Definition of Osculating Plane
Osculating plane
Let C: \( \vec{r}=\vec{r}( s )\) be a space curve of class \(\ge \) 2 and P, Q, R be three neighboring points on it. Then osculating plane at P is defined as limiting position of a plane determined by PQR as \( Q, R \to \) P.
It is written as
Osculating plane at P \(=\displaystyle \lim_{Q,R \to P}\) PQR
NOTE
- The osculating plane at P contains the tangent line at P.
- Osculating plane at P has three point contact with the curve at P.
- Space curve मा कुनै तिनवटा point हरु ले limiting form मा बनाउने plane लाई osculating plane भनिन्छ ।
Equation of Osculating Plane
Equation of Osculating PlaneLet C: \( \vec{r}=\vec{r}( s )\) be a space curve of class \(\geq 2\).
Also let P be a point on C whose unit tangent at P is
\( \vec{t}=\vec{r}'( s )\)
Assume, Q is neighborhood of P and arc length s of C is measured from suitable base point A such that
arcAP = s, arcAQ = s+\(\delta \)s
Then, position vectors of P and Q are
\( \vec{OP}=\vec{r}( s )\) and \( \vec{OQ}=\vec{r}( s+\delta s )\)
Now, osculating plane at P is defined as
Osculating plane at P = \( \displaystyle \lim_{Q\to P}\) plane determined by \( \vec{t}\) and Q
Osculating plane at P = \(\displaystyle \lim_{\delta s\to 0}\) plane determined by \(\vec{t}\) and Q
Let \( \Gamma \) be the osculating plane at P, then suppose R be the position of arbitrary point T on \( \Gamma \), then
\( \vec{PT},\vec{t}\) and \( \vec{PQ}\) are coplanar at \( \Gamma \)
Thus,
\( [ \vec{PT},\vec{t},\vec{PQ} ]=0\)
or \( [ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}( s+\delta s )-\vec{r}( s ) ]=0\)(i)
Now, equation of osculating plane at P is
\( \displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}( s+\delta s )-\vec{r}( s ) ]=0\)
Expanding \( \vec{r}\)(s+\(\delta \)s) about \( \vec{r}(s)\) by Taylor’s theorem in powers of \(\delta s \), we get
\( \displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\{ \vec{r}( s )+\delta s\vec{r}'( s )+ \frac{\delta s^2}{2!} \vec{r}''( s )+O( \delta s )^3 \}-\vec{r}( s ) ]=0\)
or \( \displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}( s )+\delta s\vec{r}'( s )+\frac{\delta s^2}{2!}\vec{r}''( s )+ O( \delta s )^3-\vec{r}( s ) ]=0\)
or \( \displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\delta s\vec{r}'( s )+\frac{\delta s^2}{2!}\vec{r}''( s )+O( \delta s )^3 ]=0\)
or \( \displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\frac{\delta s^2}{2!}\vec{r}''( s )+O{{( \delta s )}^{3}} ]=0\)
or \( \frac{\delta s^2}{2!}\displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}''( s )+O( \delta s ) ]=0\)
or \( \displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}''( s )+O( \delta s ) ]=0\)
or \( [ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}''( s )+0 ]=0\)
or \( [ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}''( s ) ]=0\)
This is required equation of osculating plane at P.
