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Unformatted text preview: Mathematics 234, Fall 2004 Lecture 3 (Wilson)
Final Exam December 22, 2004 ANSWERS Problem 1
For the helical (“corkscrew”) motion F(t) = 2 cos(t)'Z’— 2 sin(t)f+ 375k: %
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(b) Find the acceleration 6(75
( = —2 cos(t)i’+ 2 sin(t)j’+ 0/2. ANSWER: 50:) = 77' t (c) Find the unit tangent vector T(t).
ANSWER: TOE) is the unit vector in the direction of 17(t). The magnitude of 17(75), using part (a), is (17(t)l = 4sin2(t) + 4cos2(t) + 9 = V13. Hence T(t) = “Ll—31705) = —\/i1—3 sin(t)'7— “Ll—3 cos(t)j’
3 _,
+7131; (d) Find the principal unit normal vector JWt). ANSWER: 07(t) = ‘jgéj; and Eli—Z = (fag. We got [7705)] = m in part (c). Also from part (c) we have T(t)7 so we can calculate T”(t) : —\/i1—3 cos(t)?+ “ll—3 sin(t)j+ 01?. Then 2le =
VLF?) (_\/i1_3 cos(t)i’+ “ll—3f) = —12—3 cos(t)77+ % sin(t)j’. Now we can calculate ldT/dsl = 12—3, and NOS) 2 12—3 (—12—3 cos(t)i’+ % sin(t)f) = — cos(t)'Z’—l— sin(t)f+ 01;. You might also see this answer geometrically with almost no calculation7 in several ways: (i) Since
the motion in the I; direction has constant speed, the normal vector will point directly inward
from 7705) toward the zaxis. Hence its I; component will be zero and its '2' and j’ components will
be just the negatives of what appear in 77 except for normalizing to a unit vector. (ii) Since the
“speed” z7 is a constant, m. the acceleration must be entirely in the direction of the normal
vector ﬁ(t). Hence we could take Ei(t) from (b) and multiply by a constant to get unit length.
(iii) You can also realize that (i) or (ii) applies if you notice that (i and 7" are the same except
for the I; component. BUT: A common incorrect procedure involved dividing by a vector. No matter what else you
are doing7 that should raise a waving red ﬂag to say something is wrong! (e) Find the curvature Mt). ANSWER: m) = % so using what we got in (d) we have Mt) = 12—3. One error that you should not have made was to get a vector as an answer: The curvature is a
number7 and furthermore it is never negative. Problem 2 2:53}2
$2+y4 The function f(:1:,y) 2 does not have a limit as (:13,y) —> (0, 0). Show that this is true. ANSWER: Since we are told the limit does not exist, we look for paths such that approaching (0,0)
along different paths produces different apparent limits. There are many possible paths to choose.
Here is one way to do it “all at once”: Suppose we approach (0,0) along the curve 90 = ay2 for
some number a, a parabola if a 74 0 and the y—axis if a = 0. Where 90 = of, but (90,31) 74 (0,0), f(ac,y) = ﬁ%; 2 3121—1. That is a constant as (m,y) —> (0,0) and hence the limit along the path is
that constant. But choosing different values for a gives different values for the limit, e.g. a = 1 gives §=1whilea=2gives%751. Problem 3
Let f(a:, y) 2 ﬂy + cw sing. (a) What is the gradient V f as a function of :1: and y? (9 8
ANSWER: 8—f = 2mg + yew sing and 3—f = 51:2 + $6323; siny + emy cos y.
$ 3/ Thus Vf(ac, y) = (2303/ + yew sin y) 77 + ($2 + mew siny + 6:09 cos y) j’ (b) At the point (1,0), in what direction 11’ is the directional derivative D1; f largest? In what
direction is the directional derivative smallest? ANSWER: At (1,0) the gradient, from (a), is 0551— 2}: 2}. That is not a unit vector, so we
use the unit vector 71' = j’ in the same direction to specify the direction of the greatest directional
derivative at (1, 0). If we use 1? in the opposite direction, 1? = —j, we will get the least directional
derivative at the same point. (c) What is the value of the directional derivative at (1, 0), in the direction making the directional
derivative largest? ANSWER: The value of the directional derivative in the direction of the gradient at a point
is the magnitude of the gradient at that point, Vf (1, 0) = v 02 + 22 = 2. Problem 4
Set up but do not evaluate an iterated integral for the integral of f($, y, z) = 3% — 22 cos(a:y) over the region in space which is inside the cylinder $2 + 22 = 4 and between the planes y = 0 and
a: + y = 3.
