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Solution to the Final Examination
MAT1322D, Fall 2017
Part I. Multiple-choice Questions (2 12 = 24 marks)
FADAFEECBBDD
1. The area of the region above the graph of y = x^2 and under the graph of y = 8 x^2 in the first quadrant is
(A)^27
; (B)^29
; (C)^37
; (D)^41
; (E)^35
; (F)^32
Solution. (F) Let x^2 = 8 x^2. 2 x^2 = 8, x = 2. The area is
2 2 3 2 2 2 2 (^0 )
(^3) x 3
x x dx x dx x^ x
^
2. Let R be the region above the parabola y = x^2 and under the line y = 2 x. Solid B is obtained by revolving R about the line y = 4. Then the volume of B is calculated by the integral
(A)
(^2 2 2 )
0 ((4 x ) (4 2 ) ) x dx ; (B)
(^4 2 2 )
0 ((4 x ) (4 2 ) ) x dx ;
(C)
(^2 2 2 )
0 ((4 x ) (4 2 ) ) x dx ; (D)
(^4 2 2 )
0 ((4 x ) (4 2 ) ) x dx ;
(E)
(^2 2 2 )
0 ((4 2 ) x (4 x ) ) dx ; (F)
(^2 2 2 )
0 ((4 x ) (4 2 ) ) x dx.
Answer. (A)
r in = 4 – 2 x r out = 4 − x^2
y = 4
y = 2 x y = x^2
X
Y
O
3. Suppose a pool has the shape of an inverted truncated pyramid. The top of the pool at the ground level is a square with side length 20 meters, and the bottom of the pool is a square with side length 10 meters. The depth of the pool is 5 meters. The pool is filled with water with
density kg / m^3. Let x be the distance between a horizontal layer of water and the top of the
pool.
Denote the acceleration of gravity be g m / sec^2. Then the work, in Joules, needed to pump the water in the pool to a point 2 meters above the ground is calculated by the integral
(A)
(^5 )
g 0 (20 2 ) (7 x x dx ) ; (B)
(^7 )
g 0 (20 2 ) ( x x 2) dx
(C)
(^7 )
g 0 (20 2 ) (7 x x dx ) ; (D)
(^5 )
g 0 (20 2 ) ( x x 2) dx
(E)
(^5 )
g 0 (20 2 ) (7 x x xdx ) ; (F)
(^5 )
g 0 (20 2 ) ( x x 2) dx
Solution. (D) A horizontal layer of water in the pool is a square with side-length L ( x ) = 10 + 2 (5 – x ) = 20 – 2 x.
The volume of this layer with thickness dx is V ( x ) = ( L ( x ))^2 = (20 – 2 x )^2 dx.
The weight of this layer of water is w ( x ) = gV ( x ) = g (20 – 2 x )^2 dx.
The work needed to pump this layer of water to a point 2 meters above the ground is
W ( x ) = w ( x )( x + 2) = g (20 – 2 x )^2 ( x + 2) dx.
The total work is
W =
(^5 )
g 0 (20 2 ) ( x x 2) dx.
4. Recall that the length of the arc y = f ( x ), a x b , is calculated by the formula
L = 1 ( '( ))^2
b
a f x dx. Then the length of the arc y = ln x −
2 8
^ x , 1 x e , is
x
ground level
Solution. (E) The mass is m =
(^1 ) 0
x x dx . The moments are
Mx = 1 3 4 5 1 2 2 (^00)
(^2 2 3 2 5) x 2 3 2 5 60
x x dx^ x^ x^ x
^ ^ ^
My =
(^1 ) 0
x x x dx .
Then 1/12^1 1/ 6 2
x , 1/ 60^1 1/ 6 10
y . The centroid is^1 ,^1 2 10
(^) .
7. Consider improper integral
1 0 2
x (^1) dx x x
. Which one of the following argument is true?
(A) Since 2 x^1 x x
<^2
x
in (0, 1) and
1 1 0 0
(^2) dx 2 1 dx x x
converges,
1 0 2
x (^1) dx x x
converges.
(B) Since 2 x^1 x x
<^2
x
in (0, 1) and
1 1 0 0
(^2) dx 2 1 dx x x
diverges,
1 0 2
x (^1) dx x x
diverges.
(C) Since 2 x^1 x x
<^2 x x
in (0, 1) and
1 1 0 0 1/ 2
2 x dx 2 1 dx x x
converges,
1 0 2
x (^1) dx x x
converges.
(D) Since 2 x^1 x x
x x
in (0, 1) and
1 1 0 2 0 3/ 2
x dx dx x x
diverges,
1 0 2
x (^1) dx x x
diverges.
(E) Since 2 x^1 x x
>^1
2 x
in (0, 1) and
1 1 0 0
dx dx x x
diverges,
1 0 2
x (^1) dx x x
diverges.
(F) Since 2 x^1 x x
>^1
2 x
in (0, 1) and
1 1 0 0
dx dx x x
converges,
1 0 2
x (^1) dx x x
converges.
Answer. (E).
