Find the equations of the normal plane and the osculating plane of the helix r(t) = 2cos(t) i +...

Question:

Find the equations of the normal plane and the osculating plane of the helix {eq}\; \mathrm{r}(t) = 2 \cos(t) \, \mathrm{i} + 2 \sin(t) \, \mathrm{j} + t \, \mathrm{k} \; {/eq} at the point {eq}\; P(0, 2, \frac{\pi}{2}) {/eq}.

Finding An Equations of Normal and Osculating Planes:

The normal plane is orthogonal to the velocity vector and passing through the point. Equation of the normal plane is {eq}\displaystyle v(t) \cdot (x-x_{0}, y-y_{0}, z-z_{0}) =0 {/eq}.

The osculating plane is orthogonal to the cross product of velocity and acceleration vectors and passing through the point. Equation of the osculating plane is {eq}\displaystyle \left( v(t) \times a(t) \right) \cdot (x-x_{0}, y-y_{0}, z-z_{0}) =0 {/eq}.

Answer and Explanation:

The given helix is {eq}\; \displaystyle \mathrm{r}(t) = 2 \cos(t) \, \mathrm{i} + 2 \sin(t) \, \mathrm{j} + t \, \mathrm{k} \; {/eq} at the point {eq}\; \displaystyle P\left( 0, 2, \frac{\pi}{2} \right) {/eq}.

Finding the cross product of velocity and acceleration vectors at {eq}\displaystyle t=\frac{\pi}{2} {/eq}:

{eq}\begin{align*} \displaystyle \mathrm{v}(t) &= -2\sin \left(t\right) \, \mathrm{i} + 2\cos \left(t\right) \, \mathrm{j} + 1 \, \mathrm{k} \\ \displaystyle \mathrm{v}\left( \frac{\pi}{2} \right) &= -2\sin \left(\frac{\pi}{2}\right) \, \mathrm{i} + 2\cos \left(\frac{\pi}{2}\right) \, \mathrm{j} + 1 \, \mathrm{k} \\ \displaystyle \mathrm{v}\left( \frac{\pi}{2} \right) &= -2 \, \mathrm{i} + 0 \, \mathrm{j} + 1 \, \mathrm{k} \\ \displaystyle \mathrm{a}(t) &= -2\cos \left(t\right) \, \mathrm{i} -2\sin \left(t\right) \, \mathrm{j} + 0 \, \mathrm{k} \\ \displaystyle \mathrm{a}\left( \frac{\pi}{2} \right) &= -2\cos \left(\frac{\pi}{2}\right) \, \mathrm{i} -2\sin \left(\frac{\pi}{2}\right) \, \mathrm{j} + 0 \, \mathrm{k} \\ \displaystyle \mathrm{a}\left( \frac{\pi}{2} \right) &= 0 \, \mathrm{i} -2 \, \mathrm{j} + 0 \, \mathrm{k} \\ \displaystyle \mathrm{v}\left( \frac{\pi}{2} \right) \times \mathrm{a}\left( \frac{\pi}{2} \right) &=\begin{vmatrix} i & j & k\\ -2 & 0 & 1\\ 0 & -2 & 0 \end{vmatrix} \\ \displaystyle &=((0)(0)-(-2)(1))\vec{i}-((0)(-2)-(0)(1))\vec{j}+((-2)(-2)-(0)(0))\vec{k} \\ \displaystyle \mathrm{v}\left( \frac{\pi}{2} \right) \times \mathrm{a}\left( \frac{\pi}{2} \right) &= 2 \vec{i} +0 \vec{j}+ 4 \vec{k} \end{align*} {/eq}

Finding the equation of the normal plane:

{eq}\begin{align*} \displaystyle v(t) \cdot (x-x_{0}, y-y_{0}, z-z_{0}) &=0 \\ \displaystyle \left \langle -2, 0, 1 \right \rangle \cdot \left \langle x-0, y-2, z-\frac{\pi}{2} \right \rangle &=0 \\ \displaystyle (-2)(x-0)+(0)(y-2)+(1)\left( z-\frac{\pi}{2} \right) &=0 \\ \displaystyle -2x+z-\frac{\pi }{2} &=0 \end{align*} {/eq}

The equation of the normal plane is {eq}\ \displaystyle \mathbf{\color{blue}{ -2x+z-\frac{\pi }{2}=0 }} {/eq}.


Finding the equation of the osculating plane:

{eq}\begin{align*} \displaystyle \left( v(t) \times a(t) \right) \cdot (x-x_{0}, y-y_{0}, z-z_{0}) &=0 \\ \displaystyle \left \langle 2, 0, 4 \right \rangle \cdot \left \langle x-0, y-2, z-\frac{\pi}{2} \right \rangle &=0 \\ \displaystyle (2)(x-0)+(0)(y-2)+(4)\left( z-\frac{\pi}{2} \right) &=0 \\ \displaystyle 2x+4z-2\pi &=0 \\ \displaystyle x+2z-\pi &=0 \end{align*} {/eq}

The equation of the osculating plane is {eq}\ \displaystyle \mathbf{\color{blue}{ x+2z-\pi =0 }} {/eq}.


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