ANSWER: The cylinder, since 172 + Z2 = 4 makes no mention of y, is
parallel to the yaxis. It intersects the rz—plane in a circle of
radius 2, centered at the origin. The plane 1E + y = 3 extends
vertically through the line $+y = 3 in the wyplane, and cuts
through the cylinder at 45°. See the picture to the right. You could set this integral up in different orders for the variables. Using dz dy dzv: The largest value
of m, overall, is 2, and the smallest is —2. Hence the outermost integral will go from —2 to 2. For
any :1: between —2 and 2, y goes from 0 to 3 — :1: since the region is bounded by the planes y = 0 and
y = 3 — ac (rewritten version of 90 + y = 3.) (When 90 = —2, y ranges from 0 to 3 — (—2) = 5, along
the back edge. On the front edge, :1: = 2 and y ranges from 0 to 3 — 2 = 1.) At any pair of values for
ac and y, z ranges from the lower half of the cylinder to the upper half, i.e. —\/4 — :32 g 2 g V4 — 252.
Hence we can write the integral as 3 a? 4 \/—a72
/: /0 /\/H 2(3zr’Z — 22 cos(acy)) dz dy dm. Problem 5 2 vii—$2
For // :10 (/x2+ 2d dx:
0 0 y y 3/ Evaluate the integral by converting to polar coordinates and evaluating the resulting polar integral. ANSWER: The region of integration is a quartercircle, the portion of the circle of radius 2 centered
at the origin that is in the ﬁrst quadrant where lb 2 0 and y 2 0. In polar coordinates it can be
described by 0 g r g 2 and 0 g 6 g %. That takes care of the limits on the integrals. The integrand, wyx/ﬂ + y2, can be converted using :1: = rcos 0, y = rsin0, and r = «$2 + y2. We must remember
to replace dy d$ by rdr d0. Doing all of these things, we get /0— /[)2(TC030)(rsjn6)(r)(r)drdﬁ 0r /0/02,.4(cos6)(sin0)drde. We can “pull” the (cos 0)(sin0) outside of the dr integral, and also replace it by %sin(26) using an
identity you were given on the exam, to get 2 .1 2 1 5 1 .1
/21sin(20)/ r4drd62/21sin(20) T— d62/21s1 ‘ n(20)¥d6= E/2 sin26d6
0 2 0 0 2 5 0 0 2S 5 5 0 1 1
2 §[— cos20]§ = —6.
5
Problem 6
If 116,3}, and z satisfy Z3 — my + yz + 3/3 = 2:
32
F d —
(a) in 8:1: ANSWER: It is anot practical to solve the given equation for z. We ﬁnd the derivatives implicitly. Takinga am across the equation and remembering the chain rule, we get 322— oz —y + y3—i — —.0 Grouping on the left the terms that do include 3—2 and on the right those that don’t and factoring gives a720(322 + y)— — y, and dividing we get 3—8; — 3251+? ANSWER: Proceeding in the same way except that we differentiate with respezct to y, we have
(remember the product rule this time as well as the chain rule!) 3223—2 — $ + yg—Z + z + 3y2 =0, 177273312
322+y ‘ so 3—4322 +y)=$—z—3y2 and 3—5: (c) Find 3—: at the point (1,1,1). ANSWER: We just use y = 1 and 2 = 1 (:10 = 1 happens not to matter) in the expression y . . . 1
3z2+y we got 1n (a), g1v1ng Z' Problem 7
Let f($,y,2) = my + y + 2. Let C be the curve ﬁt) 2 2tf+ tf—t (2 — 2t)k for 0 < t <1. Evaluate the line integral / f(x, 3/, 2) d3.
C ANSWER: This is a straightforward line integral. Along the curve, lb 2 2t and y = t and 2 = 2 — 2t,
so f($,y,2) = (2t)(t)+t+(2—2t) = 2752—7512. 55' = 2, y' = 1, and 2' = —2, so (33’)2 + (y’)2 + (2’)2 =
1/4 + 1 + 4 = 3. Hence we can evaluate the line integral as the ordinary integral 1 2 1 1 2 1 13
2t2—t 2 3dt=3[—t3——t2 2t] =3(——— 2):—.
/0( + M) 3 2 + 0 3 2+ 2 Problem 8
One of the following two vector ﬁelds is conservative and the other is not. 151(5):, y, 2) = (2a; — 3)?— 27+ (cos z)/Z
FEW, y, 2) 2 (ex cosy + yzﬁ—l ($2 — em sing/17+ (mg + 2)]? (a) Which vector ﬁeld i_s conservative? Which one is not conservative? ANSWER: To test whether a ﬁeld E = M '7 + N j’ + P]; is conservative we check whether 8M_3N 8M_8P 8N_8P
(9y _ (917’ (92 _ 8177and (92 _ By For the ﬁeld E1 where M = 22 — 3, N = —2, and P = cos 2, these amount to 0 = 0, 0 = 0,
and —1 = 0. These are not all true so F1 is not conservative. From this we can infer F2 is
conservative, but the problem as given on the exam said to test each explicitly. So we repeat the check for E2. For E2: 83—]? and 38—]; both give 2 — em siny, 83—]? and 83—]; both give 3/, and 83—]: and 88—]; both give an. Hence F2 is a conservative ﬁeld. (b) For the vector ﬁeld F that you found to be conservative, evaluate C where C is any path leading from (0,0,0) to (—1, g, 2). Note: As printed on the exam, the integral read / F(:I:,y,2) d5.
