# How can I generate tangent vectors to any non parametric curve in in the XY plane at any arbitrary point for which the derivative exists?

It doesnt seem enough to write a vector function w respect to some time variable. If I use a linear variable t (which is the easiest choice), then I will get a linear combo of t and f(t). But when we take the derivative w respect to time, we get the tangent vectors (1, f'(t)) where the vector is always an x... show more It doesnt seem enough to write a vector function w respect to some time variable. If I use a linear variable t (which is the easiest choice), then I will get a linear combo of t and f(t). But when we take the derivative w respect to time, we get the tangent vectors (1, f'(t)) where the vector is always an x direction of 1? I realize depending where we are in time, we could simply translate/rotate the tangent vector to its corresponding location since it always begins with 1. But there has to be a better way. But then I thought if I wanted my vector valued derivative to begin w t, I could use .5t^2 but this isnt enough? Then I remembered the gradient is always normal to a level curve. So I thought we could perform rotations such that the normal becomes tangent. Could I do this? For example T(x, y)= (-y, x) transforms the corresponding vector under a 90° counterclockwise rotation. Assuming the gradient does something similar, then I thought we could undo the normal vector relationship to the curve nto a tangent one using the proper rotation transformations. But then I thought the gradient wouldn't be enough in every case, even if it was possible to undo the normal relationship..since they are already normal at a point to the curve. But then wouldn't this make normal tangents but not tangents not necessarily normal? Wow thanks for reading this

Im just a community college student w zero training. one day I'll know how to be good at problem solving. Thank you for mercy and time