hands of a clock problem

[mark22]

[FONT=Verdana,Arial,Helvetica][FONT=Verdana,Arial,Helvetica]The hour hand of a clock has length 3. The minute hand has length 4. Find the distance between the tips of the hands when that distance is increasing the most rapidly. Find the precise time on the clock.

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[FONT=Verdana,Arial,Helvetica][FONT=Verdana,Arial,Helvetica][FONT=Verdana,Arial,Helvetica][FONT=Verdana,Arial,Helvetica]d= distance between the hands
m= minute hand
h= hour hand
if angle x; between m and h known then from law of cosines
d^2= m^2+h^2 -2mh cos x
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[FONT=Verdana,Arial,Helvetica][FONT=Verdana,Arial,Helvetica][FONT=Verdana,Arial,Helvetica][FONT=Verdana,Arial,Helvetica]Distance between the tips is sqrt(25 - 24 cos x). I differentiate and get 12 dx/dt sin x / sqrt(25 - 24 cos x). I differentiate again and find the maximum is when cos x (25 - 24 cos x) = 12 (sinx)^2 hence 12 cos2x - 25 cos x + 12 = 0. Factored to solve x and find one solution of angle x between 0 and 1 is 3/4 so plugged that back into the original distance equation. sqrt(25 - 24cos 3/4) = sqrt 7.

I think my calculus is correct but I don't really know how to find the time on the clock. Also if someone could show a way of solving it with vectors that would be interesting. Thanks[/FONT][/FONT]
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[John Harris]

My first thought was that "the distance is increasing the most rapidly" is satisfied by an angle between the hands and not by a specific time, and that the hands frozen at that angle could be turned through the whole circle without damaging the relationship, and so the time can't be determined.

Being a clock though, there's an additional constraint that the hour hand has to say the exact same number of minutes between any two hours as the minute hand is showing. There's always going to be one position each hour where that's true, so my conclusion is that there'll be exactly 11 times when the clock is showing a genuine time because the final 11 hours 60 minutes overlays the original 12 hours 0 minutes.

As for the angle between the hands, the distance is increasing most rapidly at the moment they pass each other. So the 11 solutions in hours minutes and seconds are 12:00:00, 1:05:300/11, 2:10:300/11*2, 3:15:300/11*3 [...] 10:50:300/11*10, 11:55:300.



eta 1: I'm writing nonsense, it's only true if the hands are the same length and that's not the question you posed. That's what comes of logging on at 4.30am.

eta 2: Go on, I'll stick with what I wrote. Which means the distance between the tips is invariably 1 at each of the 11 solutions.

 

[MarkFL]

I think I would approach this parametrically. Let 12:00:00 (am or pm) be time \(\displaystyle t=0\) in minutes and \(\displaystyle m(t)\) represent the position of the tip of the minute hand while \(\displaystyle h(t)\) represents the position of the tip of the hour hand both with respect to the center of the clock, which is the origin of our coordinate system. Then:

\(\displaystyle m(t)=4\left\langle \cos\left(\dfrac{\pi}{30}t \right),\sin\left(\dfrac{\pi}{30}t \right) \right\rangle\)

\(\displaystyle h(t)=3\left\langle \cos\left(\dfrac{\pi}{360}t \right),\sin\left(\dfrac{\pi}{360}t \right) \right\rangle\)

Let \(\displaystyle D(t)\) represent the distance between the two tips, hence:

\(\displaystyle D^2(t)=\left(4\cos\left(\dfrac{\pi}{30}t \right)-3\cos\left(\dfrac{\pi}{360}t \right) \right)^2+\left(4\sin\left(\dfrac{\pi}{30}t \right)-3\sin\left(\dfrac{\pi}{360}t \right) \right)^2\)

Expanding and simplifying via Pythagorean identities, we find:

