- #1

Mr. Heretic

- 16

- 0

## Homework Statement

"On a certain clock, the minute hand is 8 cm long and the hour hand is 6 cm long.

How fast, in cm/min, is the distance between the tips of the hands changing

at 9 am?"

## Homework Equations

## The Attempt at a Solution

I should be able to transform this somewhat difficult differentiation problem into a simpler limits problem due to the fact that the hands are perpendicular to each other at 9AM, and the velocity of the tip of a hand is perpendicular to the length of the hand at any time.

So where the speed of the tip of the minute hand is 'm' and that of the hour hand is 'h': (sqrt((8 - ht)^2 + (6 + mt)^2) - sqrt(8^2 + 6^2))/t should be an approximation of the change in distance between the tips over period t while t is non-zero, and the limit as t goes to zero should be exactly the rate of change that's desired.

However with m as 4pi/15 cm/min and h as pi/120 cm/min, I get 23pi/150 where the answers state 22pi/150. I've been over my working for an algebra error enough times to be pretty sure there hasn't been one.

The answers use a completely different method which I do understand (cosine rule and chain rule), I just want to know why this method isn't working.