Posted by MG on Wednesday, March 26, 2014 at 6:54pm.
I followed this example, where am i missing something or going about it wrong?
If we let y be the angle between the two hands and x be the distance between the two tips, then, by the law of cosines, we have:
x^2 = 5^2 + 1.5^2 - 2*5*1.5cos(y)
x^2 = 27.5 - 15cos(y)
Take the derivative of both sides with respect to t, time.
2x*dx/dt = 15sin(y)*dy/dt
Since it is 9:00, the angle between the two hands must be y = π/2. And since there is a right triangle, x = √(5^2 + 1.5^2) = √27.5. In order to find dy/dt, consider the fact that the hours hand goes around 2π in one hour and the minutes hand goes around 2π in 1/60 hour, therefore we have dy/dt = 2π - 2π/(1/60) = -118π. Plug that all in:
2(√27.5)*dx/dt = 15sin(π/2)*(-118π)
2(√27.5)*dx/dt = 15(1)*(-118π)
2(√27.5)*dx/dt = -1770π
(√27.5)*dx/dt = -885π
dx/dt = -885π/√27.5 ≈ -530.184
let Ø be the angle between them
the angular velocity of the minute hand = 2π/60 rad/min = π/30 rad/min
the angular velicity of the hour hand = 2π/(12(60)) or π/720 rad/min
then, so dØ/dt = (π/30 - π/720) rad/min
dØ/dt = 23π/720 rad/min
let the distance between the tips of the hands be d m
d^2 = 4.5^2 + 2^2 - 2(2)(4.5)cosØ
d^2 = 24.25 - 18cosØ
differentiate with respect to t
2d dd/dt = 0 + 18sinØ dØ/dt
Now at 9:00, the angle Ø = 90, and
the angle Ø between the minute and the hour hand is increasing at 23π/720 rad/min
d^2 = 24.25 - 18cos9°0 = 24.25 - 0
d = √24.25
dd/dt = 18 sin90° (23π/720) / 2√24.25
= .1834 m/min
or
11.005 m/hr
check my arithmetic.
Thank you for the response,
i tried the method mentioned above three times and was incorrect each time, i double checked all my work to match the method above.
The correct answer is always just .3 under the answer
All of your arithmetic is right as well, so it is not that.
Could it be do/dt?
o being the angle between the clock hands?
Please any help is greatly appreciated
ive been stuck on this one for days