Alternative Form of Osculating Plane
Let C:\( \vec{r}=\vec{r}( s )\) be a space curve then
\( \vec{r}'=\frac{ \dot{\vec{r}} } {\dot{s}} \)
Next, we compute second derivative given by
\( \vec{r}''\)
or \( \frac{d}{dt}(\frac{ \dot{\vec{r}} } {\dot{s}} ) \frac{dt}{ds} \)
or \( \frac{
\ddot{\vec{r}} \dot{s} -\dot{\vec{r}} \ddot{s} } {\dot{s}^3}\)
Now, equation of the osculating plane is
\( [ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}''( s ) ]=0\)
or
\( [ R-\vec{r}( t ),\frac{
\dot{\vec{r}} } {\dot{s}},\frac{
\ddot{\vec{r}}\dot{s}-\dot{\vec{r}}\ddot{s}} {\dot{s}^3} ]=0\)
or
\( [ R-\vec{r}( t ),\dot{\vec{r}},\ddot{\vec{r}} ]=0\)
Cartesian form of Osculating Plane
Let C:\( \vec{r}=\vec{r}( s )\)be a space curve and P be a point on it, in which
\( \vec{r}\)= (x,y,z)
\( \vec{r}'\)= (x’,y’,z’)
\( \vec{r}''\)= (x’’,y’’,z’’)
Also, equation of osculation plane at P is
\( [ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}''( s ) ]=0\)
In this equation,
R=(X,Y,Z)
Hence, equation of osculating plane in Cartesian form is
\( [ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}''( s ) ]=0\)
or
\( [ (X,Y,Z)-(x,y,z),(x',y',z'),(x'',y'',z'') ]=0\)
or
\( [ (X-x,Y-y,Z-z),(x',y',z'),(x'',y'',z'') ]=0\)
or
\( \begin{bmatrix} X-x & Y-y & Z-z \\ x' & y' & z' \\ x'' & y'' & z'' \\ \end{bmatrix} =0\)
Note
If any other parameter t (instead of s) is described, then equation of osculating plane is
\( \begin{bmatrix} X-x & Y-y & Z-z \\ {\dot{x}} & {\dot{y}} & {\dot{z}} \\ {\ddot{x}} & {\ddot{y}} & {\ddot{z}} \\ \end{bmatrix} =0\)
Examples
- Find osculating plane on the curve \( \vec{r}\)= (acost, asint, bt).
Solution
Given curve is
\( \vec{r}\)= (acost, asint, bt) (i)
By successive differentiation of (i) w. r. to. t, we get
\( \dot{\vec{r}}\) =(-asint, acost, b)
\( \ddot{\vec{r}}\) =(-acost, -asint, 0)
Hence, equation of osculating plane is
\( \begin{bmatrix} X-x & Y-y & Z-z \\ {\dot{x}} & {\dot{y}} & {\dot{z}} \\ {\ddot{x}} & {\ddot{y}} & {\ddot{z}} \\ \end{bmatrix} =0\)
or \( \begin{bmatrix} X-acost & Y-asint & Z-bt \\ -asint & acost & b \\ -acost & -asint & 0 \\ \end{bmatrix} =0\)
By a bit of more calculation, we get
\(b[ Xsint-Ycost-at ]+aZ=0\)
Equation of Osculating Plane at Inflexional Point
Equation at Inflexional Point
Let C: \( \vec{r}=\vec{r}( s )\) be a space curve of class \(\geq 2\).
