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MAT1322 Fall 2017 Final Exam + Solutions, Exams of Calculus for Engineers

This is the final exam for Fall 2017.

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MAT1322D Solution to Final Examination December 2017
1
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 = x2 and under the graph of y = 8 x2 in the first
quadrant is
(A)
27
3
; (B)
29
3
; (C)
37
3
; (D)
41
3
; (E)
35
3
; (F)
32
3
.
Solution. (F) Let x2 = 8 x2. 2x2 = 8, x = 2. The area is
2
3
22
2 2 2
00 0
32
(8 ) 2 (4 ) 2 4 33
x
x
x x dx x dx x




.
2. Let R be the region above the parabola y = x2 and under the line y = 2x. Solid B is obtained by
revolving R about the line y = 4. Then the volume of B is calculated by the integral
(A)
22 2 2
0((4 ) (4 2 ) )x x dx
; (B)
42 2 2
0((4 ) (4 2 ) )x x dx
;
(C)
22 2 2
0((4 ) (4 2 ) )x x dx
; (D)
42 2 2
0((4 ) (4 2 ) )x x dx
;
(E)
22 2 2
0((4 2 ) (4 ) )x x dx
; (F)
22 2 2
0((4 ) (4 2 ) )x x dx
.
Answer. (A)
rin = 4 2x
y = 4
y = x2
y = 2x
X
Y
O
pf3
pf4
pf5
pf8
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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 ), axb , 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  xe , 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 ...

xxx   xxx .

(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 + xyyz^3 = −1 at the point (2, 1, −1).

Solution. Let F ( x , y , z ) = x^2 z + xyyz^3 + 1.

Then Fx = 2 xz + y , Fy = xz^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 yz = 4.