Two Lagrangian Toads spot my cat Pythagoras simultaneously. They both run directly away from Pythagoras immediately. The first is initially 5 m from Pythagoras and runs at a rate of 2 m/s and the second is initially 1 m from Pythagoras and runs at a rate of 3 m/s. If the angle of the line of sight from Pythagoras to each of these Lagrangian Toads is 30°, then how fast is the distance between these Lagrangian Toads increasing after 4 seconds?

Explanation:

In the triangle formed , the third side , that is distance between two toads can be calculated with the help of following formula

Cos C = a² + b² - c² / 2ab

c² = a² + b² - 2abcosC

c² = 5² + 1² - 2 x 5 x 1 cos 30

= 26 - 8.66

= 17.34

c = 4.16 m

c² = a² + b² - 2abcosC

differentiating with respect to time

2 c x dc / dt = 2a da/dt + 2 b db/dt - 2 cosC ( a db/dt + b da/dt )

c x dc / dt = a da/dt +  b db/dt -  cosC ( a db/dt + b da/dt )

4.16 dc/dt = 5 x 2 + 1 x 3 - .866 ( 5 x 3 + 1 x 2 )

= 13 - 14.722

= - 1.722

dc / dt = - 1.722 / 4.16

= -  .414 m /s .

The distance between the two will decrease at the rate above.

3.2x10^7

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.