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Unformatted text preview: MATH 237 Assignment 4 Winter 2007 Solutions 3.A1 Deﬁning ﬂat, 3/) = x2y+my3, g(t) = 3t+5 and h(t) 2 2152—10, we can write w = f(a:, y),
a: 2 g(t) and y = h(t), so that w = f(g(t), h(t)). Using the chain rule, we write dw afdm 8fdy_6fd_g 8fdh _=__+(___ dt await dydt EdtJrEdt' In this case, we ﬁnd dw a = (2333/ + y3)(3) + ($2 + 3331/2)(4t) = 3(2(3t + 5)(2t2 ~ 10) + (2t2 — 10)3) + 4t((3t + 5)2 + 3(32: + 5)(2t2 — 10?). This can be simplifed further, but since we are interested in only one value, we can just
substitute t = —2 to ﬁnd = 3(2(1)(2) + (2)3) + 4(—2)((—1)2 + 3(1)(—2)2) = 3(4) — 8(11)
= —12 + 88 = 76. Note, we could avoid the use of the chain rule by differentiating the composite w = f (g(t), h(t)) =
(3t + 5)2(2t2 — 10) + (3t + 5)(2t2 — 10)3, but that would be considerably more work. 3.A3 We suppose given some function f : R3 —> IR and are told that g : R —> IR is deﬁned as
a composition involving f as g“) = fltatzatsl 'We wish to ﬁnd g’(1). Through the vector form of the chain rule (course notes page 61), we
take x(t) = (t, t2, t3) and we can write g’(t) = Vf(X(t)) . 2—: We have j—f = (1, 2t, 37:2), so, evaluating at t = 1, we ﬁnd
g/(1)= Vf(1, l, 1)  (1, 2,3) = (5, —3, —4)  (1, 2,3) = —13. In using the chain rule we made the implict assumption that f is differentiable at (1, 1, 1). 3. A6(ii) Letting s = if and t = i we can write u = x3f(s, t). The dependence of u on 3:, y
and z is sketched below U1 8U 89 6‘9 (93 89615 a: = airaairag
_ 2 s_f 1/ 3% _Z
__ 3m f(s,t)+x as ($2)+:c at ($2)
while
.82: _ @[email protected]@+3_9ﬂ
8y _ By 8882/ 813831
6f 1
_ 3_ _
— 0+3: as (an)
and 8U 89 (99 (93 89 8t
(92 dz as dz 8t (92 _ 8f 1
— 0+x3a (E)' Combining these results, we have (93: yay 232 <<—><—>><<:>>+z<ms:—:<:>> = 3x3f<etl + (—ym2>a—f + (—28% + @962)ng + (”2% 63
= 33:3f(s, t) = 311.. We conclude that any function u of this form satisﬁes this partial differential equation. 3. A9 Given the temperature at point (13,31) as T(:I:,y) = 100 + 106‘” sin(y), to deter—
mine the rate of change of temperature (per unit distance), we ﬁnd VT 2 (TmTy) = 2 (—10e"”” sin(y),106’m cos(y)). At the point (0, g), we have VT(0, %) = (—10e0 sing), 10c0 cos(§)) =
(—%§, %§). To ﬁnd the rate of change in the direction i+j = (1, 1), we ﬁnd a unit vector
in that direction: u = %(1, 1) and compute the directional derivative: vrm, 1) . u DuT 4 II II
A  mi
i
A
p—i
"\—/ n


+
I
II
C The rate of change is greatest in the direction of the gradient (ref. Theorem 2 on page 75 of the course notes). In this case that is in the direction of the vector (—1—02—ﬂ, Log), i.e. in
the direction —i + j. The rate of change in this direction is IIVTII = II
I Jana]
2 2 _ ' 3.A12 From Theorem 3 on page 75, we know that V f (a) is orthogonal to the level
curve f(w,y) = k through a. In this case f(m,y) = 22:1; — yz, SO fm(x,y) 2 2y while
fy(a:,y) = 22: — 2y. The gradient at a = (2,1) is then Vf(2, 1) = (2(1), 2(2) — 2(1)) 2 (2,2).
We conclude that the vector (2, 2) is orthogonal to the curve 29:3; — y2 = 3 at the point (2, 1).
