If you recall, when the curviliner motion of a particle is studied in an x, y and z rectangular coordinate system,
its position is represented by position vector r, and its instantaneous velocity, v
is dr/dt, and its instantaneous acceleration, a, is the second time derivative of position r, and in general
r, v and a are all three dimensional Cartesian vectors.
And don't forget this important conclusion that the velocity of the particle at any point is always tangent to the path.
Now let's look at this 3D curved path.
It can be divided into small segments of curves with equal length.
And when the segments get small enough each one of them approaches an arc, which is a segment of a circle.
And we know that a circle always falls in a 2D plane.
And for the next small segment of the path it can also be approximated by another arc 
that belongs to another circle the falls into a another 2D plane.
And then for another segment of the path, again it can be approximated by an arc the belongs to a circle the falls in yet a different 2D plane.
This plane is known as the osculating plane, which in this case refers to the plane that contains the small arced path. As you can see
this plane changes at different location and different time.
The reason to define it is because now the 3D motion can be considered as a sequence of 
2D motions that are limited within each plane.
And we know that a circle always has a center and a radius.
For the arc of the circle these are called the radius and center of curvature.
For the particle traveling at this location we can define a pair of axes from it. The first one is the t axis being tangent to the arc, 
and the other one is the n axis, pointing towards the center of curvature.
It is also normal to the arc.
And with the  definition of the t, tangent axis, and n, normal axis, we can represent the motion vectors using the tangential and normal 
components instead of the x, y and z rectangular components.
So for a particle, in a short moment dt, it travels along this curved path from location P to P prime.
The distance travelled is the length of the arc on this path, ds.
And at any given time we can always set up a pair of axes from the particle.
Again a t, tangent axis that is tangent to the path, and always points towards the direction of the motion.
Its unit vector is u_t.
And another, n, normal axis that is normal to the path, always points towards the center of curvature.
And for the normal axis its unit vector is u_n.
And if you recall the velocity vector is always tangent to the path, therefore the velocity vector 
written in the n-t components is simply v equals to v u_t,
with the scalar v being the speed, or the magnitude of the velocity.
And it equals to ds over dt.
And the unit vector u_t naturally indicates that the velocity is always in the direction that's tangent to the path.
Here I am skipping the mathematic derivation,
but we can derive that acceleration will always have two components, a_t, the tangent component, and a_n the normal component.
Therefore the acceleration is written in normal and tangential components as a equals to a_t u_t plus a_n u_n.
The tangential acceleration a_t simply equals to dv/dt,
and the normal acceleration, a_n, equals to v squared over rou, with rou being the radius of curvature at this location.
And since a_t and a_n are perpendicular to each other, we can easily tell from the Pythegorean theorem,
that the magnitude of acceleration is the square root of a_t squared plus a_n squared.
Here are some important notes when you use the normal and tangential components to describe curvilinear motion of a particle.
First, unlike the rectangular coordinate system which is generally fixed on earth, the n-t coordinate system is fixed on 
the particle instead. Therefore it moves with the particle and is different from time to time.
Second, one of the biggest advantage of using the n-t components is that the velocity always 
only has one component along the tangential direction.
Third, the two components of the acceleration have distinct meanings. 
a_t only describes the change in the magnitude of the velocity, or the change in speed,
and a_n only describes the change in the direction of the velocity.
And because the tangential acceleration fully describes the change in the magnitude of the velocity, therefore as an important conclusion,
along the tangential direction the three basic kinematic equations apply, just like in rectilinear motion that we learned before.
Therefore similarly, these three equations for constant acceleration apply too, but again only to tangential motion.
The normal acceleration, a_n, is also known as the centripetal acceleration since it always 
points towards the concave side of the path seeking the center of curvature.
And lastly, if the curved path can be modeled by a function, y equals to f(x), 
then the radius of curvature at any given point can be calculated by this formula. 
Let's look at this example. An object is traveling along this curved path and the equation for the path is given.
If at this point shown, at x equals to 16 m, that the speed of this object is 28.8 m/s and also its speed is increasing at 8 m per second squared,
We need to determine the direction of the velocity,
and also we need to determine its acceleration at this point, both magnitude and direction.
Since we know that at any given point the object will have a velocity that is tangent to its path ,we can set up 
the t axis which is tangent to the path pointing towards the direction of motion.
And n, the normal axis that is perpendicular to the path and points towards the center of curvature.
And the velocity simply points towards the same direction as the tangent axis and it has the magnitude of 28.8 m/s.
To determine the direction of the velocity is to determine the direction of the t axis,
which is characterized by this angle theta that it makes with the horizon.
And because for the path, its equation is known, y equals to ¼ times x to the 3/2 power, therefore its slope at any given time
is dy/dx which equals to 3/8 times x to the ½ power. At x equals to 16 meter we can evaluate the slope which equals to 1.5, 
and this equals to tangent theta, therefore from here we can
solve for theta to be arctangent 1.5,
which is 56.3°. And that's the direction for the velocity which is also the direction for the tangent axis.
The acceleration written in the normal-tangential components is a_t u_t plus a_n u_n
a_t is the magnitude of the tangential acceleration,
and it describes the change in the magnitude of the velocity or the change in speed, and in this case it's given to be 8 m per second squared.
And a_n is the normal acceleration and is evaluated by v squared over rou, rou 
being the radius of curvature at this point, and since the equation of the path is given,
therefore rou can be calculated using this formula.
So since y equals to again 1/4 times x to the 3/2 power, therefore dy/dx equals 
to 3/8 x to the ½ power, and the second derivative equals to 
3/16 x to the -1/2 power. And we can evaluate these two both at x equals to 16 m,
to be dy/dx equals 1.5, second derivative equals to 3/64,
substitute both into the equation for the radius of curvature, we can calculate at this point the radius of curvature is 125 m.
Therefore if we substitute in
the radius of curvature we can calculate the normal acceleration to be 6.64 meter per second squared, and these two are the two components 
for the acceleration vector. Therefore the magnitude of the acceleration is a_t squared plus a_n squared, square root, 
which is 10.4 m per second squared, and is represented here visually.
And the direction of the acceleration is characterized by this angle made with the horizon,
and it equals 56.3 degree which is the angle of the tangential axis, plus arctangent 6.64 over 8,
and totally that is 96.0°. And these two are the answers we're looking for, the magnitude and direction of acceleration.
