972 微甲01-05班期中考解答
1. (8%) Determine whether the series converges absolutely, or converges conditionally, or diverges.
(a)
∞
X
n=1
(−1)n3 · 5 · 7 · · · (2n + 1) n! · 3n . (b)
∞
X
n=1
ln (1 + 1
√n).
Sol:
(a) Let an= (−1)n3 · 5 · 7 · · · (2n + 1)
n! · 3n , applying Ratio Test,
n→∞lim
an+1 an
= lim
n→∞
3 · 5 · 7 · · · (2n + 1) · (2n + 3)
(n + 1)! · 3n+1 · n! · 3n 3 · 5 · 7 · · · (2n + 1)
= lim
n→∞
2n + 3 3 · (n + 1)
= 2
3 < 1 So,
∞
X
n=1
(−1)n3 · 5 · 7 · · · (2n + 1)
n! · 3n is absolutely convergent.
(b) Let an= ln(1 + 1
√n), bn= 1
√n,
n→∞lim an
bn = lim
n→∞
ln(1 + √1n)
√1 n
= 1 (L’Hospital’s rule)
Since P bn is a divergent series (p-series with p = 12), by Limit Comparison Test, P an is also divergent.
Note: The Limit Comparison Test must compare two positive series. We cannot compare P an with P ln(√1n)
2. (4%) Determine the real values of p for which the series
∞
X
n=3
(−1)n 1
npln n(ln ln n)2 is absolutely convergent, conditionally convergent or divergent, respectively.
Sol:
Consider the series
∞
X
n=3
1
npln n(ln ln n)2
p < 0: lim
n→∞
1
npln n(ln ln n)2 = lim
n→∞
n−p
ln n(ln ln n)2 → ∞ 6= 0.
So the series is divergent.
p = 1: Using Integral Test,
Z ∞ 3
1
x ln x(ln ln x)2 dxu=ln ln x= Z ∞
ln ln 3
1
u2 du = −1 u
∞
ln ln 3
< ∞ So, the series is absolutely convergent.
p > 1: Using Comparison Test, comparing with p-series (p > 1)
∞
X
n=3
1
npln n(ln ln n)2 <
∞
X
n=3
1
np < ∞.
So the series is absolutely convergent.
0 ≤ p < 1: Compare with X 1 n, lim
n→∞
1 npln n(ln ln n)2
1 n
= lim
n→∞
n1−p
ln n(ln ln n)2 → ∞, Since X1
n is divergent, X 1
npln n(ln ln n)2 is divergent.
However, X
(−1)n 1
npln n(ln ln n)2 is convergent by Alternating Series Test. Hence, the series is conditionally convergent.
3. (11%)
(a) Write down the Maclaurin series of arctan x.
(b) What is the interval of convergence of the above series?
(c) Find the sum of the series
∞
X
n=0
(−1)n (2n + 1)3n. Sol:
(a) Since d
dxarctan x = 1
1 + x2, arctan x = C +
∞
X
n=0
(−1)n x2n+1
2n + 1. Because arctan 0 = 0, we have C = 0, therefore arctan x =
∞
X
n=0
(−1)n x2n+1 2n + 1.
(b) Since lim
n→∞
n
r
| x2n+1
2n + 1| = x2, this series converges absolutely when −1 < x < 1. For x = ±1, this series also converge (conditionally) by Leibniz theorem. Hence the interval of convergence of this seris is [−1, 1].
(c)
∞
X
n=0
(−1)n
(2n + 1)3n =√ 3
∞
X
n=0
(−1)n (2n + 1)( 1
√3)2n+1 =√
3 arctan 1
√3 =
√3π 6 .
4. (11%) Let r(t) be a motion governed by Newton’s Law F = ma and the Gravitational Law F = −GM m
r3 r where r = |r|.
(a) Show that r × v is a constant vector h. Deduce that the orbit of the particle is a plane curve.
