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Find out the general formula for the tangent vector and unit tangent vector to the curve specified by
r→ (t) = t2 i→ + 2 sin t j→ + 2 cos t k→.
Solution
First, by common formula we mean that we won't be plugging in a particular t and thus we will be finding out a formula that we can utilize at a later date if we would like to find the tangent at any point on the curve. Along with that said there really isn't all that much to do at this point other than to do the work.
Here below is the tangent vector to the curve.
r→′ (t) = 2t i→ + 2 cos t j→ - 2 sin t k→
To obtain the unit tangent vector we require the length of the tangent vector.
|| →r′ (t)|| = √ (4t2 + 4cos2 t + 4 sin2 t)
= √ (4t2 + 4)
After that the unit tangent vector is,
25/5(2+3)
Thus, just why do we care regarding direction fields? Two nice pieces of information are there which can be readily determined from the direction field for a differential equation.
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