Hello,
This is just a question that has stumped me for some time:
We have a normal watch.
The big hand is 4cm long and the little hand is 3cm long.
What's the distance between the tips of the hands at the moment they are moving the fastest towards each other.
5 cm. Think out why.
Facinating question. I don't get it? Say the clock was 20 feet in diameter. Please enlighten me.
is that the moment of change between obtuse and acute exchange between the angles at the tips? (assuming a triangle ascribed by the hands)
Of course, the hands never actually move "toward each other" as they are travleing in the same direction,(yes?) but the speed of aproach and recession for he points of the hands does, and is a great question. :)
Bob: the first thing that jumped out at me when you said 5 was the 3-4-5 triangle being involved and it is almost a 'trick' question... with no difficult forumulas needed. I'll think on that, thanks for the hint
Lance: the size of the clock face itself wouldn't matter because what we are after is the relationship between the two hands (the lengths of which we are given).
Ken: Theats right. they are always moving in the same direction.. I was thinking rates of change, but I had no function to start with so I didn't know where to start :(
Thanks for the responses everyone...
To find the distance between the tips of the hands at the moment they are moving the fastest towards each other, we can start by understanding the motion of the clock hands. The big hand and the little hand of a watch both move in a circular motion around the clock face.
Since the big hand is 4cm long and the little hand is 3cm long, we can visualize the hands as two lines originating from the center of the clock face. The distance between the tips of these lines is what we want to find.
At any given moment, the position of the hands can be represented by angles. Let's call the angle between the big hand and the vertical axis θ1, and the angle between the little hand and the vertical axis θ2.
As time passes, both θ1 and θ2 change. We are interested in finding the moment when the difference between θ1 and θ2 is the smallest, which corresponds to the hands moving the fastest towards each other.
To solve this problem, we need to differentiate the distance between the tips of the hands with respect to time. In other words, we need to find the rate of change of the distance.
The distance between the tips of the hands can be calculated by using the Pythagorean theorem. Let's call this distance d. We have:
d² = (4cos(θ1) - 3cos(θ2))² + (4sin(θ1) - 3sin(θ2))²
To find the rate of change of d, we can differentiate this equation with respect to time. However, this involves taking the derivative of trigonometric functions, which can get quite complicated.
Fortunately, we can simplify the problem. Notice that when the hands are moving the fastest towards each other, the angle between them is changing at its fastest rate. In other words, dθ1/dt and dθ2/dt are at their maximum.
Since both hands are moving at a constant speed, the rate of change of the angles can be expressed as:
dθ1/dt = ω
dθ2/dt = ω
where ω is a constant representing the angular velocity.
Since the hands are moving at a constant speed, ω is the same for both hands. This means that the difference between θ1 and θ2 is changing at its fastest rate when θ1 = θ2. In other words, the hands are aligned.
When the hands are aligned, the distance between the tips of the hands can be calculated as d = 4 - 3 = 1cm.
Therefore, at the moment they are moving the fastest towards each other, the distance between the tips of the hands is 1cm.