jcook735
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1. A moving particle has position (x(t),y(t)) at any time t. The position of the particle at t =1 is (2,6), and the velocity vector at any time t > 0 is given by (1 - (1/(t^2)), 2 + (1/(t^2))).
The particle approaches a line as t -> infinity. Find the slope of the line. Show the work that leads to your conclusion.
2. I found that the position vector is <t + (1/t), 2t - (1/t) + 5>
3.
I decided that the best thing to do was take the limit of y(t) over x(t), and applied L'Hôpital's rule until the equation turned into something tangible. I ended up with the limit as t approaches infinity to be 2. Is that right? Is that the slope of the line? Did I not go far enough or did I do it completely wrong?
The particle approaches a line as t -> infinity. Find the slope of the line. Show the work that leads to your conclusion.
2. I found that the position vector is <t + (1/t), 2t - (1/t) + 5>
3.
I decided that the best thing to do was take the limit of y(t) over x(t), and applied L'Hôpital's rule until the equation turned into something tangible. I ended up with the limit as t approaches infinity to be 2. Is that right? Is that the slope of the line? Did I not go far enough or did I do it completely wrong?