solution

Differentiation of vectors: If a vector R varies continuously as a scalar variable t changes, then R is said to

be a function of t and is written as ( )tFR = . We define the derivative of a vector function ( )tFR = as

( ) ( )t

tttLt0 t

+

FF

and write it as dtdR

or dtdF

or ( )t F . General rules of differentiation:

General rules of differentiation are similar to those of ordinary calculus provided

the order of factors in vector products is maintained. Thus, if is scalar and F, G, H are vector functions of a scalar variable t, then we have

1st Topic

Vector Calculus

Differentiation of vectors, Space curves (Curves in Space), Curvature, Torsion, Radius of curvature

and radius of torsion, Frenets Formulae

Prepared by: Dr. Sunil

NIT Hamirpur (HP) (Last updated on 24-10-2008)

Latest update available at: http://www.freewebs.com/sunilnit/

Vector Calculus: Differentiation of vectors, Curves in space Prepared by: Dr. Sunil, NIT Hamirpur

2

(i) ( )dtd

dtd

dtd

dtd HGFHGF +=+

(ii) ( )d d ddt dt dt

= +FF F

(iii) ( )d d d. . .dt dt dt

= +F GF G G F

(iv) ( )d d ddt dt dt

= + F GF G G F

(v) ( )

+

+

=

dtd

dtd

dtd

dtd HFGHGFGHFFGH

(vi) ( )[ ] ( )dtd

dtd

dtd

dtd HGFHGFHGFHGF +

+

= .

(iv) Prove that ( ) GFGFGF +=dtd

dtd

dtd

.

Proof: ( ) ( ) ( )t

Ltdtd

0 t ++

=

GFGGFFGF

( ) ( ) ( ) ( ) ( )t

Lt0 t

+++=

GFGFGFGFGF

( ) ( ) ( )t

Lt0 t

++=

GFGFGF

GFGFGFGFGF +=

+

+

=

dtd

dtd

tttLt

0 t

[ ]0t as 0 G

This completes the proof.

Result No.1.: If F(t) has a constant magnitude, then show that 0FF =dtd

. .

Proof.: Given F(t) has a constant magnitude ( ) constant t = F ( ) ( ) ( ) ( ) 0.

dtd

constant tt.t 2 === FFFFF

0dtd

.0dtd

.20.dtd

dtd

. ===+FFFFFFFF .


Note: FFFF =dtd0

dtd

. .


3

Result No.2.: If F(t) has a constant direction, then show that 0FF =dtd

.

Proof: Let ( ) ( )tft =F .

If we assume ( )tG be a unit vector in the direction of F(t), then ( ) ( )t)t(ft = GF .

d d dff (t)dt dt dt

= +F G G . (i)

If F(t) has a constant direction, then ( )tG has also a constant direction.

Thus, ( )tG is a constant vector ( )tdt

=G

0 .

From (i), we have d d dff (t)dt dt dt

= +F G G d df

dt dt

=F G

Now d df dff fdt dt dt

= = =

FF G G G G 0

ddt

=FF 0 .


Now let us solve few problems using the general rule of differentiation:

Q.No.1.: If

+= KJIA 32 ttt5 ,

= JIB tcostsin ,

find (i) ( )BA.dtd

, (ii) ( )BA dtd

.

Sol.: (i) ( ) BABABA .dtd

dtd

..

dtd

+=

( )

++

+=

JIKJIJIKJI tcostsin.t3t10tsincost . ttt5 232

( ) ( ) tcostsint11tcost5tcostsint10tsinttcost5 22 +=++= . Ans. (ii): ( ) BABABA +=

dtd

dtd

dtd

++

+

+=

JIKJIJIKJI tcostsint3t10tsintcosttt5 232


4

+=

IJKK tsinttcosttcosttsint5 332

+

++

IJKK tcost3tsint3tsintcost10 22

( ) ( ) ( )[ ] ++= KJI tcost11tsin1t5tsin3tcostttcost3tsint 2223 . Ans. Space curves (Curves in Space): Associated with each point on a curve, there is a set of three mutually perpendicular lines known as Tangent, Principal normal, Binormal and three mutually perpendicular planes determined by these in pairs and known as Osculating plane, Normal plane, Reflecting plane.

B

N

T Tangent

Principal Normal

Binormal

Osculating Plane

Reflecting Plane

Normal Plane

Curve O


5

(1) Tangent at a point:

Let ( ) ( ) ( ) ++= KJIR tz)t(ytxt be the position vector of a point P. We observe that, as the scalar parameter t takes different values, then the point P traces out a curve in space.

