P=2πsqrt(L/g)
dP/dL = 2π(1/2)L^(-1/2)/sqrt(g)
=2πsqrt(L/g)/(2L)
transpose appropriate terms:
2*[dP/(2πsqrt(L/g))] = [dL/(L)]
Each term in square brackets is equal to the fractional form of percentage error, thus proved.
(use Δ => differentials instead of derivative.)
need help please. the time required for one complete oscillation of a pendulum is called its period. if L is the length of the pendulum, then the period is given by P= 2 times pie multiplied by the square root of (L/g), where g is a constant called the acceleration due to gravity. use differentials to show that the percentage error in P is approximately half the percentage error in L.
1 answer
1 answer