Suggested+Solutions+MaC2NVC08+Part+II+VG MVG Level+Test

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Suggested solutions MaC2NVC08-Part II: VG/MVG-level Test NV-College

CHAPTER TEST: Ch 2 Rate of Change and Differentials

MATHEMATICS COURSE C

Fall 2009: MaCNVC08

Part II: VG/MVG-level

Warning: There are more than one versions of the test.

Instructions

Test period 11:35-12:50

Tools Formula sheet, ruler and graphic calculator.

The test For most items a single answer is not enough. It is also expected

• that you write down what you do

• that you explain/motivate your reasoning

• that you draw any necessary illustrations.

After every item is given the maximum mark your solution can receive.

[2/3] means that the item can give 2 g-points (Pass level) and 3 vg-points

(Pass with distinction level).

Items marked with give you a possibility to show MVG-quality (Pass with

special distinction quality). This means that you use generalised methods,

models and reasoning, that you analyse your results and account for a clear

line of thought in a correct mathematical language.

Try all of the problems. It can be relatively easy, even towards the end of

the test, to receive some points for partial solutions. A positive evaluation

can be given even for unfinished solutions.

Mark limits The test gives totally at the most 26 points, out of which 21 vg-points and 3

MVG-pints. To pass the test you must have at least 9 points.

To qualify for higher grade you need to have a score more than 27 points in

the G-test.

To get the test character Pass with distinction (VG) you must have at least

17 points. Pass with distinction level Excellence (MVG) requires at least 20

points out of which at least 14 VG points and excellent quality presentation

of the solutions ¤¤.

Name: ________________________ Class-School: NVC08 Sjödalsgymnasiet

Problem 1a 1b 2 3a 3b 3c 4a 4b 5 Sum 26 Total/Limits Grade

G 1 1 1 1 1 5 9 G

VG 1 2 3 1 2 2 2 4 4 21 17 VG

MVG ¤ ¤ ¤ ¤¤¤ 20 MVG+¤¤

G

VG

MVG

© [email protected] Not for sale. Free to use for educational purposes☺





1. Differentiate the following functions with respect to x , and calculate the value of the

derivative of the function at the given point. Give the answer in the surd (exact) form.

Show the necessary details of your calculations as clear as possible:

a. Find ( )1−′ y if x x

x x y 5

35

930252

⋅−

+−= . [1/1]

b. Find ( )4 f ′ if ( )1

2

−

+−=

x

x x x x x f . [0/2]

Suggested solutions:

a. x x

x x y 5

35

93025 2

⋅−

+−=

( )( )

( ) x x x x x x

x y 15255355

35

35 2

2

−=⋅−=⋅−

−=

155015225 12 −⋅=−⋅⋅=′ − x x y Answer: 1550 −⋅=′ x y [0/1]

( ) ( ) 651550151501 −=−−=−−⋅=−′ y Answer: ( ) 651 −=−′ y [1/0]

b. ( )1

2

−

+−=

x

x x x x x f

( )( ) ( ) ( ) 2

12

11

1

1

12 x x x x x x

x

x x

x

x x x x f −=−=−⋅=

−

−⋅=

−

+−⋅=

( ) ( ) x

x x x f x x x f 2

11

2

11

2

11 2

11

2

1

2

1

−=−=−=′⇔−=−−

[0/1]

( ) ( )43

411

42

1142

11 =−=−=′⇔−=′ f x

x f Answer: ( )434 =′ f [0/1]

2. Find( )

h

h

h

11lim

5678

0

−+−

→by interpreting it as the derivative of a function. [0/3]

Suggested Solutions:

If we interpret( )

h

h

h

11lim

5678

0

−+−

→as the derivative of a function, then we may

rewrite it as:

( )( )

h

h f

h

11lim1

5678

0

−+=′

−

→

Then may recognize the function is ( ) 5678−= x x f . This is due to the fact that:

( )( ) ( ) ( ) ( ) ( )

h

h

h

h

h

f h f f

hhh

11lim

11lim

11lim1

5678

0

56785678

00

−+=

−+=

−+=′

−

→

−−

→→ [0/1]