Also let P be a point on C whose unit tangent at P is
\( \vec{t}=\vec{r}'( s )\)
Assume, Q is neighborhood of P and arc length s of C is measured from suitable base point A such that
arcAP = s, arcAQ = s+\(\delta \)s
Then, position vectors of P and Q are
\( \vec{OP}=\vec{r}( s )\) and \( \vec{OQ}=\vec{r}( s+\delta s )\)
Now, osculating plane at P is defined as
Osculating plane at P = \( \displaystyle \lim_{Q\to P}\) plane determined by \( \vec{t}\) and Q
Osculating plane at P = \(\displaystyle \lim_{\delta s\to 0}\)plane determined by \(\vec{t}\) and Q
Let \( \Gamma \) be the osculating plane at P, then suppose R be the position of arbitrary point T on \(
\Gamma \), then
\( \vec{PT},\vec{t}\) and \( \vec{PQ}\) are coplanar at \( \Gamma \)
Thus,
\( [ \vec{PT},\vec{t},\vec{PQ} ]=0\)
or
\( [ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}( s+\delta s )-\vec{r}( s ) ]=0\)(i)
Now, equation of osculating plane at P is
\( \displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}( s+\delta s )-\vec{r}( s ) ]=0\)
Since P is inflexional point
\( \vec{r}''\)=0 at P with \( \vec{r}'.\vec{r}''=0\)
\( \vec{r}'.\vec{r}''=0 \)(ii)
Differentiating (ii) w. r. to. s, we get
\( \vec{r}''.\vec{r}''+\vec{r}'.\vec{r}'''=0\)
If \( \vec{r}'''\neq 0 \), we stop the process of differentiation
If\( \vec{r}'''= 0 \), we differentiate w. r. to. s to get
\( \vec{r}''.\vec{r}'''+\vec{r}'.\vec{r}''''=0\)
or
\( \vec{r}'.\vec{r}'''' =0 \)
If \( \vec{r}'''' \ne 0 \), we stop the process of differentiation
If \( \vec{r}''''\)=0, we continue the similar argument to arrive at
\( \vec{r}'.{{\vec{r}}^k}\)=0, where\({{\vec{r}}^k}; k \ge 3\) is the first non-zero derivatives of \( \vec{r}\) at P.
Now from (i), we have
\( [ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}( s+\delta s )-\vec{r}( s ) ]=0\)
Expanding \( \vec{r}( s+\delta s )\)about\( \vec{r}( s )\) by Taylor’s theorem in powers of\( \delta s\), we get
\( \displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\{ \vec{r}( s )+\delta s\vec{r}'( s )+\frac{\delta {s^{2}}}{2!}\vec{r}''( s )+...+\frac{\delta {s^k}}{k!}{{{\vec{r}}}^k}( s )+O{{( \delta s )}^{k+1}} \}-\vec{r}( s ) ]=0\)
or
\( \displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}( s )+\delta s\vec{r}'( s )+\frac{\delta {s^{2}}}{2!}\vec{r}''( s )+...+\frac{\delta {s^k}}{k!}{{{\vec{r}}}^k}( s )+O{{( \delta s )}^{k+1}}-\vec{r}( s ) ]=0\)
or
\( \displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\delta s\vec{r}'( s )+\frac{\delta {s^{2}}}{2!}\vec{r}''( s )+...+\frac{\delta {s^k}}{k!}{{{\vec{r}}}^k}( s )+O{{( \delta s )}^{k+1}} ]=0\)
or
\( \displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\frac{\delta {s^{2}}}{2!}\vec{r}''( s )+...+\frac{\delta {s^k}}{k!}{{{\vec{r}}}^k}( s )+O{{( \delta s )}^{k+1}} ]=0\)
or
\(\displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\frac{\delta {s^{3}}}{3!}\vec{r}'''( s )+...+\frac{\delta {s^k}}{k!}{{{\vec{r}}}^k}( s )+O{{( \delta s )}^{k+1}} ]=0\)
or
\( \displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\frac{\delta {s^k}}{k!}{{{\vec{r}}}^k}( s )+O{{( \delta s )}^{k+1}} ]=0\)
or
\( \frac{\delta {s^k}}{k!}\displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}^k( s ) +O( \delta s ) ]=0\)
or
\( \displaystyle \lim_{\delta s\to 0}[ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}^k( s )+O( \delta s ) ]=0\)
or
\( [ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}^k( s )+0 ]=0\)
or
\( [ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}^k( s ) ]=0, k \geq \)3.
This is required equation of osculating plane at P.
Osculating Plane has three point contact
Three point ContactShow that osculating plane at P has, in general, three point contact (contact of second order) with the curve at P.
Solution
Let C:\( \vec{r}=\vec{r}\)(s) be a space curve and P be a point on it. For simplicity, we measure arc length s of C such that s=0 at P.
