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Masteriza [31]
2 years ago
11

Please help me with the below question.

Mathematics
1 answer:
o-na [289]2 years ago
7 0

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) = \dfrac{d\vec r}{dt} = \left\langle (4-4t) e^{-t}, \dfrac8{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 <em>unit</em> tangent vector, multiply this by 1/(norm of tangent vector) :

\|\vec T(0)\| = \sqrt{4^2 + 8^2 + 8^2} = \sqrt{144} = 12

Then the unit tangent vector is

\dfrac{\vec T(0)}{\|\vec T(0)\|} = \left\langle \dfrac4{12}, \dfrac8{12}, \dfrac8{12}\right\rangle = \boxed{\left\langle\dfrac13, \dfrac23, \dfrac23\right\rangle}

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