The tangent vector to r(t) at any t in the domain is
[tex]\mathbf T(t)=\dfrac{\mathrm d\mathbf r(t)}{\mathrm dt}=2\cos t\,\mathbf i-7\sin t\,\mathbf j[/tex]
At t = π/6, the tanget vector is
[tex]\mathbf T\left(\dfrac\pi6\right)=\sqrt3\,\mathbf i-\dfrac72\,\mathbf j[/tex]
To get the unit tangent, normalize this vector by dividing it by its magnitude:
[tex]\left\|\mathbf T\left(\dfrac\pi6\right)\right\|=\sqrt{(\sqrt3)^2+\left(-\dfrac72\right)^2}=\dfrac{\sqrt{61}}2[/tex]
So the unit tangent at the given point is
[tex]\dfrac{\mathbf T\left(\frac\pi6\right)}{\left\|\mathbf T\left(\frac\pi6\right)\right\|}=2\sqrt{\dfrac3{61}}\,\mathbf i-\dfrac7{\sqrt{61}}\,\mathbf j[/tex]
Applying derivatives, the tangent vector of unit length at the point given is:
[tex]r_{u}{\prime}(\frac{\pi}{6}) = \frac{2\sqrt{3}}{\sqrt{61}}i - \frac{7}{\sqrt{61}}j[/tex]
The vector function is:
[tex]r(t) = 2\sin{(t)}i + 7\cos{(t)}j[/tex]
The tangent vector is it's derivative, which is given by:
[tex]r^{\prime}(t) = 2\cos{(t)}i - 7\sin{(t)}j[/tex]
At point [tex]t = \frac{\pi}{6}[/tex], we have that:
[tex]r^{\prime}(\frac{\pi}{6}) = 2\cos{(\frac{\pi}{6})}i - 7\sin{(\frac{\pi}{6})}j[/tex]
[tex]r^{\prime}(\frac{\pi}{6}) = \frac{2\sqrt{3}}{2}i - \frac{7}{2}[/tex]
[tex]r^{\prime}(\frac{\pi}{6}) = \sqrt{3}i - \frac{7}{2}[/tex]
The norm of the vector is:
[tex]|r^{\prime}(\frac{\pi}{6})| = \sqrt{\sqrt{3}^2 + (-\frac{7}{2})^2}[/tex]
[tex]|r^{\prime}(\frac{\pi}{6})| = \sqrt{\frac{61}{4}}[/tex]
[tex]|r^{\prime}(\frac{\pi}{6})| = \frac{\sqrt{61}}{2}[/tex]
The unit vector is given by each component divided by the norm, thus:
[tex]r_{u}{\prime}(\frac{\pi}{6}) = \frac{\sqrt{3}}{\frac{\sqrt{61}}{2}}i - \frac{7}{2\frac{\sqrt{61}}{2}}j[/tex]
[tex]r_{u}{\prime}(\frac{\pi}{6}) = \frac{2\sqrt{3}}{\sqrt{61}}i - \frac{7}{\sqrt{61}}j[/tex]
A similar problem is given at https://brainly.com/question/20733439