Hence,
\( \vec{OP}=\vec{r}( s )=\vec{r}( 0 )\)
Now, osculating plane at P is
\( [ R-\vec{r}( s ),\vec{r}'( s ),\vec{r}''( s ) ]=0\)
\( [ R-\vec{r}( 0 ),\vec{r}'( 0 ),\vec{r}''( 0 ) ]=0\)
The intersection of curve and plane is given by
\(F( s ):[ \vec{r}( s )-\vec{r}( 0 ),\vec{r}'( 0 ),\vec{r}''( 0 ) ]=0\)
Now, expanding \( \vec{r} (s) \) about \( \vec{r} (0) \) by Taylor’s theorem in powers of s, we get
F(s): \( [ \vec{r}( 0 )+s\vec{r}'( 0 )+\frac{{s^2}}{2!}\vec{r}''( 0 )+\frac{{s^3}}{3!}\vec{r}'''( 0 )+O( {s^4} ) .. -\vec{r}( 0 ),\vec{r}'( 0 ),\vec{r}''( 0 ) ]\)
F(s):\( [ s\vec{r}'( 0 )+\frac{{s^2}}{2!}\vec{r}''( 0 )+\frac{{s^3}}{3!}\vec{r}'''( 0 )+O( {s^4} ),\vec{r}'( 0 ),\vec{r}''( 0 ) ]\)
F(s):\( [ \frac{{s^3}}{3!}\vec{r}'''( 0 )+O( {s^4} ),\vec{r}'( 0 ),\vec{r}''( 0 ) ]\)
F(s):\(\frac{{s^3}}{3!}[ \vec{r}'''( 0 )+O( s ),\vec{r}'( 0 ),\vec{r}''( 0 ) ]\)
It leads that
\(F( s )=0,F'( s )=0,F''( s )=0\)
but
\( F'''( s )\ne 0\) in general at s=0
Therefore, we conclude that osculating plane at P has, in general, three point contact (contact of second order) with the curve at P.
ExerciseFind equations of osculating plane of the curve at the given point
- \( x = 5 \sin (3t), y = t, z = 5 \cos (3t) \) at \( \pi\)
Solution
Given curve is
\( \vec{r}= (5 \sin (3t), t, 5 \cos (3t)) \) (i)
At \(t=\pi\), it is
\( \vec{r}= (0, \pi, -5) \)
By successive differentiation of (i) w. r. to. t, we get
\( \dot{\vec{r}} =(15 \cos (3t), 1, -15 \sin (3t)) \)
At \(t=\pi\), it is
\( \dot{\vec{r}} =(15 , 1, 0) \)
And
\( \ddot{\vec{r}} =(-45\sin (3t), 0, -45\cos (3t)) \)
At \(t=\pi\), it is
\( \ddot{\vec{r}} =(0,0,45) \)
Hence, equation of osculating plane at \(t=\pi\) is
\( \begin{bmatrix} X-x & Y-y & Z-z \\ {\dot{x}} & {\dot{y}} & {\dot{z}} \\ {\ddot{x}} & {\ddot{y}} & {\ddot{z}} \\ \end{bmatrix} =0\)
or \( \begin{bmatrix} x-0 & y-\pi & z+5 \\ -15 & 1 & 0 \\ 0 & 0 & 45 \\ \end{bmatrix} =0\)
By a bit of more calculation, we get
\( x+15y-15 \pi= 0\)
This completes the solution.
- \(x = 5t, y = 𝑡^2, z = 𝑡^3 \) at t=1
Solution
Given curve is
\( \vec{r}= (5t, t^2, t^3) \) (i)
At \( t = 1 \), it is
\( \vec{r}= (5, 1, 1) \)
By successive differentiation of (i) w.r.t. \( t \), we get
\( \dot{\vec{r}} = (5, 2t, 3t^2) \)
At \( t = 1 \), it is
\( \dot{\vec{r}} = (5, 2, 3) \)
And
\( \ddot{\vec{r}} = (0, 2, 6t) \)
At \( t = 1 \), it is
\( \ddot{\vec{r}} = (0, 2, 6) \)
Hence, equation of osculating plane at \( t = 1 \) is
\( \begin{bmatrix} X-x & Y-y & Z-z \\ \dot{x} & \dot{y} & \dot{z} \\ \ddot{x} & \ddot{y} & \ddot{z} \end{bmatrix} = 0 \)
or \( \begin{bmatrix} x-5 & y-1 & z-1 \\ 5 & 2 & 3 \\ 0 & 2 & 6 \end{bmatrix} = 0 \)
By a bit of more calculation, we get
\( x + 5y - 6z - 4 = 0 \)
This completes the solution.