(b) Let u = r
r. Show that a × h = GM u0. Deduce that there is a constant vector c such that v × h = GM u + c. ( Hint: use the formula a × (b × c) = (a · c)b − (a · b)c. )
Sol:
(a) By F = ma = −GM m
r3 r, we obthain
a = −GM r3 r So, a is parallel to r. Thus
d
dt(r × v) = r0× v + r × v0
= v × v + r × a = 0 + 0 = 0
Therefore r × v = h is a constant vector.
(We may assume that h 6= 0; that is, r and v are not parallel.)
This means that the vector r = r(t) is perpendicular to h for all values of t, so the planet always lies in the plane through the origin perpendicular to h. Thus the orbit of the planet is a plane curve.
(b)
h = r × v = r × r0 = ru × (ru)0
= ru × (ru0 + r0u) = r2u × u0 + rr0(u × u)
= r2u × u0
Then
a × h = −GM
r2 u × (r2u × u0) = −GM u × (u × u0)
= −GM [(u · u0)u − (u · u)u0]
But u · u = kuk2 = 1, it follows from Example 4 in section 13.2 that u · u0 = 0. Therefore
a × h = GM u0
and since h is a constant vector,
(v × h)0 = v0× h = a × h = GM u0
Integrating both side of this equation, we get
v × h = GM u + c ,for some constant vector c.
5. (11%) Let r(t) = 2 sin ti + 5tj + 2 cos tk, t ∈ R.
(a) Find the unit tangent vector T(t), unit normal vector N(t) and binormal vector B(t).
(b) Find the curvature κ.
Sol:
(a) Since r0(t) = 2costi + 5j − 2 sin tk. |r0(t)| = q
4(cos t)2+ 4(sin t)2+ 25 =√ 29 We have T(t) = r0(t)
|r0(t)| = 1
√29(2 cos ti + 5j − 2 sin tk).
T0(t) = 1
√29(−2 sin ti − 2 cos tk). |T0(t)| = r 4
29((cos t)2+ (sin t)2) = r 4
29.
We have N(t) = T0(t)
|T0(t)| = − sin ti − cos tk.
B(t) = T(t) × N(t) =
i j k
2 cos t√ 29
√5 29
−2 sin t√ 29
− sin t 0 − cos t
= 1
√29(−5 cos ti + 2j + 5 sin tk).
(b) By the definition we have κ(t) =
dT dS =
T0(t) r0(t)
=
q4
√29
29 = 2 29 6. (11%) Determine whether the function is continuous at (0, 0).
(a) f (x, y) =
x2y
x2+ y2 (x, y) 6= (0, 0) 0 (x, y) = (0, 0).
(b) f (x, y) =
x2y
(x2+ y2)2 (x, y) 6= (0, 0) 0 (x, y) = (0, 0).
Sol:
(a) Since lim
(x,y)→(0,0)
x2y
x2+ y2 = lim
r→0
r3(sin θ(cos θ)2)
r2 . let x = r cos θ, y = r sin θ
= lim
r→0r(sin θ(cos θ)2) = 0 = f (0, 0). (∵ | sin θ(cos θ)2| is bounded by 1. ) By the definition we know that the function is continuous at (0, 0).
(b) Along the path y = x.
We have lim
(x,y)→(0,0)
x2y
(x2+ y2)2 = lim
(x,y)→(0,0)
x3
4x4 = lim
(x,y)→(0,0)
1
4x 6= 0 = f (0, 0).
The function isn’t continuous at (0, 0).
7. (11%) Let f (x, y) =
x2tan−1 y
x − y2tan−1 x
y xy 6= 0
0 x = 0 or y = 0.
(a) Find fx(0, 0) and fx(0, y).
(b) Find ∂2f
∂y∂x(0, 0).
Sol:
(a)
∂f
∂x(0, 0) = lim
h→0
f (h, 0) − f (0, 0)
h = lim
h→0
0 − 0 h = 0.
