Parametric equations for trajectory

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SUMMARY

The discussion focuses on deriving the parametric equations for the trajectory of a jet taking off from the point (1, 1, 0) with a constant speed vector of v = (−5, 0, 1). The correct position of the jet in three dimensions is given by the equation (1 - 5t, 1, t). The trajectory as observed from the point (1, 0, 0) is determined by the line equation (1 - 5st, 1 + s, ts), which intersects the yz-plane at x = 0, leading to the parametric equations y = 1 + (1/5)t and z = (1/5)t.

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  • Understanding of parametric equations in three-dimensional space
  • Familiarity with vector representation of motion
  • Knowledge of projection techniques in geometry
  • Basic calculus for parameterization
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  • Learn about vector calculus and its applications in physics
  • Explore projection methods in geometry, particularly parallel projections
  • Investigate the concept of trajectory analysis in flight simulations
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Homework Statement


A jet takes off from (1, 1, 0) at time t = 0 and moves with constant speed v = (−5, 0, 1).
In a flight simulator, the trajectory of the jet is displayed in the yz-plane as it would appear to an observer at the point (1, 0, 0). Find the formula (in the form y = y(t), z = z(t)) for the trajectory on the screen.

Homework Equations



The Attempt at a Solution


y(t) = 1
z(t) = t
Is that right?
 
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That would be the "parallel projection" onto the yz-plane, not "it would appear to an observer at the point (1, 0, 0)". The position of the jet, in 3 dimensions, is (1- 5t, 1, t). A straight line from (1, 0, 0) through that, in parameter s, would be (1- 5st, 1+ s, ts). That passes through the "yz-plane" (x= 0) when 1- 5st= 0 or s= 1/5t: (0, 1+1/5t, 1/5).
 

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