I need help figuring this error percentage homework problem. Question: The period of a simple pendulum is given by $T=2\pi\sqrt{\frac{L}{g}}$ where $L$ is the length of the pendulum in feet, $g$ is the constant of acceleration due to gravity, and $T$ is measured in seconds. Suppose that the length of a pendulum was measured with a maximum error of $\frac{1}{4}$%.
What will be the maximum percentage error in measuring its period?
So this is what I have so far. $$Δ{t} ≈ dT=T'dT$$ Error in period $$\frac{ΔT}{t} ≈ \frac{dT}{T}$$$$dL=\frac{1}{4}$$
$$T'=2\pi(\frac{\sqrt{g}\frac{1}{2}(L)^{-\frac{1}{2}}-\sqrt{L}\frac{1}{2}(g)^{-\frac{1}{2}}}{g})$$
Annnnnnd that's all. I have no idea how to approach this problem and I don't even know if what I have so far is right. A detailed step by step explanation would be greatly appreciated, thanks in advance. 
 A: The following computes the first order relative error. In this case one can compute the maximum relative error exactly (since $\sqrt{\cdot}$ has nice properties), but ones misses out on seeing the functional form of the sensitivities.
Think of $T$ as a function of $L$, that is, $T(L) = 2 \pi \sqrt{L \over G}$.
Then a small change in $L$, say $\lambda$, will result in a change in $T$
of $T(L+\lambda) -T(L) \approx T'(L)\lambda$.
Hence the relative error is $\approx {T'(L) \over T(L)} \lambda$.
In this case you are given $\lambda \approx {1 \over 400} L$, so the relative error becomes
$\approx {T'(L) \over T(L)} {1 \over 400} L$.
We have $T'(L) = \pi \sqrt{1 \over LG }$, so substituting gives
the relative error $\approx \pi \sqrt{1 \over LG } {1 \over 2 \pi} \sqrt{G \over L} {1 \over 400} L = {1 \over 800}$.
A quick check shows that the quantity is 'unit less', as it should be, and
as a percentage, this is ${1 \over 8}$% (as would be expected as the power of $L$ in $T(L)$ is ${1 \over 2}$).
A: Brute force:
$$
T_1 = 2\pi \sqrt{\frac{L}{g}}\sqrt{1\pm 0.0025} = T_0 \left(1\pm \frac{1}{2}.0025 +\frac{\frac{1}{2}(\frac{1}{2}-1)}{2}0.0025^2+\cdot\cdot\cdot\right)
$$
So the error is $\frac{1}{2}0.0025$
A: First you have a measure of the real period $T$ with error, call it $T'$. And  a measure of the length $L$ with error call it $L'$
$$T'=2\pi\sqrt{\frac{L'}{g}}$$
Assume that your $L'$ have an error of max $\frac{1}{4}\%$. Then
$$L'=L(1+\epsilon)$$ 
Where $\epsilon$ is the error with $\left|\epsilon\right| \leq \frac{1}{4}\%$
Replacing:
$$T'=2\pi\sqrt{\frac{L'}{g}}=2\pi\sqrt{\frac{L(1+\epsilon)}{g}}=2\pi\sqrt{\frac{L}{g}}\sqrt{1+\epsilon}=T\sqrt{1+\epsilon}$$
Therefore:
$$\left|\frac{T}{T'}-1\right|=\left|\sqrt{1+\epsilon}-1\right|\leq\left|\sqrt{1-\frac{1}{400}}-1 \right|= 0.00125078222809105=0.12507\%$$
A: Here's how an experimental physicist would compute this error.  Assuming that all the errors in a given measurement are linearly independent (that is to say, one error does not affect another error), then for any given function$$F(x_1,x_2,...,x_N)$$ the standard error $\sigma_F$ is given by 
$$
\sigma_F^2=\sum_{i=1}^N\bigg(\frac{\partial F}{\partial x_i}\bigg)^2\sigma_{x_i}^2
$$
Since you only have one source of error, for $T(L,g)$ you have
$$
\sigma_T=\frac{\partial T}{\partial L}\sigma_L=2\pi\bigg(\frac{1}{2}\bigg)\bigg(\frac{L}{g}\bigg)^{-\frac{1}{2}}\bigg(\frac{1}{g}\bigg)(.0025L)=.0025\pi\sqrt{\frac{L}{g}}=.00125T$$
This result informs you that your most accurate results will be obtained when the period is small.
