 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Like Clockwork (Posted on 2004-02-27) 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 it is increasing most rapidly?

 See The Solution Submitted by DJ Rating: 4.0000 (9 votes) Comments: ( Back to comment list | You must be logged in to post comments.) my formulae for d(t), d'(t) | Comment 26 of 29 | I started with theta_M(t)=pi/2 - 2pi t  and theta_H(t)=pi/2 - (pi/6)t    where t is in hours

d(t)=sqrt[25 - 24cos{(11/6)pi t}]

d'(t)= [22 sin{(11/6)pi t}] / d(t)

d"(t) was very tedious but setting it to zero eventually led to a quadratic expression in cos(x)   where x= (11/6)pi t

(2pi - 1) (cos(x))^2  - (25 pi/12) cos(x) + 1  =0

solving this gives cos(x)= {0.1785, 1.0603} but the cosine can't be greater than 1, so only 0.1785 can be valid.

then x= 1.3913 radians;   t=(6x)/(11 pi)=2.384 hours
or t is about 2:23
and d(t)=4.551

unless of course I made a math error.

And that's my story

 Posted by Larry on 2004-03-07 02:10:02 Please log in:

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