A particle moves with its position given by x=cos(2t) and y=sin(t), where positions are given in feet from the origin and time t is in seconds.
A)Find the speed of the particle.
Speed = ____________
(include units)
B)Find the first positive time when the particle comes to a stop.
t=_____
(include units)
C)If n is any odd integer, write a formula (in terms of n) for all positive times t at which the particle comes to a stop.
t=______
(include units)
For the first one I got sqrt((-2sin(2t))^2+(cost(t))^2) ft/s, which is wrong, and I don't know why....
I don't know ho to calculate the second and third part... please someone help...
Ok..I found the problem with my first part, I accidentally typed cost(t)^2... Ok I'm still stucked with the last two parts, I don't know what to do with them....
To find the speed of the particle, we need to find the magnitude of its velocity vector. The velocity vector can be found by taking the derivative of the position vector with respect to time.
Given: x = cos(2t) and y = sin(t)
A) Find the speed of the particle:
The velocity vector is calculated by taking the derivative of the position vector with respect to time.
vx = dx/dt = d(cos(2t))/dt = -2sin(2t)
vy = dy/dt = d(sin(t))/dt = cos(t)
The magnitude of the velocity vector is given by the formula:
Speed = sqrt(vx^2 + vy^2)
Substituting the values of vx and vy:
Speed = sqrt((-2sin(2t))^2 + (cos(t))^2) ft/s
Now, let's simplify this expression:
Using the trigonometric identity: sin^2(x) + cos^2(x) = 1
((-2sin(2t))^2 + (cos(t))^2)
= 4sin^2(2t) + cos^2(t)
= 4(1 - cos^2(2t)) + cos^2(t)
= 4 - 4cos^2(2t) + cos^2(t)
Simplifying further:
= 4 - 3cos^2(t) - 4cos^2(2t)
So, the correct expression for the speed of the particle is:
Speed = sqrt(4 - 3cos^2(t) - 4cos^2(2t)) ft/s
B) To find the first positive time when the particle comes to a stop, we need to find when the speed (magnitude of velocity) becomes zero.
Setting Speed = 0:
0 = sqrt(4 - 3cos^2(t) - 4cos^2(2t))
Squaring both sides:
0 = 4 - 3cos^2(t) - 4cos^2(2t)
This equation can now be solved to find the values of t when the particle comes to a stop.
C) If n is any odd integer, we need to find a formula in terms of n for all positive times t at which the particle comes to a stop.
Using the equation derived in part B, we can write it as:
3cos^2(t) + 4cos^2(2t) = 4
Let's let n = 2t, so 2t = n
Now, the equation becomes:
3cos^2(t) + 4cos^2(n) = 4
Therefore, the formula in terms of n for all positive times t at which the particle comes to a stop is:
3cos^2(t) + 4cos^2(n) = 4 (where n is any odd integer)