APPM 2350: Section exam 1, version A
June 15, 2012.
ON THE FRONT OF YOUR BLUEBOOK write: (1) your name, (2) your instructor’s name, and
(3) a grading table. Text books, class notes, and calculators are NOT permitted.
Problem 1: (25 points) Alvin the ant is out exploring one morning when he finds a spilled ice
cream cone located at P (−4, 3). With much delight, he takes some ice cream and heads back to
his hill (located at the origin) taking the path ~r(t) = h4t3 − 4, 3 − 3t2 i. Throughout the morning
and afternoon, Alvin and his ant friends make repeated trips to and from the ice cream cone, each
time making slight corrections to the path. After several trips, the path is essentially a straight
line.
(a) How much shorter is the distance from the ice cream cone to the hill along the new path than
the distance along Alvin’s original path? All distances are in meters.
(b) On one trip, Alvin is at the ice cream cone when he encounters an ant from an enemy colony.
Alvin is in danger and needs to get to his colony’s territory to be safe. The border of his
colony’s territory is formed by the parabola y = 2x2 + x − 1. Alvin plans to run in a straight
line from the ice cream cone to the border. In what direction should he run in order to reach
the border in the least distance?
Problem 2: (25 points) For each of the surfaces described below (a through d), list the equation
and the plot that fit the description. Note that for each description there is exactly one equation
and one plot that will fit.
(a) Elliptic Paraboloid parallel to the x-axis.
Equation:
Plot:
(b) Hyperboloid of one sheet parallel to the y-axis.
Equation:
Plot:
(c) Hyperbolic Paraboloid parallel to the z-axis.
Equation:
Plot:
(d) Ellipsoid centered at (0, 1, 1).
Equation:
Plot:
Equations:
(1) z 2 + 2y 2 + x2 − 2x = 0
(2) z 2 + 2y 2 − x = 0
(3) 2x2 + z 2 − y 2 = 1
(4) z 2 − 2y 2 − x2 = 1
(5) z 2 + 2y 2 + x2 − 2z − 4y = −2
(6) z − 2y 2 + x2 = 0
Plots:
y 1
0
-1
2
HaL
y 1
0
-1
2
HbL
y 1
0
-1
-2
2
-2
2
-2
2
1
z0
1
z0
1
z0
-1
-1
-1
-2
-2
-2
-2
-2
-2
-1
x
0
1
-1
x
2
0
1
-1
x
2
0
2
HcL
1
2
y 1
0
-1
2
HdL
2
y 1
0
-1
HeL
y 1
0
-1
-2
2
-2
2
-2
2
1
z0
1
z0
1
z0
-1
-1
-1
-2
-2
-2
-2
-2
-2
-1
x
0
-1
x
1
2
0
1
Hf L
2
-1
x
0
1
2
2
Problem 3: (25 points) Consider the plane containing the points (0, 1, 1), (− 12 , 0, 0), and (1, 12 , 1).
and the path given by ~r(t) = ht, 2t, (t − 2)2 i.
(a) Calculate the equation for the plane.
(b) Find the point along the path ~r(t) where the tangent vector of ~r(t) is parallel to the plane.
(c) Find the parametric equations for the line parallel to the tangent vector of ~r(t) at the point
found above.
(d) Calculate the distance from the line found above to the plane.
Problem 4: (25 points) Dr. Chim Richalds is a new helicopter pilot and having a little trouble
landing his helicopter. On his way to the helipad, which is located at the origin, he travels along
the path ~r(t) = he−t cos t, e−t sin t, e−t i.
(a) Calculate the tangent, normal, and binormal vectors along Chim’s path.
(b) Calculate Chim’s speed and curvature experienced at time t = π2 .
(c) It turns out that Chim’s path lies on a particular quadratic surface. Find the equation for the
quadratic surface on which Chim’s path lies.
Projections and distances
~a · ~b
proj~a~b = (
)~a = (â · ~b)â
~a · ~a
|P~S × ~v |
d=
|~v |
~n ~
d = P S ·
|~n| Arc length, TNB, curvature, torsion, and tangental and normal components
ds
= |~v |
dt
Z
b
|~v |dt
L=
a
~v
d~r/dt
T~ =
=
|d~r/dt|
|~v |
~
~ = dT /dt
N
|dT~ /dt|
dT~ |dT~ /dt|
|~v × ~a|
|f 00 (x)|
κ = =
=
=
ds
|~v |
|v|3
(1 + (f 0 (x))2 )3/2
~ + aT T~
~a = aN N
aT =
~v · ~a
|v|
τ =−
aN =
|~v × ~a|
|~v |
~ = T~ × N
~
B
~
dB
~
·N
ds