User:Egm6321.f12.team6.Bang/report1

=Report 1=

Problem Statement

 * Derive
 * a) $$\displaystyle\;$$$$\displaystyle\frac{d}{dt}f(Y^1(t),t)=\frac{\partial f(Y^1(t),t)}{\partial S}\dot Y^1+\frac{\partial f(Y^1(t),t)}{\partial t}$$
 * and
 * b) $$\displaystyle\;$$ $$\displaystyle\frac{d^2f}{dt^2}=f_{,S}(Y^1(t),t)\ddot Y^1+f_{,SS}(Y^1(t),t)(\dot Y^1)^2+2f_{,St}(Y^1(t),t)\dot Y^1+f_{,tt}(Y^1(t),t)$$,
 * and
 * c) $$\displaystyle\;$$ show the similarity with the derivation of the Coriolis force.

Given

 * $$\displaystyle f(S,t)|_{S=Y^{^{1}}(t)}=f(Y^{1}(t),t)$$

Nomenclature

 * $$\displaystyle \dot{Y}^1:=\frac{dY^1(t)}{dt}$$
 * $$\displaystyle f,_{S}(Y^1,t):=\frac{\partial f(Y^1(t),t)}{\partial S}$$
 * $$\displaystyle f,_{S\,t}(Y^1(t),t):=\frac{\partial ^2f(Y^1(t),t)}{\partial S\partial t}$$

Part a)

 * By using Chain Rule,
 * $$\displaystyle \frac{d}{d t}f(S,t)=\frac{\partial f}{\partial S}\frac{d S}{d t}+\frac{\partial f}{\partial t}\cancelto{1}\frac{d t}{d t}=\frac{\partial f}{\partial S}\frac{d S}{d t}+\frac{\partial f}{\partial t}$$.
 * Also,
 * $$\displaystyle S=Y^{^{1}}(t)$$
 * is a given condition. This implies that
 * $$\displaystyle\frac{d S}{d t}=\frac{dY^1(t)}{dt}=:\dot{Y^1}$$.
 * Therefore, when the equation is rewritten in terms of $$\displaystyle f(Y^{^{1}}(t),t)$$, then the equation for $$\displaystyle\frac{d}{dt}f(Y^1(t),t)$$ becomes following:
 * {| style="width:100%" border="0"|-

$$\displaystyle\;$$ $$\displaystyle\frac{d}{dt}f(Y^1(t),t)=\frac{\partial f(Y^1(t),t)}{\partial S}\dot Y^1+\frac{\partial f(Y^1(t),t)}{\partial t}$$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Part b)

 * By using Part a) derivation
 * $$\displaystyle \frac{d }{d t}f(S,t)=\frac{\partial f}{\partial S}\frac{d S}{d t}+\frac{\partial f}{\partial t}$$,
 * The second derivate of $$\displaystyle f(Y^1(t),t)$$, can be rewritten in following:
 * $$\displaystyle \frac{d^2 f}{d t^2}=\frac{d}{d t} \left [ \frac{d}{d t} f(S,t) \right ]=\frac{d}{d t} \left [ \frac{\partial f}{\partial S}\frac{d S}{d t}+\frac{\partial f}{\partial t}\right ]$$
 * Then, the expression above can expanded by using Chain and Product Rule.
 * $$\displaystyle \frac{d^2 f}{d t^2} = \frac{\partial f}{\partial S}\frac{d}{d t}\left ( \frac{d S}{d t} \right )+\frac{d }{d t} \left ( \frac{\partial f}{\partial S} \right )\frac{d S}{d t}+\frac{d}{d t}\left ( \frac{\partial f}{\partial t} \right )$$
 * $$\displaystyle = \displaystyle \frac{\partial f}{\partial S}\frac{d^2 S}{d t^2} + \left ( \frac{\partial^2 f}{\partial S^2} \frac{d S}{d t} + \frac{\partial^2 f}{\partial S \, \partial t} \cancelto{1}\frac{d t}{d t} \right )\frac{d S}{d t} + \left ( \frac{\partial^2 f}{\partial t \, \partial S} \frac{d S}{d t} + \frac{\partial^2 f}{\partial t^2} \cancelto{1}\frac{d t}{d t} \right ) \cancelto{1}\frac{d t}{d t}$$
 * $$\displaystyle = \frac{\partial f}{\partial S}\frac{d^2 S}{d t^2} + \frac{\partial^2 f}{\partial S^2} \left ( \frac{d S}{d t} \right )^2 + \frac{\partial^2 f}{\partial S \, \partial t} \frac{d S}{d t} + \frac{\partial^2 f}{\partial t \, \partial S} \frac{d S}{d t} + \frac{\partial^2 f}{\partial t^2}$$
 * Also, in this case,
 * $$\displaystyle\frac{\partial^2 f}{\partial S \, \partial t} \frac{d S}{d t} = \frac{\partial^2 f}{\partial t \, \partial S} \frac{d S}{d t}$$
 * Adding both terms will give
 * $$ \frac{d^2 f}{d t^2} = \frac{\partial f}{\partial S}\frac{d^2 S}{d t^2} + \frac{\partial^2 f}{\partial S^2} \left ( \frac{d S}{d t} \right )^2 + 2\frac{\partial^2 f}{\partial S \, \partial t} \frac{d S}{d t} + \frac{\partial^2 f}{\partial t^2} $$.
 * By using $$\displaystyle f(Y^{^{1}}(t),t)$$, terms and different nonmenclature technique, then equation becomes following:
 * {| style="width:100%" border="0"|-

$$\displaystyle\;$$ $$\displaystyle\frac{d^2f}{dt^2} = f_{,S}(Y^1(t),t)\ddot Y^1 + f_{,SS}(Y^1(t),t)(\dot Y^1)^2 + 2f_{,St}(Y^1(t),t)\dot Y^1 + f_{,tt}(Y^1(t),t) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

