Instructor: Frank Liou
Exam Time: 8:10 am to 10:00 am Total Score: 110 Points
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1. (20 Points) Find the radius of convergence and the interval of convergence of the power series
∞
X
n=2
xn nlnn.
You need to explain why the boundary points of the interval of convergence is di- vergent or convergent. (There is no partial credit if there is no explanation.)
Solution:
The radius of convergence is R= 1 by ratio test.
To determine the interval of convergence, we first takex= 1.The infinite series P∞
n=21/(nlnn) is divergent either by integral test or by Cauchy-condensation prin- ciple. Let us use integral test. Sincef(x) = 1/(xlnx) is nonnegative and decreas- ing on [3,∞) with f(n) = 1/(nlnn) for n ≥3,the infinite series P∞
n=31/(nlnn) (henceP∞
n=21/(nlnn) is divergent) and the improper integralR∞
3 dx/(xlnx) both diverges or converges. Using the change of variable u= lnx,
Z ∞
3
1
xlnxdx= Z ∞
ln 2
du u =∞ Therefore the improper integral R∞
3 dx/(xlnx) is divergent.
Let us consider the case when x=−1.The infinite series P∞
n=2(−1)n/(nlnn) is an alternating series such that
(a) {1/(nlnn) :n≥1} is a decreasing sequence of real numbers.
(b) limn→∞(1/(nlnn)) = 0.
By Leibnitz test, the seriesP∞
n=2(−1)n/(nlnn) is convergent.
We conclude that the interval of convergence is [−1,1).
Let us go back to the discussion about the monotonicity of the function f(x) = 1/(xlnx) on [3,∞).This can be proved by considering the first derivative of f :
f0(x) =−lnx−1
(xlnx)2, x≥3.
Forx≥3,lnx >lne= 1.Hence f0(x)<0 onx≥3.
Let us prove that limx→∞1/(xlnx) = 0.Forx≥3,lnx >1 and hence 0< 1
xlnx < 1
x, x≥3.
By Sandwich principle, we obtain the required result.
2. (20 Points) LetC be the regular curve given by r(t) = 2
3(1 +t)3/2i+2
3(1−t)3/2j+√
2tk, |t|<1.
(Present your work clearly or you will receive no partial credits.) (a) (2 Points) Find the arclength functions(t) of C.
(b) (3 Points) Find the arclength reparametrization of the curve C by α(s) = r(t(s)),wheret=t(s) is the inverse function ofs=s(t).
(c) (9 Points) Find the unit tangent T,the principal normalN and the binormal B.
(d) (6 Points) Find the curvature κ and the torsionτ.
Taking the derivative of rwith respect tot, we find r0(t) = (1 +t)1/2i−(1−t)1/2j+
√ 2k.
Hence the speed is given by kr0(t)k=p
(1 +t) + (1−t) + 2 = 2.
The arclength function
s(t) = Z t
0
2du= 2t.
Hence t=s/2.The arc length reparametrization ofC is then α(s) = 2
3(1 + s
2)3/2i+2 3(1− s
2)3/2j+
√ 2 2 sk The unit tangent is the velocity of α:
T=α0(s) = 1
2 (1 + s
2)1/2i−(1− s 2)1/2j+
√2 2 k
! . Hence the derivative of Tis
T0= 1 8
(1 +s
2)−1/2i+ (1−s 2)−1/2j
. Then the curvature of the curve is given by the formula
κ=kT0k= 1 8
s 1
1 + (s/2)+ 1 1−(s/2) =
√2 4√
4−s2. We obtain
N= T0 κ =
√2 2
(1− s
2)1/2i+ (1 + s 2)1/2j
.
By definition, B=T×N,we know B=
i j k
1
2(1 +s2)1/2 12(1−2s)1/2
√ 2
√ 2 2
2 (1− s2)1/2
√2
2 (1 + s2)1/2 0
=−1 2(1 +s
2)1/2i+1 2(1− s
2)1/2j+
√2 4 sk.
To compute the torsion, we need to compute B0 = 1
8 −(1 +s
2)−1/2i−(1− s
2)−1/2j+
√2 4 k
! . Since the torsion ofC is defined to beτ =−B0·N,we have
τ =
√2 16
2(1 + s
2)1/2(1−s 2)1/2
=
√2 16
p4−s2.
