Compute the velocity vector.

Question

Can you solve explicitly? please. I don't know how to solve. Thank you for help.

Answer 1

$$\begin {bmatrix} -2x \sin 2 t - 2y \cos 2t \\2x \cos 2t -2y \sin 2t \end {bmatrix}$$ is the derivative @B11. You have to treat x and y as constants, and derivate with respect to t. Now you can fill in t = 0 to obtain:$$\begin {bmatrix} -2y \\2x \end {bmatrix}$$

Answer 2

Hint: If you multiply the matrices, you get $c_p(t)=\begin {bmatrix} x \cos 2 t - y \sin 2t \\x \sin 2t +y \cos 2t \end {bmatrix}\begin {bmatrix} x \\ y \end {bmatrix}$. Now take the derivative of each term and evaluate it at $t=0$

Added: you have copied the matrix $c_p$, not taken the derivative. You can take the derivative first if you want, but would get $c'_{p}(t)=\begin{bmatrix}-2\sin2t & -2\cos2t \ \\2\cos2t &\ -2\sin2t \end{bmatrix}\begin {bmatrix} x \\ y \end {bmatrix}$, evaluate this at $t=0$, to get $c_p(0)=\begin {bmatrix} 0 & -2 \\ 2 & 0 \end {bmatrix}\begin {bmatrix} x \\ y \end {bmatrix}$then carry out the multiplication to get $c_p(0)=\begin {bmatrix} -2y \\ 2x \end {bmatrix}$ (The last two steps commute, but this order is easier).