Expert: Paul Klarreich - 3/26/2011

Question
After Lee gives his little sister Kara a big push on a swing, her horizontal position as a function of time is given by the equation x(t)= 3cost(e^-0.05t), where x(t) is her horizontal displacement, in metres, from the lowest point of her swing, as a function of time, t, in seconds.

a) Determine the highest speed Kara will reach
and when this occurs.

b) How long will it take for Kara’s maximum
horizontal displacement at the top of her
swing arc to diminish to 1 m? After how
many swings will this occur?

Answer
Questioner: Tasmina
Category: Calculus
Private: No
Subject: Caculus
Question: After Lee gives his little sister Kara a big push on a swing, her horizontal position as a function of time is given by the equation x(t)= 3cost(e^-0.05t), where x(t) is her horizontal displacement, in metres, from the lowest point of her swing, as a function of time, t, in seconds.

a) Determine the highest speed Kara will reach
and when this occurs.

b) How long will it take for Kara’s maximum
horizontal displacement at the top of her
swing arc to diminish to 1 m? After how
many swings will this occur?
.......................................
If x(t) = cos t (e^- kt), then  <<< skipping the useless 3 and putting k for the constant:

Diff:

v = - k cos t (e^-kt) - sin t (e^-kt)
v = - e^-kt [k cos t + sin t ]

Now to get the max, diff again:

a = - [e^-kt(-k sin t + cos t) - k e^-kt(k cos t + sin t) ]
a = - e^-kt [-k sin t + cos t - k^2 cos t - k sin t) ]
a = - e^-kt [- 2k sin t + (1 - k^2) cos t ]

and set that equal to zero:

- 2k sin t + (1 - k^2) cos t = 0

(1 - k^2) cos t = 2k sin t
1 - k^2     sin t
---------- = ----- = tan t
  2k        cos t

Now you can put k = 0.05 and compute tan t, then t.
...........................................
For your (b)

The max disp occurs at  t = 2n pi.  Solve  e^-kt = 1/3  for t, then solve 2n pi = t and you have your number of 'swings'.