Question

Find parametric equations for the tangent line at t=2 for the motion of a particle given by x(t)=8t^2+6, y(t)=8t^3. For the line, x(t)= y(t)=

Answer 1

You'll need the slope of the tangent line first.
\[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\cdots\]
Find the slope at \(t=2\).

Answer 2

Please, you have to calculate this: (I call T the tangent vector):
\[T=(dx/ds,dy/ds)\]
where:
\[(ds)^{2}=(dx)^{2}+(dy)^{2}\]