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Unformatted text preview: Vector Calculus Second Midterm Solutions 1 Find the semimajor and semiminor axes of the ellipse 2m2 + 4333/ + 5y2 = 30. Solution: To keep this short I will cut to the chase: the eigenvalue equation is Z—A 2
det[ 2 5__)\]—=O, this works out to be A2 — 7 A + (10  4 = 6) = 0. The quadratic formula tells us
that the eigenvalues are A__=§[7i¢(72—46)]=%[7i5]=1ore. [2r 531l=léil kills the (normalized) vector v+ = 715(—2, 1), and the matrix [236 SEEH'; 31] kills the (normalized) vector v : \—}g(1, 2). It follows that The matrix 62(90, 1!) = /\+(V+ X)2 + A—(V x)2 = %[(—2x + y)2 + 6(93 + 21W] ; I’ve used the usual sign convention for eigenvalues here7 which is opposite of
that used in the notes I posted. It is straightforward to check that Q(vi) 2 Ai, so Q(—2, 1) = 5 and Q(1, 2) =
30. It follows that (1,2) lies on the ellipse in question, as does (/6 (—2, 1);
these (perpendicular) vectors are its semiminor and semimajor axes. 2 Calculate the integral of g(m, y, z) = xyz over the tetrahedron deﬁned by the
intersection of the plane a: +3; + z = 1 with the ﬁrst octant x Z 07 y 2 0, z 2 0. Solution: We want to calculate ///a:yzd:cdydz,
T Where T is the tetrahedron deﬁned above. I propose to integrate ﬁrst with
respect to 2; this reduces the problem to the calculation of the double integral // éxyz2 dm dy ,
B where now B is the base of the tetrahedron, which is the region in the ct, y plane
bounded by the first quadrant (cc 2 0, y 2 O) and the line 1' + y = 1. Now
2 = 1 — a: — y, so we can rewrite the integral as m=1 y=1—a:
%/ / xy(1—x—y)2dydx.
$20 y=0
The interior integral (with respect to y) can be written 1
/[(1 — $)2xy  2(1 — $)$y2 + 962/3] dy = $0 — 90293212 — $941  90.743 + 1131/4 ; substituting in the limits y = 0 and y = 1 — :r of integration yields 1
—;~$(1 —— (8)4 — §x(1 — x)4 + ix(1 — 2:)4 = E541 — as)4 . To get the ﬁnal answer there doesn’t seem to be any simpler way than just
expanding out this mess and integrating term by term to get 1
5121 / [m—4$2+6m3—4x4+x5] dx,
0
which gives This adds up to 1 2 1
30— 0 90—4 10=—=—.
2460[ 8 + 8+ ] 2460 720 3 Find the volume of the solid enclosed between the upper surface 2 = 8—2? —y2
and the lower surface 2 = m2 + 3y2. Solution: The two surfaces intersect along the curve
$2+3y2=z=8—m2—y2,
which lies above the ellipse x2 + 23/2 g 4 in the 2:, y—plane. This is very similar to the problem in the practice test. We can write the volume
as the iterated integral 1 2 (2—§mz)l/2 8—.r2—y2
/ f / dz dy d2: .
—2 —(2—%z2)1/2 :r3+3y2 2 (2—%z2)1/z '
/ / 1 (8 — 21:2 — 4y2)dy dm . 2 _(2_§m2)1/2 This reduces ﬁrst to 2 —1m2)3/2l dm
.. _8( ._ 2 [2 [2(8 — 2m2)(2 ~32—w2)1/2
—2 which simpliﬁes to 2 1
ill/2— ] (4——z2)3/2 (maM ] (1—39?” dm.
3 ,2 3 1 to conclude that the integral equals 877\/2.
front page of the exam contained a typo, f the formula abov .
to sort out it for the problem mula on the
' instead 0 al difﬁculty] Without any re
itical points of 4 Find and classify the or my) = (as? ~ 1)2 + <wa — x — 1)?
st We need to calculate the gradient, and ﬁnd Where it vanishes. Solution: Fir We have af
5; =2(m2 ~1)2x+2(a:
and 6f
——— :2(:I:2y—m#1) 362.
3y
rI‘he second expressron equals zero only if either cc = r 9521; — :15 ~ 1 .— 0', but
if a: is zero then
ﬂf0+2(0—1)(0—1)—
8m '—
h :1: = 0 can nev
If mgy — m ~ 1 == 0, on the other hand, then
5% =4m(m2 ~1)+0
can vanish only if either a: z: 0 (which We have already seen leads to a contra—
diction), or if x2 = 1, which happens if :1: = :11. if at := 1 then
:cZy—x——1=y——2=0
soy=2,andifx:~1thenac2y—os~1=y+1—1:0soy:0. 3 There are thus two critical points, at (1, 2) (Case I) and at (—1,0) (Case II).
[Lots of people wrote them down, but without any justiﬁcation; I assume they
found them with the aid of a calculator. We didn’t give any credit for this, or
for accurately classifying them, using the second derivative test as below (unless
there was some explanation as to where they came from). We did give credit
for calculating the second derivatives correctly, even though I suspect that was
done in many cases by the use of the same calculator. But a surprising number
of people seemed to have trouble copying their computer’s output onto the exam page!] Next we need to calculate the matrix of second derivatives. Note that both ﬁrst
derivatives are multiplied by two, which clutters things up; so I’m going to omit
those factors in the calculations below. We start with 2
gx—J;22(552——1)+2x2x—l—(2wy—1)2+(a:2y—x—1)2y; this equals 0+22+(2 1 2—1)2+0 = 13 in case I, and 0+(—2)2+ (—1)2+0 = 5
in case II.
For the mixed derivatives we get (again omitting superﬂuous two’s) 82 f _
away _ 11320321; — a: — 1) + 932(2xy —~ 1) which equals 0 + 1  (2  1  2 — 1): 3 in case I, and 0 + (—1)2(0 — 1) = ——1 in
case II. Finally we have
if
8312 again omitting a factor of two. This then equals 2 in both cases. So the two Hessian matrices are
1 3 3 & 5 — 1
3 2 — 1 2 which both have positive terms in the upper left corner, as well as positive
determinants, 17 in the ﬁrst case, 9 in the second (or, if you put back the
omitted factors of two, four times that. Both critical points, remarkably enough,
are therefore local minima. [This can’t happen for a function of one variable! In
that case we would have a local maximum between them.] A few astute people
noticed that the function f is a sum of squares, which is thus bounded below
by zero, and thus must have at least some minima.] = 3:2  29:2 = 2m4 ...
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This note was uploaded on 05/23/2009 for the course MATH 110.202 taught by Professor Staff during the Fall '04 term at Johns Hopkins.
 Fall '04
 Staff
 Calculus

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