If the neighbouring point Q corresponds to tt + , then ( ) ( )t t t . = + R R R

As t is scalar, then the vector ( ) ( )t

tttt

+=

RRR

is parallel to

PQ

or directed along the chord PQ.

Now, taking the limit, when PQ , and consequently 0t , we get t

Lim0tPQ

R ,

which becomes the tangent (vector) to the curve at P, whenever it exists and is not zero.

But Q Pt 0

dLimt dt

= =

R R R

P(t)

x-axis

y-axis

z-axis

Tangent

P0(t0) Q

O

R

( )tt +R( )tR

X

s s


6

Thus, the vector ddt

=RR is a tangent to the space curve ( )tFR = .

Vector equation of the tangent: If u is the scalar parameter, R is the position vector of the point P and X is the

position vector of any point on the tangent.

Then the vector equation of the tangent at P is dtd

uRRX +=

( )t u (t) = +X F F . Formula for evaluating arc length:

Let 0P be a fixed point of this curve corresponding to 0tt = .

If s be the length of the arc PP0 , then

tPQ chordPQ arc

t.

s

ts

=

=

RR

R.

Now, as PQ along the curve QP i.e., 0t , and consequently PQ chord

PQ arc 1 .

Then Q P Q Pt 0 t 0

sLim Limt t

=

R

( )ds d tdt dt

= =R R .

If ( )tR is continuous, then by taking integration, the arc PP0 is given by ( ) ( ) ( ) dtzyxdt s 222t

t

t

t 00++== R .

This is the required formula for evaluating the length of an arc.

Remarks:

Since, we know ds d .dt dt

=

R

If we take s as the parameter in place of t, then the magnitude of the tangent vector is equal to one, i.e.

1dsd

=

R.


7

Thus, if we denote the unit tangent vector by

T , then we have dsdRT =

.

(2) Principal Normal:

Since

T is unit tangent vector, then we have 0.dsd

=

TT .

dds

T is perpendicular to

T

or 0dsd

=

T

T is constant vector w.r.t. the arc length s and so has a fixed direction

i.e. the curve is a straight line.

If we denote a unit normal vector to the curve at P by

N ,

then

dsd

T is in the direction of

N ,

which is known as the principal normal to the space curve at P.

Osculating plane:

The plane between

T and

N is called the osculating plane of the curve at P.

(3) Binormal:

A third unit vector

B defined by

= NTB , is called the binormal at P.

Since

T and

N are unit vectors

B is also a unit vector perpendicular to both

T and

N .

Hence

B

is normal to the osculating plane at P.

Final Conclusions: Thus, at each point P of a space curve, there are three mutually perpendicular unit

vectors

T ,

N ,

B which form a moving trihedral such that


8

= BNT ,

= TBN ,

= NTB .

This moving trihedral determines the following three fundamentals planes at each point of the curve:

(i) The osculating plane containing

T and

N .

(ii) The normal plane containing

N and

B .

(iii) The reflecting plane containing

B and

T .

Remarks:

(i) The tangent is parallel to the vectors ddsR

The principal normal is parallel to the vector

2

2dds

R

The binomial is parallel to the vector

2

2d dds ds

R R

.

(ii) Equation of normal plane at a point P with position vector R, is

( ) 0dtd

. =

RRX ,

where X is the position vector of any point on the plane.

(iii) Equation of osculating plane is

( ) 0dtd

dtd

. 2

2=

RRRX

2 2

2 2d d d d

.

dt dt dt dt =

R R R RX R .

B

T

N


9

Curvature:

The arc rate of rotation of the tangent (i.e. the magnitude of dsd

T ) is called

curvature of the curve and is denoted by k.

Thus, kdsd

=

T.

or

The arc rate at which the tangent changes its direction as the point moves along the curve is known as curvature.

Torsion:

The arc rate of rotation of the bi-normal (i.e. the magnitude of dsd

B ) is called

torsion of the curve and is denoted by .Thus, =

dsdB

.

or

The arc rate at which the bi-normal changes its direction as the point moves along the curve is known as torsion.

Remarks: Here, the torsion is positive or negative according as the vectors

N and

dsd

B have the same or opposite senses.

Radius of curvature and radius of torsion: The reciprocal of curvature is called radius of curvature and is denoted by .

