A clock's minute hand has length 4 and its hour hand length 3. What is the distance between the tips at the moment when the distance separating them is increasing most rapidly? Don't cheat; I'm looking for thorough explanations.
When the distance between them is one; they are pointing in the same direction. Nope that's wrong. It would be when the minute hand is approaching the hour hand and the right triangle formed by the tip distance is at 45 degrees. Not as easy as it first looks.
Why not just work in either polar or cartesian coordinates, work out the distance between the tips, say \(D\) in terms of the angle between them, then work out \(\partial_{\theta}D\). Then you have the expression for the relative velocity of the tips in terms of their angle. Then it's just a matter of finding when that function is extremised. This is basically M2 mechanics, ie working with vectors which are a function of something you then need to differentiate, so the first step is to draw a picture.
After looking at the fire and the bucket of water, AlphaNumeric exclaims "A solution exists," and goes back to sleep...Please Register or Log in to view the hidden image!
Hi AlphaNumeric; just from visualization it seems to me that the maximum velocity would occur when the tips form a right triangle with two sides equal to one. There's two places where that happens; one when the hour hand is approaching and one when it is departing. I'm not able to determine which of those would provide the greatest velocity. I just guessed it would be when minute hand is approaching the hour hand. I think you could probably work it out though Please Register or Log in to view the hidden image!
Try a phase diagram, with two variables which are the tips of each hand. The phase space is straightforward - it's isn't spherical but circular (easy-peasy)...? But it's vector arithmetic too.
The distance between the tips can be found based on thier angles of seperation, and using the cosine law: D(theta)=sqrt(25-24cos(theta)) differentiate with respect to theta, to find the rate of distance change yields: D(theta)=(25-24cos(theta))^(1/2) d(theta)=0.5(25-24cos(theta))^(-1/2) * 24sin(theta) Now punch that into a graphing calculator and you will find that the rate of change in ditance is actualy the smallest at 90 degrees (or 1 pi radians)! (edit: being awake, it is painfully obviouse that 90deg does not equal a pi radian.) Of course, we are looking for a maxima (d't=0), so time for another derrivative (this one will be ugly, and hence may be more prone to error): d'(theta)=-.25(25-24cos(theta))^(-3/2) * (24sin(theta))^2 + 0.5(25-24cos(theta))^(-1/2) * 24cos(theta) d'(theta)=0 0=-.25(25-24cos(theta))^(-3/2) * (24sin(theta))^2 + 0.5(25-24cos(theta))^(-1/2) * 24cos(theta) We can limit our range to 0<t<pi, and solve. Im going to use a GDC to solve. If your feeling particularly board, or masochistic, or perhaps know some secret that I don't to making this easy, feel free to solve it with limited (or no) mechanical aid Please Register or Log in to view the hidden image! My GDC says: at theta=0.72273425 occurs the maximum rate of distance change. So, now we can sub that back into D(theta), and calculate the distance between the hands: D(0.72273425)=2.645751318 Assuming units in inches, the distance the two hands of your clock are apart at the time when the distance is changing most rapidly is approx 2.6 inches. Now I think I got that generaly correct, but please point out any errors. Good night. -Andrew
Before I turn in: work in a (rotating) reference frame where the minute hand is stationary. Then the distance is increasing the fastest when the tip of the hour hand is moving directly away from the tip of the minute hand, or the hour hand is at right angles to the line connecting the tips of the two hands. So \(d^2 + 3^2 = 4^2\) and the distance is \(\sqrt{7} = 2.64575 \ldots\)
Correct przyk. Why is the distance increasing the fastest at that angle? andbna, I believe you have the correct approach and the correct approximate solution. The cos law is what I thought of as well. http://www.mat.itu.edu.tr/gungor/IMO/www.kalva.demon.co.uk/putnam/psoln/psol832.html
The distance between the tips of the hands can be easily written in terms of one variable: \(d^2 = \left( 4 \sin 12 \theta - 3 \sin \theta \right) ^ 2 + \left( 4 \cos 12 \theta - 3 cos \theta \right) ^ 2\) Then the naive procedure is to solve for the extremal rate of change of the distance: \({ { \partial ^ 2 d} \over { \partial \theta ^ 2} } = { { \partial ^ 2 } \over { \partial \theta ^ 2} } \sqrt{ \left( 4 \sin 12 \theta - 3 \sin \theta \right) ^ 2 + \left( 4 \cos 12 \theta - 3 cos \theta \right) ^ 2 } = 0\) But you save yourself some time by noting that the distance is periodic in \(\phi = 11 \theta\) (because as the hour hand goes around once in twelve hours, the minute hand laps it 11 times). After some manipulation we have: \(d^2 = 25 - 24 \cos \phi\) and \({ { \partial ^ 2 d} \over { \partial \phi ^ 2} } = 12 { { 25 \cos \phi - 18 - 6 \cos 2 \phi } \over { d^3 } } = 0\) which has solution \(d = \sqrt{7}\) and \(\cos 11 \theta = \cos \phi = \frac{3}{4}\) (phi is in the first quadrent). But we can learn more if we solve a more general case. If the minute hand (length a) goes around the clock k > 1 times in the time the hour hand (length b) goes around once then we have, for a > b: \(d^2 = a^2 + b^2 - 2 a b \cos ( k - 1) \theta\) and \(\phi = ( k - 1) \theta\) and \({ { \partial ^ 2 d} \over { \partial \phi ^ 2} } = a b { { ( a^2 + b^2 ) \cos \phi - \frac{3}{2} a b - \frac{a b}{2} \cos 2 \phi } \over { d ^ 3 } } = 0 \) which has solution \(d = \sqrt{a^2 - b^2}\) and \(\cos (k - 1) \theta = \cos \phi = \frac{b}{a}\) (phi is in the first quadrent). I would like to see przyk finish the post, although the geometric approach is tougher for a text web.
Okay, a more complete reply: First, we work in a coordinate system whose origin is fixed at the centre of rotation of the hands and which is "rotating with" the minute hand. The tip of the minute hand is fixed at the position vector \(\bar{r}_{0}\), and the tip of the hour hand is rotating anticlockwise (at 11 complete revolutions every 12 hours if I'm not mistaken, though all that's important is that the angular velocity is constant) with a time-dependent position vector \(\bar{r} = \bar{r}(t), \; r \, < \, r_0\) : Please Register or Log in to view the hidden image! Defining \(\bar{s} = \bar{r} - \bar{r}_{0}\), direct calculation yields: \(\frac{\mathrm{d}s}{\mathrm{d}t} = \bar{v} \cdot \bar{1}_{s}\)with the unit vector \(\bar{1}_{s} \equiv \frac{\bar{s}}{s}\) and \(\bar{v} \equiv \frac{\mathrm{d}\bar{r}}{\mathrm{d}t}\). Since the module of \(\bar{v}\) is constant, \(\bar{v} \cdot \bar{1}_{s}\) is maximal when \(\bar{s}\) and \(\bar{v}\) are parallel (intuitively, when all of \(\bar{v}\) is contributing to increasing \(\bar{s}\) and no component is lost in perpendicular motion). \(\bar{r}\) is perpendicular to \(\bar{v}\) (circular motion) and therefore also to \(\bar{s}\) at this point, which justifies the use of Pythagoras' theorem in calculating the separation, yielding: \(s_{0} = \sqrt{{r_{0}}^{2} - r^{2}}\)
