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Page 1
Edurev123
6. Curves in Space
6.1 Find ?? /?? for the curve
??? (?? )=?? ?????? ?? ??ˆ+?? ?????? ?? ??ˆ+???? ??ˆ
(2010: 12 Marks)
Solution:
Given:
?? (?? )=?? cos ?? ??ˆ+?? sin ??ˆ+???? ??ˆ
?? ?? (?? )
????
=-?? sin ?? ??ˆ+?? cos ??ˆ+?? ??ˆ
?? 2
?? (?? )
?? ?? 2
=-?? cos ?? ??ˆ-?? sin ??ˆ
?? 3
?? (?? )
?? ?? 3
=?? sin ?? ??ˆ-?? cos ??ˆ
?? ?? (?? )
????
×
?? 2
?? (?? )
?? ?? 2
=|
??ˆ ??ˆ ??ˆ
-?? sin ?? ?? cos ?? ?? -?? cos ?? -?? sin ?? 0
|
=??ˆ(0+???? sin ?? )-?? (0+???? cos ?? )+??ˆ
(?? 2
sin
2
?? +?? 2
cos
2
?? )
=???? sin ?? ??ˆ-???? cos ??ˆ+?? 2
??ˆ
?? =
|
????
????
×
?? 2
?? ?? ?? 2
|
|
????
????
|
3
=
|???? sin ?? ??ˆ-???? cos ??ˆ+?? 2
??ˆ
|
|-?? sin ?? ??ˆ+?? cos ??ˆ+?? ??ˆ3
=
(?? 2
?? 2
sin
2
?? -?? 2
?? 2
cos
2
?? +?? 4
)
1/2
(?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
)
3/2
=
(?? 2
?? 2
+?? 4
)
1/2
(?? 2
+?? 2
)
3/2
=
?? (?? 2
+?? 2
)
1/2
(?? 2
+?? 2
)
3/2
=
?? ?? 2
+?? 2
?? =
[
????
????
?? 2
?? ?? ?? 2
?? 3
?? ?? ?? 3
]
|
????
????
×
?? 2
?? ?? ?? 2
|
2
=
(
????
????
×
?? 2
?? ?? ?? 2
)·
?? 3
?? ?? ?? 3
|
????
????
×
?? 2
?? ?? ?? 2
|
2
=
(???? sin ?? ??ˆ-???? cos ?? ??ˆ+?? 2
??ˆ
)·(?? sin ?? ??ˆ-?? cos ?? ??ˆ)
|???? sin ?? ??ˆ-???? cos ?? ??ˆ+?? 2
??ˆ
|
2
Page 2
Edurev123
6. Curves in Space
6.1 Find ?? /?? for the curve
??? (?? )=?? ?????? ?? ??ˆ+?? ?????? ?? ??ˆ+???? ??ˆ
(2010: 12 Marks)
Solution:
Given:
?? (?? )=?? cos ?? ??ˆ+?? sin ??ˆ+???? ??ˆ
?? ?? (?? )
????
=-?? sin ?? ??ˆ+?? cos ??ˆ+?? ??ˆ
?? 2
?? (?? )
?? ?? 2
=-?? cos ?? ??ˆ-?? sin ??ˆ
?? 3
?? (?? )
?? ?? 3
=?? sin ?? ??ˆ-?? cos ??ˆ
?? ?? (?? )
????
×
?? 2
?? (?? )
?? ?? 2
=|
??ˆ ??ˆ ??ˆ
-?? sin ?? ?? cos ?? ?? -?? cos ?? -?? sin ?? 0
|
=??ˆ(0+???? sin ?? )-?? (0+???? cos ?? )+??ˆ
(?? 2
sin
2
?? +?? 2
cos
2
?? )
=???? sin ?? ??ˆ-???? cos ??ˆ+?? 2
??ˆ
?? =
|
????
????
×
?? 2
?? ?? ?? 2
|
|
????
????
|
3
=
|???? sin ?? ??ˆ-???? cos ??ˆ+?? 2
??ˆ
|
|-?? sin ?? ??ˆ+?? cos ??ˆ+?? ??ˆ3
=
(?? 2
?? 2
sin
2
?? -?? 2
?? 2
cos
2
?? +?? 4
)
1/2
(?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
)
3/2
=
(?? 2
?? 2
+?? 4
)
1/2
(?? 2
+?? 2
)
3/2
=
?? (?? 2
+?? 2
)
1/2
(?? 2
+?? 2
)
3/2
=
?? ?? 2
+?? 2
?? =
[
????
