Rate of Change of Clock Hands

Date: 08/11/99 at 13:43:19
From: Hoang Nguyen
Subject: Related Rates

Hi Dr. Math,

I am taking calculus and have a problem that I can't find the answer 
to. I hope that you will help me to find the solution. The problem is:

The minute hand on a watch is 8 mm long and the hour hand is 4 mm 
long. How fast is the distance between the tips of the hands changing 
at one o'clock?

I need you to show me how to solve this problem step by step, and how 
to draw a diagram of this problem.

Thanks,
Hoang

Date: 08/11/99 at 14:14:53
From: Doctor Fwg
Subject: Re: Related Rates

Dear Hoang,

Here is a solution for the clock problem.

Problem Statement: Find the rate of change (with respect to time) of 
the distance between the tips of the minute hand and the hour hand of 
a watch at one o'clock.

  

Note: this drawing shows the time a little after one o'clock so that a 
more general solution can be worked out.

The changing distance is marked "C" on the drawing and is the 
hypotenuse of the shaded triangle. The length of the minute hand is R 
and the length of the hour hand is r, where R > r.

The hypotenuse ("C") of the shaded 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

and the final solution will contain d(Theta)/dt and d(Phi)/dt terms.

I hope this is what you were after.

- Doctor Fwg, The Math Forum
http://mathforum.org/dr.math/