- \( x = 7 \sin 3t, y = t, z = 7 \cos 3t \) at \( \pi\)
Solution
Given curve is
\( \vec{r}= (7 \sin 3t, t, 7 \cos 3t) \) (i)
At \( t = \pi \), it is
\( \vec{r}= (0, \pi, -7) \)
By successive differentiation of (i) w.r.t. \( t \), we get
\( \dot{\vec{r}} = (21 \cos 3t, 1, -21 \sin 3t) \)
At \( t = \pi \), it is
\( \dot{\vec{r}} = (21 , 1, 0) \)
And
\( \ddot{\vec{r}} = (-63 \sin 3t, 0, -63 \cos 3t) \)
At \( t = \pi \), it is
\( \ddot{\vec{r}} = (0, 0, 63) \)
Hence, equation of osculating plane at \( t = \pi \) is
\( \begin{bmatrix} X-x & Y-y & Z-z \\ \dot{x} & \dot{y} & \dot{z} \\ \ddot{x} & \ddot{y} & \ddot{z} \end{bmatrix} = 0 \)
or \( \begin{bmatrix} x-0 & y-\pi & z+7 \\ 21 & 1 & 0 \\ 0 & 0 & 63 \end{bmatrix} = 0 \)
By a bit of more calculation, we get
\( x + 21y - 21 \pi = 0 \)
This completes the solution.
- \(\vec{r}(t) = (\cos t, \sin t, \log (\cos t)) \) at 0
Solution
Given curve is
\( \vec{r}= (\cos t, \sin t, \log (\cos t)) \) (i)
At \( t = 0 \), it is
\( \vec{r}= (1, 0, \log 1) \)
By successive differentiation of (i) w.r.t. \( t \), we get
\( \dot{\vec{r}} = (-\sin t, \cos t, -\tan t) \)
At \( t = 0 \), it is
\( \dot{\vec{r}} = (0, 1, 0) \)
And
\( \ddot{\vec{r}} = (-\cos t, -\sin t, -\sec^2 t) \)
At \( t = 0 \), it is
\( \ddot{\vec{r}} = (-1, 0, -1) \)
Hence, equation of osculating plane at \( t = 0 \) is
\( \begin{bmatrix} X-x & Y-y & Z-z \\ \dot{x} & \dot{y} & \dot{z} \\ \ddot{x} & \ddot{y} & \ddot{z} \end{bmatrix} = 0 \)
or \( \begin{bmatrix} x-1 & y-0 & z-0 \\ 0 & 1 & 0 \\ -1 & 0 & -1 \end{bmatrix} = 0 \)
By a bit of more calculation, we get
\( x + z - 1 = 0 \)
This completes the solution.
