posted by MG on .
Posted by MG on Wednesday, March 26, 2014 at 6:54pm.
The hands of a clock in some tower are 4.5m and 2m in length. How fast is the distance between the tips of the hands changing at 9:00? (Hint: Use the law of cosines)
The distance between the tips of the hands is changing at a rate of _______ m/hr at 9:00?
I tried several times, mathlab is stating the correct answer as a two digit integer with the following decimals round to the tenths place.
After following several different paths of approach I am getting a three digit number.
Any help would be great,
College Calculus - MG, Wednesday, March 26, 2014 at 6:58pm
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
College Calculus - Reiny, Wednesday, March 26, 2014 at 8:25pm
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
check my arithmetic.
College Calculus - MG, Thursday, March 27, 2014 at 3:31am
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
The hands of a clock in some tower are 4.5m and 2m in length. How fast is the distance between the tips of the hands changing at 9:00
at time t hours after 12:00, (at t=0)
the minute hand is at 4.5 sin(2pi*t)
the hour hand is at 2.0sin(2pi*t/12)
at 9:00, the distance d is
d^2 = 4.5^2 + 2.0^2 - 2(4.5)(2.0)cos(2pi*t - 2pi/12 * t)
= 24.25 - 18cos(11pi/6 t)
= 24.25 - 18.0cos(33pi/2)
d = 4.9
Now, 2d dd/dt = 18sin(11pi/6 t) (11pi/6)
= 33pi sin(11pi/6 t)
2*4.9 dd/dt = 33pi
dd/dt = 10.58 m/hr
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Finally got this one correct on Math Lab!
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