Mathematical Models 2 - Winter 2009 Final Exam, Exams of Mathematics

The solutions to a university-level mathematics exam covering topics such as calculus, trigonometry, logarithms, and integration. It includes both multiple-choice and free-response questions, with answers provided to the nearest decimal place.

Typology: Exams

2012/2013

Uploaded on 02/27/2013

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Mathematical Models 2
201-225 Instructor: Bob DeJean
Winter 2009
Final Exam
Answers to 4 decimal places please.
2 mark question
What is the Root Mean Square of the function y = 20 sin (60πt) ?
Is y = 3x2 – 5 a solution of xy’ – 2y = 10 ?
3 mark questions
Find the derivatives
y = 12 tan-1 (2x+3)
y = x3 ln(x - 2)
y = 173x-5
1
1
2
2
+
=x
x
e
e
y
pf3
pf4
pf5
pf8
pf9

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Mathematical Models 2 201-225 Instructor: Bob DeJean Winter 2009 Final Exam Answers to 4 decimal places please.

2 mark question

What is the Root Mean Square of the function y = 20 sin (60πt)?

Is y = 3x^2 – 5 a solution of xy’ – 2y = 10?

3 mark questions

Find the derivatives y = 12 tan-1^ (2x+3)

y = x^3 ln(x - 2)

y = 173x-

1

1 2

2

− = (^) x

x

e

e y

Integrate

∫ x^ −^ x 6 dx =

∫ −^ =

2 1 ( 4 x 3 )^2 dx

∫ +^ =

9 4

3 x dx

2

0

15 sin^4 3 cos 3

π x x dx

∫ e − dx =

e x

x 3 1

∫ x^ dx =

2 4 x^3

What is this area?

Find the area under the curve y = 4x^3 + x – 1 between x = 2 and x = 3.

Here’s an area: Approximate the area using the Trapezoidal Rule with n = 5

Solve for y xy’ + 3y^2 = 0

Find y: y’ – y^2 = 16 When x = 0, y = 0

Find y: ′^ + = 2 x

y^ y

What is the equation of the tangent line to y = 6x^2 – 24x at the point where x = 5?

Solve ln (x + 3) - 2x = 0 using Newton’s Method. Your answer should be accurate to 3 decimal places. Show your work.

7 mark question

Here is a function 

x if x

if x f x 0

and here is its graph:

I want the first few terms of its Fourier Series. What is a 0?

What is a 1?

What is b 1?

Use these answers to write the first part of the Fourier Series forf (x).

Answers

yes

( 2 3 ) 1

x +^2 +

3 ln( 2 ) 2 3 −

x

x x^ x

17 3x-5^ (ln 17) 3

2 2

2 ( 1 )

x +

x e

e

3x 4 + 2x-5^ + C

  • 1 5/3 ln | e x^ – 1 | + C

C

x

ln 4

4 3

1

3

1/5 sin 5x + C 1/3 (sin x) 3 – 1/5 (sin x)^5 + C x (^) + C

sin 1

10 tan -1^ (x + 3) + C

x^ x + x + C 25

sin 5 5

cos 5

C x y

− 1 = − 3 ln or x C

y

3 ln

y (^) = x

tan 4

(^1 1) or y = 4 tan( 4x)

y = x + C/x y = 36x – 150 x ≈ 0.