Math 227 Midterm 1 Solutions
1.
r(t) = et (cos t − sin t) , et (cos t + sin t) , et
= et cos t − sin t, cos t + sin t, 1
v(t) = et cos t − sin t, cos t + sin t, 1 + et − sin t − cos t, − sin t + cos t, 0
= et − 2 sin t, 2 cos t, 1
p
|v(t)| = et (−2 sin t)2 + (2 cos t)2 + 1
√
= 5et
1
T̂(t) = √ − 2 sin t, 2 cos t, 1
5
1
d
√ − 2 cos t, −2 sin t, 0
dt T̂(t) =
5
1
d
√ − 2, 0, 0 = κN̂(0) ds
dt T̂(0) =
dt
5
√ Rπ
√
Rπ
(a) The arc length is s = 0 |v| dt = 5 0 et dt = 5 (eπ − 1)
(b) At time 0, T̂ =
√1 (0, 2, 1),
5
N̂ = (−1, 0, 0), κ =
√2 √1
5 5
√1 (0, −1, 2)
5
(1, 1, 1) + 25 (−1, 0, 0)
= 52 , B̂ = T̂ × N̂ =
(c) The osculating circle has radius ρ = 5/2, centre rc = r(0) + κ1 N̂(0) =
= (−3/2, 1, 1)
and lies a plane spanned by T̂(0) and N̂(0). It is given parametrically by
r(t) = rc + ρ cos t T̂(0) + ρ sin t N̂(0) = −
3
2
−
5
2
√
sin t ı̂ı + 1 + 5 cos t ̂ + 1 +
√
5
2
cos t k̂
2. All of the streamlines of F~ = z ı̂ı + z ̂ − (3x + y)k̂ obey
dx
z
=
dy
z
dz
= − 3x+y
=⇒ dy = dx =⇒ y = x + C
For the field line to pass through (1, 1, 2) we need 1 = 1 + C and hence C = 0. So x = y and
dx
z
dz
= − 3x+y
=⇒
dx
z
dz
= − 4x
=⇒ z dz = −4x dx =⇒
1 2
2z
= −2x2 + C ′ =⇒ z 2 = −4x2 + 2C ′
For the field line to pass through (1, 1, 2) we need 4 = −4 + 2C ′ and hence 2C ′ = 8. So the stream line is
√
y = x, z = 8 − 4x2 .
Warnings. (a) When you use separation of variables to solve, for example,
dy
z
dz
= − 3x+y
, you must first
somehow eliminate all x’s and then move all y’s to the left hand side of the equation, and all z’s to the right
R
2
hand side of the equation, before integrating. Integrating (3x + y) dy = 3xy + y2 + C is wrong, because x
is not a constant. It is some (possibly unknown) function of y (and possibly z). In this example, x = y.
(b) An equation of the form f (x, y, z) = 0 is the equation of a surface in IR3 , not the equation of a curve in
IR3 . To get a curve in IR3 you need two equations in three unknowns, not one equation in three unknowns.
3. (a) The field is conservative only if
∂F1
∂y
=
∂F2
∂x
∂F1
∂z
=
∂F3
∂x
∂F2
∂z
=
∂F3
∂y
see over
That is,
(2x sin(πy) − ez ) =
∂
∂y
∂
∂x
ax2 cos(πy) − 3ez
∂
(2x sin(πy) − ez ) = − ∂x
(x + by) ez
∂
ax2 cos(πy) − 3ez = − ∂y
(x + by) ez
∂
∂z
∂
∂z
⇐⇒
2πx cos(πy) = 2ax cos(πy)
⇐⇒
−ez = −ez
⇐⇒
−3ez = −bez
Hence only a = π, b = 3 works.
(b) When a = π, b = 3
F~ = (2x sin(πy) − ez )ı̂ı + πx2 cos(πy) − 3ez ̂ − (x + 3y) ez k̂
= ∇ x2 sin(πy) − xez − 3yez + C
so φ(x, y, z) = x2 sin(πy) − xez − 3yez + C for any constant C.
~ = F~ + 3yez k̂, with the F~ of #3 evaluated with a = π, b = 3. Hence
4. Observe that G
Z
C
~ · d~r =
G
Z
C
Z
F~ · d~r +
z
C
3ye k̂ · d~r = φ(1, 1, ln 2) − φ(0, 0, 0) +
Z
C
z
3ye k̂ · d~r = −8 +
Z
C
3yez k̂ · d~r
1
To evaluate the remaining integral, parametrize the curve by ~r(t) = tı̂ı+t̂+ln(1+t)k̂. Then ~r′ (t) = ı̂ı+̂+ 1+t
k̂
and 3yez = 3t(1 + t)k̂ so that 3yez k̂ · d~r = 3t dt. Subbing in
Z
C
~ · d~r =
G
Z
C
F~ · d~r = −8 +
Z
0
1
3t dt = −8 +
3
2
= − 13
2
see over