Question

Please help me with the below question.

Answer 1

Given that

`\vec r(t) = \left\langle 4t e^(-t), 8 \arctan(t), 8 e^t \right\rangle`

we first differentiate with respect to t to get the tangent vector,
`\vec T(t)` :

`\vec T(t) = (d\vec r)/(dt) = \left\langle (4-4t) e^(-t), \frac8{1+t^2}, 8e^t \right\rangle`

At t = 0, the tangent vector is

`\vec T(0) = \left\langle 4, 8, 8 \right\rangle`

To get the unit tangent vector, multiply this by 1/(norm of tangent vector) :

`\|\vec T(0)\| = √(4^2 + 8^2 + 8^2) = √(144) = 12`

Then the unit tangent vector is

`(\vec T(0))/(\|\vec T(0)\|) = \left\langle \frac4{12}, \frac8{12}, \frac8{12}\right\rangle = \boxed{\left\langle\frac13, \frac23, \frac23\right\rangle}`