????
?? 2
?? ?? ?? 2
?? 3
?? ?? ?? 3
]
|
????
????
×
?? 2
?? ?? ?? 2
|
2
=
(
????
????
×
?? 2
?? ?? ?? 2
)·
?? 3
?? ?? ?? 3
|
????
????
×
?? 2
?? ?? ?? 2
|
2
=
(???? sin ?? ??ˆ-???? cos ?? ??ˆ+?? 2
??ˆ
)·(?? sin ?? ??ˆ-?? cos ?? ??ˆ)
|???? sin ?? ??ˆ-???? cos ?? ??ˆ+?? 2
??ˆ
|
2
=
?? 2
?? sin
2
?? +?? 2
?? cos
2
?? (?? 2
?? 2
+?? 4
)
=
?? 2
?? ?? 2
(?? 2
+?? 2
)
=
?? ?? 2
+?? 2
?? ?? =
?? ?? 2
+?? 2
?? ?? 2
+?? 2
=
?? ??
6.2 Derive the Frenet-Serret formulae. Define the curvature and torsion for a space
curve. Compute them for the-space curve
?? =?? ,?? =?? ?? ,?? =
?? ?? ?? ??
Show that the curvature and torsion are equal for this curve.
(2012 : 20 Marks)
Solution:
-The following three relations are known as Serret-Frenet formulae:
?? '
=???? (??)
?? '
=???? -???? (???? )
?? '
=-???? (?????? )
where, ?? is the magnitude of curvature, ?? is the magnitude of torsion, ?? ?? ?? ,?? are the unit
tangent vector, the unit principal normal vector and the unit bi-normal vector respectively
and ' } ' denotec the differentiation w.r.t. the arc length ?? .
Derivation: We know
?? 2
=1
Differentiating it w.r.t. the arc length ?? ,
?? ·?? '
=0??? '
is perpendicular to ?? .
The equation of the oscillating plane at a point ?? (??) of the curve is
[?? -?? ,?? ;?? '
)=0
This equation shows that ?? '
lies in the oscillating plane and hence ?? '
is perpendicular to
the binormal ?? (since oscillating plane is perpendicular to ?? ).
??? '
is parallel to ?? ×?? .
??? is parallel to ?? .
? ?? '
=±????
Page 3
Edurev123
6. Curves in Space
6.1 Find ?? /?? for the curve
??? (?? )=?? ?????? ?? ??ˆ+?? ?????? ?? ??ˆ+???? ??ˆ
(2010: 12 Marks)
Solution:
Given:
?? (?? )=?? cos ?? ??ˆ+?? sin ??ˆ+???? ??ˆ
?? ?? (?? )
????
=-?? sin ?? ??ˆ+?? cos ??ˆ+?? ??ˆ
?? 2
?? (?? )
?? ?? 2
=-?? cos ?? ??ˆ-?? sin ??ˆ
?? 3
?? (?? )
?? ?? 3
=?? sin ?? ??ˆ-?? cos ??ˆ
?? ?? (?? )
????
×
?? 2
?? (?? )
?? ?? 2
=|
??ˆ ??ˆ ??ˆ
-?? sin ?? ?? cos ?? ?? -?? cos ?? -?? sin ?? 0
|
=??ˆ(0+???? sin ?? )-?? (0+???? cos ?? )+??ˆ
(?? 2
sin
2
?? +?? 2
cos
2
?? )
=???? sin ?? ??ˆ-???? cos ??ˆ+?? 2
??ˆ
?? =
|
????
????
×
?? 2
?? ?? ?? 2
|
|
????
????
|
3
=
|???? sin ?? ??ˆ-???? cos ??ˆ+?? 2
??ˆ
|
|-?? sin ?? ??ˆ+?? cos ??ˆ+?? ??ˆ3
=
(?? 2
?? 2
sin
2
?? -?? 2
?? 2
cos
2
?? +?? 4
)
1/2
(?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
)
3/2
=
(?? 2
?? 2
+?? 4
)
1/2
(?? 2
+?? 2
)
3/2
=
?? (?? 2
+?? 2
)
1/2
(?? 2
+?? 2
)
3/2
=
?? ?? 2
+?? 2
?? =
[
????
