Math 241 Midterm 1 (September 25, 2014)
DO NOT OPEN!
Time: 50 minutes
Name:
NetID:
Instructions:
1. Write your name clearly and legibly in the space above.
2. Do not open until told to do so.
3. Show your work. Answers have to be justified.
4. Check that your copy contains 8 problems, some scratch paper, and a formula sheet. Extra scratch
paper is available upon request.
5. If the computations are too complicated to handle, proceed as far as you can, and leave the answer in
such a form that someone with a calculator can get the numerical result. 856 −23 cos 9 is OK; a b −2k(k−1)
is not OK, since there are some a’s, b’s and k’s.
6. Books, notes, or calculators cannot be used. Only pens, pencils, and erasers are allowed.
7. You can rely on anything result which was mentioned in class, or can be found in your textbook.
8. If an answer box is provided, write your answer there. Otherwise, circle your answer.
Good luck!
Problem Max. grade Your grade
1
6 pts.
2
3 pts.
3
4 pts.
4
4 pts.
5
6 pts.
6
6 pts.
7
5 pts.
8
6 pts.
Total:
40 pts.
1 (6 points): For each function f (x, y), label one graph that most closely matches the formula. You do not have
to justify your answer.
(a) f (x, y) = x 2 − y 2
(b) f (x, y) = e x cos(y)
(c) f (x, y) = (−x)e −(x
2
+y 2 )
2 (3 points): Circle the equation for the quadric surface shown at right. You do not have to justify your answer.
z
1. x 2 + y 2 + z 2 = 1
2. x 2 − y 2 + z 2 = 1
3. y 2 + z 2 − x 2 = 1
4. x 2 − y 2 − z 2 = 1
5. x − y 2 − z 2 = 1
x
y
3 (4 points): For each of the following statements, determine whether it is true or false, and circle T or F accordingly. You do not have to justify your answer.
• If a spacecurve’s torsion is 0 everywhere, then this curve must be planar (it must lie in a plane).
T/F
• For vectors ~
u ,~
v ∈ R2000 , we always have |~
u ·~
v | ≤ |~
u ||~
v|
T/F
~ ∈ R3 , we always have (~
~ =~
~)
• For vectors ~
u ,~
v, w
u ×~
v) · w
u · (~
v ×w
T/F
• If the vectors ~
u ,~
v ∈ R3 satisfy ~
u ·~
v = 0 and ~
u ×~
v =~0, then either ~
u =~0 or ~
v =~0
T/F
4 (4 points): Find the limit (or show it does not exist). Justify your answer!
3x y + xz 2 − y 4 z 2
lim
=
(x,y,z)→(0,0,0)
x2 + y 2 + z2
5 (6 points): Find the linearization of the function f (x, y) = x 4 e 2y at (1, 0).
Put your answer here:
6 (6 points): Determine whether the lines
L 1 : x = 3 + t , y = t − 1, z = 3t + 3
and
L 2 : x = 2t , y = 2 − 2t , z = 4t − 3
are parallel, intersecting, or skew. If the lines intersect, find the point of intersection.
Parallel / Intersecting / Skew (circle one).
Point of intersection (if it exists):
³
,
,
´
7 (5 points): We are given the points P (2, 1, 0), Q(1, 1, −1), and R(3, 3, 0).
(a) The area of the triangle PQR equals
(b) Find an equation of the plane containing P , Q, and R.
8 (7 points): The curve C is the intersection of the cylinder x 2 +y 2 = 1 and the plane z = x+y. Find the curvature
of C at the point (−1, 0, −1).
Put your answer here:
Scratch paper
Scratch paper – may be detached
Formula sheet – may be detached
2
2
Trigonometric identities: sin2 x + cos2 x = 1, tan
Z x + 1 = sec x, sin 2x = 2 sin x cos x,
1
1
cos2 x = (1 + cos 2x), sin2 x = (1 − cos 2x),
sec θ d θ = ln | sec θ + tan θ| +C .
2
2
Exponential and logarithmic functions: ln(ab) = ln a + ln b, ln(a/b) = ln a − ln b, ln(a b ) = b ln a,
¡ ¢b
e a e b , e a−b = e a /e b , e ab = e a .
e a+b =
The TNB frame of a spacecurve, and the curvature: Throughout, we consider a spacecurve ~
r =~
r (t ).
~ =~
• The unit tangent vector: T
r 0 /|~
r 0 |.
~ =T
~ 0 /|T
~ 0 |.
• The unit normal vector: N
~=T
~ ×N
~.
• The binormal vector: B
¯
¯
~ /d s ¯, where s is the arclength function. Also, κ = |T
~ 0 |/|~
• The curvature: κ = ¯d T
r 0 | = |~
r 0 ×~
r 00 |/|~
r 0 |3 .
• The torsion:
~
dB
~.
= −τN
ds
• Connections between the derivatives of frame vectors:
~
~
~
dT
dN
dB
~,
~ + τB
~.
~.
= κN
= −κT
= −τN
ds
ds
ds
Tangential and normal acceleration: If the motion of a particle is given by ~
r (t ), then for the acceleration ~
a
0
2
0
~
~
we have ~
a = a T T + a N N , with a T = v and a N = κv (here v = |~
r |).