Distance between a Clock's hands when the distance is increasing most rapidly

[WMDhamnekar]

Homework Statement

Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question)

Relevant Equations

Answer is already known.

Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question)

Answer: Making assumption that both the hands moves at constant angular velocities, the answer is $\sqrt{7} .$ But don't you think this assumption is somewhat doubtful and wrong?

 

[berkeman]

Answer: Making assumption that both the hands moves at constant angular velocities, the answer is SQRT(7) But don't you think this assumption is somewhat doubtful and wrong?

The hands move with constant and different angular velocities. At what point of their movement do you think the difference in their tip velocity vectors is the max? How did you approach this question?

 

[WMDhamnekar]

Correct answer to this question is given in the following video. 30 seconds correct answer given at the end of this video is also time and energy saving. Calculus

 

[haruspex]

Answer: Making assumption that both the hands moves at constant angular velocities, the answer is $\sqrt{7} .$ But don't you think this assumption is somewhat doubtful and wrong?

Seems reasonable to me for an ideal clock. Why do you doubt it?

 

[OmCheeto]

Fun problem. Guessing the problem involves civilian clocks(12 hr) and not military(24 hr) clocks?

 

[haruspex]

Fun problem. Guessing the problem involves civilian clocks(12 hr) and not military(24 hr) clocks?

Not that that makes any difference here.

 

[wrobel]

looks like a problem about differentiation of a function
$$f(t)=\sqrt{a_1^2+a_2^2-2a_1a_2\cos(\omega_1-\omega_2)t}$$

 

[OmCheeto]

Not that that makes any difference here.

Interesting. It doesn't.
So the clock could be broken, with the hour hand not even moving, and we would still come up with √7 .

 

[vela]

Answer: Making assumption that both the hands moves at constant angular velocities, the answer is $\sqrt{7} .$ But don't you think this assumption is somewhat doubtful and wrong?

Without that assumption, the answer would depend on the particulars of how you model the movement of the hands. How could you answer the question without that additional information?

 

[haruspex]

Interesting. It doesn't.
So the clock could be broken, with the hour hand not even moving, and we would still come up with √7 .

Yes, and that is the easiest way to solve it: take the minute hand as stationary. That makes it obvious when the relative velocity of rate of increase of the distance between the tips is greatest.

 

[256bits]

Yes, and that is the easiest way to solve it: take the minute hand as stationary. That makes it obvious when the relative velocity of the tips is greatest.

Reading the comments from the video, the obviousness seems to be not ensured. A mixup of, or distinction, between velocity and distance confused a few people. One comment went against her analysis saying that the (relative) velocity must be greatest when the hands are 180 degrees apart.
Although age, training, and experience not given.

 

[256bits]

Interesting. It doesn't.
So the clock could be broken, with the hour hand not even moving, and we would still come up with √7 .

Once every hour, 24 times a day it happens.

 

[haruspex]

Reading the comments from the video, the obviousness seems to be not ensured. A mixup of, or distinction, between velocity and distance confused a few people. One comment went against her analysis saying that the (relative) velocity must be greatest when the hands are 180 degrees apart.
Although age, training, and experience not given.

Yes, my wording was sloppy. Will edit. As stated in post #1 the intent is clear.

 

[256bits]

Yes, my wording was sloppy. Will edit. As stated in post #1 the intent is clear.

Oh sorry. I do apologize.
If what I wrote turned out to appear to be a critique, it was not meant as you are well versed, having and giving great insights as anyone can see.

 

[berkeman]

Question: A clock's minute hand has length 4 and its hour hand has length 3.

Yes, and that is the easiest way to solve it: take the minute hand as stationary.

@haruspex -- minute hand or hour hand as stationary?

 

[haruspex]

Oh sorry. I do apologize.
If what I wrote turned out to appear to be a critique, it was not meant as you are well versed, having and giving great insights as anyone can see.

Whoa, too much deference!
Nothing to apologise for. My post #10 was wrong and I am glad you picked that up.

 

[haruspex]

@haruspex -- minute hand or hour hand as stationary?