3. (20 Points) Letf(x, y, z) =exylnz.Find the directional derivative off atP(1,0, e) in the directions parallel to the line in which the planesx+y−z= 0 and 4x−y−z= 1 intersect. (There are two unit vectors in this direction. No partial credits if the final answer is wrong.)
Solution:
Let us compute the partial derivatives of f :
fx=yexylnz, fy =xexylnz, fz= exy z . Hence the gradient of f atP is
∇f(P) =fx(P)i+fy(P)j+fz(P)k=j+1 ek.
Letn1=i+j−kandn2 = 4i−j−kbe the normal vectors of the planex+y−z= 0 and 4x−y−z= 1 respectively. Compute the cross product ofn1 and n2:
i j k
1 1 −1
4 −1 −1
=−2i−3j−5k.
Take unit vectors
u=±2i+ 3j+ 5k
√ 38 .
Then u are parallel to the line in which the two planes intersection. Hence the directional derivatives off along u are
Duf(P) =∇f(P)·u=±3 + (5/e)
√ 38 .
4. (10 Points) Find the limit if it exists. If it does not exist, explain why it is divergent.
(x,y)→(0,0)lim
4−4 cosp
|xy|
|xy| .
(Hint: consider the Taylor expansion of f(θ) = cosθ at θ = 0. You can omit this hint if you have any other idea. )
Solution:
The Taylor expansion of cosθ atθ= 0 is given by cosθ= 1−θ2
2! +θ4 4! − θ6
6! +· · ·. This implies that
1−cosθ= θ2 2! −θ4
4! + θ6
6! +· · ·. Substituting θ=p
|xy|into the above equation, we obtain 1−cosp
|xy|= |xy|
2 −|xy|2
4! +· · · . For xy6= 0,
4−4 cosp
|xy|
|xy| = 2−|xy|
6 +· · ·. Hence we find
(x,y)→(0,0)lim
4−4 cosp
|xy|
|xy| = 2.
In fact, we have the following inequality 2−|xy|
6 ≤ 4−4 cosp
|xy|
|xy| ≤2, xy6= 0.
Using Sandwich principle, we also obtain the same result.
5. (9 Points) Let
f(x, y) = ( x2y
x2+y2 (x, y)6= 0 0 (x, y) = (0,0).
(a) (2 Points) Use definition to compute fx(0,0) and fy(0,0),i.e. find fx(0,0) = lim
h→0
f(h,0)−f(0,0)
h , fy(0,0) = lim
h→0
f(0, h)−f(0,0)
h .
(b) (4 Points) Compute fx(x, y) andfy(x, y) for (x, y)6= (0,0).
(c) (3 Points) Are fx, fy continuous at (0,0)?
Solution:
We know f(h,0) = 0 and f(0, h) = 0.Hence fx(0,0) = 0 and fy(0,0) = 0.
For (x, y)6= (0,0)
fx = 2xy(x2+y2)−x2y·2x
(x2+y2)2 = 2xy3 (x2+y2)2 fy = x2(x2+y2)−x2y(2y)
(x2+y2)2 = x4−x2y2 (x2+y2)2. Let us consider straight-line x= 0.For allh6= 0,
fx(0, h) = 0, fy(0, h) = 0.
Then limh→0fx(0, h) = limh→0fy(0, h) = 0. Let us consider the staightline x =y.
For allh6= 0,
fx(h, h) = 2h4 (2h2)2 = 1
2.
Hence limh→0fx(h, h) = 1/2. We find that the function fx(x, y) has two different limits when we consider the limits of the the function along x = 0 and x = y.
Therefore lim(x,y)→(0,0)fx(x, y) does not exist. Similarly, we consider the straight- line y= 2x.Forh6= 0,
fy(h,2h) = h4−4h4
(h2+ 4h2)2 =− 3 25.
Hence limh→0fy(h,2h) = −3/25. We conclude that lim(x,y)→(0,0)fy(x, y) does not exist.
Thereforefx, fy are not continuous at (0,0).
6. (20 Points) Find all first and second partial derivatives of the function f(x, y) =xey+ycosx+ tan−1(x/y), (x, y)∈R2.
(No Partial credits if the final answers are wrong. Make sure that all of your cal- culation are correct.)