C This was corrected during the exam to the verstion given above. This should not have caused
confusion, the printed version really could not mean anything but F ($,y,2)  d? in context.
The change was intended to make the problem match the formulas we had seen. In particular Theorem A on page 742 in the text exactly matches the problem in this form. If you did the
problem without the change you must have interpreted it to mean exactly this anyway! ANSWER: The ﬁeld to use is 132. We do not need to know what the curve is, if we can ﬁnd a potential function f($, y, z) for the ﬁeld. Once we have f we can simply evaluate f(—1, %, 2) —
f(0, 07 0) To ﬁnd f: We know Vf = F}, so g—f = M 2 ex cosy + yz. Thus f must be of the form
617 cosy + myz + g(y, z) where g(y, 2) denotes some part that can vary with y andz but not :16. We also have % = N = $2 — e‘” siny: From what we had so far, %— — —em siny + 5132 + 3—9, so if we let g—g— — 0 we are still OK. That means g(y, 2) might depend on 2 or might be constant, but it does not vary with y. As our last step we need g—zf = P = my + z. From what we have so far,
where f— — 6”” cos y+$yz+(some function g(z ) depending at most on Z)7a ﬂ— — $y+ 2—: , so 2—: must
be 2: We can achieve that if 9(2) 2 %z2. Putting it all together, f(a:, y,z z)— — eit cosy +$yz+ %22
is a potential function for ﬁg. (We could add a constant and get another potential function, but
we only need to ﬁnd a potential function so this is suﬁicient. If we did add a constant it would
occur with both + and — signs in the next calculation and hence it would have no effect on the answer.) Now we evaluate f(—l, %,2) = 2 — 7r and f(0,0,0) = l, and the integral is 2 — 7r — 1 = 1 — 7r. Problem 9
Evaluate j; —y2 dx + my dy
C around the square in the :ryplane with vertices (0,0), (1,0), (1, 1), and (0, 1), in that order.
ANSWER: You could parametrize each of the four legs of the curve C and do this as a line integral.
It is much easier to use Green7s Theorem and convert this to a double integral over the square that C
is the boundary of. Since C is traversed in a counter clockwise direction we don’t even need to worry
about changing a sign. Viewing ji—y2dzr+$ydy as %Md:r+Ndy
C C' where M = —y2 and N = $3], Green’s theorem says
8N 3M
ifde+Ndy=// (———> dA
In this case %_1;/ = y and 8%; 2 —2y, so we have to evaluate [ﬂy — (—2y))dA where S is the square S
0§$§1and0§y§1, 1 1 _ 1 _ 3y2 1—3
/0/0(3y)dmdy—/0 (3y)dy— [710—3 If you do it as a line integral, you really need to have m pieces, not three. The problem says “around
the square77 and squares have four sides! So your path should go back to the starting point. But this
is not an integral of a conservative ﬁeld, so going around a closed path does not imply the result is
zero, and as we saw above it is in fact not zero. Problem 10 Evaluate % F"  T ds where
C ﬁ($,y,z) = yf—l :rzj—l— 3:2]: and C is the boundary of the triangle in the plane :1: + y + z = 1 with
vertices (1,0,0), (0,1,0), and (0,0,1), traversed counterclockwise as viewed from above.
ANSWER: I will use Stokes’ Theorem: It is also possible to do the problem directly, parametrizing
the curve C in three parts and adding the results. Letting S be the triangle whose boundary is C, the
theorem tells us ifﬁTds=//(curlﬁ)~d3
C
S where 77 is a normal vector oriented consistently with the righthand—rule for traversing C in the
prescribed direction. Since C and 8' lie in the plane 1E + y + z = 1, a vector perpendicular to the plane
is 77+ j'+ 1;". Since we go around 0 counterclockwise, the right—hand—rule indicates an upwardpointing
vector, so this vector has the correct orientation. But it has length x/S. So we let 7? = Eta—F ﬁfl— $19. We compute curlﬁ and get —:Jc'Z’— 2xf+ (z — DE. Taking the dot product of that with 775 we get
%(—3m + z — 1). Now we need to compute the surface integral of i \/§
1 Wedothisas //—(—3IE+Z—1)«/f§+f§+1dA
R x/g where fl. and fy come from the expression of the surface in the form 2 = f (r, 3;), which in this case is
z = 1 — ac — 3/. So fm = —1 and fy = —1, and R is the triangle in the anyplane underneath the surface
S. Thus we have the integral //%(—3$+(1—$—y)—1)\/§dA=//(—4IE—y)d14
R R Finally, we set that up as an iterated integral. R is the triangle in the plane with vertices the origin,
(1,0), and (0, 1), which we can describe as 0 S 1E S 1 and 0 S y S 1 — :10. The integral in iterated form is
1 1—:v 1 2 921*” 1 1
/ (—4$ — y) dy d3: 2/ —4a:y — y— (13: 2/ <Z$2 — 3m — —> (13:
0 0 0 2 F0 0 2 2 (—3£E + z — 1) over the triangle. ...
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 Fall '08
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 Multivariable Calculus

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