User talk:Ramu.12317

Given
$$\displaystyle \left. f(S,t) \right|_{S=Y^1(t)} = f(Y^1(t),t) $$

Problem
Derive the first and second time derivative of $$f$$.

Also show the similarity between the second time derivative and the Coriolis acceleration expression.

Solution
Using the chain rule,

$$ \displaystyle \frac {df}{dt} =\frac {\partial f}{\partial S}.\frac{d Y^1}{ dt} + \frac {\partial f}{\partial t}. \frac{dt}{ dt}$$

This is the first time derivative.

The second time derivative is calculated as follows:

$$\displaystyle \frac {d^2f}{dt^2} = \frac {d}{dt}\left(\frac {\partial f}{\partial S}.\frac{d Y^1}{ dt} + \frac {\partial f}{\partial t}\right)$$

$$\displaystyle = \left \{\dot Y^1 \left(\frac {\partial^2f}{\partial S^2}.\dot Y^1 + \frac {\partial^2 f}{\partial t \partial S} \right ) + \frac {\partial f}{\partial S}(\ddot Y^1) \right \}+\left \{\frac {\partial^2f}{\partial S \partial t}.\dot Y^1 +\frac {\partial^2f}{\partial t^2}\right\}$$

$$\displaystyle =(\ddot Y^1).\frac {\partial f}{\partial S}+(\dot Y^1)^2\frac {\partial^2f}{\partial S^2}+ 2\dot Y^1.\frac {\partial^2 f}{\partial S \partial t}+\frac {\partial^2f}{\partial t^2}$$

Coriolis Acceleration Derivation
Angular velocity of a vector is defined as $$\displaystyle \dot \vec r = \vec \omega \times \vec r $$

Consider a point in space $$ \displaystyle \vec a = a_x \hat i+ a_y \hat j+a_z\hat k $$

Velocity of the particle as seen in an inertial frame is

$$ \displaystyle \left.\frac {d\vec a}{dt}\right|_I = \frac {d} {dt}(a_x \hat i+ a_y \hat j+a_z\hat k) $$

$$ \displaystyle = \left( \frac {da_x}{dt} \hat i+ \frac {da_y}{dt} \hat j+\frac {da_z}{dt}\hat k\right ) + \left ( a_x \frac{d\hat i}{dt}+a_y \frac{d\hat j}{dt}+a_z \frac{d\hat k}{dt} \right) $$

$$ \displaystyle = \left( \frac {da_x}{dt} \hat i+ \frac {da_y}{dt} \hat j+\frac {da_z}{dt}\hat k\right ) + \left ( a_x\vec \omega \times \hat i+a_y\vec \omega \times \hat j+a_z \vec \omega \times \hat k\right ) $$

$$ \displaystyle = \left. \frac {d\vec a}{dt}\right|_A + \vec \omega \times\vec a = \vec v $$

Where $$ \left. \frac {d\vec a}{dt}\right|_A $$ is the local velocity in the accelerating frame of reference.

Differentiating again,

$$ \displaystyle \left.\frac{d^2\vec a}{dt^2}\right |_I =\frac{d}{dt}(\vec v)+ \vec \omega \times \vec v $$

$$ \displaystyle =\frac{d}{dt} \bigg(\left.\frac{d\vec a}{dt}\right |_A +\vec \omega \times \vec a \bigg)+ \vec \omega \times\bigg(\left.\frac{d\vec a}{dt}\right |_A +\vec \omega\times \vec a\bigg) $$

$$ \displaystyle = \left.\frac{d^2\vec a}{dt^2}\right |_A +\vec \omega \times \frac{d\vec a}{dt}+ \frac{d\vec \omega}{dt}\times \vec a+ \vec \omega \times\left.\frac{d\vec a}{dt}\right |_A +\vec \omega (\vec \omega \times \vec a) $$

$$ \displaystyle = \left.\frac{d^2\vec a}{dt^2}\right |_A +2\vec \omega \times \frac{d\vec a}{dt}+ \frac{d\vec \omega}{dt}\times \vec a +\vec \omega (\vec \omega \times \vec a) $$

The above equation is the expression showing the coriolis and centrifugal accelerations.

The similarity between the second time derivative and the above expression are listed below

$$ \displaystyle 2\dot Y \frac{\partial f}{\partial S \partial t}\cong 2 \vec \omega \times \frac{d\vec a}{dt} $$

$$ \displaystyle \frac{\partial^2 f}{\partial t^2}\cong \left.\frac{d^2\vec a}{dt^2}\right|_A $$

$$ \displaystyle \ddot Y^1\frac{\partial f}{\partial S}\cong \frac{d\vec \omega}{dt}\times \vec a $$

$$ \displaystyle (\dot Y^1)^2\frac{\partial^2 f}{\partial S^2}\cong \vec \omega\times (\vec \omega\times \vec a ) $$