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Unformatted text preview: Mathematics 234, Fall 2004 Lecture 3 (Wilson)
First Midterm Exam October 4, 2004 ANSWERS Problem 1
In class we saw pictures of the helix (corkscrew) given by the position vector m) : sin(t)?+ tj+ cos(t)lZ. Calculate the arclength of one turn of this helix, from t = 0 to t : 271 ANSWER: The velocity vector 17(t) is cos(t)T+f—sin(t)E, so (17‘ is (/cos2(t) + 1 + sin2(t) : V? at any value of t. Thus the arclength is 271'
ﬁdt : 2\/§7r.
0 Problem 2
Let F(t) : 3sin(t)?+ 3cos(t)f+ 4tk describe the motion of an object along a curve in
space. Find as functions of t: (a) The velocity 17(t)
The acceleration 5(2?) The unit tangent vector )
)
d) The principal unit normal vector 1W1?)
) The curvature H(t) ) The tangential (scalar) component of acceleration (1T (g) The normal (scalar) component of acceleration aN
ANSWERS: Differentiating,
27(t) : 3 cos(t)?— 3 sin(t)f+ 4i?
and
(7(1?) 2 —3 sin(t)7— 3 cos(t)f+ 01?.
To find T, we need a vector of unit length in the direction of 17(25). The magnitude of 17(25) is
27(t) = (/9 cos2(t) + 9sin2(t) + 16 = m = 5. Thus “(t)_1_, _3 H 3. #4
_ — Eva) — ECOS(t)Z — 6811105)]... 6 IE. There are several ways to proceed to JV. One way is to go ahead and ﬁnd the components of
acceleration, CLT and cm, and since we need them anyway I will do that. For the tangential
component we have am) : cm) for) : —3sin(t) >< geese) + (—3cos(t)) >< (—gsin(t)) + 0 x g = 0. Since (7(2?) 2 aTT(t) + aNZWt) : (MIN/(t), where = 1, we have is a unit vector in
the direction of (7(1?) and (ii) cm is the magnitude of (ﬁt). Hence cm 2 9sin2(t) + 9sin2(t) 2
\/§ : 3, and = écﬂt) : —sin(t)i’— cos(t)f+ 01;.
This leaves only Mt) to be determined. Again there are several ways to ﬁnd it: Given what we
have already computed, it may be easiest to compute GN 3
H“) : W2 Z % which happens not to depend on 15. Problem 3
For each of the four descriptions, ﬁll in A or B or C or D to indicate which graph below it corresponds to. (i) 2; : y sin($)
ANSWER: Notice that ﬁgure C has a sine—wave cross—section in one direction, like .2 : sin(x), while in the other horizontal direction it has cross sections that are straight
lines, like 2 =(some constant)y. C is the answer here. (ii) 7" = 1 + cos(6), any value of z ANSWER: From the variables 7“, 6, and z, we can infer this is in cylindrical coordinates.
Since .2 does not appear in the equation, if any point is on the surface so is the entire
vertical line through that point. That alone is enough to pick out ﬁgure D. When you
also notice that a horizontal cross—section will be 7" = 1 + cos6 in polar coordinates, a
cardioid, you have conﬁrming evidence. (iii) z:3—$2—y ANSWER: This surface slopes downward as y increases, and upward in the —y direction,
and follows the downward opening parabola z = 3 — x2 as you move in the ::x directions
along the :r—axis, and surface B certainly ﬁts this description. (iv) p : sin(q§), any value of 6 ANSWER: This must be in spherical coordinates, and since 6 does not appear in the
equation it will be completely symmetric around the z axis. That picks out surface A.
In addition, for any ﬁxed 6, the cross—section will be p = sin gb: 7“ : sinH gave a circle in
polar coordinates, so looking out from the z—aXis in any direction we should see a circle:
This conﬁrms the ” donut” shape in ﬁgure A. x
is . xﬁﬁgﬁﬁgﬂuﬁ .\ . l m
N ¢ \ xx . 2
a. ___.______________________ m W, EEEESEE m x, .l e§§w§§§x a fe §§§§3. .‘AVVSR‘KS KW / ﬁr. .
an) a); 0 Ham 2 1 8 4% 3“ + 2y). 2 ll} ( = 2:1: cos Problem 4 ﬁnd: 2 sin(x2 + 2y 1193731) For ANSWER: Remembering the Chain rule7 % : cos(x2 + 2y) >< 2x + 2y) X 2 : 2(:0s(x2 + 2y). 2 : cos(:1: ﬂ
3y ANSWER: Similarly7 {if
8:162 ) C ( ANSWER: We had % = 2x cos(ar2 + 2y): we differentiate that with respect to 36, using the product and chain rules, we get % : 2 cos(x2 + 2y) — 4:152 sin(x2 + 2y).
32 f
(d)
agar ANSWER: Since this says to take the derivative first with respect to ac and then with respect
to y, we take the derivative of % = 236 cos(ac2 + 23/) with respect to 3/. Using the chain rule we get 883/28]; : —4x sin(x2 + 2y).
