Download FINAL REVIEW: KEY 1. Given the parametric equations x = 3 cos

FINAL REVIEW: KEY
1.
Given the parametric equations

√

x = 3 cos t


√
y = 3 sin t


 z = 4√t
of a curve C, where 0 6 t 6 π 2 , find the velocity and acceleration vectors as
they depend on t. Find also the speed and the length of curve C. Describe the
motion and its trajectory in words.
Key:
√
√ 2
dx dy dz
3
3
~v =
, ,
= − √ sin t, √ cos t, √ ;
dt dt dt
2 t
2 t
t
dvx dvy dvz
,
,
~a =
dt dt dt
√
√
√
√
3
3
3
3
1
√ sin t − cos t, − √ cos t − sin t, − √ ;
=
4t
4t
4t t
4t t
t t
5
k~v k = √ ;
2 t
Z π2
Z π2
5
√ dt = 5π.
length =
k~v k dt =
2
t
0
0
The motion starts at time t = 0 at point (3, 0, 0) and proceeds along the helix
5
trajectory with decreasing speed k~v k = √ , to arrive at point (−3, 0, 4π) at
2 t
time t = π 2 .
2. Given the force field F~ = ~i + 2~j − t ~k, find the work over the curve C in
the previous problem.
Key:
Z
W =
Z
−3
1 dx + 2 dy − t dz =
Z
Z
2 dy −
dx +
3
0
0
Z
0
π2
= −6 + 0 −
0
π2
√
t d(4 t)
4
4
t √ dt = −6 − π 3 .
3
2 t
3. For each of the following force fields that are path-independent, find the
corresponding potential function:
(a) F~ = −y ~i + x ~j; (b) F~ = (x3 − 3xy 2 , y 3 − 3x2 y).
Key:
(a) F~ = P ~i + Q ~j, where P = −y and Q = x. One has
∂P
= −1,
∂y
∂Q
∂P
= 1 6=
,
∂x
∂y
1
so that field F~ is not path-independent.
(b) F~ = P ~i + Q ~j, where P = x3 − 3xy 2 and Q = y 3 − 3x2 y. One has
∂P
= −6xy,
∂y
∂Q
∂P
= −6xy =
,
∂x
∂y
so that field F~ is path-independent.
If F~ = grad f , then fx = x3 − 3xy 2 and fy = y 3 − 3x2 y, so that the potential
function is
Z
Z
3x2 y 2
x4
−
+ g(y);
f (x, y) = fx dx = (x3 − 3xy 2 ) dx =
4
2
fy = −3x2 y + g 0 (y) = y 3 − 3x2 y;
g 0 (y) = y 3 ;
Z
y4
g(y) = y 3 dy =
+ C;
4
3x2 y 2
y4
x4 − 6x2 y 2 + y 4
x4
−
+
+C =
+ C.
f (x, y) =
4
2
4
4
4. For each of the force fields in Problem 4 that are not path-independent,
find the work done over the circle of radius 2 centered at point (0, 1). Do this
in two ways: (i) using Green’s theorem and (ii) without using it.
Key:
(i)
Z Z
∂F2
∂F1 −
dA = (1 − (−1)) dA = 2 Area(R) = 2 × π22 = 8π.
∂x
∂y
R
R
(ii) Parametric eqs. are x = 2 cos t and y = 1 + 2 sin t, 0 6 t 6 2π.
Z
F~ · d~r =
Z
Z
C
C
2π
−y dx + x dy =
−(1 + 2 sin t)(−2 sin t) dt + 2 cos t (2 cos t) dt
0
Z
=
2π
(2 sin t + 4) dt = 8π.
0
5. Find the following fluxes:
(a) of the field F~ = xy 2 z 2 ~i + x ~j − y ~k through the square of side 2 centered
on the x-axis in plane x = 3 and oriented in the negative x-direction;
(b) of the field F~ = −2~r through the sphere of radius 5 centered at the origin
and oriented outward.
Key:
(a) ~n = −~i;
Z
S
(F~ · ~n) dA =
Z
(−xy 2 z 2 ) dA = −3
Z
1
−1
S
2
Z
1
4
y 2 z 2 dy dz = − .
3
−1
(b) ~n = ~r/k~rk = 15 ~r;
Z
Z
(F~ · ~n) dA = (−2~r · ( 15 ~r)) dA
S
S
Z
2
=−
~r · ~r dA
5 S
Z
2
2
2
dA = − × 52 Area(S) = − × 52 4π × 52 = −1000π.
= − × 52
5
5
5
S
3