# Related rates of an pendulum workings [closed]

I have an question that I would like to see the workings of. Pendulum decreasing at rate of 0.100cm/sec. What is the time rate of change of the Period T ( in sec) of the pendulum when L = 16.0 cm , if equation related to period of length is

T = pie * sqrt( L / 245)

## closed as off-topic by Leucippus, mlc, user99914, Semiclassical, The Dead LegendMay 12 '17 at 5:24

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$$T=2\pi \sqrt{\frac{L}{g}}$$ $$\frac{dT}{dt}= \frac{\pi}{2\sqrt{\frac{L}{g}}}.\frac{dL}{dt}$$
You have $\frac{dL}{dt}=-0.001 cm/sec$ , $g=9.8 m/s^2$ and $L=16cm$
• Given your question earlier, could you start using \cdot instead of . when you mean multiplication? – epimorphic May 12 '17 at 3:19