Plz help with parametric equations
The position of a particle is given by the parametric equations
x = 1 + sin pi*t - cos pi*t
y = 3*sin pi*t + 2*cos pi*t
(a) Describe the graph of these parametric equations as t ranges over all real numbers. (In other words, find all possible positions of the particle.)
(b) Describe the motion of the particle as t ranges from 0 to 2.
(c) Find a parametrization such that the overall graph of this parametrization from t = 0 to t = 2 matches the graph of part (b), but the motion of the particle is different.
(a) To describe the graph of the parametric equations, we can eliminate the parameter t to obtain an equation in terms of x and y. From the given equations, we have:
sin pit = x - 1 + cos pit
cos pit = (y - 3sin pi*t)/2
Substituting the first equation into the second equation, we get:
cos pit = (y - 3(x - 1 + cos pit))/2
2cos pit = y - 3x + 3
cos pit = (y - 3x + 3)/2
Substituting this into the first equation, we get:
sin pit = x - 1 + (y - 3x + 3)/2
2sin pit = 2x - 2 + y - 3x + 3
sin pit = x + y - 1
Therefore, the graph of the parametric equations is the set of all points (x, y) that satisfy the equation x + y - 1 = sin pi*t, where t ranges over all real numbers.
(b) To describe the motion of the particle as t ranges from 0 to 2, we can evaluate x and y at t = 0 and t = 2, and plot the resulting points.
At t = 0, we have:
x = 1 + sin 0 - cos 0 = 1
y = 3sin 0 + 2cos 0 = 2
At t = 2, we have:
x = 1 + sin pi2 - cos pi2 = 1
y = 3sin pi2 + 2cos pi2 = -1
Therefore, the particle moves horizontally along the line y = 2, then returns to its starting point by moving horizontally along the line y = -1.
(c) To find a parametrization that matches the graph of part (b) but with a different motion, we can use a different function for x or y. For example, we can use:
x = 1 + sin pit
y = 2cos pi*t
This parametrization produces the same graph as the original parametrization for t ranging from 0 to 2, but the motion of the particle is different. In this case, the particle moves along an ellipse centered at (1, 0), starting from the point (2, 0) and ending at the point (0, 0). HEB Partner