972 微甲08-13班期中考解答
1. (12%) Determine whether the series is convergent or divergent.
(a)
∞
X
n=1
(−1)n
en1 − 1 .
(b)
∞
X
n=2
1 n(ln n)3. Sol:
(a)
∞
X
n=1
(−1)n(e1n − 1)
∵ e1n − 1 → 0 as n → ∞ and e1x − 1 is decreasing in x, ( since f (x) = ex1 − 1, f0(x) = − 1
x2ex1 < 0 for x ≥ 1) By alternating series test, the series converges.
(b)
∞
X
n=2
1 n(lnn)3 Let g(x) = 1
x(lnx)3. Then g(x) ≥ 0, is decreasing to 0, and continuous for x ≥ 2 (since the function x(lnx)3 is increasing.)
g(n) = 1
n(lnn)3 and the improper integral Z ∞
2
dx
x(lnx)3 = −1
2(lnx)−2
∞ 2
= 1
2(ln2)2 converges.
By integral test, the series converges.
2. (12%)
(a) Evaluate lim
x→0
1 − cos x x2 . (b) Evaluate lim
n→∞
1 − cos (1n) 1 − cos (n+11 ).
(c) Find the interval of convergence of the power series
∞
X
n=1
1 − cos (1 n)
xn. Sol:
(a) By L’Hopital’s rule,
x→0lim
1 − cos x
x2 = lim
x→0
sin x 2x = lim
x→0
cos x 2 = 1
2. 1
(b) By (a),
n→∞lim
1 − cos(1n)
1 − cos(n+11 ) = lim
n→∞
1 − cos(n1)
(n1)2 · (n+11 )2 1 − cos(n+11 )
= lim
n→∞
1 − cos(n1) (n1)2 · lim
n→∞
(n+11 )2 1 − cos(n+11 )
= (12) (12)
= 1.
(c) By (b), let an = (1 − cos(n1))xn,
n→∞lim
an+1 an
= lim
n→∞
1 − cos(1n) 1 − cos(n+11 )x
= |x|
Hence, the radius R = 1.
For x = −1, since 1 − cos(n1) decreases to 0, so according to alternating test,
∞
X
n=1
(−1)n(1 − cos(1
n)) converges.
For x = 1, since lim
n→∞
1 − cos(n1) (1n)2 = 1
2 and
∞
X
n=1
1
n2 converges, so by comparison test,
∞
X
n=1
(1 − cos(1
n)) converges.
Therefore, the interval of convergence is [−1, 1].
3. (10%) Let f (x) = sin (x3). Find f(15)(0).
Sol:
sin(t) =
∞
X
n=0
(−1)nt2n+1
(2n + 1)! = t − t3 3!+ t5
5! − t7 7!+ t9
9! − · · ·
⇒ sin(x3) =
∞
X
n=0
(−1)nx6n+3
(2n + 1)! = x3− x9 3! +x15
5! − x21 7! + x27
9! − · · ·
⇒ f15(0) = 15!
5! .
4. (12%) Find the curvature κ, the unit tangent vector T, and the unit normal vector N of the curve r(t) = hcos t, sin t, ln (cos t)i at r(0) = (1, 0, 0).
Sol:
r0(t) = (− sin t, cos t,− sin t
cos t ) =⇒ |r0(t)| =p
1 + tan2t = 1
cos t (because cos t > 0)
=⇒ T (t) = (− sin t cos t, cos2t, − sin t), so T (0) = (0, 1, 0)
T0(t) = (− cos2t + sin2t, −2 sin t cos t, − cos t)
=⇒ k(0) = |dT /ds|t=0 = |T0(0)|/|r0(0)| =√
2; N (0) = T0(0)
|T0(0)| = 1
√2(−1, 0, −1)
5. (10%) Where does the tangent plane of the surface z = ex−y at (1, 1, 1) intersect the z axis?
Sol:
∂z
∂x = ex−y, ∂z
∂y = −ex−y (1)
So, the tangent plane is
z − 1 = ∂z
∂x(1, 1)(x − 1) + ∂z
∂y(1, 1)(y − 1). (2)
z − axis ⇒ x = y = 0 input in equation (2), then we obtain z = 1 (3)
Therefore, the intersect point is (0, 0, 1).
6. (12%) You are wandering around in a strange desert where the temperature at the point (x, y) is given by the function T (x, y) = ey−x2.
(a) You have stopped at the point (2, 1). Suddenly you are feeling chilled and want to warm up – in what direction should you go to warm up as rapidly as possible?
(b) Figure out the coordinates of all the points where there is no increase or decrease in temperature in the h1, 1i direction.
