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Particles $P_1$ and $P_2$ have position vectors $r_1(t)=t\vec i+\frac{t^2}{2}\vec j +\vec t$ and $r_2(t)=\frac{t^2}{2}\vec i +t\vec j+t\vec k$ respectively.

Question asks us to:

Find the angle $θ$ between the paths of these two particles at times $t = 0$ and $t → \infty$.

So the derivative of the particle's path is the velocity of the particle so to find that:

\begin{align*} v_1={}& 1\vec i + t\vec j + 1\vec k \\ v_2 ={}& t\vec i + 1\vec j + 1\vec k \end{align*}

Then to find the angle I am assuming we compute the dot product:

$$v_1\cdot v_2= |v1||v2|cos θ,$$

which I then got to be:

$$\cos\theta = \dfrac{(2t+1)}{(2+t^2)},$$

and I'm not really sure where to go from here?

David K
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john
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    Put in the values of $t$ and work out the angles... – David Quinn Apr 30 '16 at 14:30
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1 Answers1

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You have $\cos \theta = \frac {2t+1}{2+t^2}$. When you take the $\arccos$ of both sides, you need to pay attention to the quadrant. At $t=0$ this gives $\cos \theta=\frac 12$. Look at the velocities to see whether $\theta=\frac \pi 3$ or $\frac {-\pi}3$. As $t \to \infty$, you need to take the limit to find $\cos \theta$

Ross Millikan
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