Question

For the following parameterized​ curve, find the unit tangent vector T​(t) at the given value of t. r​(t) = < 8 t,10,3 sine 2 t >​, for 0

Answer 1

Answer:

The tangent vector for $t = 0$ is:

$\vec T (t) = \left \langle \frac{8}{10}, 0, \frac{6}{10} \right\rangle$

Step-by-step explanation:

The function to be used is $\vec r(t) = \langle 8\cdot t, 10, 3\cdot \sin (2\cdot t)\rangle$

The unit tangent vector is the gradient of $\vec r (t)$ divided by its norm, that is:

$\vec T (t) = \frac{\vec \nabla r (t)}{\|\vec \nabla r (t)\|}$

Where $\vec \nabla$ is the gradient operator, whose definition is:

$\vec \nabla f (x_{1}, x_{2},...,x_{n}) = \left\langle \frac{\partial f}{\partial x_{1}}, \frac{\partial f}{\partial x_{2}},...,\frac{\partial f}{\partial x_{n}} \right\rangle$

The components of the gradient function of $\vec r(t)$ are, respectively:

$\frac{\partial r}{\partial x_{1}} = 8$, $\frac{\partial r}{\partial x_{2}} = 0$ and $\frac{\partial r}{\partial x_{3}} = 6 \cdot \cos (2\cdot t)$

For $t = 0$:

$\frac{\partial r}{\partial x_{1}} = 8$, $\frac{\partial r}{\partial x_{2}} = 0$ and $\frac{\partial r}{\partial x_{3}} = 6$

The norm of the gradient function of $\vec r (t)$ is:

$\| \vec \nabla r(t) \| = \sqrt{8^{2}+0^{2}+ [6\cdot \cos (2\cdot t)]^{2}}$

$\| \vec \nabla r(t) \| = \sqrt{64 + 36\cdot \cos^{2} (2\cdot t)}$

For $t = 0$:

$\| \vec r(t) \| = 10$

The tangent vector for $t = 0$ is:

$\vec T (t) = \left \langle \frac{8}{10}, 0, \frac{6}{10} \right\rangle$