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Cuve C is traced out by the vector function shown here: \vec{r}(t)=\langle 3\sin(t), 4t, 3\cos(t)...

Question:

Cuve C is traced out by the vector function shown here:

{eq}\vec{r}(t)=\langle 3\sin(t), 4t, 3\cos(t) \rangle{/eq}

(a) Find the vectors {eq}\vec{T}, \vec{N}{/eq}, and {eq}\vec{B}{/eq} at the point {eq}P(0,4 \pi, -3){/eq} on the curve.

(b) Find an equation for the normal plane to the curve {eq}C{/eq} at the point {eq}P{/eq}.

(c) Find an equation for the osculating plane to the curve {eq}C{/eq} at the point {eq}P{/eq}.

(d) Find parametric equations for the tangent line to the curve {eq}C{/eq} at the point {eq}P{/eq}.

Application of Derivatives:

We have to find some given terms by using the velocity and acceleration vectors. The normal plane is normal to the velocity vector and passing through the point.

The osculating plane is normal to the binormal vector and passing through the point.

Answer and Explanation:

Let us consider the given vector function {eq}\displaystyle \vec{r}(t)=\langle 3\sin(t), 4t, 3\cos(t) \rangle{/eq} at the point {eq}\displaystyle P(0,4 \pi, -3){/eq}.

The parameter value is {eq}\displaystyle t=\pi {/eq} which is corresponding to the point.

(a) Finding the vectors {eq}\vec{T}, \vec{N}{/eq}, and {eq}\vec{B}{/eq} at the point on the curve:

{eq}\begin{align*} \displaystyle \text{Unit tangent vector}: \\ \displaystyle \vec{T}(t) &=\frac{{\vec{r}}'(t)}{\left \| {\vec{r}}'(t) \right \|} \\ \displaystyle \vec{r}'(t) &=\langle 3\cos \left(t\right), 4, -3\sin \left(t\right) \rangle \\ \displaystyle \left \| \vec{r}'(t) \right \| &=\sqrt{(3\cos \left(t\right))^{2}+(4)^{2}+(-3\sin \left(t\right))^{2}} \\ \displaystyle \left \| \vec{r}'(t) \right \| &=5 \\ \displaystyle \vec{T}(t) &=\left \langle \frac{3\cos \left(t\right)}{5}, \frac{4}{5}, \frac{-3\sin \left(t\right)}{5} \right \rangle \\ \displaystyle \vec{T}(\pi) &=\left \langle \frac{3\cos \left(\pi\right)}{5}, \frac{4}{5}, \frac{-3\sin \left(\pi\right)}{5} \right \rangle \\ \displaystyle \vec{T}(\pi) &=\left \langle -\frac{3}{5}, \frac{4}{5}, 0 \right \rangle \\ \\ \displaystyle \text{Unit normal vector}: \\ \displaystyle \vec{N}(t) &=\frac{{\vec{T}}'(t)}{ \left \| {\vec{T}}'(t) \right \|} \\ \displaystyle \vec{T}'(t) &=\left \langle -\frac{3}{5}\sin \left(t\right), 0, -\frac{3}{5}\cos \left(t\right) \right \rangle \\ \displaystyle \vec{T}'(\pi) &=\left \langle -\frac{3}{5}\sin \left(\pi\right), 0, -\frac{3}{5}\cos \left(\pi\right) \right \rangle \\ \displaystyle \vec{T}'(\pi) &=\left \langle 0, 0, \frac{3}{5} \right \rangle \\ \displaystyle \left \| \vec{T}'(\pi) \right \| &=\frac{3}{5} \\ \displaystyle \vec{N}(\pi) &=\left \langle 0, 0, \frac{\frac{3}{5}}{\frac{3}{5}} \right \rangle \\ \displaystyle \vec{N}(\pi) &=\left \langle 0, 0, 1 \right \rangle \\ \\ \displaystyle \text{Binormal vector}: \\ \displaystyle \vec{B}(t) &=\vec{T}(t)\times \vec{N}(t) \\ \displaystyle &=\begin{vmatrix} i & j & k\\ -\frac{3}{5} & \frac{4}{5} & 0\\ 0 & 0 & 1 \end{vmatrix} \\ \displaystyle &=\left( (1)\left( \frac{4}{5} \right)-(0)(0) \right)\vec{i}-\left( (1)\left( -\frac{3}{5} \right)-(0)(0) \right)\vec{j}+\left( (0)\left( -\frac{3}{5} \right)-(0)\left( \frac{4}{5} \right) \right)\vec{k} \\ \displaystyle \vec{B}(\pi) &= \frac{4}{5} \vec{i} + \frac{3}{5} \vec{j}+ 0 \vec{k} \end{align*} {/eq}