Sol:
(a) f (2, 1) is the direction we need (you show that you known this property) f (2, 1) = e−3(−4, 1) (this answer can be replace by t(−4, 1), for any t > 0) (b) f (x, y) · (1, 1) = 0 ⇒ x = 1
2, y ∈ R
7. (10%) Let z = f (x, y) such that all the second partial derivatives of f are continuous. Let x = r cos θ and y = r sin θ.
(a) Evaluate ∂r
∂x, ∂r
∂y, ∂θ
∂x, and ∂θ
∂y. Express the results in ( functions of ) r and θ.
(b) Express fx in terms of ( functions of ) r, θ, fr, fθ. 3
(c) Express fxx in terms of ( functions of ) r, θ, fr, fθ, frr, frθ, fθθ. Sol:
(a) Since r =px2+ y2 Hence∂r
∂x = x
px2 + y2 = r cos θ
r = cos θ.
Similarly ∂r
∂y = y
px2+ y2 = r sin θ
r = sin θ.
Since θ = tan−1 y x Hence ∂θ
∂x =
−y x2 1 + (y
x)2
= −y
x2+ y2 = − sin θ r .
Similarly∂θ
∂y = 1 x 1 + (y
x)2
= x
x2+ y2 = cos θ r .
(b) Follows chain rule ⇒ fx = ∂f
∂x = ∂f
∂r
∂r
∂x +∂f
∂θ
∂θ
∂x = frcos θ + fθ− sin θ r (c)
fxx = ∂(frcos θ − fθsin θ r )
∂x
= cos θ ∂
∂xfr+ fr
∂
∂xcos θ − sin θ r
∂
∂xfθ− fθ
∂
∂x sin θ
r
= cos θ
frr∂r
∂x + frθ∂θ
∂x
− frsin θ∂θ
∂x
−sin θ r
fθr∂r
∂x + fθθ∂θ
∂x
−fθ ∂
∂r sin θ
r
∂r
∂x + ∂
∂θ sin θ
r
∂θ
∂x
= frrcos2θ + frθcos θ
−sin θ r
− frsin θ
−sin θ r
−fθr sin θ r
cos θ + fθθ sin θ r
2
−fθ − sin θ
r2 cos θ + cos θ r
−sin θ r
= frrcos2θ − frθsin 2θ
r + frsin2θ
r + fθθ sin θ r
2
+ fθsin 2θ r2
8. (12%) Find and classify the critical points of f (x, y) = xye−x2−y2.
Sol:
fx(x, y) = 0,fy(x, y) = 0 ⇒ (x, y) = (0, 0), ( 1
√2, 1
√2), (−1
√2, −1
√2), ( 1
√2, −1
√2), (−1
√2, 1
√2) (5 分) fxx(x, y) = (−6xy + 4x3y)e−x2−y2,
fyy(x, y) = (−6xy + 4xy3)e−x2−y2, fxy = (1 − 2x2)(1 − 2y2)e−x2−y2 D(x, y) = fxxfyy− (fxy)2
⇒ D(0, 0) < 0, (0, 0) is a saddle point
⇒ D( 1
√2, 1
√2) > 0, fxx( 1
√2, 1
√2) < 0, ( 1
√2, 1
√2) attains maximum
⇒ D(−1
√2, −1
√2) > 0, fxx(−1
√2, −1
√2) < 0, (−1
√2,−1
√2) attains maximum
⇒ D(−1
√2, 1
√2) > 0, fxx( 1
√2, −1
√2) > 0, ( 1
√2,−1
√2) attains minimum
⇒ D( 1
√2, −1
√2) > 0, fxx( 1
√2, −1
√2) > 0, ( 1
√2,−1
√2) attains minimum
9. (10%) Use the method of Lagrange multipliers to find the extreme values of z on the curve of intersection of x2+ z2 = 1 and y2+ z2+ z = 1.
Sol:
we want to find the extreme value of function: f (x, y, z) = z, Constrain functions: g(x) = x2+ z2− 1 and h(x) = y2+ z2+ z − 1
by the Lagrange multiplier, we have ∇f = λ∇g + µ∇h That is
−2λx = 0
−2µy = 0
λ(2z) + µ(2z + 1) = 1 Then we discuss for conditions based on λ and µ
(1) λ = 0, µ = 0 Contradiction to the equation λ(2z) + µ(2z + 1) = 1 (2) λ = 0, µ 6= 0 we can get y = 0 → z2+ z − 1 = 0 → z = −1 ±√
5
2 . By examine x2+ z2 = 1, we find that z = −1 +√
5 2
(3) λ 6= 0, µ = 0 we can get x = 0 → z = ±1. By examine y2 + z2 + z = 1, we find that 5
z = −1
(4) λ 6= 0, µ 6= 0 we can get x = 0 and y = 0 which are contradict to the constrain x2+ z2 = 1 and y2+ z2+ z = 1
Extreme values:
Maximum: z = −1 +√ 5 2 Minimum: z = −1
