- \( x = 5 \sin (3t), y = t, z = 5 \cos (3t) \) at \( \pi\)
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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
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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 \( \pi\)
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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
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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 \( \pi\)
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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 \( \pi\)
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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
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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 \( \pi\)
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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 \( t = \frac{\pi}{2}\)
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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 \( 2 \pi\)
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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 ❤️
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