Thus k1

= .

The reciprocal of torsion is called the radius of torsion and is denoted by .

Thus

=1

.


10

Frenets Formulae: We shall now establish the following important results, known as Frenets formulae.

(i)

= NT kdsd

, (ii)

= NBdsd

, (iii)

= TBN kdsd

Proof:

(i) To prove:

= NT kdsd

.

Since, we know that

N is the unit vector having the sense and direction of dsd

T.

i.e. dsd

T

N .

And moreover, we also know that kdsd

=

T.

Combining these two facts, we obtain

= NT kdsd

.

(ii) To prove:

= NBdsd

.

Since

B is a unit vector, we have 0.dsd

=

BB .

dsd

B is perpendicular to

B .

Also

TB. = 0 0.dsd

=

TB

0dsd

..

dsd

=+

TBTB

0k..dsd

=

+

NBTB . d kds

=

T N


11

0.dsd

=

TB

=

0.NB

dsd

B is perpendicular to

T .

Thus, dsd

B

B and dsd

B

T .

Hence, dsd

B is parallel to the vector

N dds

B

N .

And moreover, we also know that =

dsdB

.

Combining these two facts, we obtain

= NBdsd

.

Here, the torsion is positive or negative according as the vectors

N and dsd

Bas defined

above, have the same or opposite senses.

Remarks: If dsd

B has direction of

N , then

= NBdsd

.

(iii) To prove:

= TBN kdsd

.

Since we know that

= TBN .

Differentiate w.r.t. s, we get

d d dds ds ds

= + N B TT B .

k

= + N T B N d d, kds ds

= =

B TN N

k

= B T [ , = B T N ]. = T N B


12

Now let us solve some more problems using the general rule of differentiation:

Q.No.2.: Show that, if tcostsin += BAR , where A, B and are constants, then

RR 222

dtd

=

and BARR =dtd

.

Sol.: , A, B and are constants (i) tcostsin += BAR Differentiating w. r. t. t, we get

( )tsintcosdtd

+= BAR

Differentiating again w. r. t. t, we get

( )tcostsintcostcosdtd 222

2

2+== BABAR

. Ans.

(ii)

tsintcosdtd

= BAR

( )( )tcostsin 22 += BA

( )BA= . Ans. Q.No.3.: BAR nm tt += , where A, B are constant vectors,

show that, if R and 22

dtd R

are parallel vectors, then m + n =1, unless m = n.

Sol.: BAR nm tt +=

Differentiating w. r. t t, we get dtd

tdtd

tntmtdtd nm1n1m BABAR +++=

Differentiating again w. r. t t, we get

tcostsin += BAR

RR 222

dtd

=

tcostsin += BAR

( ) ( )tsintcostcostsindtd

= BABARR

tcostsin 22 = BABA

( ) ( ) BAR 2n2m22

t1nnt1mmdtd

+=


13

Since R and 22

dtd R

are parallel vectors, then 0dtd

2

2=

RR

( ) ( ) ( )[ ] 0t1nnt1mmtt 2n2mnm =++ BABA ( ) ( ) 0t1mmt1nn 2nm2nm =+ + ABBA ( ) ( )( ) ABBA = + 2nm2nm t1mmt1nn ( ) ( )( ) BABA = + 2nm2nm t1mmt1nn

( )( ) ( )1nn1mm = nmnmnnmm 2222 == ( )( ) 1nmnmnmnm =+=+

( ) ( )[ ] 01nmnm =+

R and 22

dtd R

are parallel if m + n =1, unless m = n.

Hence this proved the result.

Q.No.4.: If

+= KJIP ttt5 32 ,

+= KJIQ t5tcostsin2 ,

find (i) ( )QP.dtd

(ii) .

Sol.:(i) ( ) QPQPQP .dtd

dtd

..

dtd

+=

+

++

++

+=

KJIKJIKJIKJI t5tcostsin2t3t10t5tsintcos2ttt5 232

t5tcost3tsint20t5tsinttcost10 232 ++=

t10tsint20tcost7tsint 23 ++= . Ans.

(ii)

Now

t5tcostsin2

ttt5 32

=

KJI

QP

( )QP dtd

( )QP dtd


14

( ) ( ) ( )tsint2tcost5t25tsint2tcostt5 3234 ++= KJI ( )QP

dtd

( ) ( )[ ] ++ K232 t3tsintcost2tcostsin2tsint5

[ ] [ ] +++= JI tsin2t75tcost2tcostsintt20 23

[ ] ++ Ktcos10tsinttcost2t 2 . Ans. Q.No.5.: If UWU =

dtd

and VWV =dtd

, prove that ( ) ( )VUWVU =dtd

.

