Calculus III
Practice Problems 3: Answers
1. Let L
cos θ t I
sin θ t J be a unit vector -valued function of t in the xy-plane. Show that
K
dθ
L
dt
d
L
dt
Answer. L and K are unit vectors, so L K is a unit vector orthogonal to L and K such that L K L
right-handed. Thus L K
L
sin θ I cos θ J. Thus
K
d
L
dt
dθ
I
dt
cos θ
sin θ
dθ
J
dt
dθ
L
dt
K is
2. A particle moves in the plane according to the equation
Xt
t sintI
costJ
Find the speed, tangential and normal accelerations and the curvature of the trajectory at the time t
Answer. Differentiate;
V
2π , V
Evaluating at t
sint sintJ 2I J. Then
A
cost T
I
t cost I
2π I, A
ds
dt
2π
since A is clockwise from T. This gives us A
2T
N
cost
N, so a T
aN
t sint I
costJ
2 and κ
a n ds dt
J
2
2 4π 2 .
3. A particle moves in the plane according to the equation
t t 1 I 2
Xt
t 3J
Find the velocity, speed, acceleration, tangent and normal vectors of the particle at any time t
2t 1 I 3t J A 2I 6tJ Then
2t 1 I 3t J
3t I 2t 1 J
ds 2t 1 9t T N 2t 1 9t dt
2t 1 9t
2 2t 1 6t 3t a
2t 1 9t a 6t 2t 6t1 2t 9t 1 2t6t 1 6t 9t Notice that the choice of sign for N was correct, since a 0,
Answer. Differentiate:
2
V
2
2
2
T
2
4
2
2
4
2
4
2
N
N
4
2
2
4
2
2π .
4
0.
4. A particle moves in the plane according to the equation
Xt
tI
lntJ
Find the velocity, speed, acceleration, tangent and normal vectors, and tangential and normal acceleration of
the particle at any time t 0.
Answer. Differentiate:
V
ds
dt
1 I
1
J
t
1
J
t2
A
t 1 T tI J t
t 1
1
1 a
t t 1
t t aT
2
N
2
I
N
2
2
tJ
1
2
1
t2
t2
1
5. A particle moves in the plane according to the equation
eat costI
Xt
sintJ
Show that the angle between the position vector and the tangent line to the trajectory is constant.
Answer. Let that angle be β . We have
sintJ
VV RR e
We calculate:
cos β
aeat costI
V
eat
sintI
costJ
a
1 e a 1
ae2at
at
a2
at
2
so β is constant. The easiest way to do this calculation is to introduce L
eat L V eat aL L . Then, since L is a unit vector, we have
ae2at
V R
R
eat
V
eat
costI
sint
J, so thatR
a 1
2
6. A particle moves in space according to the formula
Find the tangential and normal accelerations and the curvature at t π 4.
Xt
Answer. Differentiating:
V
At t
sintI
π 4:
V
costJ
1
I
2
costI
2 sin 2t K
1
J
2
cos 2t K
costI
sintJ
2K
A
A
1
I
2
sintJ
1
J
2
4 cos 2t K
We have ds dt
5. Noting that A is orthogonal to V, we conclude that A
aN
A 1, and finally, κ a N ds dt 2 1 5.
a N N, so that aT
0 and
7. A particle moves in space according to the formula
Xt
costI
sintJ
et K
Find the tangential and normal accelerations for this motion.
Answer. Differentiate:
V
Then ds dt
1
sintI
costJ
e2t and
et K
At
costI
sintJ
et K
To find a it is probably easiest to use the formula a
A V V . This calculation gives
1 2e a
1 e
We can see this with less calculation by introducing L costI sintJ, so that
X L e K V L e K A L e K Then V A is easily computed since the system
L L1 2eK is. a right handed system of orhogonal unit vectors. We get V A e L e L K, so V A
aT
e2t
1 e2t
A T
N
N
2t
N
2t
t
t
t
t
t
2
2t
8. A particle moves in space according to the formula
Find aN κ at the point t
Xt
1 2
t I
2
1
J
t
A
tI
V
1, we obtain V
tK
1.
Answer. Differentiate
At t
I
J K A I ds
3
dt We now calculate
aN N
A
aT T
Then, aN is the magnitude of this vector:a N
1
J K
t2
2J. Then
T
I
J K
3
I J
I
I
2J
2
J
t3
aT
K
4I
3
42 3 and κ 42 9 .
9. Let U t V t be unit vectors which are always orthogonal. Let W
functions α β γ such that
dU
dt
αV
βW
dV
dt
αU
γW
dW
dt
1
3
5J
3
U
K
V. Show that there are scalar
βU
γV
Answer. Since U V are orthogonal unit vectors, W is a unit vector orthogonal to both U and V. Now, since
dU dt is orthogonal to U, it lies in the plane of V and W. Similarly, dV dt lies in the plane of U and W, and
dW dt lies in the plane of U and V. This tells us that there are scalar functions α β γ δ µ ν such that
dU
dt
αV
βW
dV
dt
But now, since U and V are orthogonal, U V
δU
γW
dW
dt
µV νV 0 so
0
U dV
dt
or δ α . The other identities, µ β ν γ are derived in the same way.
dU
V
dt
Answer. Let V
10. Let X X t represent the motion of a particle in space.
a) Show that if V is constant, then A is orthogonal to T.
c, so that V
cT. Then
dV
dt
A
c
dT
dt
cκ
ds
N
dt
so, since N is orthogonal to T, so is A. Another way to see this is to note that the hypothesis tells us that
speed is constant, so
d 2s
0
aT
dt 2
so A aN N, and the conclusion follows.
b) Show that if A is constant, the motion lies in a plane.
Answer. By the hypothesis, dV dt
C, a constant vector. Integrating, we get
V
Integrating again:
so that X t
tC
V0
t2
C tV0 X0
2
X0 is always in the plane spanned by C and V 0 .
Xt