this one is so good (and not solved) I pulled it foreward in time
Posted by quizzical on Thursday, November 16, 2006 at 6:24am.
Hello,
This is just a question that has stumped me for some time:
We have a normal watch.
The big hand is 4cm long and the little hand is 3cm long.
What's the distance between the tips of the hands at the moment they are moving the fastest towards each other.
For Further Reading
* math Puzzler.. - bobpursley, Thursday, November 16, 2006 at 9:24am
5 cm. Think out why.
o math Puzzler.. - Lance, Thursday, November 16, 2006 at 11:53am
Facinating question. I don't get it? Say the clock was 20 feet in diameter. Please enlighten me.
* math Puzzler.. - Ken, Thursday, November 16, 2006 at 12:02pm
is that the moment of change between obtuse and acute exchange between the angles at the tips? (assuming a triangle ascribed by the hands)
* math Puzzler.. - Ken, Thursday, November 16, 2006 at 12:50pm
Of course, the hands never actually move "toward each other" as they are travleing in the same direction,(yes?) but the speed of aproach and recession for he points of the hands does, and is a great question. :)
* math Puzzler.. - quizzical, Thursday, November 16, 2006 at 6:04pm
Bob: the first thing that jumped out at me when you said 5 was the 3-4-5 triangle being involved and it is almost a 'trick' question... with no difficult forumulas needed. I'll think on that, thanks for the hint
Lance: the size of the clock face itself wouldn't matter because what we are after is the relationship between the two hands (the lengths of which we are given).
Ken: Theats right. they are always moving in the same direction.. I was thinking rates of change, but I had no function to start with so I didn't know where to start :(
Thanks for the responses everyone...
Assume the clock hands are phasors. Maximum apparent frequency occurs when the displacements are ninety degrees. Frequency on the phasor plot relates to velocity on postion, time plot.
This is a very old planetary problem, originally laid out on the astrolabe. Apparent planet veloicty was always maximum when the moving planet was halfway in angular rotation between the minimum and maximum distance from the reference planet.
Analytically, this problem is a mess. It probably is easier to simulate than to solve.
They are moving toward each other
They are never moving toward each other
To solve this problem, we need to determine the distance between the tips of the clock hands at the moment they are moving the fastest towards each other.
The first clue we have is that the big hand is 4cm long, and the little hand is 3cm long.
One important concept to consider is that the hands of a clock move at different rates. The big hand moves at a constant rate, while the little hand moves at a variable rate.
To find the moment when the hands are moving the fastest towards each other, we need to determine the point where the rate of change of the distance between the tips is maximized. In other words, we want to find the point where the derivative of the distance function is maximized.
Let's define a coordinate system with the center of the clock as the origin. We can place the big hand along the positive x-axis, and the little hand at an angle theta counterclockwise from the positive x-axis.
Now, let's find the distance between the tips of the hands. We can use the Pythagorean theorem:
Distance = sqrt((3cos(theta))^2 + (4-3sin(theta))^2)
Next, we can compute the derivative of the distance function with respect to theta:
d(Distance)/d(theta) = (18sin(theta) - 16cos(theta)) / sqrt((3cos(theta))^2 + (4-3sin(theta))^2)
To find the maximum of this derivative, we can set it equal to zero and solve for theta. However, solving this equation analytically can be quite messy and challenging.
The original poster suggests that this problem is easier to simulate than to solve analytically. So one approach could be to use a computer program or a graphing calculator to plot the distance function and visually determine the point where the hands are moving the fastest towards each other. This would involve plotting the distance as a function of theta and finding the maximum value.
Alternatively, we can use numerical methods such as Newton's method or the bisection method to find the value of theta where the derivative is zero. These methods involve iteratively refining the guess for the value of theta until we find the desired solution.
In conclusion, to find the distance between the tips of the clock hands at the moment they are moving the fastest towards each other, we can either simulate the problem using a computer program or use numerical methods to solve the equation d(Distance)/d(theta) = 0.