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 8 months ago
can anyone explain problem $1I3 from problem set 3? That is: "Describe the motions given by each of the following position vector functions, as t
goes from −∞ to ∞. In each case, give the xyequation of the curve along which P travels,
and tell what part of the curve is actually traced out by P ." The first curve given is r(t) = 2*cos^2 t i + sin^2 t j. I read the solution buit don't quite get the last step
 8 months ago
can anyone explain problem $1I3 from problem set 3? That is: "Describe the motions given by each of the following position vector functions, as t goes from −∞ to ∞. In each case, give the xyequation of the curve along which P travels, and tell what part of the curve is actually traced out by P ." The first curve given is r(t) = 2*cos^2 t i + sin^2 t j. I read the solution buit don't quite get the last step

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phi
 8 months ago
Best ResponseYou've already chosen the best response.1Problem 1I3 (a) says \[ \vec{r} = < 2\cos^2 t , \ \sin^2 t> \] in other words, r is a position vector with x component = 2 cos^2 t and a y component = sin^2 t The attached graph show the locus of points we create as we increment t notice if we start with \[ x= 2 \cos^2 t \\ y= \sin^2 t \] and (because we want to use sin^2( t) + cos^2(t) = 1 to simplify the equation) we multiply the 2nd equation by 2 and add the two equations: \[ x + 2y = 2\cos^2 t + 2 \sin^2 t \\ x+2y= 2(\cos^2 t + \sin^2 t) \\ x+2y= 2 \] that is the equation of a line. In slopeintercept form \[ y =  \frac{1}{2} x +1 \] so we should expect the locus of points to lie on this line (and it does) because sin and cos are cyclical, we will get cyclical behavior... the point moves back and forth between (0,1) and (2,0)
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