Related Rates - Clock HandsDate: 04/05/99 at 20:54:33 From: Perry Subject: Related Rates 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'm confused about how to do this related rate problem since the hour and minute hand are moving at different speeds. Date: 04/22/99 at 16:27:31 From: Doctor Fwg Subject: Re: Related Rates Dear Perry, Thank you for such an interesting question. To restate the problem: one wishes 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 1:00 o'clock. Here's a diagram - it shows the time a little after one o'clock so that a more general solution can be worked out. The changing distance you are after 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 WRT 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 1:00 o'clock. Remember the following: At 1:00 o'clock (A.M. or P.M.), 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 helps - be careful with the differentiation, it is a little tricky. Best regards, - Doctor Fwg, The Math Forum http://mathforum.org/dr.math/ |
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