Find the number of vertical tangents to the graph of the motion of a particle whose position vector is vector r with components, the 2 times the cosine of t, the sine of 2 times t . (10 points)
A) One
B) Two
C) Three
D) Four
I cannot interpret your description.
x = 2cost
y = sin2t
dy/dx = (dy/dt) / (dx/dt) = (2 cos2t)/(-2sint) = -cos2t/sint
so vertical tangents occur when sint=0 and co2t≠0
t = kπ
Since x has period 2π, that's how long it takes to trace the curve, giving 2 vertical asymptotes.
To find the number of vertical tangents to the graph of the motion of a particle, we need to analyze the derivatives of the position vector.
The position vector of the particle is given by vector r = (2cos(t), sin(2t)).
To find the derivative of r with respect to t, we differentiate each component separately.
Derivative of x component:
dx/dt = d(2cos(t))/dt
= -2sin(t)
Derivative of y component:
dy/dt = d(sin(2t))/dt
= 2cos(2t)
Now, we want to find the points where the x component of the velocity is equal to zero, i.e., when dx/dt = -2sin(t) = 0.
For sin(t) = 0, we have two possibilities:
1) t = 0
2) t = π
Note that these are the values of t that make cosine equal to 1 (t = 0) or -1 (t = π), which corresponds to the extreme points of the position vector.
Thus, we have two values of t where dx/dt = 0. At these points, the particle has vertical tangents.
So, the answer is B) Two, since there are two points where the graph has vertical tangents.