????
?? 2
?? ?? ?? 2
?? 3
?? ?? ?? 3
]
|
????
????
×
?? 2
?? ?? ?? 2
|
2
=
(
????
????
×
?? 2
?? ?? ?? 2
)·
?? 3
?? ?? ?? 3
|
????
????
×
?? 2
?? ?? ?? 2
|
2
=
(???? sin ?? ??ˆ-???? cos ?? ??ˆ+?? 2
??ˆ
)·(?? sin ?? ??ˆ-?? cos ?? ??ˆ)
|???? sin ?? ??ˆ-???? cos ?? ??ˆ+?? 2
??ˆ
|
2
=
?? 2
?? sin
2
?? +?? 2
?? cos
2
?? (?? 2
?? 2
+?? 4
)
=
?? 2
?? ?? 2
(?? 2
+?? 2
)
=
?? ?? 2
+?? 2
?? ?? =
?? ?? 2
+?? 2
?? ?? 2
+?? 2
=
?? ??
6.2 Derive the Frenet-Serret formulae. Define the curvature and torsion for a space
curve. Compute them for the-space curve
?? =?? ,?? =?? ?? ,?? =
?? ?? ?? ??
Show that the curvature and torsion are equal for this curve.
(2012 : 20 Marks)
Solution:
-The following three relations are known as Serret-Frenet formulae:
?? '
=???? (??)
?? '
=???? -???? (???? )
?? '
=-???? (?????? )
where, ?? is the magnitude of curvature, ?? is the magnitude of torsion, ?? ?? ?? ,?? are the unit
tangent vector, the unit principal normal vector and the unit bi-normal vector respectively
and ' } ' denotec the differentiation w.r.t. the arc length ?? .
Derivation: We know
?? 2
=1
Differentiating it w.r.t. the arc length ?? ,
?? ·?? '
=0??? '
is perpendicular to ?? .
The equation of the oscillating plane at a point ?? (??) of the curve is
[?? -?? ,?? ;?? '
)=0
This equation shows that ?? '
lies in the oscillating plane and hence ?? '
is perpendicular to
the binormal ?? (since oscillating plane is perpendicular to ?? ).
??? '
is parallel to ?? ×?? .
??? is parallel to ?? .
? ?? '
=±????
But we choose the direction ?? so that the curvature ?? is aiways positive, i.e., we take
?? '
=???? which proves (i) Again, we know that ?? 2
=1.
Differentiating w.r.t. the are length ?? ,
?? ·?? '
=0??? '
is perpendicular to ?? .
Also, ?? ·?? '
=0 (???? )
Differentiating w.r.t. ??
?? '
·?? +?? ·?? '
=0??? '
·?? +?? ·???? =0
?? '
·?? =0
? ?? '
perpendicular to ?? (?? )
From (iv) and ( ?? ), ?? '
is perpendicular to both ?? and ?? .
??? '
is parallel to ?? ×?? .
??? '
is parallel to ??
???? ?????? ?????????? , ?? '
=????
???? ???????????????????? ,???? h?????? ?? '
=-???? , which proves (iii).
We know, ?? =?? ×??
Differentiating w.r.t. S,
?? '
=?? '
×?? +?? ×?? '
=-???? ×?? +?? ×???? (???????? (??)?????? (?????? ))
=???? -???? , which proves (ii).
Curvature: The arc rate at which the tangent changes direction as the point ?? (?? ) moves
along the curve is called the curvature vector of the curve and its magnitude is denoted
by ?? .
Torsion: The arc rate at which the bi-normal changes direction as the point ?? (?? ) moves
along the curve is called the torsion vector of the curve and its magnitude is denoted by
?? .
Given : ?? =?? ,?? =?? 2
,?? =
2
3
?? 3
Let ?? =(?? ,?? 2
,
2
3
?? 3
)
?? =(1,2?? ,2?? 2
),???
=(0,2,4?? )
Page 4
Edurev123
6. Curves in Space
6.1 Find ?? /?? for the curve
??? (?? )=?? ?????? ?? ??ˆ+?? ?????? ?? ??ˆ+???? ??ˆ
(2010: 12 Marks)
Solution:
Given:
?? (?? )=?? cos ?? ??ˆ+?? sin ??ˆ+???? ??ˆ
?? ?? (?? )
????