- \(x = 2 \sin (3t), y = t, z = 2 \cos (3t) \) at \( \pi\)
Solution
Given curve is
\( \vec{r}= (2 \sin (3t), t, 2 \cos (3t)) \) (i)
At \( t = \pi \), it is
\( \vec{r}= (0, \pi, -2) \)
By successive differentiation of (i) w.r.t. \( t \), we get
\( \dot{\vec{r}} = (6 \cos (3t), 1, -6 \sin (3t)) \)
At \( t = \pi \), it is
\( \dot{\vec{r}} = (6 , 1, 0) \)
And
\( \ddot{\vec{r}} = (-18 \sin (3t), 0, -18 \cos (3t)) \)
At \( t = \pi \), it is
\( \ddot{\vec{r}} = (0, 0, 18) \)
Hence, equation of osculating plane at \( t = \pi \) is
\( \begin{bmatrix} X-x & Y-y & Z-z \\ \dot{x} & \dot{y} & \dot{z} \\ \ddot{x} & \ddot{y} & \ddot{z} \end{bmatrix} = 0 \)
or \( \begin{bmatrix} x-0 & y-\pi & z+2 \\ 6 & 1 & 0 \\ 0 & 0 & 18 \end{bmatrix} = 0 \)
By a bit of more calculation, we get
\( x + 6y - 6 \pi = 0 \)
This completes the solution.
- \(x = 7 \sin(3t), y = t, z = 7 \cos(3t) \) at \( \pi\)
Solution
Given curve is
\( \vec{r}= (7 \sin(3t), t, 7 \cos(3t)) \) (i)
At \( t = \pi \), it is
\( \vec{r}= (0, \pi, -7) \)
By successive differentiation of (i) w.r.t. \( t \), we get
\( \dot{\vec{r}} = (21 \cos(3t), 1, -21 \sin(3t)) \)
At \( t = \pi \), it is
\( \dot{\vec{r}} = (21, 1, 0) \)
And
\( \ddot{\vec{r}} = (-63 \sin(3t), 0, -63 \cos(3t)) \)
At \( t = \pi \), it is
\( \ddot{\vec{r}} = (0, 0, 63) \)
Hence, equation of osculating plane at \( t = \pi \) is
\( \begin{bmatrix} X-x & Y-y & Z-z \\ \dot{x} & \dot{y} & \dot{z} \\ \ddot{x} & \ddot{y} & \ddot{z} \end{bmatrix} = 0 \)
or \( \begin{bmatrix} x-0 & y-\pi & z+7 \\ 21 & 1 & 0 \\ 0 & 0 & 63 \end{bmatrix} = 0 \)
By a bit of more calculation, we get
\( x + 21y - 21 \pi = 0 \)
This completes the solution.
- \(x = \log t, y = 2 t, z = t^2 \) at 1
Solution
Given curve is
\( \vec{r}= (\log t, 2t, t^2) \) (i)
At \( t = 1 \), it is
\( \vec{r}= (0, 2, 1) \)
By successive differentiation of (i) w.r.t. \( t \), we get
\( \dot{\vec{r}} = \left(\frac{1}{t}, 2, 2t\right) \)
At \( t = 1 \), it is
\( \dot{\vec{r}} = (1, 2, 2) \)
And
\( \ddot{\vec{r}} = \left(-\frac{1}{t^2}, 0, 2\right) \)
At \( t = 1 \), it is
\( \ddot{\vec{r}} = (-1, 0, 2) \)
Hence, equation of osculating plane at \( t = 1 \) is
\( \begin{bmatrix} X-x & Y-y & Z-z \\ \dot{x} & \dot{y} & \dot{z} \\ \ddot{x} & \ddot{y} & \ddot{z} \end{bmatrix} = 0 \)
or \( \begin{bmatrix} x-0 & y-2 & z-1 \\ 1 & 2 & 2 \\ -1 & 0 & 2 \end{bmatrix} = 0 \)
By a bit of more calculation, we get
\( 4x - y + z - 3 = 0 \)
This completes the solution.
