and by the chain rule dV _ 8V ctr 8V dz ._._ .___’__.__: z._ (It 33; (ale); 4: 2“; (3Hx (2) Thus when a: = ‘24 and z : 12, we have 7 dt = 2 . 24 . 12 . (—3) +242 - (—2) = -2880 cubic inches per hour; that is, —% cubic feet per hour. 013807.034: Let an be the length of each edge of the square base of the box and let z be its height. Units: centimeters and minutes. We are given dm dz dt and dt ’ and we are to ﬁnd the rate of change V’ (t) of the volume of the box and the rate of change A’ (t) of its surface area when m = z = 100 (cm). Note that V = 1:22 and A = 2:52 + 43:2. First, dV av dz av dz 2 _.._:_.__ _._:2 .2 ._3, dt (9:1: dt+82 dt 1'2 +3 ( l and thus, when m = z = 100, we have 1‘: = 40000 — 30000 = 10000 dt cubic centimeters per minute; that is, 0.01 cubic meters 'per minute. Next, clA 8A (Ix 8A dz _=__._ ___._= . .__3 dt 8;]: (n+6; dt “1+4” “4‘” )’ so that, when {r = z = 100, E = 1600 — 1200 :2 400 dt square centimeters per minute; that is, 0.04 square meters per minute. if", 013807.038: From the equation 1 __ 1 + 1 + _1__ R _ R1 R2 R3 and the given data R1 = R2 = 100, R3 = 200, we ﬁnd that R = 40 9. Also from this equation we derive (with time t in seconds) 51 1 (1121 1 11122 1 drag R2 :R—f' rlt “LIT? (1t +R_§' dt' Then substitution of the previous data and the additional information that (1 =1 and ﬁ (131 (1R2 dt ‘ alt dt =—2 yields the result that R is increasing at the rate of 0.24 9 per second at the time in question. 013807.052: Ifw : f(u) where _ 2[.2 _ y2 1" _ \$2 + y2’ then 413/2 4:523; _ mm“; + may = :nf'(u)u3, + yf/(u)uy = mf’(u) 4 - yf'(U)~Z;2—_*--:--1;§)—2 -— 0- C13SO7.060: Here we have sin¢cos€ pcosgbcosﬂ ~psin¢\$sin6 T’(p, 912, 6) = sincﬁsinﬁ pcosgﬁsinﬁ psingbcosﬂ cos qb —~pSiIl¢o 0 Therefore |T'(p, ¢, 6)! = [)2 sin3 9*) sin2 (9 + p2 sin ¢ 6082 ¢v cos2 0 + p2 cos2 ¢ sin qS sin2 6 + ,92 sin3 45 cos2 9 = p2(sin3 ¢ + sin ¢ c052 as) = p2 sin 45. C13SO7.061: Here we compute sin qb cos 9 ,0 cos 9") cos (9 ——psin 4’) sin 6 6w 0915 6w [FI Fy Fz] sin¢sin¢9 pcosqﬁsinf) psinrbcosﬂ % , cos d) -psin¢ 0 which has ﬁrst component a. 321 2 FE sinqﬁ 0039 + Fy sind) sir19 + F2: cosgﬁ, [) second component am . r 5;),- = F1. pcosd) (2056 + Fy pcosd) sm6 — Fz psm d), and third component a, 5%) = -FI psin¢ sin6 + Fy psinqﬁ c056. C13SOB.006: Given f(:r, y, z) = W and P(12, 3, 4), the“ :1: y 2—. WW”: “\/F—‘ W 3" «Fugue 1 M and therefore Vf(12, 3, 4) = < ,ﬁ 7‘5) El 013508.011: Given f(:r., y) = 272 + 2mg + 3y2, P(2, 1), and v = (1, 1), we ﬁrst compute a unit vector with the same direction as v: u=ﬁ=<wiw§y Also Vf(:z:, y) = (255+ 22, 2x + ﬂy), so Vf(P) = (6, 10). Therefore Duf(P) = (Vf(P)) -u= <6. 