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?
4 answers
I can't make out your funny symbol in 2ðt
Oh sorry, that was supposed to be pi
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.
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.
Oh, for part A I mean 2, sorry about that.
Thanks much for the help!
Thanks much for the help!
1 answer