- \(x = 7 \sin 3t, y = t, z = 7 \cos 3t\) at \( \pi\)
Solution
Given curve is
\( \vec{r}= (7 \sin 3t, t, 7 \cos 3t) \) (i)
At \( t = \pi \), it is
\( \vec{r}= (0, \pi, -7) \)
By successive differentiation of (i) w.r.t. \( t \), we get
\( \dot{\vec{r}} = (21 \cos 3t, 1, -21 \sin 3t) \)
At \( t = \pi \), it is
\( \dot{\vec{r}} = (21 , 1, 0) \)
And
\( \ddot{\vec{r}} = (-63 \sin 3t, 0, -63 \cos 3t) \)
At \( t = \pi \), it is
\( \ddot{\vec{r}} = (0, 0, 63) \)
Hence, equation of osculating plane at \( t = \pi \) is
\( \begin{bmatrix} X-x & Y-y & Z-z \\ \dot{x} & \dot{y} & \dot{z} \\ \ddot{x} & \ddot{y} & \ddot{z} \end{bmatrix} = 0 \)
or \( \begin{bmatrix} x-0 & y-\pi & z+7 \\ 21 & 1 & 0 \\ 0 & 0 & 63 \end{bmatrix} = 0 \)
By a bit of more calculation, we get
\( x + 21y - 21 \pi = 0 \)
This completes the solution.
- \(x=\sin \ t,y= - \cos \ t ,z= t^2 \) at \( t = \frac{\pi}{2}\)
Solution
Given curve is
\( \vec{r}= (\sin t, -\cos t, t^2) \) (i)
At \( t = \frac{\pi}{2} \), it is
\( \vec{r}= (1, 0, \left(\frac{\pi}{2}\right)^2) \)
By successive differentiation of (i) w.r.t. \( t \), we get
\( \dot{\vec{r}} = (\cos t, \sin t, 2t) \)
At \( t = \frac{\pi}{2} \), it is
\( \dot{\vec{r}} = (0, 1, \pi) \)
And
\( \ddot{\vec{r}} = (-\sin t, \cos t, 2) \)
At \( t = \frac{\pi}{2} \), it is
\( \ddot{\vec{r}} = (-1, 0, 2) \)
Hence, equation of osculating plane at \( t = \frac{\pi}{2} \) is
\( \begin{bmatrix} X-x & Y-y & Z-z \\ \dot{x} & \dot{y} & \dot{z} \\ \ddot{x} & \ddot{y} & \ddot{z} \end{bmatrix} = 0 \)
or \( \begin{bmatrix} x-1 & y-0 & z-\left(\frac{\pi^2}{4}\right) \\ 0 & 1 & \pi \\ -1 & 0 & 2 \end{bmatrix} = 0 \)
By a bit of more calculation, we get
\( x + 2y - \pi z + \frac{\pi^3}{4} = 0 \)
This completes the solution.
- \(x = \sin 2 t, y = - \cos 2 t, z = 4 t\) at \( 2 \pi\)
Solution
Given curve is
\( \vec{r}= (\sin 2t, -\cos 2t, 4t) \) (i)
At \( t = 2\pi \), it is
\( \vec{r}= (\sin 4\pi, -\cos 4\pi, 8\pi) \) \)
By successive differentiation of (i) w.r.t. \( t \), we get
\( \dot{\vec{r}} = (2\cos 2t, 2\sin 2t, 4) \)
At \( t = 2\pi \), it is
\( \dot{\vec{r}} = (2\cos 4\pi, 2\sin 4\pi, 4) \) \)
And
\( \ddot{\vec{r}} = (-4\sin 2t, 4\cos 2t, 0) \)
At \( t = 2\pi \), it is
\( \ddot{\vec{r}} = (-4\sin 4\pi, 4\cos 4\pi, 0) \)
Hence, equation of osculating plane at \( t = 2\pi \) is
\( \begin{bmatrix} X-x & Y-y & Z-z \\ \dot{x} & \dot{y} & \dot{z} \\ \ddot{x} & \ddot{y} & \ddot{z} \end{bmatrix} = 0 \)
or \( \begin{bmatrix} x-0 & y+1 & z-8\pi \\ 2 & 0 & 4 \\ 0 & 4 & 0 \end{bmatrix} = 0 \)
By a bit of more calculation, we get
\( x + 2z - 16\pi = 0 \)
This completes the solution.
Thank you so much Guru ❤️
ReplyDelete