The tangent line at this point consists of all pairs (m,y) so that (a: — 2,1; — 1) is orthogonal
to the vector (2, 2). That is: (SC2,9 1)'(2,2) =0
which is equivalent to 2x—4+2y—2=0, or y=—:r+3. The curve and tangent line are sketched below. The curve is a transformed hyperbole. If you are unfamiliar with such curves, you could
use graphing software to generate the curve. Alternatively, one can rewrite the curve as
—(y — m)2 + $2 = 3. This is a “standard” hyperbola in the mu—plane. To transform to the
:ry—plane, one can observe that the asymptote at a: = 11, becomes an asymptote at y = 211:
While the asymptote at u = —a: becomes an asymptote at y = 0. We will spend some time
discussing these sort of transformations when we reach Chapter 12 in the course notes. 3‘ B1 a) The derivative of the space curve x(t) = (6" cos(t), 26t sin(t)) is
g; : (gie‘cosa), ﬁ2e‘sin(t) = (e‘(cos(t) — sin(t)), 2e‘(sin(t) + cos(t))). At time t = 0 the
curve is at the point x(0) = (60 cos(0), 280 sin(0)) = (1,0). Its tangent vector at time t = O
is (60(COS(0) — 8111(0)), 2e”(sin(0) + cos(0))) = (1, 2). The unit vector in this direction is
u = %(1, 2). To ﬁnd the rate of change of w = 3:2 +y2 in this direction at (1, 2) we calculate
the directional derivative: Du Vw(1,0)‘u (1,2) (233, 2y)l(a:.y)=(1,0) ' L
x/S
1 2
—— 2,0  1, 2 = —.
\/—5( ) ( ) \/5
b) Applying the chain rule, with x = (93,31), we have d_w _ 8_wd_w+a_wa
dt 83: dt 8y dt = 2$((et(cos(t) — sin(t)) + 2y2(et(sin(t) + cos(t)))
= Ze‘cos(t)((et(cos(t) — sin(t)) + 4etsin(t)(e‘(sin(t) + cos(t))). At time t = 0, we have a = 260 cos(0)((eo(cos(0) — sin(0)) + 460 sin(0)(e°(sin(0) + cos(0))) = 2.
t=0 c) Both of these results (% and 2) represent rates of change of m at the point (0, 1) in the direction (1, 2). They are different since the result in (a) is the normalized directional
derivative, while the result in (b) is the rate of change along the curve. Since the tangent
vector along the curve has a length of x/E, motion along the curve progresses faster than a
unity rate, and so the rate of change is accelerated relative to the normalized value. Both
of these calculations provide meaningful results. The directional derivative is normalizaed
to allow a common frame for comparing rates of change at different points or of different
functions. On the other hand, the rate of change along a particular spacecurve may be
relevant in a speciﬁc context. 3. B2. We are told that at a point a, a given function f : R2 —> R has directional
derivatives of 3 and 2 in the directions i + j and i — j respectively. Those two pieces of
information are enough to specify the two unknowns fz(a) and fy(a), that is, to specify the
gradient of f at a. Deﬁning unit vectors in these directions: u : fiﬂ, 1) and v = ﬁﬂ, —1)
we have Duf(a) = (fxfal,fy(a))u
1
= 5(fx(a)+fy(a))=3
while
vafa) = (fx(a)7fy(a))'v $02.02!) — ma» = 2. Adding these equations gives fz(a )= 5—~2‘/—_ and subtracting gives fy(a a:) 325 Thus V f ( )2
(5J9, Q) This is the direction of maximal rate of change in f ata . That maximal rate of change is IIVf(a)l = megs)” = —é—W52+12 = M k ‘/1:2—i~‘142+z2 constant of proportionality .13. Given that the temperature at the point (1, 2, 2) is 120, we ﬁnd k3 through. W: 5 — _‘,120 that IS k7_ — 360. a) To ﬁnd a rate of change of T, we ﬁnd the gradient: Extra 7. The temperature at point (any, 2) is given by T(m,y, z) = for some MIN
NIW VT=360 __i_,——y3,——z _
(w2+y2+z2) (332+y2+22)5 (m2+y2+z2) 5 Evaluating at the point (2,1,3) we ﬁnd VT(1,2, 2) = 4—3?(—1,—2,—2). The unit vector in
the direction from (11 2,2) to (2, 1, 3) is (2—1,1—2,3—2) _ (1,—1,1) 11 =H(2—1,1—2,3—2)H w The directional derivative of T in this direction at (1, 2, 2) is thus DUT(1,2,2) = VT(1,2,2)u 40 40
= — —1,—2,—2  1, —1,1 = ——.
3\/§( ) ( ) 3J3
b) At any point, the direction of greatest increase in temperature is given by the gradient
360
VT 2 ———(—£II, —y, —z). NIB) (252 + .742 + 22) This is a vector pointing from (as, y, 2) toward the origin. ...
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