Differentiation

[Tien1992]

The period of a pendulum is given by

T=2pi(sqrroot(L/g))
where L is the length of the pendulum in feet, g is the acceleration due to gravity, and T is the time in seconds. Suppose that the pendulum's length has increased by .5%; what is the approx. percent change in the period?

I've tried deriving the equation while plugging in 9.8 for g. The answer was kinda right- .25%. I got 1/400L after that. The fraction was right, but I still got that L there. Can someone show me how to do it correctly?

 

[galactus]

The percentage error in T is approximately half the percentage error in L.

Let's show in general. Remember, .5% is .005.

\(\displaystyle T=2{\pi}\sqrt{\frac{L}{g}}\)

\(\displaystyle dT=\frac{2\pi}{\sqrt{g}}\cdot\frac{1}{2\sqrt{L}}dL\)

\(\displaystyle =\frac{\pi}{\sqrt{gL}}dL\)

\(\displaystyle \frac{dP}{P}=\frac{1}{2}\frac{dL}{L}\).

So, the relative error in \(\displaystyle P\approx \frac{1}{2}\) the realtive error in L. Thus, the percentage error in \(\displaystyle P\approx \frac{1}{2}\) the percentage error in L.

 

[Tien1992]

How did you post those pictures? I can't find anything on the forum for that. o_o

 

[stapel]

It's LaTeX. You can "quote" another user's post, or (in some set-ups) hover your mouse over the formatting, to see the coding. You might also want to review an article or two on LaTeX formatting. :wink: