Selected Solutions to MAM1000W Homework Test 3 Samples – 2011 ISCIMA002 – Imaad Isaacs

The content below offers solutions and/or guidelines to questions 1 through 15 for the

Homework Test Sample Questions for HW Test 3 of 2011.

Although these methods may me be memorized, it should not be seen as a substitute for the understanding of the concepts. It should be used as a means to check for any errors in

your calculations or method after you have attempted the questions on your own.

“jusqu'ici tout va bien...”

any good derived from this project, is by the grace of the Creator, and any error is by myself...

Selected Solutions to MAM1000W Homework Test 3 Sample Questions – 2011 ISCIMA002 – Imaad Isaacs

QUESTION 1

f x( ) = ln x45 + 2( )

f | 1( ) = ?

f | x( ) =

45x!

15"

#$

%

&'

x45 + 2

f | 1( ) = 0,267

QUESTION 2

f x( ) = e 5x+3 ln 2x2 +3( )f | 1( ) = ?

f | x( ) = e 5x+3 52 5x +3!

"#

$

%&ln 2x2 +3( )+ e

5x+3 4x( )2x2 +3

f | 1( ) = 37,60

QUESTION 3

f x( ) = arctan 7x2 + 24

f | 2( ) = ?

f | x( ) = 1

1+ 7x2 + 24

!

"

####

$

%

&&&&

127x2 + 24

!

"#

$

%&

'12!

"

##

$

%

&&4.14x16

!

"#

$

%&

f | 2( ) =1,504

QUESTION 4

f x( ) = 3x2 + 4( )x+1( )

f | 1( ) = ?

y = f x( ) = 3x2 + 4( )x+1( )

ln y = x +1( ). ln 3x2 + 4( )

derive : ln y = x +1( ). ln 3x2 + 4( )1ydydx!

"#

$

%&= ln 3x2 + 4( )+

x +1( ) 9x2( )3x2 + 4

dydx!

"#

$

%&= y ln 3x2 + 4( )+

x +1( ) 9x2( )3x2 + 4

'

())

*

+,,

f | 1( ) = 221,35

QUESTION 5

6+ 5t +3 1+ x ln t( ) =14tetx

dxdt

!

"#

$

%&= ?

t =1; x = 0

derive : 6+ 5t +3 1+ x ln t( ) =14tetx

5+3 dxdt

!

"#

$

%&ln t +

xt

!

"#

$

%&=14tetx +14tetx x + t dx

dt!

"#

$

%&

!

"#

$

%&

QUESTION 6

4y3 + 5xy+ 4x2 =13

d 2ydx2!

"#

$

%&= ?

x =1; y =1

derive : 4y3 + 5xy+ 4x2 =13

12y2 dydx!

"#

$

%&+ 5y+ 5x

dydx!

"#

$

%&+8x = 0

dydx!

"#

$

%&=

'5y'8x12y2 + 5x

substitute : x =1; y =1dydx!

"#

$

%&=

'1317

QUESTION 7

x!0lim 1" e4x

2

xsin5x=

x!0lim

ddx1" e4x

2( )ddx

xsin5x( )=

x!0lim "e4x

2

.8xsin5x + 5xcos5x

=x!0lim

"e4x2

. 8x( )2 "8e4x2

5cos5x + 5cos5x " 25xsin x="810

= "0,800

QUESTION 8

substitute : t =1; x = 0

5=14+14 dxdt

!

"#

$

%&

dxdt

!

"#

$

%&=

'914

= '0,643

derive :12y2 dydx!

"#

$

%&+ 5y+ 5x

dydx!

"#

$

%&+8x = 0

d 2ydx2!

"#

$

%& 12y2 + 5x( )+ dy

dx!

"#

$

%& 24y

dydx!

"#

$

%&+ 5

!

"#

$

%&+ 5

dydx!

"#

$

%&+8 = 0

d 2ydx2!

"#

$

%&=

'8' 5 dydx!

"#

$

%&'

dydx!

"#

$

%& 24y

dydx!

"#

$

%&+ 5

!

"#

$

%&

12y2 + 5x( )

substitute : dydx!

"#

$

%&=

'1317; x =1; y =1

d 2ydx2!

"#

$

%&= '0,85

x!"lim 36x2 + 4x # 6x

=x!"lim 36x2 + 4x # 6x( )$ 36x2 + 4x + 6x

36x2 + 4x + 6x

%

&''

(

)**

=x!"lim36x2 + 4x #36x2

36x2 + 4x + 6x=

x!"lim 4x

x 36+ 4x+ 6x

=x!"lim 4x

x 36+ 4x+ 6

%

&'

(

)*

=412

= 0,33

QUESTION 9

Once we find the second derivative, we only need to solve for the second factor since we are told in the question that x=1 satisfies f||(x)=0. [NB: zero-product theorem says A.B=0 therefore either A=0 or B=0, where A and B are factors.]. If x=1 satisfies f||(x)=0, we know that ln(x)=0 at x=1 and therefore only need to solve for the second factor.

!

f x( ) = x15 ln x( )15

f | x( ) =15x14 ln x( )15 +x15 .15 ln x( )14

x=15x14 ln x( )15 +15x14 ln x( )14

f || x( ) =15 x14 ln x( )15 + x14 ln x( )14[ ]|

=15 14x13 ln x( )15 +x14 .15 ln x( )14

x+14x13 ln x( )14 +

x14 .14 ln x( )13

x

"

# $

%

& '

=15 14x13 ln x( )15 + 29x13 ln x( )14 +14x13 ln x( )13[ ]=15x13 ln x( )13 14 ln x( )2 + 29 ln x( ) +14[ ]

f || x( ) = 0

0 =15x13 ln x( )13 14 ln x( )2 + 29 ln x( ) +14[ ]

0 = 14 ln x( )2 + 29 ln x( ) +14[ ]t = ln x( )

0 =14t 2 + 29t +14t = (1,31t = (0,77

(1,31 = loge xx = e(1,31

x = 0,27

(0,77 = loge xx = e(0,77

x = 0,46

0,27 < 0,46) x = 0,27

QUESTION 10

Consider the given properties (i) – (iii) of f x( ) = ax2 + bx + cx + d

.