The unit tangent vector is {eq}\ \displaystyle \mathbf{\color{blue}{ \vec{T}(\pi)=\left \langle -\frac{3}{5}, \frac{4}{5}, 0 \right \rangle }} {/eq}, the unit normal vector is {eq}\ \displaystyle \mathbf{\color{blue}{ \vec{N}(\pi) =\left \langle 0, 0, 1 \right \rangle }} {/eq} and the binormal vector is {eq}\ \displaystyle \mathbf{\color{blue}{ \vec{B}(\pi)= \frac{4}{5} \vec{i} + \frac{3}{5} \vec{j} }} {/eq}.


(b) Finding an equation for the normal plane to the curve {eq}C{/eq} at the point {eq}P{/eq}:

{eq}\begin{align*} \displaystyle \vec{r}'(t) &=\langle 3\cos \left(t\right), 4, -3\sin \left(t\right) \rangle \\ \displaystyle \vec{r}'(\pi) &=\langle 3\cos \left(\pi\right), 4, -3\sin \left(\pi\right) \rangle \\ \displaystyle \vec{r}'(\pi) &=\langle -3, 4, 0 \rangle \\ \displaystyle \vec{r}'(t) \cdot \left \langle x-x_{0}, y-y_{0}, z-z_{0} \right \rangle &=0 \\ \displaystyle \langle -3, 4, 0 \rangle \cdot \left \langle x-0, y-4\pi, z-(-3) \right \rangle &=0 \\ \displaystyle \langle -3, 4, 0 \rangle \cdot \left \langle x-0, y-4\pi, z+3 \right \rangle &=0 \\ \displaystyle (-3)(x-0)+(4)(y-4\pi)+(0)(z+3) &=0 \\ \displaystyle -3x+4y-16\pi &=0 \end{align*} {/eq}

An equation of the normal plane is {eq}\ \displaystyle \mathbf{\color{blue}{ -3x+4y-16\pi =0 }} {/eq}.


(c) Finding an equation for the osculating plane to the curve {eq}C{/eq} at the point {eq}P{/eq}:

{eq}\begin{align*} \displaystyle \vec{B}(t) \cdot \left \langle x-x_{0}, y-y_{0}, z-z_{0} \right \rangle &=0 \\ \displaystyle \left \langle \frac{4}{5}, \frac{3}{5}, 0 \right \rangle \cdot \left \langle x-0, y-4\pi, z+3 \right \rangle &=0 \\ \displaystyle \left( \frac{4}{5} \right)(x-0)+\left( \frac{3}{5} \right)(y-4\pi)+(0)(z+3) &=0 \\ \displaystyle \frac{4x}{5}+\frac{3y}{5}-\frac{12\pi }{5} &=0 \end{align*} {/eq}

An equation of the osculating plane is {eq}\ \displaystyle \mathbf{\color{blue}{ \frac{4x}{5}+\frac{3y}{5}-\frac{12\pi }{5} =0 }} {/eq}.


(d) Finding parametric equations for the tangent line to the curve {eq}C{/eq} at the point {eq}P{/eq}:

{eq}\begin{align*} \displaystyle \vec{r}'(\pi) &=\langle -3, 4, 0 \rangle \\ \displaystyle x &=x_{0}+at, \ y=y_{0}+bt, \ z=z_{0}+ct \\ \displaystyle x &=0-3t, \ y=4\pi+4t, \ z=-3+(0)t \\ \displaystyle x &=-3t, \ y=4\pi+4t, \ z=-3 \end{align*} {/eq}

Parametric equations for the tangent line to the curve is {eq}\ \displaystyle \mathbf{\color{blue}{ x=-3t, \ y=4\pi+4t, \ z=-3 }} {/eq}.


Learn more about this topic:

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Vector Components: The Magnitude of a Vector

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