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Unformatted text preview: Your name: TA(circle one): Matt Davis Hongnian Huang Boian Popunkiov
Math 234 Lecture 3(Feldman)
February 24, 2006 Midterm Exam 1 Write your answers in the spaces provided. If you must continue an answer some—
where other than immediately after the problem statement, be sure (a) to tell where
to look for the answer, and (b) to label the answer Wherever it winds up. In any case,
be sure to make clear what is your ﬁnal answer to each problem. . _ , 7r
Wherever appllcable7 leave your answer 1n exact forms (usmg 5, x/S, cos1.2, 62, etc.) ‘ BE SURE TO SHOW YOUR WORK, AND EXPLAIN WHAT YOU DID. YOU MAY RECEIVE REDUCED OR ZERO CREDIT FOR UN—
SUBSTANTIATED ANSWERS You may use some of the following formulas: For curve r = r(t) with velocity v and acceleration a: Unit tangent vector: T = Tv—l
. v
dT ’
Curvature: n = —— = ﬂ
d8 r’ In plane, if m = f (t), y 2 g(t) then curvature is K): In plane, if y = g(a:) then curvature is K = W
[1 + (WP/2’
In space, curvature
_ lvxal __ lr’ xr”
""” “ lvl3 ‘ lr’l3
Principal normal vector
N _ T, _ a — (LTT
— lT’l — aN
Binormal vector
B = T x N iﬂﬁﬂﬂL=JﬂijﬁL
[CE/)2 + (y/)2]3/2 Kit/)2 + (g/)2]3/27 Components of acceleration: a = aTT + aNN, where aT=a T—gjg, aN=a N=I~c<%:)2=nv2,
lﬁ=é+ﬁ
Spherical coordinates:
o Spherical to Cartesian
x = psinqﬁ‘cosﬂ, y = psinqﬁsinﬂ, z = pcosqﬁ, Cartesian to Spherical p =1/zr2 + y2 + 22, tan6 = y/x, 1. (17 points) Find the curvature, the unit tangent vector, the principal normal, and
the binormal of the curve r(t) = ezti —— 6—2tj + «gtk att=0. 3. (18 points) Assume that f (m, y) is differentiable at (1,3). Given fw(1, 3) = —5 and
fy(1, 3) = 1, I (a) ﬁnd the directional derivative of f at (1,3) in the direction toward (3, —4);
(b) in what direction u does the function f decrease most rapidly at (1,3)? (0) Let L be the level curve of f that passes through (1,3). Find parametric
equations of the line tangent to L at (1,3). 4. (17 points) Find the limit; or Show that it does not exist lim :0 + y
(x,y)—>(0,0) a: — y 5. (16 points) Find the slope of the tangent line to the curve of intersection of the
surface 2 = \/4 —— $2 + 63/ and the plane a: = 1 at the point (1,0,2). 6. (16 points) Find the equation of the tangent plane to the surface
902 + me” + y3z = 0 at the point (—1,2, 0). ...
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 Spring '08
 DICKEY
 Multivariable Calculus

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