matb 113 tst2 sem2 09-10 (answer)
TRANSCRIPT
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8/18/2019 MATB 113 Tst2 Sem2 09-10 (Answer)
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No. Solution Marks
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1.(a)
b)
;k 2
t ji
++−+
+= 1))(1(tan12ln)( t t t Find the angle between the
velocity and acceleration vectors at time 0=t .
k jirv
11
1
1
2)()( 222
+
++
++
==
t
t
t t
t t t
jrv == )0()0(
( )
( )k jiva
+
+−++
+−+
+
−+==
−
1
212
11
1
2
1
)2(2)1(2)()(
2
2!122
222
2
t
t t t t
t
t
t
t t t t t
k iva +== 2)0()0(
20cos
"
2cos
11 π θ ==
+•= −−
k i j
dt t t ∫
+++
1
0
)1(#$
k ji
k jik ji 2
$
#%
1
2#%
1
0
2%
++=
+++ t t
t
t
EXAMINATION SOLUTION FORM
SUBJECT CODE: MATB 113 –TEST 2 SEMESTER: 1 2!!"#1!
&age '1' o ''
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8/18/2019 MATB 113 Tst2 Sem2 09-10 (Answer)
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No. Solution Marks
2a) ct z t a yt a x === )sin()cos(
k jir ct t at at ++= )sin()cos()(
k jivr ct at at t ++−== )cos()sin()()(
2222)cos(sin)( ct t at ++=v
22 ca +=
dt ca L
t 22
0
0
+= ∫ ] 0
0
22
t
ca +=
022 t ca += units
1 1
*
*
*
1
1 1
1
1
+otal 1"
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No. Solution Marks
2.(b)
(c)
+o ind the unit tangent vector T and the unit normal vector N or the
given circular heli,.
)(
)()(
t
t t
r
rT =
22
)cos()sin()(
ca
ct at at
+
++−=
k jiT
22
0)sin()cos()(
ca
t at at
+
−−
=T
2222
2)(
ca
a
ca
at
+
=
+
=T
0)sin()cos()(
)()( t t
t
t t −−==
T
TN
+o ind the curvature o the given circular heli,
222222
1)(
)(
1
ca
a
ca
a
cat
t +=
++== T
vκ
* *
1
*
**
**
* * *
*
*
1
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No. Solution Marks
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2
* *
1
*
**
*
*
* * *
*
*
1
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No. Solution Marks
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$a) +o ind the tangential and normal com-onents o acceleration or thecurve
k j$ir )$(2()()( t t t t ++= at t 1.
k jak jiv t t t t t /2)($2)( 2 +=++=
( ) ( )$2!1%2%2
$/%121%1)( t t t t t t
dt d t
dt d a
T +++=++== −v
%2
2
%1
1%
t t
t t
++
+=
]1%
221 ==t T a
%2
2
%1
$2)(
t t
t t t
++
++=
k jiT
( ) ( )
( )%2
$2!1%2%2
%1
$/%121%1/20
)(t t
t t t t t t t
t ++
+++−++=
−
T
( ) ( )
( ) 2!$%2
2$%
%1
$211%%1/20)(
t t
t t t t t t t t
++
+−++=T
( ) 2!$1%
11/22)1(
−−−=T
$%$
1$$
2#%%
$2%2"/%%)1( =
++=T
1//%.0$%$
1$$
1%
11≈==
dt
d T
vκ
$$.2)1%(1//%.0)( 2
=== t a N vκ
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No. Solution Marks
$a)
b)
c)
r k jak jiv t t t t t /2)($2)( 2 +=++=
]1%
22
1%
/20$21
)1(
)1()1(1
=•
=•
==v
avt T
a
] k ji
k ji
avv
av2//
/20
$21)1()1()1(
)1()1(1
+−==×
×
==t N
a
] $$.21%
#/1
===t N
a
+o ind the e3uation o the tangent -lane and normal line to the surace
y x z 2= at the -oint (21%).
4et % be the normal vector to the tangent -lane.
] ( ) k ji%k ji% −+=−+= %%2 )%122 x xy
+hus e3uation o the tangent -lane at (21%) is
%%
0)%()1(%)2(%
=−+
=−−−+−
z y xor
z y x
+o identiy and sketch the unction5s level surace222
)( z y x z y x f −+= at (000)
4et222)( z y x z y x f k −+== 6t (000) 0=k
0222 =−+ z y x 6 cone symmetrical about the z 7a,is
z
x y
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No. Solution Marks
%a) +ake a -ath along 2mx y = or 2!1my x =
(i) 2%
2%
)00()(lim
y x
y x
y x +
−→
%2%
%2%
)00()(lim
xm x
xm x
y x +−
=→
22
2%
2%
)00()( 11
)1()1(lim
mm
m xm x
y x +−=
+−=
→
+ake 0lim1 2 === it x yalong m
"
$lim22
2 −=== it x yalong m
(ii)
( )
+
+→ 22
22
)00()(
sin
lim y x
y x
y x
( )
1coslim2
cos2lim
sinlim 2
0
2
02
2
0===
=
→→→r
r
r r
r
r
r r r
*
*
1
* *
1
*
* *
*
1
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+otal81"
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