( ) ( ) ( ) ( ) 56781567815678567856795679156785678 −=⋅−=′⇔−=−=′⇔=

−−−−− f x x x f x x f [0/1]

Therefore:

Answer: ( )( )

5678

11

lim1

5678

0 −=

−+

=′

−

→ h

h

f h [0/1]





3. Atmospheric pressure ( ) mBar xP as a function of km x above the sea level, decreases

according

( ) mBar e xPx145.0

1013 −⋅=

a. At what height the atmospheric pressure is mBar .600 ? [1/1]

b. Find ( )00.4P′ . What does ( )00.4P′ mean?[1/2]

c. At what height above the sea level does the atmospheric pressure decreases at the

rate of kmmbar /0.50− ? [0/2/¤]

Suggested Solution:

P Atmospheric pressure vs altitude

0

200

400

600

800

1000

1200

0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,0

x Altitude km

P [ m B a r ]

( ) mBar e xPx145.01013 −⋅=

a. Answer: The atmospheric pressure is mBar .600 at the height

km61.3 above the sea level. x ≈

( ) mBar e xPx145.01013 −⋅=

xe145.01013600 −⋅= ⇔ 6001013 145.0 =⋅ − xe [1/0]

5923.0

1013

600145.0 ==− xe ⇔ ( ) ( )5923.0lnln

145.0 =− xe

( ) ( )5923.0lnln145.0 =⋅− e x ( ) 1ln =e , and ( ) ( )k k ak

loglog ⋅= are used

When evaluating your work, in problems 3-5 I will take intoconsideration:

• how well you perform your investigation.

• how relevant, clear and complete your solutions are.• how generalized your solutions are.• if your calculations are correct.• how well you analyze and evaluate your results.





( )kmkm x 612.3

145.0

5923.0ln=

−= Answer: km x 61.3≈ [0/1]

b. Answer: ( ) kmmBar P /2.8200.4 −≈′ .This means that the atmospheric

pressure decreases at the rate of kmmBar /2.82− at 4.00 km above the

sea level. [0/1] ( ) mBar e xPx145.01013 −⋅=

( ) ( ) kmmBar e xPx /145.01013

145.0−⋅−⋅=′ ( ) ( ) xk xk ek A x f e A x f ⋅⋅ ⋅⋅=′⇔⋅= is used

( ) kmmBar e xPx /89.146

145.0−⋅−=′ [1/0]

( ) ( )kmmBar kmmBar eP /2.82/89.14600.4

00.4145.0 −≈⋅−=′ ⋅− [0/1]

c. Answer: At km x 43.7= above the sea level the atmospheric pressure

change at the rate of km/ .mbar 0.50−

( ) kmmBar e xPx /89.146 145.0−⋅−=′ ⇔ kmmBar kmmBar e

x /0.50/89.146 145.0 −=⋅− −

kmmBar kmmBar ex

/3404.0/89.146

50145.0 ==− [0/1]

( ) ( )3404.0lnln145.0 =− x

e

( ) ( )3404.0lnln145.0 =⋅− e x ( ) 1ln =e , and ( ) ( )k k ak loglog ⋅= are used

( )kmkm x 43.74321.7

145.0

3404.0ln≈=

−= [0/1/¤]

-160

-140

-120

-100

-80

-60

-40

-20

0

20

0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,0

X Altitude [km]

P ' [ m B a r / k m ]

( ) kmmBar e xPx /89.146 145.0−⋅−=′





( )4. The figure below is the graph of the function x

x10

= and that of the tangent to it at

a

f

x = . Calculate the area of the triangle AOB

a. if 9=a . [1/2]

b. as a function of a . [1/4/¤]

Suggested solutions:The function ( )

x x f

10= and the tangent to it, mkx y += at a x = share two important

properties

a. : The function and its tangent share the point a( ) ( )a f a y = x = .

b. ( )a f : The slope of the tangent to the function at ak ′= x = is equal to

the rate at which the function is changing at the point. [0/1]

( ) x

x f 10

=

( )a

a f a y10

)( == Answer: ( )a

a f a y10

)( == [1/0]

( ) 2

1

2

110

1010 −

⋅=== x

x x

x f

( ) x x

x x x f 5

52

102

31

2

1

−=⋅−=⋅−=′−−−

Answer ( )aa

a f 5

−=′=k [0/1]

Therefore equation of the tangent to ( ) x

x f 10

= at a x = may be written as

m xaa

y +−= 5 .





