1322Final Solution S 19 PDF

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Solution to the Final ExaminationMAT1322X, Summer 2019Part I. Multiple-choice Questions (2  12 = 24 marks)DBCDEF CCFAAD1. The area of the region between the graphs of f ( x ) = − x 2 + x and g ( x ) = 2 x 2 −3 x in the interval [−1, 1] is(A) 1; (B) 2; (C) 3; (D) 4; (E) 5; (E) 6.Solution. (D) Let – ...

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MAT1322X

Solution to Final Examination

Summer 2019

Solution to the Final Examination MAT1322X, Summer 2019 Part I. Multiple-choice Questions (2  12 = 24 marks) DBCDEF CCFAAD 1. The area of the region between the graphs of f (x) = −x2 + x and g(x) = 2x2 −3x in the interval [−1, 1] is (A) 1;

(B) 2;

(C) 3;

(D) 4;

(E) 5;

(E) 6.

Solution. (D) Let –x2 + x = 2x2 – 3x. 3x2 – 4x = 0. x = 0, 4 / 3. The area of the region between the graphs of f (x) and g(x) is



0

1

2 2 (3 x  4 x) dx  0 ( 3 x  4 x )dx  3  1  4 . 1

2. Let R be the region in x-y plane bounded by the graphs of f (x) = −x2 + 2x and g(x) = x. A solid S has R as its base, and the cross sections perpendicular to the x-axis are squares. Then the volume of S is calculated by the integral 1

(C)

 ((  x  2 x)  x )dx ;   ((  x  2 x)  x )dx ;

(E)



(A)

2

2

2

0

1

2

2

2

0

1 0

x((  x 2  2 x)  x )dx ;

(B)



1 0

(( x2  2 x)  x)2 dx ;



(D) 2

1 0

((  x2  2 x)2  x2 )dx ;

1

(F)   (( x2  x)  x)2 dx . 0

Solution. (B) Let −x2 + 2x = x. x2 – x = 0, x = 0, x = 1. The side length of a cross section perpendicular to x-axis at a value x is L(x) = (–x2 + 2x) – x. The area of the cross section is A(x) = (L(x))2 = ((−x2 + 2x) − x)2. The volume of the solid is calculated by V=



1 0

1

A( x) dx   ((  x2  2 x)  x)2 dx . 0

3. A pool of the shape of an inverted truncated pyramid as shown in the following figure is filled with water. The top of the pool is a square of side-length 10 meters, the bottom of the pool is a square of side-length 2 maters, and the depth of the pool is 4 meters. The weight of a horizontal layer of water x meters above the bottom of the tank with thickness dx is w(x) = g(2x + 2)2dx. 3 Assume the density of the water is  km/m , and let the gravity of acceleration be g.

1

MAT1322X

Solution to Final Examination

Summer 2019

10

4

2

The work, in Joules, needed to pump all the water in the tank to a nozzle 1 meter above the top of the tank is calculated by the integral 5

(A)  g  (2 x  2)2 (5  x) dx ; 0

4

(C)  g  (2 x  2)2 (5 x) dx ; 0

5

(E)  g  (2 x  2) 2 ( x  1) dx ; 1

5

(B)  g (2 x  2)2 (5 x) dx; 1

5

(D)  g (2 x  2)2 ( x  1) dx ; 0

4

(F)  g  (2 x  2) 2 ( x  1) dx . 0

Solution. (C) The distance between this layer and the height of the nozzle is D(x) = (1 + 4) – x = 5 – x. The work needed to pump this layer of water to the height of the nozzle is W(x) = w(x)D(x) 2 = g(2x + 2) (5  x)dx. The layer of water at the bottom of the pool has x = 0, and the layer of water at the top of the pool has x = 4. Hence, the total work needed is calculated by W = 4

 g  (2 x  2) 2 (5  x) dx . 0

4. A trapezoidal surface is submerged vertically into water with density kg/m3 such that the top is 1 meter under the water surface as shown in the following figure. The area of a horizontal 1 stripe of the surface x meters under the water surface of height dx is A(x) = ( x  5) dx. Let g be 2 the gravity of acceleration. The force, in Newtons, acting on this surface can be calculated by the integral water surface x

