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

The tangent vector for
`t = 0` is:

`\vec T (t) = \left \langle (8)/(10), 0, (6)/(10) \right\rangle`

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) = (\vec \\abla r (t))/(\|\vec \\abla r (t)\|)`

Where
`\vec \\abla` is the gradient operator, whose definition is:

`\vec \\abla f (x_(1), x_(2),...,x_(n)) = \left\langle (\partial f)/(\partial x_(1)), (\partial f)/(\partial x_(2)),...,(\partial f)/(\partial x_(n)) \right\rangle`

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

`(\partial r)/(\partial x_(1)) = 8`,
`(\partial r)/(\partial x_(2)) = 0` and
`(\partial r)/(\partial x_(3)) = 6 \cdot \cos (2\cdot t)`

For
`t = 0`:

`(\partial r)/(\partial x_(1)) = 8`,
`(\partial r)/(\partial x_(2)) = 0` and
`(\partial r)/(\partial x_(3)) = 6`

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

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

`\| \vec \\abla 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 (8)/(10), 0, (6)/(10) \right\rangle`