A big clock has a minute hand 3 ft long and a second hand 4 ft long. let x be the distance between their tips. find max value of (dx)/(dt)
To find the maximum value of (dx)/(dt), we first need to determine the relationship between the minute hand and the second hand.
Let's consider the circular path that the second hand follows. The circumference of the path is 2πr, where r is the length of the second hand. Similarly, the minute hand traces a circular path with a circumference of 2πr, where r is the length of the minute hand.
Since both hands complete one full rotation around their respective axes in one minute, we can use this information to calculate dt, the change in time.
For the second hand:
Circumference = 2πr (where r = 4 ft)
dt = 1 minute
For the minute hand:
Circumference = 2πr (where r = 3 ft)
dt = 1 minute
Now, let's find the relationship between dx and dt. Since both hands are rotating on separate axes, we can treat each path separately.
For the second hand:
dx = 2πr (change in arc length)
dt = 1 minute
For the minute hand:
dx = 2πr (change in arc length)
dt = 1 minute
Now, let's substitute the values and simplify:
For the second hand:
(dx)/(dt) = (2πr)/(dt) = (2π*4)/(1) = 8π ft/min
For the minute hand:
(dx)/(dt) = (2πr)/(dt) = (2π*3)/(1) = 6π ft/min
To find the maximum value, we compare the rates of change for both hands:
(dx)/(dt) = 8π ft/min (for the second hand)
(dx)/(dt) = 6π ft/min (for the minute hand)
Therefore, the maximum value of (dx)/(dt) is 8π ft/min, which occurs when considering the second hand.