3m

1m

2m

4m 3 1  g  ( x  1)( x  5)dx ; 1 2 3 1 (C)  g  ( x 1)( x  5)dx ; 0 2

(A)

2 1  g  ( x  1)( x  5)dx ; 0 2 3 1 (D)  g  x( x  5) dx ; 1 2

(B)

2

MAT1322X

(E)

Solution to Final Examination

2 1  g  x( x  5) dx ; 0 2

(F)

Summer 2019

3 1  g  x( x  5) dx . 0 2

Solution. (D) The depth of a horizontal stripe of the surface x meters under the water surface with height dx is x. The pressure acting on this stripe is P(x) = gD(x) = gx. The force acting 1 on this stripe is F(x) = P(x)A(x) = gx(x + 5)dx. The stripe at the top of the trapezoid has x = 1, 2 and the stripe at the bottom of the trapezoid has x = 1 + 2 = 3. Hence, the total force is calculated 3 1 by the integral F =  g  x( x  5)dx . 1 2 5. The average value of the function y = x 4  x2 on interval [0, 2] is (A)

8 ; 3

(B)

3 ; 8

5 ; 3

(C)

(D)

3 ; 5

(E)

4 ; 3

(F)

3 . 4

Solution. (E) Use variable substitution u = 4 – x2. Then u' = −2x. When x = 0, u = 4, and, when x = 2, u = 0. 2



2

0

x 4  x2 dx 1 8 4 1 0 1 2 3/ 2 0 8  2 0 x 4  x dx    udu   = [ u ]u 4  . The average value is  . 2 0 2 4 23 3 23 3

6. The length of the arc y =

(A)

93 ; 48

(B)

91 ; 48

1 3 1 , 1  x  2, is x + 4 3x 47 ; 24

(C)

(D)

49 ; 24

(E)

19 ; 12

(F)

23 . 12

1 1 1 3 2 9 4 1 9 4 1 x − 2 , (y')2 = x −  4 , 1 + (y')2 = x +  4 3x 2 9x 2 9x 4 16 16 2 1  3 1 3 =  x 2  2  . Then 1  ( y') 2  x2  2 . 3x  4 3x 4 Solution. (F) y' =

The length of the arc is 2

1  1   1 1 23 3 2 1 3 1  1  4 x  3x 2  dx   4 x  3x  x1   2  6    4  3  12 . 2

7. Consider improper integral 

1

0

1 dx . Which one of the following statement is true? x x

3

MAT1322X

Solution to Final Examination

Summer 2019

1 1 when 0 < x < 1, and  x x 2 x



1

1

0

2 x

1 1 when 0 < x < 1, and  x x 2 x integral is convergent.



1

1

0

2 x

1 dx is convergent, this improper integral is x

(A) Because

dx is divergent, this improper integral

is divergent. (B) Because

(C) Because

1 1 when 0 < x < 1, and  x x x



1

1 1 when 0 < x < 1, and  x x x



1

1 1 when 0 < x < 1, and  x  x 2x



1 1  when 0 < x < 1, and x x x

1

0

dx is convergent, this improper

convergent. (D) Because

0

1 x

dx is divergent, this improper integral is

divergent. (E) Because

1 0

1 dx is divergent, this improper integral is 2x

divergent. (F) Because



0

1 dx is convergent, this improper integral is x

convergent. Solution. (C) When x is close to 0, x is smaller than x , which can be omitted. Hence, the 1 1 1 1 is similar to function . The integral  behaviour of dx converges, so does the 0 x x x x 1 1 integral  dx . Then (A), (D), and (E) are all false. (B) is false because the convergence 0 x x of the integral of the smaller function does not imply the convergence of the integral of the 11 bigger function. (F) is false because integral  dx diverges. 0 x 8. Suppose Euler's method with step size h = 0.05 is used to estimate y(0.1), where y(t) is the solution to the initial-value problem y' = (t + 1)y2, y(0) = 1. Which one of the following values is closest to the result that you obtained? (A) 1.015;

(B) 1.050;

(C) 1.108;

(D) 1.153;

(E) 1.175;

(F) 1.205.