Sol.: Since ( ) VUVUVU +=dtd

dtd

dtd

Given VWV =dtd

, UWU =dtd

Then ( ) ( ) ( ) VUWVWUVU +=dtd

Now using the formulae

( ) ( ) ( )ACBBCACBA .. =

( ) ( ) ( )CBABCACBA .. = ( ) ( )[ ] ( ) ( )[ ]WWUUVWVWUWVU .... +=

( ) ( )VWUUVW .. =

( ) ( )VUWVU =dtd

Hence this proved the result.

Q.No.6.: If

+= KJIA 232 xzxz2yzx and

+= KJIB 2xyz2 , find ( )BA

yx

2

at ( )2 0, ,1 Sol.: Given

+= KJIA 232 xzxz2yzx ,

+= KJIB 2xyz2 .

( )[ ] ( )[ ]JI 23 t75tsintcost2tcostsintt20 +++=


15

( )2

232

xyz2

xzxz2yzx

=

KJI

BA ( ) ( ) ( ) ++= KJI 42234233 xz4zyxxz2yzxxyzzx2

( ) ( ) ( ) ( ) ++= KJIBA 4233232 z4zxy2z2yzx4yzzx6x

( ) ++=

KJIBA xyz4zx4zyx

322

At ( )2 0, ,1 , we get ( ) ( ) ( )( ) ( )( )( ) ++=

KJIBA 10142142

yx2

2

+==

JIJI 2484 . Ans.

Problem on angle between the tangents:

Q.No.7.: Find the angle between the tangents to the curve

+= KJIR 32 tt2t at the point 1t = .

Sol.: Let 21 , TT be two tangents at t = +1 and 1t = are respectively.

To find: Angle between the tangents 1 2 and T T .

Since we know 1 2 1. . cos= 2T T T T .

i.e. we have to find:

Now given vector equation of the curve is 2 3t 2t t .

= + R I J K

Then the tangent at any point t is given by 2d 2t 2 3t .dt

= + R I J K

The tangents 21 , TT at t = +1 and 1t = are respectively, given by

+= KJIT 3221 1 17. =T

+= KJIT 3222 2 17. =T

Then the required is given by the relation


16

( ) ( )( ) =++= cos1717332.222cos. . 121 2TTTT i. e. = cos17179

=

179

cos 1 . Ans.

Problems for evaluating unit tangent vector:

Q.No.8.: Find the unit tangent vector at any point on the curve 2tx 2 += , 5t4y = , t6t2z 2 = , where t is any variable.

Also determine the unit tangent vector at any point t = 2.

Sol.: The vector equation of curve is

++= KJIR zyx

( ) ( ) ( )2 2t 2 4t 5 2t 6t . = + + + I J K To find: Unit tangent vector

T.

Since unit tangent vector

T

ddt

.

ddt

=

R

R

( ) ( ) ( ) ++= KJIR 6t44t2dtd

and 52t48t20t4836t1616t4dtd 222 +=+++=R 13t12t52 2 +=

( ) ( )2 2

dt 2 2t 31dtThus . 2t 4 4t 6

d 2 5t 12t 13 5t 12t 13dt

+ + = = + + =

+ +

RI J K

T I J KR

. Ans

2nd Part: To find: The unit tangent vector at any point t = 2. At point t = 2, we get


17

dtddtd

R

R

T =

322

++=

KJI, Ans.

which is the required unit tangent vector at any point t = 2.

Q.No.9.: If ( ) ( ) ( ) ++= KJIR tanattsinatcosa , find the value of

(i) 22

dtd

dtd RR

(ii) .

Also find the unit tangent vector at any point t of the curve.

