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Equation of osculating plane

1.Equation of osculating plane-

Equation of osculating plane, before we know it, let us know about the osculating plane and the normal line and the normal plane. Also learn about infexional tangents.

(1.) Infexional Tangent-

Definition:A straight line which meets the surface S in three coincident points i.e.,it has a second order point of contact is called Infexional tangent to the surface at that point.

(2.) Normal line and Normal plane (Perpendicular lines and planes)-

Two straight lines which intersect so as to form a pair of equal adjacent angles are Perpendicular (each line is said to be Perpendicular to the other).The condition (in analytic geometry) that two lines be Perpendicular is:(1.)In a plane,that the slope of one of the lines be the negative reciprocal of that of the other,or that one be horizontal and the other vertical;(2)in space,that the sum of the products of the corresponding direction numbers (or direction cosines) of the two lines be zero (two lines in space are Perpendicular if there exist intersecting perpendicular lines,each of which is parallel to one of the given lines).A common perpendicular to two or more lines is a line which is perpendicular to each of them.In a plane,the only lines that can have a common perpendicular are parallel lines,and they have any number.In space any two lines have any numbers of common perpendiculars (only one of which intersects both lines,unless the lines are parallel).A line perpendicular to a plane is a line which is perpendicular to every line through its intersection with the plane.It is sufficient that it be Perpendicular to two nonparallel lines in the plane.The condition (in analytic geometry) that a line be Perpendicular to a plane is that its direction numbers be proportional to those of the normal to the plane;or,what amounts to the same thing,that its direction numbers be proportional to the coefficient of the corresponding variables in the equation of the plane.The foot of the perpendicular to a line (or plane) is the point of intersection of the perpendicular with the line (or plane). Two perpendicular planes are two planes such that a line in one, which is perpendicular to their line of intersection,is perpendicular to the other;i.e.,planes forming a right dihedral angle.The condition (in analytic geometry)that two planes be Perpendicular is that their normals be Perpendicular,or that the sum of the products of the coefficients of like variables in their two equations be zero.

(3.)Normal line-

Any line perpendicular to the tangent at a point P on a curve is called a normal line at that point.Obviously in a three dimensional space curve,there will be an infinitive number of normal lines.

(4.)Normal plane-

The plane passing through a point P on a curve and perpendicular to the tangent at the point P is called the normal plane at the point P.
There the normal plane at a point P on a curve C is the locus of the normal lines to the curve at that point.

2.Equation of the normal plane-

Let r=r(t) be the curve C on which for any point P, whose position vector is r ,\dot { r } is a vector along the tangent at P.Hence the equation of the plane through P perpendicular to \dot { r } is

\left( R-r \right) .\dot { r } =0………………(1)

Where R being the position vector of a current point.
This is the equation of the normal plane to the curve C at P.
Aliter:In rectangular coordinate axes,the equation to the normal plane at a point (x,y,z) on a curve is

\left( \xi -x \right) \dot { x } +\left( \eta -y \right) \dot { y } +\left( \varsigma -z \right) \dot { z } =0
where \dot { x } ,\dot { y } ,\dot { z } are direction-ratios of the tangent at (x,y,z) and dot denotes differentiation with respect to parameter t.
Class of a vector valued function:
A vector valued function r=x\left( t \right) \hat { i } +y\left( t \right) \hat { j } +z\left( t \right) \hat { k } defined on I (real interval) is a said to be of class p,if r has a continuous derivative of pth order at each point of I.

3.The osculting plane-

(1.) The osculating plane at a point P of a curve C of class greater than or equal to two is the limiting and a neighbouring point Q on the curve C as Q\rightarrow P
Aliter:If P,Q,R be three points on a curve C, the limiting positing of the plane PQR,when Q and R tend to P,is called the osculating plane at the point P.
(2.) The osculating plane of a curve C at a point P is the plane that contains the unit tangent vector T at P and contains the principal normal vector dT/ds,where s is distance along the curve (the osculating plane does not exist if dT/ds=0,e.g.,if the curve is a straight line).The osculating plane is the plane in the limiting position,if this exists,of the plane through the tangent to C at the point P,and through a variable point P’ on C,as P’\rightarrow P along C.

