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NATIONAL UNIVERSITY OF SINGAPORE
SEMESTER 2, 2013/2014
MA1521 Calculus for Computing Tutorial Solution 7
1. (a) This is a geometric series with common ratiox 14
:
n=1
(x 1)n3 + (x 1)n14n + 22n1
=1
6(x 1)2n=1
(x 14
)n1.
It converges if and only if
x 14 < 1; that is, |x 1| < 4.
So its radius of convergence is R = 4.
(b) Let cn =n2n(2)n4n(2n+ 1)!
. Then
limn
cn+1cn = limn
((n+ 1)2n+22n+1
4n+1(2n+ 3)! 4
n(2n+ 1)!
n2n2n
)= lim
n(n+ 1)2
2(2n+ 2)(2n+ 3)
(1 +
1
n
)2n= lim
n(1 + 1
n)2
2(2 + 2n)(2 + 3
n)
(limn
(1 +
1
n
)n)2=e2
8.
Then the radius of convergence is R =
(e2
8
)1=
8
e2.
2. (a) f(x) =(x 1)2x+ 1
=(x 1)2
2 + (x 1) =(x 1)2
2 11 + x1
2
=(x 1)2
2n=0
(x 12
)n=n=0
(1)n(x 1)n22n+1
=n=2
(1)n2n1
(x 1)n.
(b) f(x) =1
x2 4x+ 5 =1
1 + (x 2)2 =n=0
(1)n(x 2)2n.
3. (a)z
x= 3x2 and
z
y= 3y2. Then z
x
(3,2)
= 27 andz
y
(3,2)
= 12.The equation of the tangent plane is given by
z = 19 + 27(x 3) 12(y 2) = 27x 12y 38.
1
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MA1521 CALCULUS FOR COMPUTING TUTORIAL SOLUTION 7 2
(b)z
x= y and
z
y= x. Then
z
x
(1,1)
= 1 and zy
(1,1)
= 1.
The equation of the tangent plane is given by
z = 1 1(x 1) + 1(y + 1) = x+ y + 1.
(c) Let f(x, y, z) = x2 + 4y2 z2. Then
f(x, y, z) = 2xi+ 8yj 2zk and f(3, 2, 5) = 6i+ 16j 10k.
It follows that the tangent plane is given by
6(x 3) + 16(y 2) 10(z 5) = 0.
That is, 6x+ 16y 10z = 0; or equivalently 3x+ 8y 5z = 0.
4. (a) Let f(x, y) = exy2 + x sin(xy). Then
f
x= exy2 + sinxy + xy cosxy,
f
y= 2yex + x2 cosxy,
2f
x2=
x
f
x= exy2 + 2y cosxy xy2 sinxy,
2f
y2=
y
f
y= 2ex x3 sinxy
2f
x y=
x
f
y= 2yex + 2x cosxy x2y sinxy.
(b) Let f(x, y) =x+ y
x2 + y. Then
f
x=
(x2 + y) (x+ y)2x(x2 + y)2
=x2 2xy + y
(x2 + y)2,
f
y=
(x2 + y) (x+ y)(x2 + y)2
=x2 x
(x2 + y)2,
2f
x2=
(2x 2y)(x2 + y)2 (x2 2xy + y) 4x(x2 + y)(x2 + y)4
=2x3 2y2 + 6xy(1 x)
(x2 + y)3
2f
y2=(x2 x) 2(x2 + y)
(x2 + y)4=
2(x x2)(x2 + y)3
2f
x y=
(2x 1)(x2 + y)2 (x2 x) 4x(x2 + y)(x2 + y)4
=2x3 + 3x2 + (2x 1)y
(x2 + y)3.
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MA1521 CALCULUS FOR COMPUTING TUTORIAL SOLUTION 7 3
5.dx
dt= 2t,
dy
dt= 6 cos 2t,
u
x= 2x and
u
y= 2y. Then
du
dt=u
x
dx
dt+u
y
dy
dt= 2x 2t+ (2y) 6 cos 2t
= 4t3 4t 36 sin 2t cos 2t = 4t(t2 1) 18 sin 4t.
6. (a) f(x, y) = 2xi+ 2yj. Let v = u|u| =113
(2i 3j). Then
Dvf(1, 2) = f(2, 4) v = (2i+ 4j) 113
(2i 3j) = 813.
(b) f(x, y, z) = tan1(y + z)i+ x1 + (y + z)2
j +x
1 + (y + z)2k.
Let v =u
|u| =13(i+ j k). Then
Dvf(1, 0, 1) = f(1, 0, 1) v =(pi
4i+
1
2j +
1
2k
) 1
3(i+ j k) = pi
43.
7. (a) Let c2n =42n
9nn5/4. Then
limn
c2n+2c2n
= limn
42
9
(n
n+ 1
)5/4= lim
n16
9
1
(1 + 1n)5/4
=16
9.
Therefore, the radius of convergence is R =
9
16=
3
4.
(b) Let c2n1 =1
32n14nn2. Then
limn
c2n+1c2n1
= limn
32n14nn2
32n+14n+1(n+ 1)2= lim
n1
36(1 + 1n)2
=1
36.
Then the radius of convergence is R =36 = 6.