Sol.: Given ( ) ( ) ( ) ++= KJIR tanattsinatcosa ( ) ( ) ( ) ++= KJIR tanatcosatsina

dtd

++= 222222 tanatcosatsinadtdR

=+= secatan1a 2

Also ( ) ( ) += JIR tsinatcosadtd

2

2

(i) 22

dtd

dtd RR

( ) ( ) ( ) ( ) +

+= KJIJI tanatsinatcostcosatsina

0tsinatcosa

tanatcosatsina

=

KJI

( ) ( ) ( ) ( ) =+=+= seca1tanaatantcosatsintana 224222 KJI

(ii) ( ) ( )= JIR tcosatsinadtd

3

3

= 3

3

2

2

dtd

dtd

.

dtd RRR

dtd

,

dtd

,

dtd

3

3

2

2 RRR

dtd

,

dtd

,

dtd

3

3

2

2 RRR


18

Now ( ) ( )

++=

=

KJI

KJIRR 23

3

2

2a00

0tcosatsina

0tsinaacostdtd

dtd

Putting the values, we get

( ) ( ) ( ) ( ) ( )

++

++=

KJIKJIRRR 233

2

2a00.tanatcosatsina

dtd

dtd

.

dtd

= tana3

3rd Part: Tangent vector at any point of curve is dtdR

= ( ) ( ) ( ) ++ KJI tanatcosatsina .

Unit tangent vector is

dtddtd

R

R

= ( ) ( ) ( )

++

=

KJI tanatcosatsinaseca

1

( ) ( ) ( ) ++= KJI sincos.tcoscos.tsin This is the required unit tangent vector at any point of the curve .

Problems on curvature and torsion:

Q.No.10.: Find the curvature and torsion of the curve btz asint,y ,tcosax === . Sol.:

*

x-axis

y-axis

z-axis

O

P0(t=0)

P


19

This curve is drawn on a circular cylinder cutting its generators at a constant angle and is known as circular helix.

The vector equation of the curve is a cos t a sin t bt .

= + +R I J K d

a sin t a cos t b .dt

= + +R I J K

To find: (i) ( ) dk curvatureds

=

T

(ii)

d (torsion) .

ds

=B

Consequently, we have to evaluate:

(i) d

d dtdsdsdt

=

TT

,

dd dt

dsdsdt

= =

RRT ,

t

0

ds dt

dt=

R

(ii) d

d dtdsdsdt

=

BB

,

= B T N ,

dd dt k ?.dsds

dt

= = =

TT N N

Its arc length from )0t(P0 = to any point P(t) is given by

( )t 2 20 ds dt a b tdt= = +R

( )2 2ds a b .dt = +

Then ( )22 babtcosatsina

dtdsdtd

dsd

+

++===

KJI

RRT

+==

+

+

==

JINNJIT

Ttsintcosk

ba

tsintcosa

dtdsdtd

dsd

22

Thus ( ) 22 baa

dsd

curvature k+

==

T.


20

Also ( )

+

+

++==

JIKJINTB tsintcosba

btcosatsina22

( )22 baatcosbtsinb

+

+

=

KJI

+==

+

+==

JINJIB

Btsintcostsintcos

bab

dtdsdtd

dsd

22

Hence 22 bab+

= . Ans.

Q.No.11.: A circular helix is given by the equation ( ) ( ) ++= KJIR tsin2tcos2)t( . Find the curvature and torsion of the curve at any point and show that they are constant.

Sol.: The vector equation of circular helix is ( ) ( ) ++= KJIR tsin2tcos2)t( ( ) ( )+= JIR tcos2tsin2

dtd

2tcos4tsin4dtd

dtds 22

=+==R

The unit tangent vector = ( )1

costsin2

tcos2tsin2

dtdsdtd

dsd

+

=

+===

JIJIR

RT

Now

= JIT tsintcosdtd


21

2tsintcos

dtdsdtd

dsd

==JI

TT

=

== JIJI

T

T

N tsintcos

212

tsintcos

dtd

dtd

21

dsdk ==

T , which is constant. (i)

( )

=

++=

== KKJI

KJI

NTB tcostsin)0()0(

0tsintcos

0tcostsin21 22

0dtd

=

B

Hence 020

dtdsdtd

dsd

====

BB

, which is constant. (ii)

From (i) and (ii), we get Curvature k and Torsion are constant. Hence this proves the result.

Remarks: Another way to calculate curvature and torsion

Curvature

2

2

3

d ddt dt

kddt

=

R R

R, Torsion

2 3

2 3

22

2

d d d.

dt dt dt

d ddt dt

=

R R R

R R.


22

Q.No.12.: Show that the curve ( ) ( ) ( ) +++= KJIR 323 tt3aat3tt3a , the curvature equals torsion.

