Posted by **Aaron** on Friday, December 19, 2008 at 3:53pm.

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?

- Check your typing...Calculus -
**Reiny**, Friday, December 19, 2008 at 4:29pm
I can't make out your funny symbol in 2ðt

- Calculus -
**Aaron**, Saturday, December 20, 2008 at 11:26am
Oh sorry, that was supposed to be pi

- Calculus -
**Damon**, Saturday, December 20, 2008 at 2:55pm
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.

- Calculus -
**Anonymous**, Tuesday, December 23, 2008 at 5:35pm
Oh, for part A I mean 2, sorry about that.

Thanks much for the help!

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