The Game of Finishing Oversoul's homework (fun calculus problem!)
- Thread starter Oversoul
- Start date
S
sageridder
Guest
Umm ...every minute.Or never if it's at a fixed position (1 o'clock) when you look, is this a calculus problem or a logic problem.Ah nevermind ...the answer is probally a seal balancing a dogdish on it's nose don't mind me I'm just an old fart move along.
L
Limited
Guest
I got 0,6515 mm per second...
Does that make any sense?
Does that make any sense?
Mooseman
Isengar Tussle
I haven't done this stuff in 13 years........ But I found this on the web.
1.18345Pi mm/min
A right triangle is created using the line between the "tips" of the clock hands as the hypotenuse.
The hypotenuse ("C") of the right triangle may be expressed as a
function of its base (X) and height (Y), where:
X = r Sin(Theta) - R Sin(Phi)
and
Y = R Cos(Phi) - r Cos(Theta)
Since the shaded triangle is a right triangle, one may write:
C^2 = X^2 + Y^2
or
C^2 = [r Sin(Theta)-R Sin(Phi)]^2 + [R Cos(Phi)-r Cos(Theta)]^2
or
C^2 = r^2 + R^2 - 2rR[Sin(Theta)Sin(Phi) + Cos(Theta)Cos(Phi)]
So, using the Theorem of Pythagoras, C may be written as:
C = {r^2 + R^2 - 2rR[Sin(Theta)Sin(Phi) + Cos(Theta)Cos(Phi)]}^(1/2)
Using the expression above for C, take the full derivative of C with
respect to time [i.e., (dC/dt)] to find the rate of change (WRT time)
of the distance between the tips of the minute hand and the hour hand
of a watch at one o'clock.
Remember the following, at one o'clock:
Theta = 0 Radians
Phi = (2 Pi r)/(12 r) Radians
= (Pi/6) Radians
d(Theta)/dt = (2 Pi) Radians/12 hr
d(Phi)/dt = (2 Pi) Radians/min
1.18345Pi mm/min
A right triangle is created using the line between the "tips" of the clock hands as the hypotenuse.
The hypotenuse ("C") of the right triangle may be expressed as a
function of its base (X) and height (Y), where:
X = r Sin(Theta) - R Sin(Phi)
and
Y = R Cos(Phi) - r Cos(Theta)
Since the shaded triangle is a right triangle, one may write:
C^2 = X^2 + Y^2
or
C^2 = [r Sin(Theta)-R Sin(Phi)]^2 + [R Cos(Phi)-r Cos(Theta)]^2
or
C^2 = r^2 + R^2 - 2rR[Sin(Theta)Sin(Phi) + Cos(Theta)Cos(Phi)]
So, using the Theorem of Pythagoras, C may be written as:
C = {r^2 + R^2 - 2rR[Sin(Theta)Sin(Phi) + Cos(Theta)Cos(Phi)]}^(1/2)
Using the expression above for C, take the full derivative of C with
respect to time [i.e., (dC/dt)] to find the rate of change (WRT time)
of the distance between the tips of the minute hand and the hour hand
of a watch at one o'clock.
Remember the following, at one o'clock:
Theta = 0 Radians
Phi = (2 Pi r)/(12 r) Radians
= (Pi/6) Radians
d(Theta)/dt = (2 Pi) Radians/12 hr
d(Phi)/dt = (2 Pi) Radians/min
B
BigBlue
Guest
Hmm... I started doing the problem (thinking I was doing it differently than mooseman)... but in the end I am doing it the same way... One thing which I believe matters, is whether the minute hand is sweeping, or if it's one where it only moves at the end of the minute... perhaps it doesn't matter... but I think it does...
I didn't bother to check his calculations, but I came up with the same angles... it's been forever since I did any trig & calculus... What is this for Oversoul? and please do give another problem... hopefully without an internet solution before I work it out...
I will say this - I'm very worried about your speeds... it shouldn't be even close to a mm a minute... the circumference of the clock is a little more than 50mm - in 1 minute the minute hand is not moving 1 mm... and the hour hand is slowly moving as well - much slower, but it is moving...
I didn't bother to check his calculations, but I came up with the same angles... it's been forever since I did any trig & calculus... What is this for Oversoul? and please do give another problem... hopefully without an internet solution before I work it out...
I will say this - I'm very worried about your speeds... it shouldn't be even close to a mm a minute... the circumference of the clock is a little more than 50mm - in 1 minute the minute hand is not moving 1 mm... and the hour hand is slowly moving as well - much slower, but it is moving...
L
Limited
Guest
Thats exactly how I did it (well almost, I wasn't smart enough to turn the watch around to match the sinusoide) but I still got another answer..
I'll blame it on a typo
I'll blame it on a typo
Mooseman
Isengar Tussle
Remember that this is not how fast is the minute hand moving around the dial, but the relationship between the minute and hour hands as they get closer.BigBlue said:
It's rate of change between two moving object, not rate of movement around a fixed point.
or something like that....... Damn you Oversoul, I am not ready for calculus discussions...... hehe
B
BigBlue
Guest
Correct mooseman... but, if it isn't even moving a mm, it isn't getting further than a mm apart... also, I realize we need to take into consideration that this is radial speed, not straight vectors...
