## University Calculus: Early Transcendentals (3rd Edition)

a) $dT=-\pi\frac{L^{\frac{1}{2}}}{g^{\frac{3}{2}}}dg$ b) As T decreases, the clock pendulum will move more frequently $dg=-0.977 \frac{cm}{s^2}$ $g=979\frac{cm}{s^2}$
Given that $T=2\pi(\sqrt{\frac{L}{g}})$ on differentiating the above equation: a) taking L as constant ${dT}=2\pi\frac{1}{2}\frac{1}{\sqrt{\frac{L}{g}}}\frac{-L}{g^2}dg$ $dT=-\pi\frac{L^{\frac{1}{2}}}{g^{\frac{3}{2}}}dg$ b) If g increases then $dg\gt0$ and $dT\lt0$ As T decreases, the clock pendulum will move more frequently. c) L=100,g=980 and dT=0.001 then $dT=-\pi\frac{L^{\frac{1}{2}}}{g^{\frac{3}{2}}}dg$ $0.001=-\pi\frac{100^{\frac{1}{2}}}{980^{\frac{3}{2}}}dg$ $dg=-0.977 \frac{cm}{s^2}$ $g=980-0.977=979\frac{cm}{s^2}$ so the final answer is a) $dT=-\pi\frac{L^{\frac{1}{2}}}{g^{\frac{3}{2}}}dg$ b) As T decreases, the clock pendulum will move more frequently $dg=-0.977 \frac{cm}{s^2}$ $g=979\frac{cm}{s^2}$