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August 29, 2014

August 29, 2014

Posted by **MG** on Thursday, March 27, 2014 at 3:36am.

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 Ų 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 dŲ/dt = (π/30 - π/720) rad/min

dŲ/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)cosŲ

d^2 = 24.25 - 18cosŲ

differentiate with respect to t

2d dd/dt = 0 + 18sinŲ dŲ/dt

Now at 9:00, the angle Ų = 90, and

the angle Ų 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.

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

- Calculus -
**Steve**, Thursday, March 27, 2014 at 5:34amThe 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 -
**MG**, Saturday, March 29, 2014 at 4:34pmThanks 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!!!

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