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ME5331 MIDTERM EXAMINATION
4:00-5:20pm, October 17, 2022
Closed book/note/no calculator
2
1. If u = x2 + y2 + z2 and v = x y z, compute ∇·(∇u ×∇v).


2. Differentiate
y=sin (xx).


3. Does
∞
∑
n=2 
 n ln ⎛
⎝ 		n2 + 1
n2 − 1
⎞
⎠
converge or diverge ? Why ?


4.
f(x) ≡ ⎧
⎪
⎨
⎪
⎩
cos x − 1
ex − 1
x ≠ 0
0
x = 0

    (a) Is f(x) continuous at x = 0 ? Why ?
    (b) Is f(x) differentiable at x = 0 ? Why ? If so, find f′(0).


5. Exchange the order of integrations for
⌠
⌡ 		4

0 
 ⌠
⌡ 		y/2

0 
 f(x, y) dx d y = ⌠
⌡ 		?

? 
 ⌠
⌡ 		?

? 
 f(x, y) dy dx.


6. Given y3 − 3 y − x = 1, compute y"(x).


7. How can you distinguish curves, surfaces and volumes in 3-D ? (3 lines max. 50% penalty if not followed.)


8. Compute
⌠
(⎜)
⌡
 1

√
a2 x2 + b2 y2 + c2 z2
 dS,
where the integral range, S, is the surface of an ellipsoid given by
a x2 + b y2 + c z2 = 1,     a, b, c > 0.


9. Compute I′(t) where
I(t) ≡ ⌠
⌡ 		t3

t 
 ln (1 + t  x2) d x.
It is not necessary to evaluate the integral.


10. Compute a normal vector perpendicular to both
a = (1, 2, −1),     b = (1, 1, −1).
Solution


1.
∇·(∇u ×∇v) = ∇v ·( ∇×∇u ) = 0.


2.
xx (log(x)+1) cos (xx).


3.
ln ⎛
⎜
⎜
⎜
⎝
1 + 1
n2
1 − 1
n2
⎞
⎟
⎟
⎟
⎠ 		= ln⎛
⎝ 		1 + 1
n2
⎞
⎠ 		− ln ⎛
⎝ 		1 − 1
n2
⎞
⎠ 		 ∼  2
n2
 ,
so

n ln ⎛
⎝ 		n2 + 1
n2 − 1
⎞
⎠ 		 ∼  2
n
 ,
which is divergent (p = 1).


4. (a)

lim
x→ 0 
 f(x) =
lim
x→ 0 
 cos x − 1
ex − 1
 =
lim
x→ 0 
 − sin x
ex
 = 0 = f(0).
Yes f(x) is continuous at x = 0.
(b)


lim
h→ 0 
 f(0 + h) − f(0)
h
 =
lim
h→ 0 
 cos h − 1
h (eh − 1)
 =
lim
h→ 0 
 − sin h
eh − 1 + h eh
 =
lim
h→ 0 
 − cos h
eh + eh + h eh
 = − 1
2
 .
Yes f(x) is differentiable at x = 0.


5.
⌠
⌡ 		4

0 
 ⌠
⌡ 		y/2

0 
 f(x, y) dx d y = ⌠
⌡ 		2

0 
 ⌠
⌡ 		4

2x 
 f(x, y) d y dx.


6.
3 y2 y′− 3 y′− 1 = 0 → y′ = 1
3 (y2 − 1)

y" = − 2 y y′
3 (y2 −1)2
 = − 2 y
9 (y2 − 1)3
 .

y"=− 2 y (y′)2
y2−1
is also OK.


7. Position vectors with one parameter define curves, two parameters define surfaces and three parameters define volumes.


8.
n =
⎛
⎜
⎜
⎜
⎝
a x
b y
c z
⎞
⎟
⎟
⎟
⎠

√
a2 x2+b2 y2+c2 z2
Choose v as
v = ⎛
⎜
⎜
⎜
⎝
x
y
z
⎞
⎟
⎟
⎟
⎠
so that
n·v = 1

√
a2 x2+b2 y2+c2 z2
 .
Hence,
⌠
(⎜)
⌡
 1

√
a2 x2+b2 y2+c2 z2
 dS = ⌠
⌡ 		⌠
⌡ 		⌠
⌡ 		∇·v dV = 3 ×(volume of a x2+b y2+c z2=1)= 4π

√
a b c
 .


9.
I′(t) = ⌠
⌡ 		t3

t 
 x2
1 + t x2
 dx + 3 t2 ln (1 + t7) − ln (1+ t3).


10.
(− 1
√2
 , 0, − 1
√2
 ).


File translated from TEX by TTH, version 4.03.
On 05 Oct 2023, 12:47.