The value of may be determined usingm ( )a

a f a y10

)( == .

maaaa

+/⋅/

−=510

aaam 15510 =+= [0/1]

Answer: The equation of the tangent to the curve ( ) x

x f 10

= at a x = is

a x

aa y

155+−=

9=a

[0/1]

If : The equation of the tangent to the curve at 9= x is 527

5+−= x y :

527

5

3

15

39

5

9

15

99

5

+−=+⋅−=+−= x x x y .

Using the general case and recognizing the fact that the height of the

triangle, OA, is the y-intercept of the tangent:a

OA15

= , and the base of

the triangle, OB , is x-intercept of the tangent:a

OBaa

1550 +−=

Multiply both sides of the equation by5

aa : aOB 3= [0/1]

And the area of the triangle is:

( ) 25.223

15

2

1

2

1lua Aa

aOBOA A ⋅=⇔⋅⎟

⎠

⎞⎜⎝

⎛ ⋅=⋅= [0/1]

For the case :9=a lua

OA 59

1515=== and luaOB 27933 =⋅==

Answer: The area of the triangle is: AOB2

5.67 lu A = : [1/0]

225.6735.2295.225.225.67275

2

1

2

1lua AluOBOA A =⋅=⋅=⋅=⇔=⋅⋅=⋅=





5. Use the definition of the derivatives, i.e.( ) ( )

h

x f h x f

h

−+→0

lim , to show that the derivative

of the function is:( ) ( ) xg x x f ⋅= 2 ( ) ( ) ( ) xg x xg x x f ′⋅+⋅=′ 22 . [0/4/¤]

Suggested solutions:

( ) ( ) ( ) ( ) ( ) ( )h

xg xh xgh x

h

x f h x f x f

hh

⋅−+⋅+=−+=′→→

22

00limlim

( )( ) ( ) ( ) ( ) ( ) ( ) ( )

h

xg xh xghh xg xhh xg x

h

xg xh xgh xh x x f

hh

⋅−+⋅++⋅++⋅=

⋅−+⋅++=′

→→

222

0

222

0

2lim

2lim

( )( ) ( ) ( ) ( )

h

h xgh

h

h xg xh

h

xg xh xg x x f

hhh

+⋅+

+⋅+

⋅−+⋅=′

→→→

2

00

22

0lim

2limlim

( )( ) ( )[ ]

( ) ( )h xghh xg xh

xgh xg x x f

hhh+⋅++⋅+

−+⋅=′

→→→ 00

2

0lim2limlim

( )( ) ( )

( ) xg xh

xgh xg x x f h ⋅+⎥⎦

⎤

⎢⎣

⎡ −+

⋅=′ → 2lim0

2

( ) ( ) ( ) xg x xg x x f ⋅+′⋅=′ 22 QED

Answer: ( ) ( ) ( ) xg x xg x x f ′⋅+⋅=′ 22

Note that in the calculations above the following steps are taken:

( ) ( ) ( ) ( ) ( ) ( )( ) xg x

h

xgh xg x

h

xg xh xg x

h

xg xh xg x

hhh′⋅=⎥⎦

⎤⎢⎣

⎡ −+⋅=

⋅−+⋅=

⋅−+⋅→→→

2

0

222

0

22

0limlimlim

( ) ( )( ) ( ) ( xg xh xg xh xg x

h

h xgh x

h

h xg xh

hhhh⋅=+⋅=+⋅=

/

+⋅)

/=

+⋅→→→→

2lim22lim2

lim2

lim0000

( )

( ) 0limlim 0

2

0=+⋅=

+⋅

→→ h xghh

h xgh

hh

Suggested+Solutions+MaC2NVC08+Part+II+VG MVG Level+Test

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