Differentials find maximum percentage error

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naspek
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Homework Statement


The period T of a simple pendulum with small oscillations is calculated from the
formula T = 2pie (L / g)^1/2 . Suppose that measurements of L and g have errors of at
most 0.5% and 0.1% respectively. Use differentials to approximate the maximum
percentage error in the calculated value of T.

The Attempt at a Solution


dT = (dT/dL)dL + (dT/dg)dg
= 2pie (1/2g(L/g)^1/2) dL + 2pie (1/2g(L/g^2)^1/2) dg
i don't know how am i going to proceed.. :confused:
 
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naspek said:

Homework Statement


The period T of a simple pendulum with small oscillations is calculated from the
formula T = 2pie (L / g)^1/2 . Suppose that measurements of L and g have errors of at
most 0.5% and 0.1% respectively. Use differentials to approximate the maximum
percentage error in the calculated value of T.


The Attempt at a Solution


dT = (dT/dL)dL + (dT/dg)dg
= 2pie (1/2g(L/g)^1/2) dL + 2pie (1/2g(L/g^2)^1/2) dg
i don't know how am i going to proceed.. :confused:
You're not using the chain rule correctly. Also, the name of the Greek letter is pi, not pie.
dT/dt = d/dt(2 pi sqrt(L/g)) = 2 pi d/dt( L^(1/2)/g^(1/2))
Now use the quotient rule to complete the differentiation on the right. After you get that, you can multiply both sides of your equation by dt to get an equation that involves dT, dL, and dg.
 
ok.. got it already.. maximum percentage is 1.2% ^_^