Sol.: The vector equation of the curve is ( ) ( ) ( ) +++= KJIR 323 tt3aat3tt3a ( ) ( ) ( ) +++= KJIR 22 t1a3at6t1a3

dtd

++= KJIR at6a6at6dtd

2

2

+= KIR a6a6dtd

3

3

Now ( ) ( )at6a6at6

t1a3at6t1a3dtd

dtd 22

2

2

+=

KJIRR ( ) ( )

++=

KJI 222 t1t21ta18

2

2

dtd

dtd RR

( ) ( ) ( )2222 t1t21ta18 +++= ( )222242 t2t1t4t21ta18 +++++=

( )242 t21t2a18 ++= ( ) ( )1ta2181ta 218 22222 +=+= And ( ) ( )22222 t1t4t1a3

dtd

+++=R 24224 t2t1t4t2t1a3 +++++=

( ) ( )2224 1ta 231t2t2a3 +=++= ( )2t1a 23 += ( ) ( )

+

++=

KIKJIRRR a6a6.t1t21ta18dtd

.

dtd

dtd 222

3

3

2

2

( ) 3222 a216t1t1a6.a18 =++=

Since we know that curvature ( )32322

3

2

2

t1a 254

)t1(a 218

dtd

dtd

dtd

k+

+=

=

R

RR


23

Curvature ( )22t1a31k+

= (i)

Also ( )2243

2

2

2

3

3

2

2

t1a2.18.18

a216

dtd

dtd

dtd

.

dtd

dtd

+=

=RR

RRR

Torsion ( )22t1a31

+=

(ii)

From (i) and (ii), we have =k

This shows that curvature equals torsion

Hence this proves the result.

Q.No.13.: Find the curvature of the (i) ellipse

+= JIR tsinbtcosa)t(

(ii) Parabola

+= JIR 2tt2)t( at the point t = 1.

Sol.: (i) The vector equation of the ellipse is

+= JIR tsinbtcosa)t(

+= JIR tcosbtsinadtd

Also ( ) 2/12222 tcosbtsinadtd

dtds

+==R

tcosbtsina

tcosbtsina

dtdsdtd

dtdsdtd

dsd

2222 +

+====

JI

RRRT

( ) ( )( )tcosbtsina

tsintcosb2tcostsina2tcosbtsina21

tcosbtsina

tsinbtcosatcosbtsina

dtd

2222

222/12222

2222

+

+

+

+

=

JI

JI

T


24

( )

( ) ( )( )tcosbtsina

tsintcosb2tcostsina2tcosbtsina 2

tcosbtsina


dtd

2222

222/12222

2/12222

+

+

+

+

=

JI

JI

T

( )( )

( ) 2/3222222

2222

tcosbtsina

tsintcosbtcostsina tcosbtsina


dtd

+

+

+

=

JI

JI

T

( ) 2/3222222

tcosbtsina

tsinbatcosabdtd

+

=

JIT

dsdk

=

T, where

dtdsdtd

dsd

=

TT

( ) ( )( )22222

222222

tcosbtsina

t2sinbt2sinatcosbtsinatsinbtcosatcosbtsina

dsd

+

+

+

=

JIJIT

( ) 2/4222222

tcosbtsina

tsinbatcosab

+

=

JI.

dsdk

=

T ( )( ) ( ) 2/3222242222

222222

tcosbtsina

ab

tcosbtsina

tsinatcosbba

+=

+

= . Ans.

(ii) The vector equation of parabola is

+= JIR 2tt2)t(

dtdsdtd

dsd

=

TT


25

+= JIR t22dtd

Now 22 t12t44dtd

dtds

+=+==R

2t12

t22

dtddtd

dtdsdtd

dsd

+

+====

JI

R

RRRT

( ) ( )2

2/122

t1

t2t121

tt1

dtd

+

+

+

+

=

JIJT

( )2

2/122

t1

t1t tt1

+

+

+

+

=

JIJ

( ) 2/322

t1

t tt1

+

++

=

JIJ

( ) 2/32t1t

+

+

=

JI

( )2

2

t12

t1t

+

+

+

=

JI

( ) ( )222

22 t12

1t

t12

t

+

+=

+

+=

JI

Put t = 1, we get ( ) 241

422

1122

dsd

2 ==

+=

T. Ans.

241

dsdk ==

T. Ans.

Problems on equations of the tangent line, the osculating plane and binormal:

Q.No.14.: Find the equation of the tangent line to the curve = cosax , = sinay ,

= tanaz at .

