HW6 - C13SOG.017: Ifw = f(:c, y) :: «$2 + 2 , then :zrdx...

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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. ...
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This note was uploaded on 03/01/2008 for the course MATH 51 taught by Professor Staff during the Spring '07 term at Stanford.

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HW6 - C13SOG.017: Ifw = f(:c, y) :: «$2 + 2 , then :zrdx...

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