For y 6= 0
∂f
∂x(0, y) = lim
h→0
f (h, y) − f (0, y)
h = lim
h→0
h2tan−1 yh − y2tan−1 hy − 0 h
= lim
h→0(h tan−1 y
h −y2tan−1 hy
h ) = −y2lim
h→0
tan−1 hy h ...0
0
= − y2lim
h→0
1 y
(hy)2+1
1 = −y 1
(0y)2+ 1 = −y.
(b)
∂2f
∂y∂x(0, 0) = ∂
∂y(∂f
∂x)|(0,0)
= lim
h→0
∂f
∂x(0, h) − ∂f∂x(0, 0) h
= lim
h→0
−h − 0
h = −1.
8. (11%) Find the parametric equation for the tangent line to the curve of intersection of z = x2+y2 and 4x2+ y2+ z2 = 9 at the point (−1, 1, 2).
Sol:
Let f (x, y, z) = x2+ y2 − z, g(x, y, z) = 4x2+ y2+ z2
This problem is to find the tangent line to the curve of intersection of
f (x, y, z) = 0 and g(x, y, z) = 9
At (-1,1,2), the normal vectors of the tangent planes to the surface functions are Of = (2x, 2y, −1) = (−2, 2, −1), Og = (8x, 2y, 2z) = (−8, 2, 4) So the direction of the tangent line is
(−2, 2, −1) × (−8, 2, 4) = (10, 16, 12)
The parametric equation for the tangent line
x = −1 + 10t y = 1 + 16t z = 2 + 12t
9. (11%) Let f (x, y) = x4+ y4− 4xy + 1.
(a) Find and classify all the critical points of f (x, y).
(b) Find the absolute maximal and minimal values of f (x, y) on the disk x2+ y2 ≤ 1.
Sol:
(a) It’s easy to see that ∇f (x, y) = (4x3− 4y, 4y3− 4x). Solve ∇f (x, y) = 0. We get x3 = y, and y3 = x. Thus y9 = y, y = 1, −1, or 0. Using x = y3, we have x = 1, −1, or 0. So the critical points are (1, 1), (−1, −1), and (0, 0). To classify the three points, consider the Hessian matrix
Hf(a, b) =
fxx(a, b) fxy(a, b) fxy(a, b) fyy(a, b)
=
12a2 −4
−4 12b2
Since Hf(1, 1) is positive definite, (1, 1) is a local minimum. By the symmetric property, (−1, −1) is also a local minimum. Now det Hf(0, 0) < 0, therefore (0, 0) is a saddle point.
(b) Since there are no maximum and minimum inside D : x2 + y2 ≤ 1, it must occur on the boundary. Let x = cos θ, y = sin θ, where θ ∈ [0, 2π]. f (cos θ, sin θ) = cos4θ + sin4θ − 4 sin θ cos θ + 1 = 2 − 2 sin 2θ − sin2θ/2. By squaring f , we have
f (cos θ, sin θ) = −1
2(sin 2θ + 2)2+ 4 , where θ ∈ [0, 4π]. Thus f has maximum 7/2 and minimum −1/2.
10. (11%) If the ellipse x2 a2 + y2
b2 = 1 (a, b > 0) is to enclose the circle x2+ y2 = 2y, what values of a and b minimize the area of ellipse?
Sol:
x2 = 2y − y2 ⇒ 2y − y2 a2 +y2
b2 = 1 ⇒ (a2− b2)y2 + 2b2y − a2b2 = 0 y has only one solution ⇒ 4b2+ 4a2(a2− b2) = 0 ⇒ a4− a2b2+ b2 = 0.
Let f (a, b) = πab and g(a, b) = a4− a2b2+ b2 = 0, then Of = (πb, πa) and Og = (4a3− 2ab2, −2a2b + 2b).
Use Lagrange multiplier, we have
πb = 4λa3− 2λab2 πa = −2λa2b + 2λb
a4− a2b2+ b2 = 0 Solve this equation, we get a =
√6
2 , b = 3√ 2
2 , λ = −
√3
3 π, and the minimum of f is f (
√6 2 ,3√
2
2 ) = 3√ 3 2 π.
