Question

Hello! I need help with this problem, its been irritating me for days! Any help would be greatly appreciated!
Here’s the question..
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 tip of the hands changing at 10:00 (Hint use the law of cosines).
So far this is what I have..
Min hand= a = 8
Hr hand= b = 4
Distance between these two points = c and I am trying to find dc/dt = ?
If you use the law of cosines (c^2=a^2 + b^2 -2ab cos (c) ) and set this up than it would be c^2=8^2 + 4^2 – 2 (8) (4) cos (c). Simplify and you end this c^2= 80-64cos(c). Also at 10:00 the angle is 300 degrees. What am I missing and where should I go from here? I not sure what the next step would be to get the answer. Again, any help would be greatly appreciated!

Answer 1

This problem is discussed in many ways. You might want to start here, and use your numbers:

http://math.stackexchange.com/questions/867482/the-rate-of-change-of-the-distance-between-the-tips-of-the-hands-of-a-clock

Answer 2

Okay this is what I got...

Minute hand = a = 8
Hour hand = b = 4
Distance between a---b = c

And we are looking for dc/dt = ?

The law of cosine states c^2 = a^2 + b^2 2ab cos (c)

So…
C^2 = 8^2 +4^2 – 2(8)(4) cos (c)

Take derivative
2c * dc/dt = 0 + 0 + 2(8)(4) sin (c) * dc/dt

2c * dc/dt = 2 * 8 * 4 * sin (5pi/6) * 11pi/360

To find side “c” use the Pythagorean theorem = c^2 = 8^2 + 4^2 which equals square root of 80.

So…

2 * sqrt(80) * dc/dt = 2 * 8* 4* (1/2) * 11pi/360

2 * sqrt(80)* dc/dt= 44pi/45

Simplify…get dc/dt on its own..

Dc/dt = 11pi sqrt(5)/450 meters per minute

Answer 3

Also, just got the news that we can not use the Pythagorean theorem b/c its not a right triangle that is why we are using the law of cosines...UGH!! If there is any other way to go about this please PLEASE HELP!