Mathematical Models II - Winter 2010 Final Examination, Exams of Mathematics

The questions and answers for the winter 2010 final examination of the mathematical models ii course. The examination covers various topics in calculus, including derivatives, integrals, and differential equations.

Typology: Exams

2012/2013

Uploaded on 02/27/2013

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Mathematical Models II
201-225-AB
Final Examination
Winter 2010
Instructor: Bob DeJean
Please answer any questions with decimal answers to 4 decimal places.
2 mark questions
The current through an 20 H inductor is given by I = 25 sin(4πt). What is
the Root-Mean-Square of this current ?
Does 1+
=
x
x
ysatisfy the equation yyxx
=
+
')1( ?
3 mark questions
Find the derivatives of the following functions:
x
x
y3sin
=
))3tan(ln( xy =
pf3
pf4
pf5
pf8
pf9

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Mathematical Models II 201-225-AB Final Examination Winter 2010

Instructor: Bob DeJean

Please answer any questions with decimal answers to 4 decimal places.

2 mark questions

The current through an 20 H inductor is given by I = 25 sin(4πt). What is the Root-Mean-Square of this current?

Does

  • 1

x

y x satisfy the equation x ( x + 1 ) y '= y?

3 mark questions

Find the derivatives of the following functions:

x

y =sin^3^ x

y =tan(ln( 3 x ))

Find the derivatives y =sec x

y = tan −^1 ( x − 5 )

y = x^3 ln( x − 1 )

5 y = ex − 2

4 mark questions

Differentiate implicitly: sin x – ln y = xy

Find the equation of the line that is normal to y = sin-1^ x at the point (0.6, 0.6435)

Integrate the following:

∫ + =

x −^33 xdx

( )

∫ −

2

1 3 x 14

dx

∫ (^) + xdx =

x 1 tan

sec 2

∫ −^ =

4 1

e^2 x^1 dx

x^ xe dx = sec tan sec x

Integrate

∫ 4 sin^2 (^3 x ) dx^ =

∫ (^9) − 4 x 2 =

dx

x^ sin(^4 x ) dx =

Find the area of this region:

6 mark question

Consider the function that is 1 for 2

0 ≤ x ≤^ π and 0 for the rest of - π to π.

Here is its graph:

I am interested in its Fourier Expansion.

What is a 0 =

What is a 1 =

What is b 1 =

Use these values to write the beginning of the Fourier Expansion of the function.

Answers

yes

2

3 cos 3 sin 3 x

y ′= x xx

x

y ′ =sec^2 (ln^3^ x )

x

y x x 2

′ =sec tan

1 ( 5 )^2

x

y

3 ln( 1 ) 2 3 −

x

y x x^ x

y’ = 5 (e x^ – 2)^4 e x

x y

y x y

cos

y= - 0.8x + 1.

4 m by 8 m

x C x

1 4 /^3

2

ln(1 + tan x) + C

e secx^ + C

2x – 1/3 sin 6x + C x (^) + C

sin^2 2

x^ x + sin 4 x + C 16

cos 4 1 4