anonymous
  • anonymous
can anyone explain problem $1I-3 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 xy-equation 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
OCW Scholar - Multivariable Calculus
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
phi
  • phi
Problem 1I-3 (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 slope-intercept 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)
1 Attachment

Looking for something else?

Not the answer you are looking for? Search for more explanations.