For the conic helix of Problem 11.95, determine the angle that the osculating plane forms with the y axis.

Question

Accepted Answer

First note that the vectors v and a lie in the osculating plane.

Now

[latex]\mathbf{r}= \left(R t \cos \omega_{n} t ight) \mathbf{i}+c t \mathbf{j}+\left(R t \sin \omega_{n} t ight) \mathbf{k}[/latex]

Then

[latex]\mathbf{v}= \frac{d r}{d t}=R\left(\cos \omega_{n} t-\omega_{n} t \sin \omega_{n} t ight) \mathbf{i}+c \mathbf{j}+R\left(\sin \omega_{n} t+\omega_{n} t \cos \omega_{n} t ight) \mathbf{k}[/latex]

and

[latex]\begin{aligned}\mathbf{a}= & \frac{d u}{d t} \= & R\left(-\omega_{n} \sin \omega_{n} t-\omega_{n} \sin \omega_{n} t-\omega_{n}^{2} t \cos \omega_{n} t ight) \mathbf{i} \& +R\left(\omega_{n} \cos \omega_{n} t+\omega_{n} \cos \omega_{n} t-\omega_{n}^{2} t \sin \omega_{n} t ight) \mathbf{k} \= & \omega_{n} R\left[-\left(2 \sin \omega_{n} t+\omega_{n} t \cos \omega_{n} t ight) \mathbf{i}+\left(2 \cos \omega_{n} t-\omega_{n} t \sin \omega_{n} t ight) \mathbf{k} ight]\end{aligned}[/latex]

It then follows that the vector [latex](\mathbf{v} imes \mathbf{a})[/latex] is perpendicular to the osculating plane.

[latex](\mathbf{v} imes\mathbf{a})=\omega_{n}R{\left|\begin{array}{l l l}{\quad\quad\quad\quad\quad\mathbf{i}}&{\mathbf{j}}&{\quad\quad\quad\quad\quad\mathbf{k}}\ {R(\cos\omega_{n}t-\omega_{n}t\sin\omega_{n}t)}&{c}&{R(\sin\omega_{n}t+\omega_{n}t\cos\omega_{n}t)}\ {-(2\sin\omega_{n}t+\omega_{n}t\cos\omega_{n}t)}&{0}&{(2\cos\omega_{n}t-\omega_{n}t\sin\omega_{n}t)}\end{array} ight|}[/latex]

[latex]=\omega_{n}R\{c(2\cos\omega_{n}t-\omega_{n}t\sin\omega_{n}t){\bf i}+R[-(\sin\omega_{n}t+\omega_{n}t\cos\omega_{n}t)(2\sin\omega_{n}t+\omega_{n}t\cos\omega_{n}t)[/latex]

[latex]-(\cos\omega_{n}t-\omega_{n}t\sin\omega_{n}t)(2\cos\omega_{n}t-\omega_{n}t\sin\omega_{n}t)]\mathbf{j}+c(2\sin\omega_{n}t+\omega_{n}t\cos\omega_{n}t)\mathbf{k}[/latex]

[latex]=\omega_{n}R\biggl[c(2\cos{\omega_{n}t}-\omega_{n}t\sin{\omega_{n}t}){\mathbf{i}-{R}}\biggr(2+\omega_{n}^{2}t^{2}\biggr)\mathbf{j}+c(2\sin{\omega_{n}t}+\omega_{n}t\cos{\omega_{n}t}){\mathbf{k}}\biggr][/latex]

The angle [latex]\alpha[/latex] formed by the vector [latex](\mathbf{v} imes \mathbf{a})[/latex] and the y axis is found from

[latex]\cos \alpha=\frac{(\mathbf{v} imes \mathbf{a}) \cdot \mathbf{j}}{|(\mathbf{v} imes \mathbf{a}) \| \mathbf{j}|}[/latex]

Where

[latex]\begin{aligned}&\quad\quad\quad\quad\quad\quad |\mathbf{j}|=1\& (\mathbf{v} imes \mathbf{a}) \cdot \mathbf{j}=-\omega_{n} R^{2}\left(2+\omega_{n}^{2} t^{2} ight)\& \quad\quad |(\mathbf{v} imes \mathbf{a})|=\omega_{n} R\left[c^{2}\left(2 \cos \omega_{n} t-\omega_{n} t \sin \omega_{n} t ight)^{2}+R^{2}\left(2+\omega_{n}^{2} t^{2} ight)^{2} ight.\& \quad\quad\quad\quad\left.+c^{2}\left(2 \sin \omega_{n} t+\omega_{n} t \cos \omega_{n} t ight)^{2} ight]^{1 / 2}\& \quad\quad =\omega_{n} R\left[c^{2}\left(4+\omega_{n}^{2} t^{2} ight)+R^{2}\left(2+\omega_{n}^{2} t^{2} ight)^{2} ight]^{1 / 2}\end{aligned}[/latex]

Then

[latex]\begin{aligned}\cos \alpha & =\frac{-\omega_{n} R^{2}\left(2+\omega_{n}^{2} t^{2} ight)}{\omega_{n} R\left[c^{2}\left(4+\omega_{n}^{2} t^{2} ight)+R^{2}\left(2+\omega_{n}^{2} t^{2} ight)^{2} ight]^{1 / 2}} \& =\frac{-R\left(2+\omega_{n}^{2} t^{2} ight)}{\left[c^{2}\left(4+\omega_{n}^{2} t^{2} ight)+R^{2}\left(2+\omega_{n}^{2} t^{2} ight)^{2} ight]^{1 / 2}}\end{aligned}[/latex]

The angle [latex]\beta[/latex] that the osculating plane forms with y axis (see the above diagram) is equal to

[latex]\beta=\alpha-90^{\circ}[/latex]

Then [latex]\quad\cos \alpha=\cos \left(\beta+90^{\circ} ight)=-\sin \beta[/latex]

[latex]-\sin \beta=\frac{-R\left(2+\omega_{n}^{2} t^{2} ight)}{\left[c^{2}\left(4+\omega_{n}^{2} t^{2} ight)+R^{2}\left(2+\omega_{n}^{2} t^{2} ight)^{2} ight]^{1 / 2}}[/latex]

Then [latex]\quad an \beta=\frac{R\left(2+\omega_{n}^{2} t^{2} ight)}{c \sqrt{4+\omega_{n}^{2} t^{2}}}[/latex]

or [latex]\quad\quad\quad\quad\quad\quad\beta= an ^{-1}\left[\frac{R\left(2+\omega_{n}^{2} t^{2} ight)}{c \sqrt{4+\omega_{n}^{2} t^{2}}} ight]\blacktriangleleft[/latex]

Question 11.180: For the conic helix of Problem 11.95, determine the angle th...

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