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Solution:

Given:

r(t) = (1 + t^{2})i – te^{-t}j + sin 2tk

To find the derivative, we will differentiate with respect to t. So derivative is

r^{‘}(t) = 2ti – [e^{-t} + te^{-t}]j + 2cos 2tk

= 2ti – e^{-t}(1 + t) + 2cos 2tk

The unit tangent vector is obtained using the formula r'(0)/|r(0)|. r(0) is the value of r'(t) at t = 0, and |r(0)| is the modulus of r(0).

now,

r'(0) = 2 . 0 . i – e^{0} (1 + 0)j + 2 cos 2.0.t

= -j + 2k

r(0) = (1 + 0)i – 0 . e^{0} . j + sin0 k

= i

|r(0)| = √(1)^{2} = 1

So, Unit vector is

\(\frac{r'(0)}{|r(0)|} = \frac{-j + 2k}{1}\)

= -i + 2k

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