Question

For constants $\alpha ,\beta$, and $\gamma$, the graph of a vector-valued function of the form

$\mathbf{r}\left(t\right)=\alpha e^{\gamma t}\left(\cos t\right)\mathbf{i}+\alpha e^{\gamma t}\left(\sin t\right)\mathbf{j}+\beta e^{\gamma t}\mathbf{k},t\ge 0$

Short answer

Ans:

Step-by-step solution

Step 1. GIven information: 

Consider

$r\left(t\right)=\alpha e^{\gamma t}\left(\cos t\right)i+\alpha e^{\gamma t}\left(\sin t\right)j+\beta e^{\gamma t}k,t\ge 0$, known as a concho-spiral.

Step 2. Finding the value of k:

The tangent vector to $r\left(t\right)$is $r^{\text{'}}\left(t\right)$.

Vector $k=⟨0,0,1⟩$

Step 3. Proving:

Let the angle between the vectors $r^{\text{'}}\left(t\right)$and vector $k$be $\theta$. Then

$\cos \theta =\frac{r^{\text{'}}\left(t\right)·k}{||r^{\text{'}}\left(t\right)||\Vert k\Vert }$

$\cos \theta =\frac{\beta \gamma e^{\gamma t}}{\sqrt{{\alpha }^2{e}^{2\gamma t}(-\sin t+\gamma \cos t{)}^2+{\alpha }^2{e}^{2\gamma t}(\cos t+\gamma \sin t{)}^2+{\beta }^2{\gamma }^2{e}^{\gamma t}}}$

$\cos \theta =\frac{\beta \gamma e^{yt}}{\sqrt{{\alpha }^2{e}^{2\gamma t}\left({\sin }^{2}t+{\gamma }^2{\cos }^{2}t-2\gamma \sin t\cos t+{\cos }^{2}t+{\gamma }^2{\sin }^{2}t+2\gamma \sin t\cos t\right)+{\beta }^2{\gamma }^2{e}^{2rt}}}$

$\cos \theta =\frac{\beta \gamma e^{\gamma t}}{\sqrt{{\alpha }^2{e}^{2\gamma t}\left(1+{\gamma }^2\right)+{\beta }^2{\gamma }^2{e}^{2\gamma t}}}$

$\cos \theta =\frac{\beta \gamma e^{\gamma t}}{\sqrt{e^{2\gamma t}\left({\alpha }^2+{\alpha }^2{\gamma }^2+{\beta }^2{\gamma }^2\right)}}$

$=\frac{\beta \gamma }{\sqrt{{\alpha }^2+{\alpha }^2{\gamma }^2+{\beta }^2{\gamma }^2}}$

$\theta ={\cos }^{-1}\left(\frac{\beta \gamma }{\sqrt{{\alpha }^2+{\alpha }^2{\gamma }^2+{\beta }^2{\gamma }^2}}\right)$which is a constant.

Thus the angle between the tangent vector to a concho-spiral and the vector $\mathrm{k}$is a constant.