Calculus III
Exam 1, Summer 2003, Answers
1. V = 3I J; W = 2I + 5J are two vectors in the plane.
a) Find the angle between V and W.
Answer. Let α be the angle. Then
cos α
so α
= 1:512 radians, or
=
VW
3(2) 5
1
jVjjWj = p10p29 = p290 =
0587 ;
:
86.63 Æ .
b) Find the vector which is orthogonal to V and counterclockwise from W.
Answer. Since W is counterclockwise from V, and the angle α is less than a right angle, the vector we seek
is V? = I + 3J.
c) Find the area of the parallelogram spanned by V and W.
Answer. The area is V? W = 2 + 15 = 17.
2. A particle moves in the plane according to the equation
1
X(t ) = lntI + J
t
Find the velocity, speed, acceleration, tangent and normal vectors, and normal acceleration of the particle at
any time t.
Answer. Differentiate X(t ):
1
V= I
t
1
1
J = 2 (tI J) :
2
t
t
1
2
A = 2I+ 3J :
t
t
ds
1
tI
J
I + tJ
=
1 + t 2; T = p
;
N= p
2
dt
t2
1 +t
1 +t2
1
1
2
1
aN = A N = p
(
+ )= p
:
2
2
2
2
t
t
1 +t
t 1 +t2
p
:
3. Find the equation of the plane through the point (0,-1,3) which is parallel to the vectors I
3I 2J + K.
Answer. The normal to the plane is the cross product of the two given vectors. This is
0I
N = det @ 1
3
Taking X0 = (0)I
J + 3K, the equation is (X
J K
2 2
2 1
1
A = 2I + 5J + 4K
:
X 0 ) N = 0, which comes to 2x + 5y + 4z = 7.
2J + 2K and
4. Find the distance of the point (2; 0; 1) from the line whose symmetric equations are
2
x
3
=
y+1
4
=
z
1
2
Answer. Let Q be the given point. P = 2I J + K is a point on the line, and L = 3I + 4J
the direction of the line. Then Q P = J, and the desired distance is
j
J Lj
jLj
=
r 13
29
:
5. A particle moves in space according to the formula
X(t ) = et I + e2t J
tK :
Find the normal acceleration at the point t = 0.
Answer. Differentiate;
V(t ) = et I + 2e2t J
K;
A(t ) = et I + 4e2t J :
Now, evaluate at t = 0 (don’t work with the general formulas!) to find
Then jVj =
p
V = I + 2J
K; A = I + 4J :
6 and since T = V=jVj we have
AT =
so aN
=
jA
(A
T)Tj =
p7 2
=
A
:
(A
T)T =
p9
6
;
1
3
I+J+ K ;
2
2
2K is a vector in