Order of Contact

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Videos on Order of contact
Videos on Order of contact
Example 1
Example 2

Let C be a space curve given by
\(x=f(t), y=g(t),z=h(t) \) (1)
Also let S be a surface given by
\(F(x,y,z) =0 \) (2)
Solving (i) and (ii) [eliminating x,y,z by t] we get
\(F(t)=0 \) (3)
Let \(t_o\) be a solution given by (iii), then we have
\(F(t_o)=0 \) (4)
Expanding \(F(t)\) about \(F(t_o)\) by Taylor’s theorem in powers of \((t-t_o)\) we get
\( F( t )=F( t_o )+( t-t_o )F'( t_o )+\frac{( t-t_o )^2}{2!}F''( t_o )+ \ldots +\frac{( t-t_o )^n}{n!}F^n(t_o )+\ldots \)
Since, \(F(t)=0\) and \(F(t_o)=0\), we have
\( ( t-t_o)F( t_o)+\frac{( t-t_o )^2}{2!}F''( t_o )+\ldots +\frac{( t-t_o )^n}{n!}F^n( t_o )+\ldots =0\)
Now,

  1. If \(F’(t_o) \neq 0 \) then \(t_o\) is simple zero.
    C and S have simple intersection. [One point contact]
  2. If \( F’(t_o)=0 \) , but \( F''(t_o) \neq 0 \) then \(t_o\) is double zero.
    C and S have two point contact. [Contact of first order]
  3. If \(F'(t_o)=0 , F''(t_o)=0 \) , but \(F'''(t_o) \neq 0 \) ; then \(t_o\) is triple zero.
    C and S have three point contact. [Contact of second order]

In similar manner
If \(F'(t_o)=0 , F''(t_o)=0 ,…., F^r(t_o)=0 \) but \(F^{r+1}(t_o) \neq 0 \) , then C and S have (r+1) point contact [Contact of rth order]

Examples
  1. Show that a space curve \( x=1+t, y=t^3, z=t^2 \) has one point contact with the surface \( x^2+y^2-z=0 \) at \( t=1 \)
    Solution
    Given the surface and space curve are respectively
    \(F:x^2+y^2-z=0 \) and (1)
    \(x=1+t, y=t^3, z=t^2 \) (2)
    Thus from (i) and (ii), eliminating Cartesian coordinates x,y,z we get
    \(F(t)=(1+t)2+t^6-t^2 =0 \) (3)
    Since \(F(-1)=0 \) is true for (iii), t=-1 is one a solution given by (i) and (ii).
    Therefore, diff. (iii) w. r. to. t at t=-1, we have
    \(F’ (t) = 2(1+t)+6t^5-2t\)
    at t=-1, we see that
    \( F’ (t) \neq 0 \)
    Thus, surface (i) and curve (ii) have one point contact at t=-1.
  2. Find equation of a plane that has three point contact with a curve \( x=t^4-1, y=t^3-1, z=t^2-1 \) at origin
    Solution
    The equation of curve is
    \(x=t^4-1, y=t^3-1, z=t^2-1 \) (1)
    As we know, equation of plane through origin is
    \(ax+by+cz=0 \) (2)
    Solving (i) and (ii) for t, [eliminating x,y,z] we get
    \(F(t)= a(t^4-1)+b(t^3-1)+c(t^2-1)=0 \) (3)
    Since, plane (ii) has three point contact with the curve (i) at origin [at t=1], we must have
    \(F(t)= 0, F’(t)= 0, F''(t)= 0 \) at t=1 …(iv)
    By successive diff. of (iii) w. r. to. t, we get
    \(F’(t)= 4at^3+3bt^2+2ct \)
    \(F''(t)= 12at^2+6bt+2c \)
    By assumption (iv), we have
    \(F’(t)=0, F''(t)= 0 \)
    or \(4at3+3bt2+2ct=0, 12at2+6bt+2c=0 \)
    or \(4a+3b+2c=0, 12a+6b+2c=0 \) at t=1 (5)
    Solving (v) for a,b,c, we get
    \( \frac{a}{6-12}=\frac{b}{24-8}=\frac{c}{24-36}\)
    or \( \frac{a}{3}=\frac{b}{-8}=\frac{c}{6}\)
    Substituting a,b,c in (ii), equation of required plane is
    \(3x-8y+6z=0. \)
    This completes the solution.
  3. If a circle \( lx+my+nz=0; x^2+ y^2 +z^2 = 2cz \) and paraboloid \( ax^2+ by^2 = 2z \) has three point contact at origin then show that \( C=\frac{l^2+m^2}{bl^2+am^2}\)
    Solution
    Given the circle is
    \(lx+my+nz=0; x^2+ y^2 +z^2 = 2cz\) (1)
    Diff. (i) w r. to. t, we get
    \( l\dot{x}+m\dot{y}+n\dot{z}=0;x\dot{x}+y\dot{y}+z\dot{z}=c\dot{z}\) (2)
    At origin, (ii) reduces to
    \( l\dot{x}+m\dot{y}=0;\text{ }\dot{z}=0\)
    or\( \frac{{\dot{x}}}{m}=\frac{{\dot{y}}}{-l};\dot{z}=0\) (A)
    Again diff. (ii) w. r. to. t, we get
    \( {{\dot{x}}^{2}}+x\ddot{x}+{{\dot{y}}^{2}}+y\ddot{y}+{{\dot{z}}^{2}}+z\ddot{z}=c\ddot{z}\) (3)
    At the origin, (iii) reduces to
    \( {{\dot{x}}^{2}}+{{\dot{y}}^{2}}=c\ddot{z}\) (B)
    Next, the paraboloid is
    \( F:ax^2+ by^2 = 2z \)
    Since, paraboloid has three point contact with the circle at origin,
    \( F’=0, F''=0 \) at origin.
    Thus,
    \( 2ax\dot{x}+2by\dot{y}-2\dot{z}=0;2a{{\dot{x}}^{2}}+2ax\ddot{x}+2b{{\dot{y}}^{2}}+2by\ddot{y}-2\ddot{z}=0\)
    At origin, this equation reduces to
    \( a{{\dot{x}}^{2}}+b{{\dot{y}}^{2}}=\ddot{z}\) (C)
    Eliminating \( \ddot{z}\) between (B) and (C) we get
    \( C=\frac{\dot{x}^2+\dot{y}^2}{a \dot{x}^2+b \dot{y}^2}\)
    From (A), substituting proportional values of Image we get
    \( C=\frac{l^2+m^2}{bl^2+am^2} \)

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