Solution:
fx =ey−ysinx+ 1
1 + (x/y)2 ·1
y =ey−ysinx+ y x2+y2 fy =xey+ cosx+ 1
1 + (x/y)2 ·
−x y2
=xey+ cosx− x x2+y2 fxx =−ycosx− 2xy
(x2+y2)2
fxy =ey−sinx+(x2+y2)−y(2y)
(x2+y2)2 =ey−sinx+ x2−y2 (x2+y2)2 fyx=ey−sinx−(x2+y2)−x(2x)
(x2+y2)2 =ey−sinx− y2−x2 (x2+y2)2 fyy =xey+ 2xy
(x2+y2). We also obtain fxy =fyx.
7. (10 Points) Letz= ln(x2+y2) andx= 1−t2 andy = 2t.Use chain rule to compute dz
dt.(There is no credit if you use other methods.) Solution: By Chain rule,
dz dt = ∂z
∂x dx
dt +∂z
∂y dy dt.
Sincedx/dt=−2tanddy/dt= 2 and∂z/∂x= 2x/(x2+y2) and∂z/∂y= 2y/(x2+ y2).We find that
dz
dt = 2(1−t)2
(1−t2)2+ (2t)2 ·(−2t) + 2·(2t)
(1−t2)2+ (2t)2 ·2 = 4t 1 +t2. 8. (10 Points) LetC be a parametrized curve in R3 with parametrization
r(t) =x(t)i+y(t)j+z(t)k,
where x(t), y(t), z(t) are differentiable functions on an interval I containing [a, b].
Suppose r(t1)6=r(t2) for any t1 6=t2 inI. (Such a curve is called a simple curve.) Show that
kr(b)−r(a)k ≤ Z b
a
kr0(t)kdt.
(Hint: consider r(t)·v for some specific v∈R3)
Proof. Let us takev= (r(b)−r(a))/kr(b)−r(a)k. Thenv is a unit vector. Using fundamental theorem of calculus,
r(b)·v−r(a)·v= Z b
a
(r0(t)·v)dt.
Notice that
r(b)·v−r(a)·v= (r(b)−r(a))·v
= (r(b)−r(a))· (r(b)−r(a)) kr(b)−r(a)k
= kr(b)−r(a)k2 kr(b)−r(a)k
=kr(b)−r(a)k.
On the other hand, using Cauchy-Schwarz inequality, r0(t)·v≤ kr0(t)kkvk, a≤t≤b.
Usingkvk= 1 and combining all of the above results, we obtain kr(b)−r(a)k=
Z b a
(r0(t)·v)dt≤ Z b
a
kr0(t)kkvkdt= Z b
a
kr0(t)kdt.
9. (20 Points) Let us consider the formal power series
f(x) = 1 +
∞
X
k=1
m k
xk.
(a) (5 Points) Find the radius of convergence R off(x).
(b) (10 Points) Show that
f0(x) = m
1 +xf(x), |x|< R.
(c) (5 Points) Define g(x) = (1 +x)−mf(x). Show that g(x) = 1 for all |x|< R.
(This implies thatf(x) = (1 +x)m for|x|< R.)
Solution: The radius of convergence isR= 1 by ratio test.
By definition off(x),we find f(0) = 1 and f(x) = 1 +mx+m(m−1)
2 x2+ m(m−1)(m−2)
3! x3+· · ·= 1 +
∞
X
k=1
m k
xk. Taking the derivative of f(x) with respect tox, we obtain
f0(x) =m+m(m−1)x+m(m−1)(m−2)
2! x2+· · ·=m
∞
X
k=0
m−1 k
xk. Multiplyingf0(x) byx,
xf0(x) =mx+m(m−1)x2+m(m−1)(m−2)
2! x3+· · ·=m
∞
X
k=1
m−1 k−1
xk.
Summing up the above two equations, (1 +x)f0(x) =m+m
∞
X
k=1
m−1 k−1
+
m−1 k
xk. Using the Pascal’s triangle:
m−1 k−1
+
m−1 k
= m
k
. This implies
(1 +x)f0(x) =m+m
∞
X
k=1
m k
xk =mf(x).
This proves (b).
For the last question, we take the derivative of g(x) with respect tox: g0(x) =−m(1 +x)−m−1f(x) + (1 +x)−mf0(x).
Using (b), we know f0(x) =m(1 +x)−1f(x) and hence
g0(x) =−m(1 +x)−m−1f(x) + (1 +x)−m·m(1 +x)−1f(x)
=−m(1 +x)−m−1f(x) +m(1 +x)−m−1f(x)
= 0.
This shows that g is a constant function. Then
g(x) =g(0) = (1 + 0)−mf(0) =f(0) = 1.
