Here is your first taste of using all the techniques taught thus far to handle a range of equations on a given position vector. I suggest you get your analytical mind on because this needs something thinking.
Just some advice. In vectors, there is a range of paths to get to the intended result. Some are easier, some are harder, which is why its best that you choose the easier one.
VECTOR DIFFERENTIAL CALCULUS >
MULTIPLE APPROACHES TO A PROBLEM.1
May want to check out:
MULTIPLE APPROACHES TO A PROBLEM.2
ACCELERATION COMPONENTS
FINDING UNIT TANGENT AND UNIT NORMAL
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VIDEO LECTURE part 1
MAIN CONCEPTS
There are 2 ways to get to  , one by finding  recommended if curve expressed in arc length. Another by Pythagorean's theorem recommended if curve expressed in t. Get the Printer Friendly Version COMMENTS
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CONCEPT
Using all that we know, we will now analyze a position vector of a curve and see the different approaches to get the vectors and quantities we want. Suppose the position vector of a particle at time t > 0 is
Then
We find the acceleration by
Using our previous definitions, the tangential component of accleration is
This is where we have multiple options to proceed with the problem. We are now concern with finding the normal component which is finding We first notice that the vectors T and N are perpendicular and so we apply the Pythagorean’s theorem to the parallelogram sum
Using the lengths
This allows us to express the acceleration vector as
From this value of
We learn from this question is that it pays to recognize whether it is easier to find the curvature All information presentated, less questions and exercises, is original content of Donny, with slight references to various books.
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. This may prove tedious because we need the curvature which in turn means we need to the unit tangent vector. The problem comes by expressing the positive vector in terms of the arc length. Try to fit in 
into the bunch of trigonometry functions and you’ll get a headache. Nonetheless, not all is lost. We are fortunate to have another approach to the problem.
and 
, we can find the magnitude, or the length of 
.
, we can easily obtain the curvature of the trajectory.
or from 
. In this case, we pick the former because making the substitution 
is tedious.