=-?? sin ?? ??ˆ+?? cos ??ˆ+?? ??ˆ
?? 2
?? (?? )
?? ?? 2
=-?? cos ?? ??ˆ-?? sin ??ˆ
?? 3
?? (?? )
?? ?? 3
=?? sin ?? ??ˆ-?? cos ??ˆ
?? ?? (?? )
????
×
?? 2
?? (?? )
?? ?? 2
=|
??ˆ ??ˆ ??ˆ
-?? sin ?? ?? cos ?? ?? -?? cos ?? -?? sin ?? 0
|
=??ˆ(0+???? sin ?? )-?? (0+???? cos ?? )+??ˆ
(?? 2
sin
2
?? +?? 2
cos
2
?? )
=???? sin ?? ??ˆ-???? cos ??ˆ+?? 2
??ˆ
?? =
|
????
????
×
?? 2
?? ?? ?? 2
|
|
????
????
|
3
=
|???? sin ?? ??ˆ-???? cos ??ˆ+?? 2
??ˆ
|
|-?? sin ?? ??ˆ+?? cos ??ˆ+?? ??ˆ3
=
(?? 2
?? 2
sin
2
?? -?? 2
?? 2
cos
2
?? +?? 4
)
1/2
(?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
)
3/2
=
(?? 2
?? 2
+?? 4
)
1/2
(?? 2
+?? 2
)
3/2
=
?? (?? 2
+?? 2
)
1/2
(?? 2
+?? 2
)
3/2
=
?? ?? 2
+?? 2
?? =
[
????
????
?? 2
?? ?? ?? 2
?? 3
?? ?? ?? 3
]
|
????
????
×
?? 2
?? ?? ?? 2
|
2
=
(
????
????
×
?? 2
?? ?? ?? 2
)·
?? 3
?? ?? ?? 3
|
????
????
×
?? 2
?? ?? ?? 2
|
2
=
(???? sin ?? ??ˆ-???? cos ?? ??ˆ+?? 2
??ˆ
)·(?? sin ?? ??ˆ-?? cos ?? ??ˆ)
|???? sin ?? ??ˆ-???? cos ?? ??ˆ+?? 2
??ˆ
|
2
=
?? 2
?? sin
2
?? +?? 2
?? cos
2
?? (?? 2
?? 2
+?? 4
)
=
?? 2
?? ?? 2
(?? 2
+?? 2
)
=
?? ?? 2
+?? 2
?? ?? =
?? ?? 2
+?? 2
?? ?? 2
+?? 2
=
?? ??
6.2 Derive the Frenet-Serret formulae. Define the curvature and torsion for a space
curve. Compute them for the-space curve
?? =?? ,?? =?? ?? ,?? =
?? ?? ?? ??
Show that the curvature and torsion are equal for this curve.
(2012 : 20 Marks)
Solution:
-The following three relations are known as Serret-Frenet formulae:
?? '
=???? (??)
?? '
=???? -???? (???? )
?? '
=-???? (?????? )
where, ?? is the magnitude of curvature, ?? is the magnitude of torsion, ?? ?? ?? ,?? are the unit
tangent vector, the unit principal normal vector and the unit bi-normal vector respectively
and ' } ' denotec the differentiation w.r.t. the arc length ?? .
Derivation: We know
?? 2
=1
Differentiating it w.r.t. the arc length ?? ,
?? ·?? '
=0??? '
is perpendicular to ?? .
The equation of the oscillating plane at a point ?? (??) of the curve is
[?? -?? ,?? ;?? '
)=0
This equation shows that ?? '
lies in the oscillating plane and hence ?? '
is perpendicular to
the binormal ?? (since oscillating plane is perpendicular to ?? ).
??? '
is parallel to ?? ×?? .
??? is parallel to ?? .
? ?? '
=±????
But we choose the direction ?? so that the curvature ?? is aiways positive, i.e., we take
?? '
=???? which proves (i) Again, we know that ?? 2
=1.
Differentiating w.r.t. the are length ?? ,
?? ·?? '
=0??? '
is perpendicular to ?? .
Also, ?? ·?? '
=0 (???? )
Differentiating w.r.t. ??
?? '
·?? +?? ·?? '
=0??? '
·?? +?? ·???? =0
?? '
·?? =0
? ?? '
perpendicular to ?? (?? )
From (iv) and ( ?? ), ?? '
is perpendicular to both ?? and ?? .