4.Equation of osculating plane-

Let r=r(s) be the given curve C of class greater than or equal to two with respect to parameter s,i.e.,the arc length.Let P and Q be two neighbouring points on the curve C with r({ s }_{ 0 }) and r(s) be their position vectors where { s }_{ 0 }and s are lengths at points P and Q from a point A, respectively.Let R be the position vector of current point T on the plane containing the tangent line at P and the point Q.Then the vector

\overrightarrow { PT } =R-r ({ s }_{ 0 }), \hat { t } =r'({ s }_{ 0 }) and \overrightarrow { PQ } =(r)s-r({ s }_{ 0 }) lie in this plane.
Hence their scalar triple product must be zero i.e.,

\left[ R-r({ s }_{ 0 }) \right] .r'({ s }_{ 0 })\quad \times \quad \left[ (r)s-r({ s }_{ 0 }) \right] =0.........(1)

which is the equation of the plane TPQ.Now expanding r(s) lie in powers of by Taylor’s theorem,we have

(r)s=r({ s }_{ 0 })+\left( s-{ s }_{ 0 } \right) r'({ s }_{ 0 })+\frac { { \left( s-{ s }_{ 0 } \right) }^{ 2 } }{ 2! } r\prime \prime ({ s }_{ 0 })+..........(2)
Substituting the value of r(s) from (2) in (1),we get

\left[ R-r({ s }_{ 0 }) \right] .r'({ s }_{ 0 })\quad \times \quad \left[ \left( s-{ s }_{ 0 } \right) r'({ s }_{ 0 })+\frac { { \left( s-{ s }_{ 0 } \right) }^{ 2 } }{ 2! } r\prime \prime ({ s }_{ 0 })+......... \right] =0\\ \left[ R-r({ s }_{ 0 }) \right] .r'({ s }_{ 0 })\quad \times \quad \left[ r\prime \prime ({ s }_{ 0 })+terms\quad of\quad order\quad \left( s-{ s }_{ 0 } \right) and\quad higher \right] =0\quad [\because r'({ s }_{ 0 })\times r'({ s }_{ 0 })=0]
Therefore, the limiting position of the plane as Q \rightarrow P i.e.,s\rightarrow { s }_{ 0 } is

\left[ R-r({ s }_{ 0 }) \right] .r'({ s }_{ 0 })\times r\prime\prime({ s }_{ 0 })=0..........(3)
Provided the vectors are r\prime ({ s }_{ 0 })\quad and\quad r\prime \prime ({ s }_{ 0 }) linearly independent.
Equation (3) can be written as

\left[ R-r({ s }_{ 0 }) \right] ,r'({ s }_{ 0 }),r\prime\prime({ s }_{ 0 })=0..........(4)
which is the equation of the osculating plane at P.
Note: From equation (4), we observe that the vectors r'({ s }_{ 0 }) and r\prime\prime({ s }_{ 0 })  lie in the osculating plane and therefore vector r'({ s }_{ 0 }) \times  r\prime\prime({ s }_{ 0 }) is normal to the osculating plane at P.

5.Equation of osculating plane in terms of general parameter t:

Let P(t) and Q\left( t+\delta t \right) be two neighbouring points with position vectors r(t)\quad and\quad r\left( t+\delta t \right) with respect to origin, respectively,on the curve C.
The tangent at these two points will be parallel to the vectors \dot { r } (t)and\quad \dot { r } \left( t+\delta t \right)  respectively.
Hence the plane passing through the tangent at P(t) and Q\left( t+\delta t \right) is perpendicular to the curve
As the osculating plane at P is perpendicular to the vector

\dot { r } (t)\times \dot { r } \left( t+\delta t \right) =\dot { r } (t)\times \left[ \dot { r } \left( t+\delta t \right) -\dot { r } (t) \right] \\ or\quad \dot { r } (t)\quad \times \frac { \left[ \dot { r } \left( t+\delta t \right) -\dot { r } (t) \right] }{ \delta t } \quad \quad \left[ \because \dot { r } (t)\times \dot { r } (t)=0 \right]

