Calculus

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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,
Thanks
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

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.
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.

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

• Calculus -

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)
= 24.25
d = 4.9

Now, 2d dd/dt = 18sin(11pi/6 t) (11pi/6)
= 33pi sin(11pi/6 t)

at t=9,

2*4.9 dd/dt = 33pi
dd/dt = 10.58 m/hr

Solutions are like standards -- there are so many to choose from!

• Calculus -

Thanks Steve!

Finally got this one correct on Math Lab!
I am eternally grateful, and have made an account here so I can get help and I have already tried helping others as well. I am very pleased with the service of this site and am glad to have found it :)

Hooray!!!