Answer:
The tangent vector for [tex]t = 0[/tex] is:
[tex]\vec T (t) = \left \langle \frac{8}{10}, 0, \frac{6}{10} \right\rangle[/tex]
Step-by-step explanation:
The function to be used is [tex]\vec r(t) = \langle 8\cdot t, 10, 3\cdot \sin (2\cdot t)\rangle[/tex]
The unit tangent vector is the gradient of [tex]\vec r (t)[/tex] divided by its norm, that is:
[tex]\vec T (t) = \frac{\vec \nabla r (t)}{\|\vec \nabla r (t)\|}[/tex]
Where [tex]\vec \nabla[/tex] is the gradient operator, whose definition is:
[tex]\vec \nabla f (x_{1}, x_{2},...,x_{n}) = \left\langle \frac{\partial f}{\partial x_{1}}, \frac{\partial f}{\partial x_{2}},...,\frac{\partial f}{\partial x_{n}} \right\rangle[/tex]
The components of the gradient function of [tex]\vec r(t)[/tex] are, respectively:
[tex]\frac{\partial r}{\partial x_{1}} = 8[/tex], [tex]\frac{\partial r}{\partial x_{2}} = 0[/tex] and [tex]\frac{\partial r}{\partial x_{3}} = 6 \cdot \cos (2\cdot t)[/tex]
For [tex]t = 0[/tex]:
[tex]\frac{\partial r}{\partial x_{1}} = 8[/tex], [tex]\frac{\partial r}{\partial x_{2}} = 0[/tex] and [tex]\frac{\partial r}{\partial x_{3}} = 6[/tex]
The norm of the gradient function of [tex]\vec r (t)[/tex] is:
[tex]\| \vec \nabla r(t) \| = \sqrt{8^{2}+0^{2}+ [6\cdot \cos (2\cdot t)]^{2}}[/tex]
[tex]\| \vec \nabla r(t) \| = \sqrt{64 + 36\cdot \cos^{2} (2\cdot t)}[/tex]
For [tex]t = 0[/tex]:
[tex]\| \vec r(t) \| = 10[/tex]
The tangent vector for [tex]t = 0[/tex] is:
[tex]\vec T (t) = \left \langle \frac{8}{10}, 0, \frac{6}{10} \right\rangle[/tex]