From (i) the equation becomes f x( ) = ax2 + bx + cx ! 4

Using long division one finds that the slant asymptote is y = ax + 4a+ b , applying one’s knowledge of property (iii) above one finds that a = 3 .

The equation now becomes f x( ) = 3x2 + bx + cx ! 4

.

Using property (ii), we can derive the function f and solve for both b and c simultaneously.

f | x( ) =6x + b( ) x ! 4( )! 3x2 + bx + c( ) 1( )

x ! 4( )2

Now use f|(x)=0 at x=1

f | x( ) =6x + b( ) x ! 4( )! 3x2 + bx + c( ) 1( )

x ! 4( )2

0 =6+ b( ) !3( )! 3+ b+ c( ) 1( )

!3( )2

0 = 6+ b( ) !3( )! 3+ b+ c( )0 = !18!3b!3! b! c"c = !21! 4b

Substitute c = !21! 4b and P(1, 3) into f(x).

f x( ) = 3x2 + bx +!21! 4b

x ! 4

3=3 1( )2 + b 1( )! 21! 4b

1! 4"b = !3"c = !21! 4b = !21! 4 !3( ) = !9

Thus f(x) becomes f x( ) = 3x2 !3x ! 9x ! 4

.

Now calculate the second derivative and check to see which type of critical point P is.

f | x( ) = 3x2 ! 24x + 21x ! 4( )2

f || x( ) =6x ! 24x( ) x ! 4( )2 ! 3x2 ! 24x + 21( ) 2( ) x ! 4( )

x ! 4( )4

f || 1( ) = !2" f || 1( ) < 0

Since f||(x)<0, P is a local maximum and 2+ f 0( ) = 4,250 .

QUESTION 11

Given: y = Ax32 +Bx2 +C and f | 1( ) = 6

P 1;7( ) is on the graph and a point of inflection

dydx

=32Ax

12 + 2Bx

substitute : f | 1( ) = 6

6 = 32A+ 2B

This yields eqn.1 12 = 3A+ 4B  

and d 2ydx2

=34Ax!

12 + 2B

substitute : f || 1( ) = 0

0 = 34A+ 2B

this yields eqn.2 B = ! 38A  

solving eqn.1 and eqn.2 simultaneously we obtain: A = 8 and B = !3

y = Ax32 +Bx2 +C

C = y! Ax32 !Bx2

substitute : A,B,P 1;7( )

C = 7! 8( ) 1( )32 ! !3( ) 1( )2

C = 2

"y = 8x32 !3x2 + 2

y 4( ) =18

QUESTION 12

y = x3 +3x2 + 5dydx

= 3x2 + 6x

0 = x x + 2( )x = 0; x = !2

check endpoints as well

y 0( ) = 5  ;     y !2( ) = 9  ;     y 1( ) = 9  ;     y !1( ) = 7

sum of global maximum and minimum = 14

QUESTION 13

Area of shaded region is 2400cm2. Using the graphical interpretation with appropriate label we derive the following relationships.

Atotal = xy

Ashaded = y!8( ) x !12( )2400 = y!8( ) x !12( )

y = 2400x !12

+8

 

substitute y = 2400x !12

+8  into   Atotal = xy , derive and find turning points.

A = x 2400x !12

+8"

#$

%

&'=2400xx !12

+8x

dAdx

=2400( ) x !12( )! 2400x

x !12( )2+8

0 =2400( ) x !12( )! 2400x

x !12( )2+8

0 = !12 2400( )+ x !12( )2

0 = x2 ! 24x !34560 = x + 48( ) x ! 72( )x ( !48x = 72

y

x

4cm

4cm

6cm 6cm A=2400cm2

QUESTION 14

First draw a diagram of the situation and label it accordingly. Once you have this setup determine what the surface area will be and derive an equation for the costs involved for the base and sides. Once you have this relationship use the differentiation rules to solve for the best x-value that will yield a minimum cost and back substitute into the cost formula. See figure 3 in appendix 1.

Base material @ R35/m2 S Note: Volume is a constant and allows us express h in terms of r

Side material @ R25/m2 V = 10/m3

QUESTION 15

!

Atotal = 3x 2 + 2xh + 2.3xh

= 3x 2 + 2x 103x 2"

# $

%

& ' + 6x

103x 2"

# $

%

& '

= 3x 2 +203x

+20x

C = 35( )3x 2 + 25( ) 203x

+20x

(

) * +

, -

=105x 2 +20003x

dCdx

= 210x . 20003x 2

0 = 210x . 20003x 2

0 = 210x 3 . 20003

x =20003.210

3

=1,47

C 1,47( ) = R680,41

!

Atotal = 3x 2 + 2xh + 2.3xh

= 3x 2 + 2x 103x 2"

# $

%

& ' + 6x

103x 2"

# $

%

& '

= 3x 2 +203x

+20x

C = 35( )3x 2 + 25( ) 203x

+20x

(

) * +

, -

=105x 2 +20003x

dCdx

= 210x . 20003x 2

0 = 210x . 20003x 2

0 = 210x 3 . 20003

x =20003.210

3

=1,47

C 1,47( ) = R680,41!

V = Abase( )h= 3x 2h

10 = 3x 2h

h =103x 2