selected solutions to mam1000w homework test 3 samples – 2011
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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
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