AsQ\rightarrow P, \delta t\rightarrow 0  the osculating plane at P is perpendicular to the vector \dot { r } \left( t \right) \times \ddot { r } \left( t \right)

Now if R be the position vector of a current point on the plane, the vector (R-r(t)) lies in the plane.
Therefore
\left( R-r\left( t \right) \right) \dot { r } (t)\times \ddot { r } (t)\quad =0\\ \left[ R-r\left( t \right) ,\dot { r } ,\ddot { r } \right] =0.........(5)
Which is the equation of  osculating plane.
Aliter: since

r'=\frac { dr }{ ds } =\frac { dr/dt }{ ds/dt } =\frac { \dot { r } }{ \dot { s } } .........(6)
Therefore, r\prime\prime=\frac { \ddot { r } \dot { s } -\dot { r } \ddot { s } }{ { \dot { s } }^{ 2 } } \times \quad \frac { 1 }{ \dot { s } } ………(7)
Substituting (6) and (7) into (4),we have
\left[ R-r,\frac { \dot { r } }{ \dot { s } } ,\frac { \ddot { r } \dot { s } -\dot { r } \ddot { s } }{ { \dot { s } }^{ 3 } } \right] =0\\ or\quad \left[ R-r,\dot { r } ,\ddot { r } \right] =0\quad ,\left[ \because \dot { r } \times \dot { r } =0 \right] ........(8)
Remark:Osculating plane is perpendicular to the normal plane.
The vector \dot { r } \times \ddot { r } is normal to  Osculating plane and is\dot { r } normal plane to normal plane.But\dot { r } .\left( \dot { r } \times \ddot { r } \right) =0 Therefore is perpendicular to \left( \dot { r } \times \ddot { r } \right) .
Hence the normal plane is perpendicular to osculating plane.

6.Equation of osculating plane in cartesian co-ordinates:

Let the coordinates of a point P on a given curve be (x,y,z) and these are functions of a parameter t.Let

a\xi +b\eta +c\varsigma +d=0
be any plane,where a,b,c,d are unknown constants.If this plane has contact of the second order at (x,y,z) with the curve,then

ax+by+cz+d=0\\ a\dot { x } +b\dot { y } +c\dot { z } =0\\ a\ddot { x } +b\ddot { y } +c\ddot { z } =0
On elimination a,b,c,d,we obtain the required equation to osculating plane,as

\left| \begin{matrix} \xi -x & \eta -y & \varsigma-z \\ \dot { x } & \dot { y } & \dot { z } \\ \ddot { x } & \ddot { y } & \ddot { z } \end{matrix} \right| =0
Which is the equation of osculating plane at a point P(x,y,z).

7.Osculating plane at a point of inflexion:

A point P where r\prime\prime=0 is called a point of inflexion and the tangent line at P is called inflexional tangent.

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Also Read This Article:- Conservation of Momentum and Energy

8.Theorem

To show that when the curve is analytic,there exists a definite osculating plane at a point of inflexion, provided the curve is not a straight line.
Proof:Since r\prime =\hat { r } is a unit tangent vector, therefore { r' }^{ 2 }=1 and differentiating with respect to s,we get
r'.r\prime\prime=0 ………..(10)
Again differentiating,we get
r".r\prime\prime+r'.r\prime\prime\prime=0 (r\prime\prime=0,at the point of inflexion)
Or r'.r\prime\prime=0 …………(11)
This shows that r’ is linearly independent of r\prime\prime\prime except when
r\prime\prime\prime=0
Differentiating successively m times and applying above argument,we get

r'.{ r }^{ (m) }=0...............(12)
Where{ r }^{ (m) }(m\ge 2)  is the first non-zero derivative of r at P.Therefore if,