Sol.: The vector equation of the curve is

++= KJIR tanasinacosa

dtdsdtd

dsd

=

TT

4pi

=


26

++=

KJIR tanacosasinadd

( ) ( )[ ] =+=++=

=

secatan1atanacosasina

dd

dds 2/1222/1222222R

++=

==

seca

tanacosasina

ddsdd

dsd KJI

RRT

At 4pi

= , 2

ax = ,

2ay = , pi= tan

4a

z

Now

+

+

= KJITsec

tan

sec21

sec21

D. C. of tangent line are

sec

tan ,

sec21

,

sec21

and it passes through the point a a a, , tan .42 2pi

Then the equation of the tangent line is

a a ax y z tan

2 2 41 1 tan

sec secsec2 2

pi

= =

pi

=

=

tan2

tan4

az

2ay

2a

x . Ans.

Q.No.15.: Find the equation of the osculating plane and binormal to the curve

(i)

=

2t

cosh2x ,

=

2t

sinh2y , t2z = at t = 0.

(ii) tcosex t= , tsiney t= , tez = at t = 0.

Sol.: (i) The vector equation of the curve is

+

+

= KJIR t2

2t

sinh22t

cosh2

+

+

=

KJIR t22t

sinh22t

cosh2dtd

dtd

+

+

= KJI 2

21

2t

cosh221

2t

sinh2


27

+

+

= KJI 2

2t

cosh2t

sinh

42t

cosh2t

sinhdtd

dtds 22 ++== R

42t

cosh2t

sinh

22t

cosh2t

sinh

dtdsdtd

dsd

22 ++

+

+

===

KJIR

RT

222

2/122

22

42t

cosh2t

sinh

21

2t

sinh2t

cosh221

2t

cosh2t

sinh242t

cosh2t

sinh21

.

22t

cosh2t

sinh21

2t

sinh21

2t

cosh42t

cosh2t

sinh

dtd

++

+

++

+

+

+

++

=

KJIJI

T

++

+

+

+

+

++

=

42t

cosh2t

sinh

2t

cosh2t

sinh242t

cosh2t

sinh21

.

22t

cosh2t

sinh21

2t

sinh2t

cosh42t

cosh2t

sinh

22

2/122

22 KJIJI

2/322

22

42t

cosh2t

sinh

2t

cosh2t

sinh.

22t

cosh2t

sinh21

2t

sinh2t

cosh42t

cosh2t

sinh

++

+

+

++

=

KJIJI

At t = 0, we have 52

+

=

KJT


28

dtdsdtd

dsd

=

TT

222

22

42t

cosh2t

sinh

2t

cosh.2t

sinh. 22t

cosh2t

sinh

2t

sinh2t

cosh42t

cosh2t

sinh

++

+

+

+

++

=

KJI

JI

At time t =0, we have

( )

==

= IIIT5

5

5

05dsd

2 .

=== IIT

N1

dtdsdtd

+

== IKJNTB52

0015

25

10

=

KJI

( )

+

+=

51

520 KJI

= KJ

51

52

52

=KJB . Ans.

which is required equation of binormal. Since we know that any vector r in the plane

containing two vectors a and b is given by bar += , where , are arbitrary constants. Thus

++=+=

KJITNR 25

qpqp

, where 5

qq = ,

which is the required equation of osculating plane.

(ii): Given tcosex t= , tsiney t= , tez = .

++=

KJIR tsintcoset

++= KJIR q2qp


29

( ) ( ) ++= KJIR ttt etsintcosetsintcosedtd

( ) t22t e 31tsintcos2edtd

dtds

=++==R

( ) ( )3

tsintcostsintcos

dtdsdtd

dsd

++

===

KJIR

RT

( ) ( )3

tsintcostcostsindtd

++=

JIT

( ) ( )te3

tsintcostcostsin

dtdsdtd

dsd

++=

JIT

T.

At t = 0, we get

3dtdsdtd

dsd

+==

JIT

T.

23

113

dsd

dsd

+=

+

+

==JI

JI

T

T

N .

Thus, at t = 0, we get 3

++

=

KJIT and 2

+

=

JIN

011

1116

1

==

KJI

NTB ( ) ( ) ( )

+++=

KJI 1101106

1


30

62

+

=

KJIB . Ans.

Also equation of plane through

NT and is

+

+

++

=+=

2'q

3'p'q'p JIKJINTR .

( ) ( ) +++= KJIR pqpqp . which is the required equation of osculating plane.

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solution

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