Solution. (C) The iteration formula is yn+1 = yn + h(tn + 1)yn2 with t 0 = 0, and y0 = 1. t0 = 0 t1 = 0.05 t2 = 0.1

y(0) = y0 = 1 y(0.05)  y1 = 1 + 0.05  (0 + 1)  12 = 1.05 y(0.1)  y2 = 1.05 + 0.05  (0.05 + 1)  1.052 = 1.108. 4

MAT1322X

Solution to Final Examination

Summer 2019

9. Which one of the following statements is true? 

(A) If

a

n 1

n



converges, then

a

n 1

n

converges.



(B) If

a

n



is an alternating series and lim a n = 0, then n 

n 1

a

n

converges.

n 1



(C) If the limit lim n a n = c > 0 exists, then series n 



(D) If

a

n

n 1

a

n

is absolutely convergent.

n 1



is an alternating series and | an+1 | < | an | for all n, then an converges. n 1



(E) If lim a n = 0, then n

a

converges.

a

diverges.

n

n 1 

(F) If lim a n 0, then n

n

n 1

Answer. (F) 

1 n x . Then the 5-th n 0 2  1

10. Suppose the Maclaurin series of a function f (x) is f (x) =  derivative of this function at x = 0 is f (5)(0) = (A)

40 ; 11

(B)

40 ; 33

(C)

1 ; 33

(D)

1 ; 11

(E)

n

120 ; 11

(F)

122 . 33

(5) 1 1 f (0) . Hence,   Solution. (A) The coefficient of x in the series is c5 = 5 2  1 33 5!

5

f (5)(0) =

5! 120 40 .   33 33 11

1 1 3  (1  x)  1/ 2  1 x x2 ... . 2 8 1 x

11. By the binomial series formula we find Using this result and the fact that



x

0



1 1 t



2





dt  ln x  1  x2 , we can find the Maclaurin 

series of the function y = ln x  1  x 2   cn x n . Then c5 =

(A)

3 ; 40

(B)

1 ; 8

(C)

3 ; 8

n 0

(D)

3 ; 32

(E)

1 ; 40

(F)

1 . 32

5

MAT1322X

Solution to Final Examination

Solution. (A) Substitute t2 for x in the Maclaurin series We have



1 1 3  (1  x)  1/ 2  1 x x2 ... . 2 8 1 x

1

1 2 3 4  1  t  t  ... . Hence, 2 8 1 t 2



ln x  1  x 2  

c5 =

Summer 2019

x 0

x 1 3 1 3 5  dt    1  t 2  t 4  ... dt  x  x 3  x  ... . 0 8 6 40  2  1 t 2

1

3 . 40

1 1 12. Recall that the Maclaurin series of the function ln (1 + x)  x  x 2  x3  ... . Then 2 3 3 3 ax  ln(1  ax ) = lim x 0 x6 (A)

1 a; 2

1 2 (C) − a ; 2

(B) 2a;

(D)

1 2 a; 2

(E) 2a2;

(F) a2.

1 1 3 Solution. (D) Use the Maclaurin series of ln (1 + t)  t  t 2  t 3  ... Let t = ax . We have 2 3 1 8 1 8 ln(1 + ax3)  ax 3  a 2 x 6  a3 x9  ... . Then ax3 − ln(1 + ax3) = a2x6 − a3 x9 + … . 2 3 2 3 1 2 6 8 3 9 a x  a x  ... ax 3  ln(1  ax 3) 1 2 2 3 lim lim   a . 6 6 x 0 x 0 x x 2

Part II. Long Answer Questions (26 marks) 1. (6 marks) Find function y(t), where y(t) is the solution of the initial-value problem y' = (cos t)(1 – y2), y(0) = 2. Solution. The equilibrium solutions are y = 1. When y  1, separate the variables: 1  1  y 2 dy   cos tdt . Integrating both sides,



1 1  1 1  1 1 y   sin t  C . dy    dt  ln  2 1 y 2  1 y 1 y  2 1 y

6

MAT1322X

Solution to Final Examination

Summer 2019

1 y 1 y 1 y = Ke 2sin t, where K =  2sin t  2C,  K1e2sin t , where K1 = e2C > 0. Then 1 y 1 y 1 y 1 y = −3e2sin t. K1  0. Use the initial condition, when t = 0, K = −3. 1 y ln

Solve this equation for y: y =

2sin t 3e  1 . 3e2sin t  1

2. (6 marks) Use an appropriate test method to determine whether each of the following series is convergent or divergent. Name the test method that you are using and the conditions why this test method applies. 