82
(e) f
amay ANSWER: We can cite the theorem that says if the partials are all continuous then the
mixed second partials will be equal, so the answer should be the same as for Or we
can directly compute the derivative of 2cos(ar2 + 2y) with respect to 1:. Either way, we get a2 
aw]; : —4:E s1n(a:2 —I— 2y). 2
(f) ANSWER: Differentiating % : 2 cos(:v2+2y) with resepect to y we get 227’; = —4 sin(:v2+2y).
Problem 5
(a) Evaluate lim $2 cos(2y). ($7y)—>(2,7T)
ANSWER: This function is constructed by multiplying two functions (362 and cos(2y))
which are each continuous at all input values, so it is continuous at all points (1:, 3/). Hence
we can evaluate the limit by “plugging in77 and ( lian ) $2 cos(2y) : 22 cos(27r) : 4.
J},y H 771' (b) Show that lim L does not exist. ($7y)—>(0,0) V $2 + y2 ANSWER: If we approach (0, 0) along the y—axis, where 36 = 0, the function is constantly
0 for all values of y 7E 0. Hence the limit is 0 as we approach the origin along the y—axis. If we instead approach along the x—axis, where y : 0, the function values are For 2
:1: > 0 this evaluates to 1, so the limit coming in along the postive :r—axis is 1, vxfhich is
not the same as the 0 we got along the y—aXis, and so the limit must not exist. In fact
the limit coming in along the negative m—aXis is actually —1, since for :v < 0 we have
V? : : —x, so we even get different limits coming to zero along the $—aXlS from opposite sides, and so we did not actually need the y—aXis result. Problem 6
Let f(a:, y) : $2 69.
Let P be the point (1, 0). (a) Find the gradient Vf at the point P.
ANSWER: The gradient at any point (36,31) is Vf(yc, y) = 236 eyﬂ— 362 69f, so when ac : 1
and y = 0 we have Vf(1,0) : 27+f (b) Find the directional derivative of f at P in the direction of an arbitrary vector 17 : via—112 f. ANSWER: First ﬁnd a unit vector in the direction of 17: Let 73 ﬁr? = 1 A v17+
1 mi. Now the directional derivative Dgf(1,0) : Vf(1, 0)  U m Um; W ‘
(c) Find the directional derivative of f at P in the direction from (1, 0) to (47 4).
ANSWER: In this case the vector 17 2 374—41“, where v1 : 3 and v2 : 4, is in the correct direction. Then MD? + 21% 2 V9 + 16 : 5 and the answer to (b) becomes “2+4 : 2. (d) In what direction is the directional derivative of f at P largest? What is the directional
derivative in that direction? ANSWER: The direction of the gradient, 2T+ f, gives the direction in which the direc—
tional derivative is greatest. In the form of a unit vector this is %?+ if. As we have
seen, the directional derivative in that direction will be the magnitude of the gradient, x/E. Problem 7 Suppose w : sin(:1:y) +3: Sin(y), where :1: : u2+v2 and y : 2u—l—v — 2.
Using the Chain rule: (a) 2—: = (y 008(xy) + Sin(y)) >< 2U + (:16 COS(:Uy) + x cos(y)) X 2. (b) 66—1: = (y 008(xy) + Sin(y)) X 22} + (x cos(xy) + :U COS(y)) X 1. Problem 8 EDGE}
983
The north—south direction is measured as 869
y and the east—west direction as x. The 500 we Suppose you are walking over some hills. altitude in feet is given by , my a:
, = 1000 — 100 — — 
ﬁx 9) 8m 3000 100 If you are at the point where a: : 100 and y : —100, and you start
walking toward the center where :1: : O and y : 0, will you begin by
walking uphill, downhill, or horizontally? Be sure to show how you
determine your answer. ANSWER: We need to ﬁnd the directional derivative for the given
altitude function f (:10, y) in the direction of a vector from (100, —100)
toward (0,0). A unit vector in that direction i —%2_’+ if The partial derivatives of f are % : —§/—0 cos (ﬂ) — {iv—0 and % : 6
(a) — At (100—100) we get = — 2a and 2—5 = —13—0 cos(—%) + 1. Thus the directional derivative we seek is —%(% cos(—%) — 2) + %(—13—0cos(—%) —— 1) which simpliﬁes to
1 20 _ﬁ (3 cos(—13—0) — 3). With a calculator you can approximate that as +6.748975979, but all we need to know is whether it is positive or negative, and we can tell that without a calculator: m : 3% is just a 3
bit larger than 7r, so the point on the unit circle determined by —%
is just above the x—axis and nearly at :10 = —1. Thus cos(—E) is 3
20 10 negative (nearly —1) so both the terms in g cos(—§) — 3 are negative, so their combination must be < O, and multiplying by —% will give us a positive number.
Hence the directional derivative is positive and our path leads uphill. ...
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This note was uploaded on 05/15/2008 for the course MATH 234 taught by Professor Dickey during the Fall '08 term at Wisconsin.
 Fall '08
 DICKEY
 Multivariable Calculus

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