Differentials find maximum percentage error

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The discussion revolves around calculating the maximum percentage error in the period T of a simple pendulum, using the formula T = 2π(L/g)^(1/2). Participants clarify the use of differentials and the correct application of the chain rule for differentiation. The maximum errors in measurements of length (L) and acceleration due to gravity (g) are noted as 0.5% and 0.1%, respectively. After applying the correct differentiation techniques, it is concluded that the maximum percentage error in T is approximately 1.2%. This highlights the importance of accurate differentiation in error analysis.
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% ^_^
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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