Either will do in principle, but I think it is more obvious if you take the longer hand as stationary. The tip of the shorter hand can move directly away from the tip of the longer hand, but the converse is never true. You cannot take a tangent to a circle from a point inside it.

 

[JimWhoKnew]

Correct answer to this question is given in the following video. 30 seconds correct answer given at the end of this video is also time and energy saving. Calculus

There is an additional benefit to the short way: if we are given that the lengths of the hands are $a$ and $b$ such that $~0<a<b~$ and $\theta$ is the angle between them, it follows immediately that the distance increases most rapidly when $~\cos{\theta}=a/b~$. To get this result via the long way, we'll need to carry out the entire calculation as done in the video.

 

[OmCheeto]

There is an additional benefit to the short way: if we are given that the lengths of the hands are $a$ and $b$ such that $~0<a<b~$ and $\theta$ is the angle between them, it follows immediately that the distance increases most rapidly when $~\cos{\theta}=a/b~$. To get this result via the long way, we'll need to carry out the entire calculation as done in the video.

I will have to work out why she said the solution was as simple as the hands making a 90° angle with the hour hand's traced perimeter.
Is this fact common knowledge? Or is this another one of my 'SENILITY NOW' moments?

 

[vela]

I will have to work out why she said the solution was as simple as the hands making a 90° angle with the hour hand's traced perimeter.

Consider which way the velocity of the hour hand's tip points at that moment.

 

[OmCheeto]

Consider which way the velocity of the hour hand's tip points at that moment.

Say that to a non-mathematician friend tomorrow, and tell me the look you got.

 

[JimWhoKnew]

I will have to work out why she said the solution was as simple as the hands making a 90° angle with the hour hand's traced perimeter.
Is this fact common knowledge? Or is this another one of my 'SENILITY NOW' moments?

Had she bothered to justify it, it would have taken another minute or two. Not as cool.

Spoiler: Hint

The "new" distance is a vectorial sum of the "old" and the "change". The direction of the change vector becomes tangent to the traced perimeter when $~dt~$ is infinitesimal. Moreover, note that the relative angular velocity between the hands is constant, so the magnitude of the infinitesimal change vector is the same for equal intervals of $~dt~$.

 

[haruspex]

the hands making a 90° angle with the hour hand's traced perimeter.

I don’t even know what that means.
Treat the minute hand as fixed (i.e. work in the frame of the minute hand).
The tip of the hour hand moves at constant speed. It will be moving away from that of the minute hand fastest when its motion is directly away from it, so the line joining the tips is tangential to the locus of the tip of the hour hand and make a right angle to the hour hand.

 

[OmCheeto]

I don’t even know what that means.

It looks as though that statement came out of my head before I finished my doodle of the problem.

I should get some kind of award for having never even considered going into teaching as I am such a bad communicator in maths, that I will look back at my old posts and claim; "This guy is an idiot."

Treat the minute hand as fixed (i.e. work in the frame of the minute hand).
The tip of the hour hand moves at constant speed. It will be moving away from that of the minute hand fastest when its motion is directly away from it, so the line joining the tips is tangential to the locus of the tip of the hour hand and make a right angle to the hour hand.

Except for the reversal of the hands, that looks very much like what I meant to say.

With the exception of the "It will be moving away from that of the minute hand fastest when its motion is directly away from it" statement. It sounds like 'a priori' knowledge to me. Perhaps having never been a science and or math instructor, I never ran across this particular problem.

In any event, after thinking about it for a few seconds, I now get it. Thanks!

 

[Ibix]

With the exception of the "It will be moving away from that of the minute hand fastest when its motion is directly away from it" statement. It sounds like 'a priori' knowledge to me.

You can write the velocity as a component directly away from the other hand's tip and a component perpendicular to that. The second component doesn't result in a change in distance, because it's tangential motion. Thus, given fixed speed, the maximum rate of change of distance happens when the velocity is directly away from the other tip, not "wasting" any of its speed on tangential motion.

 

[haruspex]

With the exception of the "It will be moving away from that of the minute hand fastest when its motion is directly away from it" statement.

You are being chased by a bear. You can run at the same speed in any direction. Which way do you run?
The a priori knowledge comes from millions of years of evolution.