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Unformatted text preview: C13SOG.017: Ifw = f(:c, y) :: «$2 + 2 , then :zrdx +ydy
ﬁzz0212 . 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.
5025 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 = [IllEl) 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 “Hee 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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