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JeeJee
Canada5652 Posts
I'll give a shot at one more numerical puzzle, but if it gets solved in less than an hour again, I shall have to take a different approach from then on, since clearly TL is too smart for their own good. Here we go! Consider an accurate analog clock (let's pretend they exist >.>) with a minute hand of length 4 and an hour hand of length 3. What's the distance between the hand tips when it is increasing the fastest? GL!  | ||
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Caller
Poland8075 Posts
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ArmChairCritic
Sweden36 Posts
dO/dt=pi/30-pi/360 H is given by H=sqrt(a^2+b^2-2abcos(O) dH/dO=+2absin(O) Now check for maxima Amirite? Gonna plug the number in now, will update EDIT: Distance should be approx 2.30 units if I havent done anything horribly wrong EDIT2: Forgot about the square root brb EDIT3: when the distance is 0.57 units, is that correct? Oooooh figured out what I was doing wrong but saw that naonao solved it  | ||
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naonao
United States847 Posts
2.65 + Show Spoiler + We can express the distance between the hands as D = sqrt(4^2 + 3^2 - 24cos(x)) = sqrt(25-24cos(x)) where x is the angle measure between the two hands. The rate of change is d(D) = 24sin(x)(1/2)(1/(sqrt(25-24cos(x)))(dx). Dx is the change in the angle which is +- .03(pi) To maximize this function we take the derivative again and set it equal to 0. (1.13097 cos(x))/sqrt(25-24 cos(x))-(13.5717 sin^2(x))/(25-24 cos(x))^(3/2) = 0, solving for X you get x = 0.72279, or 5.56046. Plugging these value of x into the equation for rate you get d(D) = + .28274334 and -.28274334 respectively, so the first value is when the hands are increasing the fastest. Now plug the value of x back into the initial equation to find the distance and you get D = 2.65 | ||
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Aim Here
Scotland672 Posts
the distance between the tips is increasing the fastest when the sum of the velocity vectors of the two hands gives the largest net vector, which will be when the hands are perpendicular, so it'll be the hypotenuse of a right-handed triangle of sides 3 and 4. Also your nicely chosen units gives another clue. My intuitive guess is that the distance is 5. | ||
Daigomi
South Africa4316 Posts
What you are asking for is when do the two hands move away from each other at the fastest speed. Since the two hands always move at the same speed, the question boils down to when the movement from the one hand is perfectly perpendicular to that of the other hand, since that would be the most effective movement away from the other hand. Since the movement is always relative to each other, you can set assume that the movement will go either way, allowing you to set a (in the image) to 90 degrees. ![[image loading]](/staff/Daigomi/Mathproblems.jpg) Once this is done, it's a case of pythagoras, and you find that the line is sqrt of 7, or 2.64575 something. Battery is going to die now, but I hope that makes sense. No idea if it is right though, but it seems to work for me! | ||
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niyade
Canada3 Posts
On March 04 2010 07:59 Daigomi wrote: I have no idea how to write this mathematically, but hopefully you can figure out what I'm trying to say: + Show Spoiler + What you are asking for is when do the two hands move away from each other at the fastest speed. Since the two hands always move at the same speed, the question boils down to when the movement from the one hand is perfectly perpendicular to that of the other hand, since that would be the most effective movement away from the other hand. Since the movement is always relative to each other, you can set assume that the movement will go either way, allowing you to set a (in the image) to 90 degrees. Once this is done, it's a case of pythagoras, and you find that the line is sqrt of 7, or 2.64575 something. Battery is going to die now, but I hope that makes sense. No idea if it is right though, but it seems to work for me! wow, a very nice solution | ||
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qrs
United States3637 Posts
On March 04 2010 07:59 Daigomi wrote: I have no idea how to write this mathematically, but hopefully you can figure out what I'm trying to say: + Show Spoiler + What you are asking for is when do the two hands move away from each other at the fastest speed. Since the two hands always move at the same speed, the question boils down to when the movement from the one hand is perfectly perpendicular to that of the other hand, since that would be the most effective movement away from the other hand. Since the movement is always relative to each other, you can set assume that the movement will go either way, allowing you to set a (in the image) to 90 degrees. Once this is done, it's a case of pythagoras, and you find that the line is sqrt of 7, or 2.64575 something. Battery is going to die now, but I hope that makes sense. No idea if it is right though, but it seems to work for me! Very nice solution. I thought about doing this the calculus way, decided it would be a lot of messy calculation without much creative thinking, so what was the point--but your answer is simple and elegant. I wish I'd thought of it. | ||
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eshlow
United States5210 Posts
On March 04 2010 07:59 Daigomi wrote: I have no idea how to write this mathematically, but hopefully you can figure out what I'm trying to say: + Show Spoiler + What you are asking for is when do the two hands move away from each other at the fastest speed. Since the two hands always move at the same speed, the question boils down to when the movement from the one hand is perfectly perpendicular to that of the other hand, since that would be the most effective movement away from the other hand. Since the movement is always relative to each other, you can set assume that the movement will go either way, allowing you to set a (in the image) to 90 degrees. Once this is done, it's a case of pythagoras, and you find that the line is sqrt of 7, or 2.64575 something. Battery is going to die now, but I hope that makes sense. No idea if it is right though, but it seems to work for me! beautiful dude just beautiful | ||
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