Unit Tangent Vector Example 1

Find the unit tangent vector given the vector equation \(r(t) = (1 + {t^3})i + t{e^{ - t}}j + \sin (2t)k\) at t=0. The components of the vector is given by i, j, and k, i is NOT the imaginary number here.

First find the derivative, which is just the derivative of each component separately.

\[r'(t) = 3{t^2}i + (1 - t){e^{ - t}}j + 2\cos (2t)k\]

Now substitute t=0 into the original and the derivative.

\[r(0) = 1i + 0j + 0k\]

\[r(0) = i\]

\[r'(0) = 0i + 1j + 2k\]

\[r'(0) = j + 2k\]

And the unit tangent vector will be given by:

\[T(0) = \frac{{r'(0)}}{{\left| {r'(0)} \right|}}\]

Note that for the denominator above, you can remove the components and sum them up because the denominator is just the magnitude or norm. Keep the components in the numerator though!

\[T(0) = \frac{{j + 2k}}{{\sqrt {1 + 4} }}\]

\[T(0) = \frac{1}{{\sqrt 5 }}j + \frac{2}{{\sqrt 5 }}k\]

Or you can write it in the other vector notation:

\[T(0) = \left\langle {\frac{1}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }},0} \right\rangle \]

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