{ r }^{ \left( m \right) }\neq 0from equation (2),we get

r(s)-r({ s }_{ 0 })=\frac { { \left( s-{ s }_{ 0 } \right) }^{ m } }{ m! } { r }^{ m }({ s }_{ 0 })+0\left[ { \left( s-{ s }_{ 0 } \right) }^{ m+1 } \right] ..........(13)
Hence the equation of osculating plane at P is

\left[ R-r({ s }_{ 0 }),r'({ s }_{ 0 }),{ r }^{ m }({ s }_{ 0 }) \right] =0......(14)

Again if,for all m\ge 2the derivative{ r }^{ \left( m \right) }=0.we concluder'\left( =\hat { t } \right)
is constant (since the curve is analytic) i.e., the tangent vector is same at each point of the curve and hence the curve is a straight line.
Hence the equation (14) is the equation of osculating plane at a point of inflexion when the curve is not a straight line.

9.Osculating plane at a point of the curve of intersection of two surfaces-

Let the equation of the surface be
f(r)=0 and\left( \phi \right) r=0...........(1)
The equations of the tangent of these surfaces are given by

\left( R-r \right) .\nabla f=0\quadand \left( R-r \right) .\nabla \phi =0…………(2)
Where \nabla f\quad and\quad \nabla \phi are normal vectors to f(r)=0 and\left( \phi \right) r=0, respectively and R be the position vector of current point on the plane.
Therefore equation of any plane through the line of intersection of two tangent planes i.e., plane through the tangent line to the curve of intersection of the two surfaces is F=\left( R-r \right) .\nabla f-\lambda \left( R-r \right) .\nabla \phi =0..........(3)
Where\lambda is a scalar
The plane (3)will be the osculating plane at P(r) if it has three point contact with the curve at point P(r) .The conditions are F=0,\dot { F } =0\quad and\quad \ddot { F } =0..........(4), at R=r, where a dot denotes differentiation with respect to parameter ‘t’

Thus from above conditions at P(r),(3)gives

\dot { r } .\nabla f-\lambda \dot { r } .\nabla \phi =0,\quad \ddot { r } .\nabla f-\lambda \ddot { r } .\nabla \phi =0.................(5)
Since\dot { r } is the tangent vector to the curve and \dot { \nabla f.\nabla \phi }
are normal vectors,thus we have
\dot { r } .\nabla f=0....................(6)\\ and\quad \dot { r } .\nabla \phi =0............(7)
Thus we observe that the first equation of (5) is an identity.The second equation gives

\lambda =\frac { \ddot { r. } \nabla f }{ \ddot { r. } \nabla \phi } .................(8)
Differentiating (6) and (7) with respect to t,we have

\ddot { r } .\nabla f+\dot { r } .{ \left( \nabla f \right) }^{ \bullet }\quad and\quad \ddot { r } .\nabla \phi +\dot { r } .{ \left( \nabla \phi \right) }^{ \bullet }=0.................(9)\quad
Therefore,

\frac { \dot { r } .{ \left( \nabla f \right) }^{ \bullet } }{ \dot { r } .{ \left( \nabla \phi \right) }^{ \bullet } } =\frac { \ddot { r. } \nabla f }{ \ddot { r. } \nabla \phi } =\lambda ........................(10)
Substituting the value of from (10) into (3),we get

\frac { \left( R-r \right) .\nabla f }{ \left( R-r \right) .\nabla \phi } =\frac { \dot { r } .{ \left( \nabla f \right) }^{ \bullet } }{ \dot { r } .{ \left( \nabla \phi \right) }^{ \bullet } } \\ or\frac { \left( R-r \right) .\nabla f }{ \dot { r } .{ \left( \nabla f \right) }^{ \bullet } } =\frac { \dot { \left( R-r \right) .\nabla \phi } }{ \dot { r } .{ \left( \nabla \phi \right) }^{ \bullet } } ........(11)
This is the required equation of osculating plane at a point P(r) of the curve of intersection of two surfaces.