====Part c) ====
 * Consider a two frames of reference centered at O, a fixed frame OXY and frame Oxy which rotates about the fixed axis OA; let $$\displaystyle \Omega$$ denote the angular velocity of the frame Oxy at a give instant.
 * $$ \displaystyle\boldsymbol{r}$$ can be expressed in vector form:
 * $$ \displaystyle\boldsymbol{r} = r_{x}\boldsymbol{i} + r_{y}\boldsymbol{j} $$
 * Differentiating with repect to t and considering the unit vectors as fixed, rate of change of $$\displaystyle\boldsymbol{r}$$ with respect to the rotating frame Oxy can be obtained.
 * $$ \displaystyle \left (\boldsymbol{\dot{r}}\right )_{Oxy} = \dot{r}_{x}\boldsymbol{i} + \dot{r}_{y}\boldsymbol{j} $$
 * To obtain the rate of change of $$\displaystyle\boldsymbol{r}$$ with respect to the fixed frame OXY, we must consider the unit vectors as variable when differentiating the above equation. Therefore the equation becomes
 * $$ \displaystyle \left (\boldsymbol{\dot{r}}\right )_{OXY} = \dot{r}_{x}\boldsymbol{i} + \dot{r}_{y}\boldsymbol{j} + r_{x}\frac{d\boldsymbol{i}}{dt} + r_{y}\frac{d\boldsymbol{j}}{dt} $$
 * Following equation can be simplified into
 * $$ \displaystyle \left (\boldsymbol{\dot{r}}\right )_{OXY} = \left (\boldsymbol{\dot{r}}\right)_{Oxy} + \boldsymbol{\Omega \times r} $$
 * or
 * $$ \displaystyle \left (\boldsymbol{\dot{r}}\right )_{OXY} = \boldsymbol{\Omega \times r} + \left (\boldsymbol{\dot{r}}\right)_{Oxy} $$
 * From above equation, rate of change with respect to OXY of the terms will give absolute acceleration.
 * $$ \displaystyle \frac{d}{dt}\left [ \left (\boldsymbol{\dot{r}}\right )_{OXY} \right ] = \left (\boldsymbol{\ddot{r}}\right)_{OXY} = \frac{d}{dt} \left [ \boldsymbol{\Omega \times r}\right ] + \frac{d}{dt}\left [ \left (\boldsymbol{\dot{r}}\right)_{Oxy}\right ] $$
 * $$ = \boldsymbol{\dot{\Omega} \times r} + \boldsymbol{\Omega \times \dot{r}} + \frac{d}{dt}\left [ \left (\boldsymbol{\dot{r}}\right)_{Oxy}\right ] $$
 * Similary with previous part, we can investigate that
 * $$ \displaystyle \frac{d}{dt}\left [ \left (\boldsymbol{\dot{r}}\right )_{Oxy} \right ] = \left (\boldsymbol{\ddot{r}}\right)_{Oxy} + \boldsymbol{\Omega \, \times} \left (\boldsymbol{\dot{r}}\right)_{Oxy} $$
 * Therefore, the equation for coriolis acceleration is
 * {| style="width:100%" border="0"|-

$$ \displaystyle \frac{d^2 \left (\boldsymbol{r}\right )_{OXY}}{dt^2} = \boldsymbol{\dot{\Omega} \times r} + \boldsymbol{\Omega \, \times} \left (\boldsymbol{\Omega \times r}\right ) + 2\boldsymbol{\Omega \, \times} \left (\boldsymbol{\dot{r}}\right)_{Oxy} + \left (\boldsymbol{\ddot{r}}\right)_{Oxy} $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }
 * The table below summarizes the similarity between the equation from Part b) and Part c)
 * {| class="prettytable"


 * colspan="3" | Comparison Table
 * Part b) Expression
 * Part c) Expression
 * 1st Term
 * $$f_{,S}(Y^1(t),t)\ddot Y^1 \,$$
 * $$\boldsymbol{\dot{\Omega} \times r} \,$$
 * 2nd Term
 * $$f_{,SS}(Y^1(t),t)(\dot Y^1)^2 \,$$
 * $$\boldsymbol{\Omega \, \times} \left (\boldsymbol{\Omega \times r}\right ) \,$$
 * 3rd Term
 * $$2f_{,St}(Y^1(t),t)\dot Y^1 \,$$
 * $$2\boldsymbol{\Omega \, \times} \left (\boldsymbol{\dot{r}}\right)_{Oxy} \,$$
 * 4th Term
 * $$\rho (f_{,tt}(Y^1(t),t) \,$$
 * $$\left (\boldsymbol{\ddot{r}}\right)_{Oxy} \,$$
 * }
 * $$2\boldsymbol{\Omega \, \times} \left (\boldsymbol{\dot{r}}\right)_{Oxy} \,$$
 * 4th Term
 * $$\rho (f_{,tt}(Y^1(t),t) \,$$
 * $$\left (\boldsymbol{\ddot{r}}\right)_{Oxy} \,$$
 * }
 * }