Re: Trajectory of an arrow

The angle through which the arrow rotates is the angle between the tangent vector evaluated at t=3 and t= 4. And you can find that from $\displaystyle \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)$. Since these are **unit** tangent vectors, that is just $\displaystyle cos(\theta)= \vec{u}\cdot\vec{v}$.

A linear transformation in two dimensions can be represented as a two by two matrix. The linear transformation that maps $\displaystyle \begin{bmatrix}x_1 & y_1\end{bmatrix}$ to $\displaystyle \begin{bmatrix}x_2 & y_2\end{bmatrix}$ is the matrix, $\displaystyle \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ satisfying $\displaystyle \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}x_1 \\ y_1\end{bmatrix}= \begin{bmatrix}x_2 \\ y_2\end{bmatrix}$. That reduces to the two equatons $\displaystyle a x_1+ by_1= x_2$ and $\displaystyle cx_1+ dy_1= y_2$. Solve those equations for a, b, c, d (using **your** values for $\displaystyle x_1, y_1, x_2, y_2$, of course).

Re: Trajectory of an arrow

so for the angle i did

$\displaystyle cos(\theta) = \frac{36\sqrt{2}*36\sqrt{2}}{\sqrt{2592}\sqrt{2736 }}$

and theta = inverse of that but it said that was wrong?

Re: Trajectory of an arrow

**Please**, do not use 'unattached' pronouns! What does "it" refer to?