Say we have a car going around a circular track... the track has a radius of 8 miles and the car goes around the track once an hour... how fast is the car going? roughly 50 mph (16pi miles/hr). There is another car going on a 4 mile radius circular track, but it takes a day for it to make it around the track... how fast is it moving? roughly 1 mph (1/3 pi miles/hr)... If they both started at the same time (from a starting line scribed across both tracks evenly), 1 hour later they would be at the "1 o' clock position"... same question in other words... just stated differently
I think it would be interesting to chart out those distances over a typical hour - they will be completely cyclic based on each time the hands are over one another... That happens once per hour... and 24 times per day... since we have that weird 12 hour clock... actually that cycle can be further simplified since it doesn't matter whether it's noon or midnight...
What is the period of that cycle? A little more than an hour... (11 times per 12 hours) if you can build a nice mathematical model... then it get's really interesting if you can map how it changes as the radii change... What are the times of day when that happens?
The only constant we know about with time, is that at 4:45, time stops for about 15 minutes until we get to go home... And at 7:30am, while trying to get ready for work, time speeds up to make it a mad dash to arrive on time...
OK, my name is Jason. I'm addicted to theoretical math. I've haven't thought about math for 5 seconds (correction I just did it again).
Say we have a car going around a circular track... the track has a radius of 8 miles and the car goes around the track once an hour... how fast is the car going? roughly 50 mph (16pi miles/hr). There is another car going on a 4 mile radius circular track, but it takes a day for it to make it around the track... how fast is it moving? roughly 1 mph (1/3 pi miles/hr)... If they both started at the same time (from a starting line scribed across both tracks evenly), 1 hour later they would be at the "1 o' clock position"... same question in other words... just stated differently
I think it would be interesting to chart out those distances over a typical hour - they will be completely cyclic based on each time the hands are over one another... That happens once per hour... and 24 times per day... since we have that weird 12 hour clock... actually that cycle can be further simplified since it doesn't matter whether it's noon or midnight...
What is the period of that cycle? A little more than an hour... (11 times per 12 hours) if you can build a nice mathematical model... then it get's really interesting if you can map how it changes as the radii change... What are the times of day when that happens?
The only constant we know about with time, is that at 4:45, time stops for about 15 minutes until we get to go home...
And at 7:30am, while trying to get ready for work, time speeds up to make it a mad dash to arrive on time...OK, my name is Jason. I'm addicted to theoretical math. I've haven't thought about math for 5 seconds (correction I just did it again).
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BigBlue
Guest
I haven't had time to calculate it yet... but, I don't think it is constant... and I am still pretty sure it isn't 1mm, though the more I think about it, you could be correct - that it could exceed the 1mm threshhold...
T
train
Guest
Without working it out... and it would take a refresher for me...
The rate of change is not constant... it does matter if sweeping or "ticking" (barely, and only for extreme specifics...)
But I must state, that at 1, the minute hand would be moving towards the hour hand (clockwise direction, the hour hand is already on the 1, and the minute hand is on the 12 at this point in time...), so... the distance between them would be decreasing...
seemingly a fractional rate of change until the minute hand passed the hour hand...
then the distance would be increasing...
The rate of change is not constant... it does matter if sweeping or "ticking" (barely, and only for extreme specifics...)
But I must state, that at 1, the minute hand would be moving towards the hour hand (clockwise direction, the hour hand is already on the 1, and the minute hand is on the 12 at this point in time...), so... the distance between them would be decreasing...
seemingly a fractional rate of change until the minute hand passed the hour hand...
then the distance would be increasing...
Oversoul
The Tentacled One
I was wondering about that myself. But this morning my teacher was giving us hints on some of the problems and he indicated that the hands are to be treated as moving at constant rates themselves. I don't know what effect it would have on the scope of the problem if they did not. This stuff is confusing me enough as it is...BigBlue said:
I'm approaching the end of my first quarter of calculus. It's been fun so far, except that my teacher is not very good and this section has my whole class running around like crazy trying to work these problems out. Once I got going, I finished all but the last two, which use more trigonometry than the others. The one I just posted is the very last one, and I did get an answer, but it seemed wrong and I found out I did it the wrong way to begin with. I still don't know what the right answer is yet.
I have the answer to the problem before it, but have yet to make my calculations match that answer. So if anyone wants to, here's that one...
And since BB is addicted to math, here's the one I thought was hardest so far (except for maybe the previous two with their confusing trigonometry). I already finished it and am pretty sure I have the right answer, but I'm not 100% certain about this...Fun calculus problem said:
B
BigBlue
Guest
This is bizarre calculus if you ask me... when I took calculus I recall it being much more theoretical... unless this is applied calculus or something...
I'l work on these tonight and get back to you in the am... it'll give me something to do while doing laundry... plus I can look up stuff in my calculus books... (Yeah, I haven't been in school for close to 12 years and I still have my calculus books... I'm truly a math geek... and proud of it...)