10>- (ya w?) = N2“. (313808.019: Given f(\$, 1, z) = exp(a:yz), the point P(4, 0, --3), and the vector v = (0, 1, —1), we ﬁrst construct a unit vector with the same direction as v: Next, Vf(;c, y, z) = exp(:m z)(yz, zz, my), and so Vf(P) = (O, —12, 0). Therefore 013808.023: Given f(:z:, y) = 111(a:2 + 3/2) and the point P(3, 4), We ﬁrst compute lvl= and its rate of increase in that direction is \ 013508.026: We are to ﬁnd the direction in which f(2:, y, z) = exp(\$ —— y —- z) is increasing the most rapidly at the point P(5, 2, 3) and its rate of increase in that direction. Such computations can easily be carried out with computer algebra systems such as Mathematica 3.0. We ﬁrst deﬁne f: f[x_, y_, 2-] := Exp[x - y - 2] Then we compute the gradient of f: {D[f[x,y,z],x], letx,y,z],y], Diflx.y,z],23} <ez—y—z __e:r:—y—:) _€z—y~z> 7 Then we evaluate the last output at P: %/. {x~5,y—§2,z—>3} (1, —1, —1) The last output, Vf(P), gives the direction in which f is increasing the most rapidly at P. Its magnitude is the rate of increase of f in that direction: Sqrt [76.70] \/3~ C13808.033: Let f(a.-, y, z) = H3 + y1/3 + 21/3 — 1. Then Vf(\$, y, z) = l 1 1 312/3’ 3y2/3’ 332/3 ’ and thus a vector normal to the surface f(x, 1 , z) = O at the point P(l, -1, 1) is aa- Therefore an equation of the plane tangent to the surface at P is §(m—1>+5(y+1)+3(z—1)=0; 3 3 that is, z+y+z= (313508.035: Hit and 11 are differenti able functions of :r: and y and a. and b are constants, then a vote», 3;) + was, 31)) = <§;(uu(x. y) + M1220): glam v) +va 10)) (aux. any) + (170:: {my} ll ll ( (m; + b’vz, (my + 5’01, ) u (um, uy) + Hot, vy) = aVu(:r, y) + va(a:, C13808.044: Because v is not a unit vector, w e must replace it with a unit vector having the same direction before we can use the formulas of this section. 80 we take “He-e The gradient vector of f is Vf = (y+z)i+(:c+z)j+(r+y)k, so Vf(1, 2, 3) = 5i+4j+3k. Hence Duf(P): <5i4v3)'( Z_ 3’ who ): grees per kilometer) for the desired range of change of temperature with respect to distance. WIN 1 39 (de C13SOB.045: In the solution of Problem 44 we calculated Vf(P) 2: (5, 4, 3), and the unit vector in the direction from P to Q is Then DunP) = Vf(P)-u = (5, 4, 3>- <%, i, %> = 7 (degrees per kilometer). Hence (111/ deg dt _dw (IS_ 7 — d5 (1". _ min as the hawk’s rate of change of temperature at P. deg km _ In) (2 — 14 C13SOS.055: The surface is the graph of the equation G(:z:, y, z) = 0 where C(Jt, y, z) = try/z —- 1, so that VG(I, y, z) = (3/2, 1:2, my). Suppose that P(a, b, c) is a point strictly within the ﬁrst octant (so that a, b, and c are all positive). Note that abc : 1. A vector normal to the surface at P is n 2 (be, ac, ab), and hence the plane tangent to the surface at P has equation bcm + acy + obz = d for some constant it. Moreover, because P is a point of the surface, ficu + (to!) + (the : d; that is, d = 30.50. Hence an equation of the tangent plane is bar: + (mg + abz occur at (3a, 0, 0), (0, 311, 0) has volume = 3ubc. The intercepts of the pyramid therefore , and (U, 0, 3c). Therefore, because of the right angle at the origin, the pyramid V = é(3u)(3b)(3c) NICO 7 —2Z;—abc = ’ independent of the choice of P, as we were to show. ...
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