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+(cos(t))^2) ft/s.
I don't know ho to calculate the second and third part... please someone help...
A) The formula you provided for the speed of the particle is correct: √((-2sin(2t))^2+(cos(t))^2) ft/s.
B) To find the first positive time when the particle comes to a stop, we need to find when the speed of the particle equals zero.
Setting the expression for speed equal to zero:
0 = √((-2sin(2t))^2+(cos(t))^2)
Squaring both sides, we get:
0 = (-2sin(2t))^2+(cos(t))^2
Since the expression inside the square root must be non-negative, we can ignore the square root and solve for when the expression inside equals zero:
(-2sin(2t))^2+(cos(t))^2 = 0
Simplifying further:
4sin^2(2t) + cos^2(t) = 0
Using trigonometric identities, we can rewrite this equation as:
4(1-cos^2(2t)) + cos^2(t) = 0
Expanding and rearranging:
4 - 4cos^2(2t) + cos^2(t) = 0
Substituting cos^2(2t) = 2cos^2(t) - 1 (double-angle identity), we have:
4 - 4(2cos^2(t) - 1) + cos^2(t) = 0
Simplifying:
9cos^2(t) - 4 = 0
Solving for cos(t):
cos^2(t) = 4/9
Taking the square root of both sides, we have:
cos(t) = ±2/3
Since the particle comes to a stop at these values of cos(t), the first positive time it comes to a stop is when cos(t) = 2/3. However, we need to check if sin(t) is positive or negative at this time.
Using the equation y = sin(t), we have:
y = sin(t) = sin(arccos(2/3))
Since sin(arccos(x)) = √(1-x^2), we have:
y = sin(t) = √(1 - (2/3)^2) = √(1 - 4/9) = √(5/9)
Hence, the particle comes to a stop at t = arccos(2/3) and y = √(5/9) ft.
C) To find a formula for all positive times at which the particle comes to a stop, we need to find when the expression inside the square root equals zero.
Setting the expression inside the square root equal to zero:
(-2sin(2t))^2+(cos(t))^2 = 0
4sin^2(2t) + cos^2(t) = 0
Using the double-angle identity sin(2t) = 2sin(t)cos(t), we can rewrite it as:
4(2sin(t)cos(t))^2 + cos^2(t) = 0
16sin^2(t)cos^2(t) + cos^2(t) = 0
Factoring out cos^2(t), we have:
cos^2(t)(16sin^2(t) + 1) = 0
Since cos^2(t) cannot be zero, we only need to solve for when 16sin^2(t) + 1 = 0.
16sin^2(t) + 1 = 0
16sin^2(t) = -1
But sine squared cannot be negative, so there are no positive times t at which the particle comes to a stop for any odd integer n.
B) To find the first positive time when the particle comes to a stop, we need to find when the velocity of the particle becomes zero.
The velocity of the particle is given by the derivatives of x and y with respect to time:
vx = d/dt (cos(2t)) = -2sin(2t)
vy = d/dt (sin(t)) = cos(t)
To find when the particle comes to a stop, both vx and vy need to be zero at the same time:
-2sin(2t) = 0 (equation 1)
cos(t) = 0 (equation 2)
From equation 2, we can see that when cos(t) = 0, t = π/2 + nπ/2, where n is any integer. And since we're looking for the first positive time, we can take n = 0.
Substituting t = π/2 into equation 1:
-2sin(2(π/2)) = 0
-2sin(π) = 0
0 = 0
Since 0 = 0 is always true, it means that at t = π/2, the particle comes to a stop.
Therefore, the first positive time when the particle comes to a stop is t = π/2 seconds.
C) To find a formula for all positive times t at which the particle comes to a stop, we need to consider equation 2: cos(t) = 0.
Since cos(t) = 0 when t = π/2 + nπ/2, where n is any odd integer, we can write a formula:
t = π/2 + nπ/2, where n = 1, 3, 5, ...
This formula gives all the positive times at which the particle comes to a stop.
For Part A, to find the speed of the particle, you need to calculate the magnitude of the velocity vector. The velocity vector is the derivative of the position vector with respect to time.
Given that x = cos(2t) and y = sin(t), we can calculate the velocity vector by taking the derivative of each component with respect to time.
dx/dt = -2sin(2t)
dy/dt = cos(t)
To find the speed, we need to calculate the magnitude of the velocity vector, which is the square root of the sum of the squares of the components.
Speed = √((dx/dt)^2 + (dy/dt)^2)
= √((-2sin(2t))^2 + (cos(t))^2) ft/s
For Part B, to find the first positive time when the particle comes to a stop, we need to find when the speed equals zero. Set the speed equation equal to zero and solve for t.
0 = √((-2sin(2t))^2 + (cos(t))^2)
To solve this equation, you need to square both sides and isolate the trigonometric terms. This will involve applying some trigonometric identities and solving for t. You may need to use a calculator or computer algebra system to find an exact solution.
For Part C, if n is an odd integer, we need to find a formula (in terms of n) for all positive times t when the particle comes to a stop. This means finding values of t where the speed is zero.
You can set the speed equation equal to zero and solve for t.
0 = √((-2sin(2t))^2 + (cos(t))^2)
Again, you will need to apply trigonometric identities and solve for t. The formula you obtain will be dependent on the value of n.
x=cos(2t) and y=sin(t)
x'(t)=-2sin(2t),
y'(t)=cos(t)
A.
Speed at time t (in ft/sec)
= sqrt(x'(t)²+y'(t)²)
= sqrt(4sin²(2t)+cos²(t))
= ... simplify as you wish
B.
x'(t)=-2sin(2t)=0
occurs when t=kπ/2, k∈Z
and y'(t)=cos(t)=0
occurs when t=(k+1/2)π, k∈Z
So what is the smallest t when x'(t)=0 AND y'(t)=0?
C.
Work out from B above.
Post if you more hint.