Describe the motion of a particle with position (x, y) as t varies in the given interval. (For each answer, enter an ordered pair of the form x, y.)
x = 2 + sin(t), y = 4 + 5 cos(t), 𝜋/2 ≤ t ≤ 2𝜋
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The motion of the particle takes place on an ellipse centered at
(x, y) =
As t goes from 𝜋/2 to 2𝜋, the particle starts at the point
(x, y) =
and moves clockwise three-fourths of the way around the ellipse to
(x, y) =
the center of
x = a sint
y = b cost
is at (0,0) so your ellipse has center (2,4)
for the other questions, just plug in the values for t. How hard is that?
extra credit: where are the foci?
omg i got it now i just had the worlds massive brainfart
The center of the ellipse is given by (x, y) = (2, 4).
When t = 𝜋/2, we have:
x = 2 + sin(𝜋/2) = 2 + 1 = 3
y = 4 + 5cos(𝜋/2) = 4
So, the particle starts at the point (x, y) = (3, 4).
When t = 2𝜋, we have:
x = 2 + sin(2𝜋) = 2
y = 4 + 5cos(2𝜋) = 4
So, the particle ends at the point (x, y) = (2, 4).
Therefore, the particle starts at (3, 4) and moves clockwise three-fourths of the way around the ellipse to (2, 4).
To describe the motion of the particle, we will substitute the given values of t into the equations x = 2 + sin(t) and y = 4 + 5cos(t) within the interval 𝜋/2 ≤ t ≤ 2𝜋.
Let's start by substituting 𝜋/2 for t:
x = 2 + sin(𝜋/2) = 2 + 1 = 3
y = 4 + 5cos(𝜋/2) = 4 + 0 = 4
So at t = 𝜋/2, the particle is at the point (x, y) = (3, 4).
Next, let's substitute 𝜋 for t:
x = 2 + sin(𝜋) = 2 + 0 = 2
y = 4 + 5cos(𝜋) = 4 - 5 = -1
So at t = 𝜋, the particle is at the point (x, y) = (2, -1).
Finally, let's substitute 2𝜋 for t:
x = 2 + sin(2𝜋) = 2 + 0 = 2
y = 4 + 5cos(2𝜋) = 4 + 5 = 9
So at t = 2𝜋, the particle is at the point (x, y) = (2, 9).
Therefore, the motion of the particle can be described as follows:
The particle starts at the point (x, y) = (3, 4) and moves clockwise three-fourths of the way around the ellipse to the point (x, y) = (2, 9).