Let f be the function given by f(t) = 2ðt + sin(2ðt)
a) Find the value of t in the open interval (0,20 for which the line tangent at (t, f(t)) is parallel to the line through (0,0) and (2,4ð)
b) Suppose the given function describes the position of a particle on the x-axis for time (0,2). What is the average velocity of the particle over that interval?
c) Determine the velocity and acceleration of the particle at t=1. Is the object speeding up or slowing down?
f(t) = 2 pi t + sin 2 pi t
line through (0,0) and (2 , 4 pi) has slope 4 pi/2 = 2 pi
Where does our function have slope = 2 pi?
df/dt = 2 pi + 2 pi cos 2 pi t
so when is df/dt = 2 pi?
2 pi = 2 pi + 2 pi cos 2 pi t
cos 2 pi t = 0
2 pi t = pi/2 , 3 pi/2 , 5 pi/2 , 7 pi/2 etc
so
t = 1/4 , 3/4 , 5/4, 7/4 , ......
are you sure you mean 20 and not 2?
Part b
f(2) = 2 pi*2 + sin 4 pi = 4 pi +0 = 4 pi
f(0) = 0 + 0 = 0
F(2) - F(0) = 4 pi
4 pi/2 = 2 pi = distance / time = av speed.
part c
v = df/dt = 2 pi + 2 pi cos 2 pi t
a = d^2f/dt^2 = -4 pi^2 sin 4 pi t
at t = 1
v = 2 pi + 2 pi = 4 pi
a = 0 so it is not speeding up or slowing down.
I can't make out your funny symbol in 2ðt
Oh sorry, that was supposed to be pi
Oh, for part A I mean 2, sorry about that.
Thanks much for the help!
a) To find the value of t for which the line tangent to the curve is parallel to the line passing through (0, 0) and (2, 4π), we need to find the equation of the tangent line.
First, let's find the derivative of the function f(t) with respect to t to get the slope of the tangent line:
f'(t) = 2π + 2πcos(2πt)
Next, we find the slope of the line passing through (0, 0) and (2, 4π):
m = (4π - 0) / (2 - 0) = 2π
Since the tangent line has to be parallel to this line, its slope should be equal to 2π. Therefore, we set f'(t) equal to 2π:
2π + 2πcos(2πt) = 2π
Now, solve for t:
2πcos(2πt) = 0
For cosine to be zero, the angle must be π/2, 3π/2, 5π/2, etc. So we have:
2πt = π/2, 3π/2, 5π/2, ...
Solving for t:
t = 1/4, 3/4, 5/4, ...
However, we only want the values of t in the open interval (0, 20). So the solution is:
t = 1/4, 3/4, 5/4, ... , 79/4, 81/4, ..., 399/4
b) The average velocity of a particle over an interval is given by the change in position divided by the change in time. In this case, the position of the particle is given by the function f(t), and the interval is from t = 0 to t = 2.
The change in position is:
f(2) - f(0) = 2π(2) + sin(4π) - 2π(0) - sin(0)
= 4π + 0 - 0 - 0
= 4π
The change in time is:
2 - 0 = 2
Therefore, the average velocity of the particle over the interval (0, 2) is:
Average velocity = (change in position) / (change in time)
= (4π) / (2)
= 2π
So, the average velocity of the particle is 2π.
c) To determine the velocity and acceleration of the particle at t = 1, we first find the derivative of f(t) to get the velocity function, and then take the derivative of the velocity function to get the acceleration function.
The derivative of f(t) = 2πt + sin(2πt) with respect to t is:
f'(t) = 2π + 2πcos(2πt)
So, the velocity of the particle at t = 1 is:
f'(1) = 2π + 2πcos(2π(1))
= 2π + 2πcos(2π)
= 2π + 2π(1)
= 4π
Next, let's find the acceleration of the particle. We take the derivative of the velocity function:
f''(t) = -4π^2sin(2πt)
So, the acceleration of the particle at t = 1 is:
f''(1) = -4π^2sin(2π(1))
= -4π^2sin(2π)
= -4π^2sin(0)
= 0
The acceleration of the particle at t = 1 is 0. Since the acceleration is zero, the object is neither speeding up nor slowing down at t = 1.