(a)

1  sin n ; 2 1 n 1

 2n

n



(b)

 2n  1    ;  n  1  3n  1 

(c)

( 1)



n 1

n

 2n  1  .  3n 1 

Solution. (a) Since the series is a positive series, we can use the comparison test. Since    2 1 1  sin n 1 sin n 2   2 , and the series converges by p -series test, series    2 2 2 2 2 2n  1 n n 0 n n 0 n n 1 2n  1 converges by comparison test. (b) Use the root test to test absolute convergence. Since this series is a positive series, it is convergent if and only if it is absolutely convergent. n

2 n 1  2  2n  1  Let an =  . Then lim n | an |  lim    < 1. This series is convergent by the    n n  3n  1   3n  1  3 root test. 2n 1  2 (c) Since lim    0, by the test of divergence, this series is divergent. n 3n  1    3 

1 ( x  2) n . For which value(s) of x is this series n 1 2 (2 n  1) absolutely convergent, for which value(s) of x is this series conditionally convergent, and for which value(s) of x is this series divergent?

3. (7 marks) Consider power series 

n

7

MAT1322X

Solution to Final Examination

Summer 2019

Solution. Use the ratio test. 1 x  2  2n  1 x 2 ( x  2)n 2n (2n  1)  lim   1. Then | x + 2 | < 2, −2 < x + 2 <    1 n n n  2 n (2n  1) (x  2) 2  2n  1 2 2, or −4 < x < 0. This series is absolutely convergent when −4 < x < 0, and it is divergent when x < −4, or x > 0.

Let lim



When x = 0, this power series becomes

2n   n n 1 2 (2n  1)



1

 2n  1 .

Let an =

n 1

1 , and let bn = 2n  1

 a n 1 1 1  > 0. Since the harmonic series  diverges, . Then nlim n  lim n  b  2n  1 n 2 n 1 n n diverges.



1

 2n  1 n 1



When x = −4, this power series becomes

 ( 2) n 1 1 . Let an = (−1)n ( 1)n  .   n 2n  1 2n  1 n 1 2 (2 n  1) n 1

Since | an | is decreasing and approaches 0, this series converges. Since series    1 1 | an |   2 n  1 diverges. (1) n 2n  1 is conditionally convergent. n 1 n 1 n 1 Summarizing: This power series is absolutely convergent when −4 < x < 0. It is conditionally convergent when x = 4. It is divergent when x  0 or x < −4. 4. (7 marks) Suppose a function z = f (x, y) is defined implicitly by the equation x2y + xz – y3z + 1 = 0 near the point p = (2, −1, 1). (a) (3 marks) Find the gradient vector of f (x, y) at the p. Solution. Let F(x, y, z) = x2y + xz – y3z + 1. Fx = 2xy + z, Fy = x2 – 3y2z, Fz = x – y3. F (2, 1,1) 3 Fx(2, −1, 1) = −3, Fy(2, −1, 1) = 1, Fz(2, −1, 1) = 3. Then, zx (2, 1,1)   x   1, Fz (2, 1,1) 3 F (2, 1,1) 1 and z y (2, 1,1)   y   . The gradient vector of function f (x, y) at point (2, −1, 1) is Fz (2, 1,1) 3

1  f ( x, y)   1,   . 3  (b) (2 marks) Find the equation of the tangent plane of the graph of this equation at point p. Solution. The equation of the tangent plane of this equation at point p is 8

MAT1322X

Solution to Final Examination

Summer 2019

−3(x – 2) + (y + 1) + 3(z – 1) = 0, or 3x − y − 3z = 4. (c) (2 marks) Find the directional derivative of function f (x, y) in the direction of vector v = (4, 3) at point p. Solution. The unit vector in the direction of v is u =

v 4  , |v |  5

3  . The directional derivative of 5

function f (x, y) in the direction of v at point p is Dv(f ; 2, −1, 1) = f ( x, y )  u 

4 1 3 3    . 5 3 5 5

9...