??? '
is parallel to ?? ×?? .
??? '
is parallel to ??
???? ?????? ?????????? , ?? '
=????
???? ???????????????????? ,???? h?????? ?? '
=-???? , which proves (iii).
We know, ?? =?? ×??
Differentiating w.r.t. S,
?? '
=?? '
×?? +?? ×?? '
=-???? ×?? +?? ×???? (???????? (??)?????? (?????? ))
=???? -???? , which proves (ii).
Curvature: The arc rate at which the tangent changes direction as the point ?? (?? ) moves
along the curve is called the curvature vector of the curve and its magnitude is denoted
by ?? .
Torsion: The arc rate at which the bi-normal changes direction as the point ?? (?? ) moves
along the curve is called the torsion vector of the curve and its magnitude is denoted by
?? .
Given : ?? =?? ,?? =?? 2
,?? =
2
3
?? 3
Let ?? =(?? ,?? 2
,
2
3
?? 3
)
?? =(1,2?? ,2?? 2
),???
=(0,2,4?? )
???
×?? =|
?? ?? ???
1 2?? 2?? 2
0 2 4?? |
=?? (8?? 2
-4?? 2
)+?? (0-4?? )+???
(2)
=(4?? 1
2
-4?? 1
+2)
? |?? ×?? |=v16?? 4
+4?? 2
+4=2(2?? 2
+1)
[?? ,?? ,?? ]=?? ×?? ×??
=(4?? 1
2
-4?? 1
2
)(0,0,4)
=8
? Torsion, ?? =
|?? ×?? |
|?? |
=
2(2?? 2
+1)
(1+4?? 2
+4?? 4
)
3/2
=
2
(2?? 2
+1)
3
?????????????????? , ?? =
[?? ,?? ,?? ]
|???×??¨|
2
=
8
4(2?? 2
+1)
2
=
2
(2?? 2
+1)
2
6.3 Show that the curve ???? (?? )=?? ??ˆ+(
?? +?? ?? )??ˆ+
(?? -?? ?? )
?? ??ˆ
lies in a plane.
(2013 : 10 Marks)
Solution:
For the curve to lie in a plane the binormal ?? must point in a constant direction as one
moves along the curve.
??.?? .,
????
????
=0
? -???? =0??? =0
i.e., torsion must be zero.
?? =
[?????¨???]
(???×??¨)·(???×??¨)
?????? , ??? =
?? ??
????
=??ˆ+(-?? -2
)??ˆ-(?? -2
+1)??ˆ
??¨ =
?? 2
??
?? ?? 2
=2?? -3
??ˆ+2?? -3
??ˆ
??? =
?? 3
??
?? ?? 3
=-6?? -4
??ˆ-6?? -4
??ˆ
??¨×??? =|
?? ?? ?? 0 2?? -3
2?? -3
0 -6?? -4
-6?? -4
|=0
[?????¨ ???] =0
Page 5
Edurev123
6. Curves in Space
6.1 Find ?? /?? for the curve
??? (?? )=?? ?????? ?? ??ˆ+?? ?????? ?? ??ˆ+???? ??ˆ
(2010: 12 Marks)
Solution:
Given:
?? (?? )=?? cos ?? ??ˆ+?? sin ??ˆ+???? ??ˆ
?? ?? (?? )
????
=-?? sin ?? ??ˆ+?? cos ??ˆ+?? ??ˆ
?? 2
?? (?? )
?? ?? 2
=-?? cos ?? ??ˆ-?? sin ??ˆ
?? 3
?? (?? )
?? ?? 3
=?? sin ?? ??ˆ-?? cos ??ˆ
?? ?? (?? )
????
×
?? 2
?? (?? )
?? ?? 2
=|
??ˆ ??ˆ ??ˆ
-?? sin ?? ?? cos ?? ?? -?? cos ?? -?? sin ?? 0
|
=??ˆ(0+???? sin ?? )-?? (0+???? cos ?? )+??ˆ
(?? 2
sin
2
?? +?? 2
cos
2
?? )
=???? sin ?? ??ˆ-???? cos ??ˆ+?? 2
??ˆ
?? =
|
????
????
×
?? 2
?? ?? ?? 2
|
|
????
