Download Past Exam 2 Practice Problems - Calculus III--Multivariable | MATH 113 and more Exams Advanced Calculus in PDF only on Docsity!
Math 113 – Calculus III Exam 2 Practice Problems Spring 2004
- A train is traveling northwest at 10 miles per hour. A person in the train walks at 2
miles per hour from a window on the left side to a window directly across the train on
the right side. (Left and right refer to the sides relative to a person facing the front of
the train.)
Assume that
i points east, and
j points north. Express your answers in terms of these
unit vectors.
(a) What is the velocity vector of the train?
(b) What is the velocity vector of the person relative to the train?
(c) What is the velocity vector of the person relative to the ground?
(d) What is the speed of the person relative to the ground?
- (a) TRUE or FALSE? For any vectors ~v and w~, (~v + w~) · (~v − w~) = ‖~v‖
2
− ‖ w~‖
2
.
(Briefly explain.)
(b) TRUE or FALSE? For any vectors ~v and w~, ‖~v + w~‖ = ‖~v‖ + ‖ w~‖. (Briefly
explain.)
- Let
~v = 3
i + 2
j +
k, w~ =
i −
j +
k, ~p = a~i −
j +
k, ~q = (3 − a)
i − 4
k.
Find:
(a) the cosine of the angle between ~v and w~;
(b) the values of a for which the vectors ~p and ~q perpendicular;
(c) the values of a for which ~p is a unit vector;
(d) the vector that is the component of ~v parallel to w~.
- Given the plane
x + y + z = 1,
find the point in the plane that is closest to the point P = (3, 3 , 2).
- Simplify the following expression as much as possible:
((~v · ~u)~u) · (~v × w~) − ( w~ × ~v) · (~v − (~u · ~v)~u).
- Let P = (1, 1 , 1), Q = (1, − 3 , 0) and R = (2, 2 , 2).
(a) Find the equation of the plane that contains the points P , Q, and R.
(b) Find the area of the triangle formed by the three points.
(c) Find the distance from the plane found in (a) to the point (3, 4 , 5).
- We say two planes are perpendicular if their normal vectors are perpendicular. Given
the following two planes (which are not perpendicular):
x + 2y + 4z = 1, −x + y − 2 z = 5,
find the equation of a plane that is perpendicular to both of these planes, and that
contains the point (3, 2 , 1).
- Suppose the motion of a particle is given by
x = 4 cos t, y = sin t.
(a) Describe the motion of the particle, and sketch the curve along which the particle
travels.
(b) Find the velocity and acceleration vectors of the particle.
(c) Find the times t and the points on the curve where the speed of the particle is
greatest.
(d) Find the times t and the points on the curve where the magnitude of the acceler-
ation is greatest.
- Find the coordinates of the points where the line
x = t, y = 1 + t, z = 5t
intersects the surface
z = x
2
2
.
(c) ~p is a unit vector if ‖~p‖ = 1. Thus we want
a
2
= 1 =⇒ a
2
= 1 =⇒ a
2
=
=⇒ a = ±
~p is a unit vector if a = −
11 /4 or a =
(d) Let ~u =
w~
‖ w~‖
i −
j +
k). Then
~vpar = (~v · ~u)~u =
i −
j +
k)
i −
j +
k)
- Let C = (x c
, y c
, z c
) be the point in the plane closest to P. We will find the vector
−→
CP ,
from which we can find C.
Let P 0
= (0, 0 , 1); this is a point in the plane. A vector normal to the plane is
~n =
i +
j +
k, and a unit vector in the same direction as ~n is
~u =
~n
‖~n‖
i +
j +
k).
Let ~v =
−→
P
0
P = 3
i + 3
j +
k. Then
−→
CP is the component of ~v parallel to ~u (i.e.
−→
CP is
the projection of ~v on ~u). Thus
−→
CP = ~vpar = (~v · ~u)~u =
i +
j +
k) =
i +
j +
k.
Since
−→
CP is the displacement vector from C to P , we also know
−→
CP = (3 − x c
i + (3 − y c
j + (2 − z c
k.
Thus
3 − x c
=⇒ x c
, 3 − y c
=⇒ y c
, 2 − z c
=⇒ z c
so C =
((~v · ~u)~u) · (~v × w~) − ( w~ × ~v) · (~v − (~u · ~v)~u)
= ((~v · ~u)~u) · (~v × w~) + (~v × w~) · (~v − (~v · ~u)~u)
= [(~v · ~u)~u + ~v − (~v · ~u)~u] · (~v × w~)
= ~v · (~v × w~)
- (a) We have a point in the plane (in fact, we have three). All we need is a normal
vector. This is given by
~n =
P Q ×
P R
i + (− 3 − 1)
j + (0 − 1)
k
×
i + (2 − 1)
j + (2 − 1)
k
j −
k
×
i +
j +
k
i −
j + 4
k.
By using ~n and the point P , we find the equation of the plane to be
−3(x − 1) − (y − 1) + 4(z − 1) = 0 or − 3 x − y + 4z = 0.
(b)
‖~n‖
2
√
9+1+
2
√
26
2
(c) Let A be the point (3, 4 , 5). Let ~u =
~n
‖~n‖
1 √
26
i−
j +
k
. Then d = |
P A·~u| =
1 √
26
i + 3
j + 4
k
i −
j + 4
k
∣ =^
7 √
26
- Normal vectors for the two given planes are ~n 1
i + 2
j + 4
k and ~n 2
i +
j − 2
k,
respectively. Then ~n 3
= ~n 1
× ~n 2
i − 2
j + 3
k is perpendicular to both ~n 1
and ~n 2
and therefore a plane with normal vector ~n 3
will be perpendicular to the given planes.
We are told that (3, 2 , 1) is a point in the plane, so the equation of the plane is
−8(x − 3) − 2(y − 2) + 3(z − 1) = 0.
- (a) Note that (x/4)
2
2
= cos
2
t + sin
2
t = 1, so the path of the particle is the ellipse
(x/4)
2
2 = 1. The motion is counter-clockwise around the ellipse.
y
x
–1.
–0.
0
1
–4 –2 2 4
(b) The path is ~r(t) = (4 cos t)
i + (sin t)
j, so the velocity is
~v(t) = ~r
′
(t) = (−4 sin t)
i + (cos t)
j
and the acceleration is
~a(t) = ~r
′′
(t) = (−4 cos t)
i + (− sin t)
j.