8. Suppose Euler's method with step size h = 0.05 is used to find an approximation of y (0.1), where y ( t ) is the solution to the initial-value problem y' = (2 t – 1)( y + 1), y (0) = 1. Which one of the following is closest to the answer? y (0.1)
(A) 0.905; (B) 0.845; (C) 0.815; (D) 0.742; (E) 0.707; (F) 0.685.
Solution. (C)
i ti yi 0 0 1 1 0.05 1 + 0.05 (2 0 −1) (1 + 1)= 0. 2 0.10 0.900 + 0.05 (2 0.05 – 1) (0.9 + 1) = 0.
9. The sum of the series
1 0 2
n n n n n
is S =
(A)^43
; (B)^45
; (C)^29
; (D)^25
; (E)^13
; (F)^23
Solution. (B)
1 1 0 2 0 2 0 2
n n n n n n n n^ n n^ n n
^
. The first series is a geometric series
with first term a 1 = 2 and common ratio r 1 =^2 9
. The second series is also a geometric series
with first term a 2 = 1 and common ratio r 2 =^5 9
. Hence,
S =^2 2 1 5 18 9
10. The interval of convergence of the series 0
n n n n^ x
is
(A) −1.5 x 1.5; (B) −1.5 < x < 1.5; (C) −1.5 < x 1.5; (D) −1.5 x < 1.5; (E) −∞ < x < ∞; (F) only at x = 0.
Solution. (B)
1 1 1 lim 2 3 lim^2 3 2 3 3
n n n n n^ n^ n n
x x (^) x x
(^) ^ ^ . When | x | <
, this series is absolutely
convergent. When | x | >^3 2
, this series is divergent. When x = −^3 2
, this series becomes
0 0
n^ n n n n n
^ ; when^ x^ =
, this series becomes 0 0
n^ n n n n
. In both cases, the
series diverges. Hence, the interval of convergence is^3 ,^3 2 2
11. If f ( x , y ) = x^2 y^2 + 3 xy + y^3 , then the gradient vector of f ( x , y ) at the point x = 2, y = −1 is
Solution. (a) Since this series is positive, we can use the limit comparison test. Let an =
2
n n n
, and let bn =^1 n
. Then
2 2 2 lim lim 2 lim 2 1/ 2 1 1 1/
n n (^) n n n
a n n n n (^) b (^) n n
^
. Since series 1
n n
diverges, series 2 1
n^1
n n n
diverges.
This question can also be solved by comparison test.^2 2 1 2 2
n n n n n n
. Since series
1 1
n^2 n^^2 n n
^ diverges, series 2
1
n^1
n n n
diverges.
(b) Since function f ( n ) = 21 n 1
is decreasing, andlim 21 n (^) n 1 = 0, By the alternating series
test, this series converges.
(c) Use the root test. Let an =^1 3
n^ n n
lim lim^1 3 3
n (^) n n n a n n . This series is convergent.
This question may also be solved by the ratio test.
Since 1 (^1 ) 1
lim lim 3(^ 1)^ lim 2 3 1 lim 1 1 lim 1 1 1 1 3(^ 1)^1 3 1 3
n n (^) n n n n n n (^) n n n n n
n a n n n a (^) n n n n n n
(^)
^ ^ ^ ^ ^ ^ ^
This series is (absolutely) convergent.
4. (6 marks) The Maclaurin series of the function y = sin x is
sin x =
2 1 3 5 0
n^ n n
x (^) x x x n
^
(a) (4 marks) Find the first three non-zero terms of the Maclaurin series of the function 2 ( ) 0 sin(2 )
x
F x t dt.
(b) (2 marks) Find The fifth and the seventh derivative of function F ( x ) at x = 0, i.e., F (5)(0) and F (7)(0).
Solution. (a) sin (2 t^2 ) =
2 2 1 2 3 6 5 10 0
( 1) (2^ )^2 2 ...
n^ n n
t (^) t t t n
^
2 2 2 1 2 3 6 5 10 0 0 0 0 ( ) sin(2 ) ( 1) (2^ )^2 2 ... (2 1)! 3! 5!
x x (^) n^ n x n
F x t dt t^ dt t t^ t dt n
^
^
=^2 3 8 7 32 11 ... 2 3 4 7 4 11 ...
x x x x x x .
(b) F (5)(0) = 0, F (7)(0) = 7!^4 21
5. (4 marks) Find the equation of the tangent plane of the graph of the equation x^2 z + xy − yz^3 = −1 at the point (2, 1, −1).
Solution. Let F ( x , y , z ) = x^2 z + xy − yz^3 + 1.
Then Fx = 2 xz + y , Fy = x − z^3 , and Fz = x^2 − 3 yz^2. Hence, Fx (2, 1, −1) = −3, Fy (2, 1, −1) = 3, and Fz (2, −1, 3) = 1.
The equation of the tangent plane at the point (2, 1, −1) is −3( x – 2) + 3( y – 1) + ( z + 1) = 0, or 3 x − 3 y − z = 4.