????
|
3
=
|???? sin ?? ??ˆ-???? cos ??ˆ+?? 2
??ˆ
|
|-?? sin ?? ??ˆ+?? cos ??ˆ+?? ??ˆ3
=
(?? 2
?? 2
sin
2
?? -?? 2
?? 2
cos
2
?? +?? 4
)
1/2
(?? 2
sin
2
?? +?? 2
cos
2
?? +?? 2
)
3/2
=
(?? 2
?? 2
+?? 4
)
1/2
(?? 2
+?? 2
)
3/2
=
?? (?? 2
+?? 2
)
1/2
(?? 2
+?? 2
)
3/2
=
?? ?? 2
+?? 2
?? =
[
????
????
?? 2
?? ?? ?? 2
?? 3
?? ?? ?? 3
]
|
????
????
×
?? 2
?? ?? ?? 2
|
2
=
(
????
????
×
?? 2
?? ?? ?? 2
)·
?? 3
?? ?? ?? 3
|
????
????
×
?? 2
?? ?? ?? 2
|
2
=
(???? sin ?? ??ˆ-???? cos ?? ??ˆ+?? 2
??ˆ
)·(?? sin ?? ??ˆ-?? cos ?? ??ˆ)
|???? sin ?? ??ˆ-???? cos ?? ??ˆ+?? 2
??ˆ
|
2
=
?? 2
?? sin
2
?? +?? 2
?? cos
2
?? (?? 2
?? 2
+?? 4
)
=
?? 2
?? ?? 2
(?? 2
+?? 2
)
=
?? ?? 2
+?? 2
?? ?? =
?? ?? 2
+?? 2
?? ?? 2
+?? 2
=
?? ??
6.2 Derive the Frenet-Serret formulae. Define the curvature and torsion for a space
curve. Compute them for the-space curve
?? =?? ,?? =?? ?? ,?? =
?? ?? ?? ??
Show that the curvature and torsion are equal for this curve.
(2012 : 20 Marks)
Solution:
-The following three relations are known as Serret-Frenet formulae:
?? '
=???? (??)
?? '
=???? -???? (???? )
?? '
=-???? (?????? )
where, ?? is the magnitude of curvature, ?? is the magnitude of torsion, ?? ?? ?? ,?? are the unit
tangent vector, the unit principal normal vector and the unit bi-normal vector respectively
and ' } ' denotec the differentiation w.r.t. the arc length ?? .
Derivation: We know
?? 2
=1
Differentiating it w.r.t. the arc length ?? ,
?? ·?? '
=0??? '
is perpendicular to ?? .
The equation of the oscillating plane at a point ?? (??) of the curve is
[?? -?? ,?? ;?? '
)=0
This equation shows that ?? '
lies in the oscillating plane and hence ?? '
is perpendicular to
the binormal ?? (since oscillating plane is perpendicular to ?? ).
??? '
is parallel to ?? ×?? .
??? is parallel to ?? .
? ?? '
=±????
But we choose the direction ?? so that the curvature ?? is aiways positive, i.e., we take
?? '
=???? which proves (i) Again, we know that ?? 2
=1.
Differentiating w.r.t. the are length ?? ,
?? ·?? '
=0??? '
is perpendicular to ?? .
Also, ?? ·?? '
=0 (???? )
Differentiating w.r.t. ??
?? '
·?? +?? ·?? '
=0??? '
·?? +?? ·???? =0
?? '
·?? =0
? ?? '
perpendicular to ?? (?? )
From (iv) and ( ?? ), ?? '
is perpendicular to both ?? and ?? .
??? '
is parallel to ?? ×?? .
??? '
is parallel to ??
???? ?????? ?????????? , ?? '
=????
???? ???????????????????? ,???? h?????? ?? '
=-???? , which proves (iii).
We know, ?? =?? ×??
Differentiating w.r.t. S,
?? '
=?? '
×?? +?? ×?? '
=-???? ×?? +?? ×???? (???????? (??)?????? (?????? ))
=???? -???? , which proves (ii).
Curvature: The arc rate at which the tangent changes direction as the point ?? (?? ) moves
along the curve is called the curvature vector of the curve and its magnitude is denoted
by ?? .
Torsion: The arc rate at which the bi-normal changes direction as the point ?? (?? ) moves
along the curve is called the torsion vector of the curve and its magnitude is denoted by
?? .
