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?
(In reply to
re: why calculus by Steve)
In addition to analytical and spreadasheeet methods there is a third one, a “common sense” solution.
Assume that:
-the hour hand A is stationary and rests on 12 o’clock.
-The minute hand B is of the same length as A and, beginning at 12 rotates with the speed Vt.
The distance between tips of A and B, (D) changes at the max rate at t=0, when D is perpendicular to A.
For B>A the above occurs when:
A=cos(a)*B, (a) is the angle between A and B
We have:
D=sin(a)*B
And
D=SQRT(B^2-A^2)
For our example
D=sqrt7
Cos(a)=3/4
QED !!!!!!!!!!!!!!!!!
Note that D is independent of Vt.
I move to post the three solutions.
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Posted by joe
on 2004-03-03 20:06:00 |
KISS, keep it simple!!!!!!!