The first one is in my opinion the hardest certainly... because both points are moving... unlike the second one... where I get myself into trouble is that I try to solve it with variables - to get a general answer which I can apply to multiple situations... usually before I even try it with just your example... It is more challenging, more useful, and more interesting I guess...
Train - good to see you man! I can remember the days when I could do this sort of thing faster, but my trig is definately rusty... Heck, I was having a hard time remembering the distance formula earlier... have to keep referencing back to pythagorean theorom and such... Sine = Opp over hyp... Cos = Adj over Hyp...
I'l work on these tonight and get back to you in the am... it'll give me something to do while doing laundry... plus I can look up stuff in my calculus books... (Yeah, I haven't been in school for close to 12 years and I still have my calculus books... I'm truly a math geek... and proud of it...)
The first one is in my opinion the hardest certainly... because both points are moving... unlike the second one... where I get myself into trouble is that I try to solve it with variables - to get a general answer which I can apply to multiple situations... usually before I even try it with just your example...
It is more challenging, more useful, and more interesting I guess...Train - good to see you man! I can remember the days when I could do this sort of thing faster, but my trig is definately rusty... Heck, I was having a hard time remembering the distance formula earlier... have to keep referencing back to pythagorean theorom and such... Sine = Opp over hyp... Cos = Adj over Hyp...
B
BigBlue
Guest
Last night I didn't spend my entire evening working these out - sorry Oversoul...
But, I can't recall exactly how to derive f(t)= (g(t))^n
I think it's ng(t)^(n-1) * g'(t)... but it's been forever and I didn't feel like researching it all that much... a first semester student (unless you're taking differential calculus first - which I doubt) wouldn't be doing multivariable functions... I wrote my function in terms of a & t... where a is the starting angle and t is time elapsed...
I plotted out some values for it, it is a nice curve.... and as train says it has negative slope at 1 o'clock...
anywho..
Let X be the length of the Minute hand.
Let n be the ratio of the length of the hour hand to the minute hand.
Let a (would be theta but I'm too lazy to look for those codes anywhere) be the initial angle inscribed by the hands.
f(a,t) is a function to calculate the change in distance between the tips of the hands after t hours. (I used hours since it keeps the formula nicer - and can be converted later anyways)...
f(a,t) = sqrt((Xcos(a) - nX)^2 + (Xsin(a))^2) *****(distance at time 0)
- sqrt((Xcos(a-11*t*pi/6) - nX)^2 + (Xsin(a-11*t*pi/6))^2) ***** (distance at time t)
for your example we can set a as a constant (pi/6)... which eliminates the first number since it's just a constant and the derivative makes it vanish... then it just get's ugly... But, since my trig is rusty, maybe there is a better way to solve it (looking at mooseman's method)...
But, I can't recall exactly how to derive f(t)= (g(t))^n
I think it's ng(t)^(n-1) * g'(t)... but it's been forever and I didn't feel like researching it all that much... a first semester student (unless you're taking differential calculus first - which I doubt) wouldn't be doing multivariable functions... I wrote my function in terms of a & t... where a is the starting angle and t is time elapsed...
I plotted out some values for it, it is a nice curve.... and as train says it has negative slope at 1 o'clock...
anywho..
Let X be the length of the Minute hand.
Let n be the ratio of the length of the hour hand to the minute hand.
Let a (would be theta but I'm too lazy to look for those codes anywhere) be the initial angle inscribed by the hands.
f(a,t) is a function to calculate the change in distance between the tips of the hands after t hours. (I used hours since it keeps the formula nicer - and can be converted later anyways)...
f(a,t) = sqrt((Xcos(a) - nX)^2 + (Xsin(a))^2) *****(distance at time 0)
- sqrt((Xcos(a-11*t*pi/6) - nX)^2 + (Xsin(a-11*t*pi/6))^2) ***** (distance at time t)
for your example we can set a as a constant (pi/6)... which eliminates the first number since it's just a constant and the derivative makes it vanish... then it just get's ugly... But, since my trig is rusty, maybe there is a better way to solve it (looking at mooseman's method)...
Killer Joe
New member
"France"?
N
Nightstalkers
Guest
Is this a mechanical watch or a quartz watch?
If mechanical and a newer style movement it'll be probably at 28,800 impulses a minute. So you have around 5 minutes minus probably around 8 milliseconds added to it if the watch is at full wind, and perhaps minus 5 milliseconds if the watch is almost run down.
Also take into account your hour hand will move (even slightly) by the time your minute hand reaches it.
EDIT Sorry, I'm going stupid again...
If mechanical and a newer style movement it'll be probably at 28,800 impulses a minute. So you have around 5 minutes minus probably around 8 milliseconds added to it if the watch is at full wind, and perhaps minus 5 milliseconds if the watch is almost run down.
Also take into account your hour hand will move (even slightly) by the time your minute hand reaches it.
EDIT Sorry, I'm going stupid again...