Rate of Change Problem with Clock Hand Position

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SUMMARY

The discussion centers on solving the rate of change problem involving the positions of clock hands at 9 AM. The minute hand, measuring 8 cm, and the hour hand, measuring 6 cm, are analyzed using differentiation and limits. The initial attempt yielded an incorrect rate of change due to a miscalculation of the hour hand's velocity, which was incorrectly set to one rotation per 24 hours instead of 12. After correcting this error, the correct rate of change was confirmed to be 22π/150 cm/min, validating the alternative method of using the cosine rule and chain rule.

PREREQUISITES
  • Understanding of basic calculus concepts, particularly differentiation.
  • Familiarity with the cosine rule in trigonometry.
  • Knowledge of angular velocity and its application in circular motion.
  • Proficiency in limits and their use in calculus problems.
NEXT STEPS
  • Study the application of the cosine rule in related rate problems.
  • Learn about angular velocity and its calculation for rotating objects.
  • Explore advanced differentiation techniques, including implicit differentiation.
  • Practice solving similar rate of change problems involving circular motion.
USEFUL FOR

Students studying calculus, particularly those focusing on related rates and trigonometry, as well as educators seeking to enhance their teaching methods for these concepts.

Mr. Heretic
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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.
 
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Check your value for h.
 
Oh my god, I can't believe it was that... I accidentally did one rotation of the hour hand per 24 hours instead of per 12. Now I'm getting the right answer and I can know my side-step of a method is valid. Thanks, Voko.
 

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