Given : ?? =?? ,?? =?? 2
,?? =
2
3
?? 3
Let ?? =(?? ,?? 2
,
2
3
?? 3
)
?? =(1,2?? ,2?? 2
),???
=(0,2,4?? )
???
×?? =|
?? ?? ???
1 2?? 2?? 2
0 2 4?? |
=?? (8?? 2
-4?? 2
)+?? (0-4?? )+???
(2)
=(4?? 1
2
-4?? 1
+2)
? |?? ×?? |=v16?? 4
+4?? 2
+4=2(2?? 2
+1)
[?? ,?? ,?? ]=?? ×?? ×??
=(4?? 1
2
-4?? 1
2
)(0,0,4)
=8
? Torsion, ?? =
|?? ×?? |
|?? |
=
2(2?? 2
+1)
(1+4?? 2
+4?? 4
)
3/2
=
2
(2?? 2
+1)
3
?????????????????? , ?? =
[?? ,?? ,?? ]
|???×??¨|
2
=
8
4(2?? 2
+1)
2
=
2
(2?? 2
+1)
2
6.3 Show that the curve ???? (?? )=?? ??ˆ+(
?? +?? ?? )??ˆ+
(?? -?? ?? )
?? ??ˆ
lies in a plane.
(2013 : 10 Marks)
Solution:
For the curve to lie in a plane the binormal ?? must point in a constant direction as one
moves along the curve.
??.?? .,
????
????
=0
? -???? =0??? =0
i.e., torsion must be zero.
?? =
[?????¨???]
(???×??¨)·(???×??¨)
?????? , ??? =
?? ??
????
=??ˆ+(-?? -2
)??ˆ-(?? -2
+1)??ˆ
??¨ =
?? 2
??
?? ?? 2
=2?? -3
??ˆ+2?? -3
??ˆ
??? =
?? 3
??
?? ?? 3
=-6?? -4
??ˆ-6?? -4
??ˆ
??¨×??? =|
?? ?? ?? 0 2?? -3
2?? -3
0 -6?? -4
-6?? -4
|=0
[?????¨ ???] =0
? ?? =0
So, the curve lies in a plane.
6.4 Find the curvature vector at any point of the curve ??? (?? )=?? ?????? ?? ??ˆ+?? ?????? ?? ??ˆ,?? =
?? =?? ?? . Give its magnitude also.
(2014 : 10 Marks)
Solution:
The position vector ?? of any point on the given curve is ??
?? =?? cos ?? ??ˆ+??sin ?? ??ˆ
i.e., ?? =(?? cos ?? ,?? sin ?? )
?
?? ??
????
=(cos ?? -?? sin ?? +sin ?? +?? cos ?? )
?????? ,
????
????
=|
?? ??
????
|
=vcos
2
?? +?? 2
sin
2
?? -2?? cos ?? sin ?? +sin
2
?? +?? 2
cos
2
?? +2?? cos ?? sin ?? =v1+?? 2
?????????? ,?? =
?? ??
????
=
?? ?? /????
???? /????
=
1
v1+?? 2
(cos ?? -?? sin ?? ,sin ?? +?? cos ?? )
Differentiating this w.r.t. ?? , we get
????
????
=???? =
???? /????
???? /????
=
1
v1+?? 2
·
????
????
=
1
v1+?? 2
[(sin ?? -sin ?? -?? cos ?? )v1+?? 2
-
1
2v1+?? 2
(2?? )(cos ?? -?? sin ?? )
(cos ?? +cos ?? -?? sin ?? )v1+?? 2
-
1
2v1+?? 2
(2?? )(sin ?? +?? cos ?? )·
1
(1+?? 2
)
]
=
1
v1+?? 2
·
1
(1+?? 2
)
[
(-2sin ?? -?? cos ?? )(1+?? 2
)-?? (cos ?? +sin ?? )
v1+?? 2
(2cos ?? -?? sin ?? )(1+?? 2
)-?? (?? sin ?? +?? cos ?? )
v1-?? 2
]
kN=
1
(1+?? 2
)
2
(-(2+?? 2
)(sin ?? +cos ?? ),(2+?? 2
)(cos ?? -sin ?? ))
kN=
(2+?? 2
)
(1+?? 2
)
2
(sin ?? +cos ?? )??ˆ+(
2+?? (1+?? )
2
·cos ?? -sin ?? )??ˆ
which is the required curvature vector
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