Cartesian form:The osculating plane at a point of the curve of the of the surfaces

f\left( \xi ,\eta ,\varsigma \right) =0\quad and\quad \phi \left( \xi ,\eta ,\varsigma \right) =0..........(i)
The equation to the tangent planes at (x,y,z) are

\left( \xi -x \right) \frac { \partial f }{ \partial x } +\left( \eta -y \right) \frac { \partial f }{ \partial y } +\left( \varsigma-z \right) \frac { \partial f }{ \partial z } =0........(ii)\\ \left( \xi -x \right) \frac { \partial \phi }{ \partial x } +\left( \eta -y \right) \frac { \partial \phi }{ \partial y } +\left( \varsigma-z \right) \frac { \partial \phi }{ \partial z } =0........(iii)
and any plane through above tangent planes is

\lambda \left\{ \left( \xi -x \right) \frac { \partial f }{ \partial x } +\left( \eta -y \right) \frac { \partial f }{ \partial y } +\left( \varsigma-z \right) \frac { \partial f }{ \partial z } \right\} =\mu \left\{ \left( \xi -x \right) \frac { \partial \phi }{ \partial x } +\left( \eta -y \right) \frac { \partial \phi }{ \partial y } +\left( \varsigma-z \right) \frac { \partial \phi }{ \partial z } \right\} .......(iv)
If this plane is the osculating plane,then it should have a contact of the second order with the curve at (x,y,z), therefore
\lambda \left\{ \dot { x } \frac { \partial f }{ \partial x } +\dot { y } \frac { \partial f }{ \partial y } +\dot { z } \frac { \partial f }{ \partial z } \right\} =\mu \left\{ \dot { x } \frac { \partial \phi }{ \partial x } +\dot { y } \frac { \partial \phi }{ \partial y } +\dot { z } \frac { \partial \phi }{ \partial z } \right\} .......(v)\\ and\quad \lambda \left\{ \ddot { x } \frac { \partial f }{ \partial x } +\ddot { y } \frac { \partial f }{ \partial y } +\ddot { z } \frac { \partial f }{ \partial z } \right\} =\mu \left\{ \ddot { x } \frac { \partial \phi }{ \partial x } +\ddot { y } \frac { \partial \phi }{ \partial y } +\ddot { z } \frac { \partial \phi }{ \partial z } \right\} .......(vi)
Where a dot denotes differentiation with respect to’t’
But\frac { \partial f }{ \partial x } \dot { x } +\frac { \partial f }{ \partial y } \dot { y } +\frac { \partial f }{ \partial z } \dot { z } =0\\ and\frac { \partial \phi }{ \partial x } \dot { x } +\frac { \partial \phi }{ \partial y } \dot { y } +\frac { \partial \phi }{ \partial z } \dot { z } =0.........(vii)

Therefore equation (v) is an identity.It is true,since any plane through the tangent to a curve has contact of the first order with the curve.
Now differentiating the equations of (vii),we have
{ \dot { x } }^{ 2 }\frac { { \partial }^{ 2 }f }{ { \partial x }^{ 2 } } +................+2\dot { y } \dot { z } \frac { { \partial }^{ 2 }f }{ { \partial x\partial z } } +.................=-(\ddot { x } \frac { \partial f }{ \partial x } +........),\\ and\quad { \dot { x } }^{ 2 }\frac { { \partial }^{ 2 }\phi }{ { \partial x }^{ 2 } } +................+2\dot { y } \dot { z } \frac { { \partial }^{ 2 }\phi }{ { \partial x\partial z } } +.................=-(\ddot { x } \frac { \partial \phi }{ \partial x } +........)\quad ...........(viii)
Hence eliminating (iv) and (vi),and using (viii),we get

\frac { \sum { \left( \xi -x \right) \frac { \partial f }{ \partial x } } }{ \sum { { \dot { x } }^{ 2 }\frac { { \partial }^{ 2 }f }{ { \partial x }^{ 2 } } +2\left( \sum { \dot { y } \dot { z } \frac { { \partial }^{ 2 }f }{ { \partial x\partial z } } } \right) } } =\frac { \sum { \left( \xi -x \right) \frac { \partial \phi }{ \partial x } } }{ \sum { { \dot { x } }^{ 2 }\frac { { \partial }^{ 2 }\phi }{ { \partial x }^{ 2 } } +2\left( \sum { \dot { y } \dot { z } \frac { { \partial }^{ 2 }\phi }{ { \partial x\partial z } } } \right) } } .........(ix)
Question-1.Find the equation of osculating plane at any point"\theta " on the helix.

