Trajectory of a body falling at the Equator during a period of 10 seconds.
If you drop a pebble down a mine-shaft, it will not fall vertically, but will be deflected slightly to the East by the Coriolis force, an effect of the Earth’s rotation. We can solve the equations to calculate the amount of deflection; for a ten-second drop, the pebble falls about 500 metres (air resistance is neglected) and is deflected eastward by about 25 cm. The figure on the left shows the trajectory in the vertical xz-plane (scales are not the same).
We derive the equations after making some simplifying assumptions. We assume the mine-shaft is at the Equator; we assume the meridional or north-south motion is zero; we neglect variations in the gravitational force; we neglect the sphericity of the Earth; we neglect air resistance. We can still get accurate estimates provided the elapsed time is short. However, carrying the analysis to the extreme, we obtain results that are completely unrealistic. The equations predict that the pebble will reach a minimum altitude and then rise up again to its initial height a great distance east of its initial position. Then this up-and-down motion will recur indefinitely.
to the Weierstrass CND function.
that is differentiable at a point 
is continuous there, and if differentiable on an interval ![{[a, b]}](https://s0.wp.com/latex.php?latex=%7B%5Ba%2C+b%5D%7D&bg=ffffff&fg=000000&s=0&c=20201002)
, is continuous on that interval. However, the converse is not necessarily true: the continuity of a function at a point or on an interval does not guarantee that it is differentiable at the point or on the interval.
ThatsMaths 