This graph is easily drawn with the PlotPositionVector command in the Student VectorCalculus package.
Since r(t) is a planar curve, the graph would also be planar, not 3D.
There is no command that returns either the tangential or normal components of the acceleration vector. However, these are easily obtained since, if t is time, then the second derivative of r(t) is a linear combination of the tangent and principal normal vectors. The coefficient of the unit tangent vector is the derivative of the speed; the coefficient of the (unit) principal normal is kappa*speed^2, where kappa is the curvature of r(t) and the speed is the length of dr/dt.
Since r(t) defines a circle, the curvature is constant. So is the speed constant. So, the acceleration vector lies along the principal normal. There is no tangential component of the acceleration in this case.
The speed is 72, kappa=1/36, so the length of the acceleration vector is 144. The radius of the circle is 36, so each acceleration vector points inward, passes through the center of the circle, and goes beyond the circumference by a considerable amount. The unit tangent vector is so small in comparison, that it is nearly invisible on a graph of the circle.
Some command that might be helpful:
PlotPositionVector(R,t=0..Pi,pvdiff=[t$2]) (Draws circle with the acceleration vector that necessarily lies along the principal normal)
PlotPositionVector(R,t=0..Pi,normal=true) (Draws circle with principal normal vectors. They are more visible than tangent vectors, but barely so.)
Change "normal" to "tangent" and you get a graph of the circle with unit tangent vectors, but they are so small as to be just about invisible. The visualization problem stems from taking a circle with such a large radius. Change that 36 to something smaller like 5 and both the tangent and normal vectors along the circle will assume more reasonable dimensions.