\(\displaystyle D^2(t)=25-24\left(\cos\left(\dfrac{\pi}{30}t \right)\cos\left(\dfrac{\pi}{360}t \right)+\sin\left(\dfrac{\pi}{30}t \right)\sin\left(\dfrac{\pi}{360}t \right) \right)\)

Using the angle-difference identity for cosine, there results:

\(\displaystyle D^2(t)=25-24\cos\left(\dfrac{11\pi}{360}t \right)\)

Let \(\displaystyle \theta=\dfrac{11\pi}{360}t\)

\(\displaystyle D^2(t)=25-24\cos(\theta)\)

Differentiating, we find:

\(\displaystyle 2D(t)D'(t)=24\sin(\theta)\dfrac{d\theta}{dt}\)

\(\displaystyle D'(t)=12\dfrac{d\theta}{dt}\dfrac{\sin(\theta)}{D(t)}\)

Let \(\displaystyle k_1=12\dfrac{d\theta}{dt}\), differentiate again and equate to zero:

\(\displaystyle D''(t)=k_1\dfrac{D(t)\cos(\theta)\dfrac{d\theta}{dt}-\sin(\theta)D'(t)}{D^2(t)}=0\)

This implies:

\(\displaystyle D(t)\cos(\theta)\dfrac{d\theta}{dt}-\sin(\theta)D'(t)=0\)

\(\displaystyle \sqrt{25-24\cos(\theta)}\cos(\theta)\dfrac{d\theta}{dt}-\sin(\theta)k_1\dfrac{\sin(\theta)}{D(t)}=0\)

\(\displaystyle (25-24\cos(\theta))\cos(\theta)\dfrac{d\theta}{dt}-k_1\sin^2(\theta)=0\)

\(\displaystyle 25\cos(\theta)-24\cos^2(\theta)-12(1-\cos^2(\theta))=0\)

\(\displaystyle 25\cos(\theta)-24\cos^2(\theta)-12+12\cos^2(\theta)=0\)

\(\displaystyle 12\cos^2(\theta)-25\cos(\theta)+12=0\)

\(\displaystyle (4\cos(\theta)-3)(3\cos(\theta)-4)=0\)

Discarding the invalid root, we find:

\(\displaystyle \cos(\theta)=\dfrac{3}{4}\)

Hence, the tips of the hands are moving away from one another at the greatest rate when their distance apart is:

\(\displaystyle \sqrt{25-18}=\sqrt{7}\)

This coincides with times of:

\(\displaystyle \dfrac{11\pi}{360}t=\cos^{-1}\left(\dfrac{3}{4} \right)+2k\pi\) where \(\displaystyle k\in\mathbb{Z}\)

\(\displaystyle t=\dfrac{360}{11\pi}\cos^{-1}\left(\dfrac{3}{4} \right)+\dfrac{720k}{11}\)

So, to find the times this corresponds to, use \(\displaystyle 0\le k\le 10\) to find the 11 such times in minutes after 12:00:00.

 

[John Harris]

EDITED: I've edited this post after MarkFL's change noted below, to retain coherence in the thread.

It's interesting, Mark, that our result's the same except for this phase difference.

I note that the tips of the hands each move at constant speed. I still think the rate of separation is at a maximum when the tips are closest, which happens when the hands overlap. I'll spend a bit of this evening plotting the rate of separation against time for 12/11ths of an hour and post the result, I'm keen to see why my thinking's awry.

 

[MarkFL]

I made a faulty assumption, and have corrected my work above.

 

[John Harris]

If I'm allowed to post a graphic, I reckon this says the hands are at right angles which is what Mark originally posted. The blue line is the distance between tips, the orange is the rate of change of the distance multiplied by a scaling factor to get the same magnitude for the graph. That maximum is when the minute hand leads the hour hand by 90 degrees.

 

[mark22]

@MarkFL

Parametric equations is very clever, well done. I will remember that to model these kind of systems.

@JohnHarris

Thanks for the graphic