x=a\quad cos\theta \quad ,\quad y=a\quad sin\theta ,\quad \quad z=c\theta
Solution-Equation of the curve are

x=a\quad cos\theta \quad ,\quad y=a\quad sin\theta ,\quad \quad z=c\theta
Therefore

\dot { x } =-a\quad sin\theta \quad \quad \quad \quad \quad \dot { y } =a\quad cos\theta \quad \quad \dot { z } =c\\ \ddot { x } =-a\quad cos\theta \quad \quad \quad \quad \ddot { y } =-a\quad sin\theta \quad \quad \ddot { z } =o
The equation of the osculating plane at the point \theta is given by

\left| \begin{matrix} x-a\quad cos\theta & \quad \quad \quad y-a\quad sin\theta & \quad \quad \quad z-c\theta \\ -a\quad sin\theta & \quad \quad \quad a\quad cos\theta & \quad \quad c \\ -a\quad cos\theta & \quad \quad -a\quad sin\theta & \quad \quad 0 \end{matrix} \right| =0\\ -c\begin{vmatrix} x-a\quad cos\theta & \quad \quad y-a\quad sin\theta \\ cos\theta & sin\theta \end{vmatrix}+\left( z-c\theta \right) \begin{vmatrix} -a\quad sin\theta & \quad \quad a\quad cos\theta \\ cos\theta & sin\theta \end{vmatrix}=0\\ -xc\quad sin\theta +ac\quad sin\theta \quad cos\theta +cy\quad cos\theta -ac\quad sin\theta \quad cos\theta +\left( z-c\theta \right) \left( -a\quad { sin }^{ 2 }\theta -a\quad cos^{ 2 }\theta \right) =0\\ -xc\quad sin\theta +cy\quad cos\theta -az+ac\theta =0\\ c\left( x\quad sin\theta -y\quad cos\theta -a\theta \right) +az=0

This is the required equation of osculating plane.
Question-2.Show that the osculating plane at t=1 of the curve C given by r =\left( 3at,3a{ t }^{ 2 },c{ t }^{ 3 } \right) is \frac { x }{ a } -\frac { y }{ b } +\frac { z }{ c } =1
Solution-Equation of the curve are

x=3at,\quad y=3a{ t }^{ 2 },\quad z=c{ t }^{ 3 }
Therefore

\dot { x } =3a\quad \quad \quad \quad \quad \dot { y } =6bt\quad \quad \dot { z } =3c{ t }^{ 2 }\\ \ddot { x } =0\quad \quad \quad \quad \quad \quad \ddot { y } =6b\quad \quad \ddot { z } =6ct
The equation of osculating plane at t=1 is given by

\left| \begin{matrix} x-3a & \quad \quad \quad y-3b & \quad \quad z-c \\ 3a & \quad \quad 6b & \quad \quad 3c \\ 0 & \quad \quad 6b & \quad \quad 6c \end{matrix} \right| =0\\ \left( x-3a \right) \begin{vmatrix} 6b & \quad \quad 3c \\ 6b & \quad \quad 6c \end{vmatrix}-3a\begin{vmatrix} y-3b & z-c \\ 6b & 6c \end{vmatrix}=0\\ \left( x-3a \right) \left( 36bc-18bc \right) -3a\left[ 6yc-18bc-6zb+6bc \right] =0\\ 18\quad \left( x-3a \right) \left( bc \right) -3a\left( 6yc-12bc-6zb \right) =0\\ 18xbc-54abc-18yac+18zab+36abc=0\\ 18xbc-18yac+18zab=18abc\\ xbc-yac+zab=abc\\ \frac { x }{ a } -\frac { y }{ b } +\frac { z }{ c } =1

This is equation of osculating plane.

Also Read This Article:- Linear differential equation of second order

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