# Given a control polygon, how do you find a tangent vector at a given point (CAGD, Bezier Curve)?

I am having a hard time with my Computer Aided Graphic Design class. I am presented with the bezier control polygon (in a Mathematica statement):

P = Table[{x^2, x^3, 0}, {x, 4,7}]

And the questions:

Given the curve P, display the point on the curve and the tangent vector there for t=[0,1]. Use the manipulate function.

My question isn't how to do this in Mathematica (yet). I know I need to come up with a parametric equation, but I am not sure how to go from the table to a parametric equation. Then I need to calculate the tangent, which should just be the derivative...?

I am desperately confused on what this question is asking. Hopefully someone can kind of explain the steps needed to display a tangent vector at a given point on a Bezier curve.

The given table expands to $$(4^2,4^3,0), \; (5^2, 5^3,0), \; (6^2,6^3,0), \; (7^2,7^3,0)$$ which is $$(16,64,0), \; (25,125,0), \; (36, 216, 0), \; (49,343,0)$$ You said this table of four points is supposed to be used as the control polygon of a Bézier curve. So, the parametric equation of this Bézier curve is $$\mathbf{P}(t) = (1-t)^3(16,64,0) +3t(1-t)^2(25,125,0) +3t^2(1-t)(36,216,0) + t^3(49,343,0)$$ The tangent direction at the parameter value $t$ is in the direction of the first derivative vector, which you can get by differentiating the equation above. To draw a tangent, just draw a line from $\mathbf{P}(t)$ to $\mathbf{P}(t)+k\mathbf{P}'(t)$. Choose the value of $k$ to give the tangent lines the right length for a good-looking display. Using $k=1/3$ would probably work ok.
Code all this in Mathematica, and use the Manipulate function to animate the tangent line as $t$ varies from $0$ to $1$.