# Trajectory of an arrow

• Jun 15th 2013, 02:21 AM
linalg123
Trajectory of an arrow
An arrow shot from a bow travels along a curved trajectory given by the set of points satisfying

$P(x,y) = (0,22) + r(t)$

where $r(t) = <36\sqrt{2}t , (36t - 6t^2)>$

a. find the unit tangent vector T to this curve as a function of t
b. find unit nomrla vector N to this curve as function of t
c. taking {T,N} as an orthonormal basis defininf the orientation of the arrow, find the angle through which the arrow rotates from t=3 to t=4
d. using this angle, or otherwise, find a transformation matrix that maps the basis at time t=3 to the basis at time t=4

k so $r'(t) = <36\sqrt{2} , (36 - 12t)$

so a. $T = \frac{<36\sqrt{2} , (36 - 12t)>}{\sqrt{3888-864t+144t^2}}$

b. $N=\frac{<12t-36 , 36\sqrt{2}>}{\sqrt{3888-864t+144t^2}}$

c. i tried evaluating the values
$T(3) = \frac{<36\sqrt{2},0>}{\sqrt{2592}}$
$N(3)= \frac{<0,36\sqrt{2}>}{\sqrt{2592}}$

$T(4) = \frac{<36\sqrt{2},-12>}{\sqrt{2736}}$
$N(4) = \frac{<-12,36\sqrt{2}>}{\sqrt{2736}}$

but i don't know if this is usefull or where to go from there.

d. No idea.
• Jun 15th 2013, 07:45 AM
HallsofIvy
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 $\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)$. Since these are unit tangent vectors, that is just $cos(\theta)= \vec{u}\cdot\vec{v}$.

A linear transformation in two dimensions can be represented as a two by two matrix. The linear transformation that maps $\begin{bmatrix}x_1 & y_1\end{bmatrix}$ to $\begin{bmatrix}x_2 & y_2\end{bmatrix}$ is the matrix, $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ satisfying $\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}x_1 \\ y_1\end{bmatrix}= \begin{bmatrix}x_2 \\ y_2\end{bmatrix}$. That reduces to the two equatons $a x_1+ by_1= x_2$ and $cx_1+ dy_1= y_2$. Solve those equations for a, b, c, d (using your values for $x_1, y_1, x_2, y_2$, of course).
• Jun 15th 2013, 10:37 AM
linalg123
Re: Trajectory of an arrow
so for the angle i did

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

and theta = inverse of that but it said that was wrong?
• Jun 22nd 2013, 11:45 AM
HallsofIvy
Re: Trajectory of an arrow
Please, do not use 'unattached' pronouns! What does "it" refer to?