Help my friend in calc .. because I can't :P

[TheVrolok]

Alright, basic related rates problem.

Small wristwatch has a minute hand with a length of 8mm and an hour hand of 4 mm, at what rate is the distance of the tips of the hands changing at 1 o'clock?

I really tried to help him out but related rates were three years ago for me, last year was my last calc year and I can barely remember what I did then. As much as I did really enjoy calc at the time .. I'm not too upset I'll never use it again .. much. 🙂

 

[TuxDave]

Originally posted by: TheVrolok
Alright, basic related rates problem.

Small wristwatch has a minute hand with a length of 8mm and an hour hand of 4 mm, at what rate is the distance of the tips of the hands changing at 1 o'clock?

I really tried to help him out but related rates were three years ago for me, last year was my last calc year and I can barely remember what I did then. As much as I did really enjoy calc at the time .. I'm not too upset I'll never use it again .. much. 🙂

You could do it related rates, or do it the el-cheapo no skill way.

The minute hand tip has a tangental velocity equal to 8mm*2*pi/60 minutes
The hour hand tip has a tangental velocity equal to 4mm*2*pi/24 hours

The angle of the minute velocity is 0 degrees (0 degrees being directly right)
The angle of the hour velocity is is -30degrees

So solve for:
Magnitude of (Vmin-Vhour) where Vmin and Vhour are vectors....

 

[TheVrolok]

Impressively done, I wouldn't have thought of that, but he does need it done via calculus. It's for his class tomorrow, I'm going back to try and help him.. I do recall not enjoying related rates 😛

 

[glugglug]

This isn't a calculus problem.

It's more of a trig problem.

 

[dighn]

you can do it the calc way

eg have a triangle made up by hand, minute and distance, you hae the angle inbteween. use cosine lawn to find csitance then diferenatiate with respect to angle difference which is a function of teh angles of handla an dminutes

 

[DrPizza]

I'm not sure I'm in agreement with tuxdave... I'd really have to think about what he's talking about...
Yes, I disagree, because there's 12 hours on a clock, not 24.

besides, the calculus way is easier...

the two hands (and the distance between the hands) form a triangle. Use the law of cosines to express the distance as a function of the angle between the hands.

c^2 = a^2 + b^2 - 2ab cos C
where c is the distance between the hands.
c is a variable, C (the angle) is a variable, and a and b are constants.

take the derivative of both sides (implicitly is fine, since he'll have covered it by now)

2c dc/dt = 2absinC dCdt

solve dor dc/dt = 2absinC/2c dC/dt = (ab/c) sinC dC/dt
and simply fill in a,b, the angle at that moment is 1/12 of 360 = 30 degrees (teacher probably prefers pi/6)
Use the law of sines again to find side c at that moment, and as Tuxdave pointed out, the rate of change of the angle is the difference between the angular velocities of each (angle is decreasing, minute hand is moving 6 degrees per minute (or 1/10 degree per second) Hour hand is moving 360 degrees per 12 hours = 30 degrees per hour = .5 degrees/min = .5/60 degrees/second) Simply subtract to get dC/dt.

 

[TuxDave]

hahaha.. oops... change the 24 to 12.. there, it should work.

 

[dighn]

Originally posted by: TuxDave
hahaha.. oops... change the 24 to 12.. there, it should work.

yeah i also believe that would work

 

[TuxDave]

Originally posted by: DrPizza
I'm not sure I'm in agreement with tuxdave... I'd really have to think about what he's talking about...
Yes, I disagree, because there's 12 hours on a clock, not 24.

besides, the calculus way is easier...

the two hands (and the distance between the hands) form a triangle. Use the law of cosines to express the distance as a function of the angle between the hands.

c^2 = a^2 + b^2 - 2ab cos C
where c is the distance between the hands.
c is a variable, C (the angle) is a variable, and a and b are constants.

take the derivative of both sides (implicitly is fine, since he'll have covered it by now)

2c dc/dt = 2absinC dCdt

solve dor dc/dt = 2absinC/2c dC/dt = (ab/c) sinC dC/dt
and simply fill in a,b, the angle at that moment is 1/12 of 360 = 30 degrees (teacher probably prefers pi/6)
Use the law of sines again to find side c at that moment, and as Tuxdave pointed out, the rate of change of the angle is the difference between the angular velocities of each (angle is decreasing, minute hand is moving 6 degrees per minute (or 1/10 degree per second) Hour hand is moving 360 degrees per 12 hours = 30 degrees per hour = .5 degrees/min = .5/60 degrees/second) Simply subtract to get dC/dt.

Haha... i guess I like this way better. I've gotten lazy and do whatever comes to mind first.

 

[DrPizza]

Originally posted by: TuxDave
hahaha.. oops... change the 24 to 12.. there, it should work.

Ahhhhhhhhhhhhhhh, now I see what you meant... I read it a little more carefully. Yes, your way works just fine