calculus problem

Question

The period T of oscillation (in seconds) of a simple pendulum of length L (in feet) is given by T=2pi sqrt L/32. What is the rate of change of T with respect to L when L= 4 ft?

Answer 1

T=2pi √(L/32) = 2pi (L/32)^1/2

 

Differentiate using the Chain Rule and Power Rule:

 

dT/dL = 2pi·(1/2)(L/32)^-1/2 ·(1/32)

dT/dL = (pi/32)·(32/L)^1/2

 

Plug in L = 4 to get the answer