MATH 227 MIDTERM 1 – SOLUTIONS
Wednesday, February 4, 2009
1. (a) Sketch the flow lines of the vector field
F(x, y, z) =
x2
y
x
i− 2
j.
2
+y
x + y2
The flow lines are circles centered at the origin, oriented clockwise.
(b) Calculate the curl of the vector field in (a).
i
j
k −(x2 + y 2 ) + x(2x) (x2 + y 2 ) − y(2y)
∂y
∂z =
∇ × F = ∂x
−
k
(x2 + y 2 )2
(x2 + y 2 )2
2 y 2 2−x 2 0 x +y
x +y
−2(x2 + y 2 ) + 2x2 + 2y 2
k = 0.
=
(x2 + y 2 )2
2. Prove that if f and g are two C 2 functions on an open set X ⊂ R3 , then ∇ · (∇f × ∇g) = 0.
We first compute
i
j
k
∇f × ∇g = ∂x f ∂y f ∂z f = (fy gz − fz gy )i − (fx gz − fz gx )j+)fx gy − fy gx )k,
∂x g ∂y g ∂z g so that
∇ · (∇f × ∇g) = (fxy gz + fy gxz − fxz gy − fz gxy )
− (fxy gz + fx gyz − fyz gx − fz gxy )
+ (fxz gy + fx gyz − fyz gx − fy gxz )
= 0.
3. The length of a curve x(t), t ∈ [0, ∞), is defined as the limit of the length of x(t) between t = 0 and
t = R as R → ∞. Prove that the curve x(t) = (e−at cos t, e−at sin t), t ∈ [0, ∞), in R2 has finite length for
any a > 0, and find all values of a for which the curve has length 2.
We have x0 (t) = (−ae−at cos t − e−at sin t, −ae−at sin t + e−at cos t) = e−at (−a cos t − sin t, −a sin t + cos t),
so that
1/2
kx0 (t)k = e−at a2 cos2 t − 2a cos t sin t + sin2 t + a2 sin2 t + 2a cos t sin t + cos2 t
√
= e−at a2 + 1,
and the length of the curve is
Z R
Z
√
−at
2
L(a) = lim
e
a + 1dt =
R→∞
0
0
∞
√
√
e−at ∞
−at
2
2
e
a + 1dt = a + 1
=
−a 0
√
a2 + 1
,
a
which is finite. We used here that limx→∞ e−at = 0 if a > 0.
√
√
Suppose that√L(a) = 2, then a2 + 1 = 2a, a2 + 1 = 4a2 , 3a2 = 1, a = 1/ 3. (We discard the negative
root a = −1/ 3 because we assumed that a > 0. Also, when a < 0, the curve has infinite length.)
4. Find the unit tangent and principal normal vectors, and the curvature at a general point on the curve
3
2
x(t) = ti − t2 j + t3 k.
√
We have x0 (t) = (1, −t, t2 ) and s0 (t) = kx0 (t)k = 1 + t2 + t4 . Hence
1
1
1
1
2
T= √
(1, t, t ) = √
,√
,√
,
1 + t2 + t4
1 + t2 + t4
t−2 + 1 + t2
t−4 + t−2 + 1
−4t3 − 2t)
−2t−3 + 2t
4t−5 + 2t−3
,
,
T (t) =
2(1 + t2 + t4 )3/2 2(t−2 + 1 + t2 )3/2 2(t−4 + t−2 + 1)3/2
−2t3 − t
−1 + t4
2t + t3
=
,
,
(1 + t2 + t4 )3/2 (1 + t2 + t4 )3/2 (1 + t2 + t4 )3/2
0
T0 (t)
=
κN = T (s) = 0
s (t)
0
κ = kT0 (s)k =
−1 + t4
2t + t3
−2t3 − t
,
,
(1 + t2 + t4 )2 (1 + t2 + t4 )2 (1 + t2 + t4 )2
1
4t6 + 4t4 + t2 + t8 − 2t4 + 1 + t6 + 4t4 + 4t2
,
1/2
(1 + t2 + t4 )2
1/2
1
t8 + 5t6 + 6t4 + 5t2 + 1
=
2
4
2
(1 + t + t )
1/2
(1 + t2 + t4 )(t4 + 4t2 + 1)
=
(1 + t2 + t4 )2
(t4 + 4t2 + 1)1/2
=
.
(1 + t2 + t4 )3/2
5. Recall the Frenet-Serret formulas:
T0 (s) = κN,
N0 (s) = −κT + τ B,
B0 (s) = −τ N.
Find a vector W such that these formulas can be written as T0 (s) = W × T, N0 (s) = W × N, B0 (s) =
W × B.
Let W = aT + bN + cB, where a, b, c are unknown parameters to be found. Then
W × T = bN × T + cB × T = −bB + cN = κN,
W × N = aT × N + cB × N = aB − cT = −κT + τ B,
W × B = aT × B + cN × B = −aN + bT = −τ N,
so that a = τ , b = 0, c = κ, W = τ T + κB.