Ask a question
0 0

calculus problem

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?
Tutors, please sign in to answer this question.

1 Answer

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