# User:Egm6321.f11.team3.chang

## Problem1. Second total time derivative

### Given

From lecture note 2-4, first total time derivative is given by,

${\displaystyle \displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}f(Y^{1}(t),t)={\frac {\partial f(Y^{1}(t),t)}{\partial S}}{\dot {Y^{1}}}+{\frac {\partial f(Y^{1}(t),t)}{\partial t}}}$

with

${\displaystyle \displaystyle {\dot {Y^{1}}}:={\frac {\mathrm {d} Y^{1}(t)}{\mathrm {d} t}}}$

### Find

Show that

${\displaystyle \displaystyle {\frac {\mathrm {d^{2}} f}{\mathrm {d} t^{2}}}=f_{,S}(Y^{1},t){\ddot {Y}}^{1}+f_{,SS}(Y^{1},t)({\dot {Y}}^{1})^{2}+2f_{,St}(Y^{1},t){\dot {Y}}^{1}+f_{,tt}(Y^{1},t)}$

with

${\displaystyle \displaystyle f_{,S}(Y^{1},t):={\frac {\partial f(Y^{1},t)}{\partial S}}}$
${\displaystyle \displaystyle f_{,St}(Y^{1},t):={\frac {\partial ^{2}f(Y^{1},t)}{\partial S\partial t}}}$

Similarly for the other partial derivatives

### Solution

Let

${\displaystyle \displaystyle {\frac {\mathrm {d} ^{2}f}{\mathrm {d} t^{2}}}={\frac {\mathrm {d} }{\mathrm {d} t}}({\frac {\mathrm {d} }{\mathrm {d} t}}f(Y^{1}(t),t))={\frac {\mathrm {d} }{\mathrm {d} t}}({\frac {\partial f(Y^{1}(t),t)}{\partial s}}{\dot {Y}}^{1}+{\frac {\partial f(Y^{1}(t),t)}{\partial t}})}$
${\displaystyle \displaystyle ={\frac {\partial f(Y^{1}(t),t)}{\partial S}}{\frac {\mathrm {d} ^{2}Y^{1}(t)}{\mathrm {d} t^{2}}}+{\frac {\partial }{\partial s}}({\frac {\partial f(Y^{1}(t),t)}{\partial s}}{\frac {\mathrm {d} Y^{1}(t)}{\mathrm {d} t}}+{\frac {\partial f(Y^{1}(t),t)}{\partial t}}){\frac {\mathrm {d} Y^{1}(t)}{\mathrm {d} t}}}$
${\displaystyle \displaystyle +{\frac {\partial }{\partial t}}({\frac {\partial f(Y^{1}(t),t)}{\partial S}}{\frac {\mathrm {d} Y^{1}(t)}{\mathrm {d} t}}+{\frac {\partial f(Y^{1}(t),t)}{\partial t}})}$
${\displaystyle \displaystyle ={\frac {\partial ^{2}f(Y^{1}(t),t)}{\partial S^{2}}}({\frac {\mathrm {d} Y^{1}(t)}{\mathrm {d} t}})^{2}+2{\frac {\partial ^{2}f(Y^{1}(t),t)}{\partial S\partial t}}{\frac {\mathrm {d} Y^{1}(t)}{\mathrm {d} t}}+{\frac {\partial ^{2}f(Y^{1}(t),t)}{\partial t^{2}}}}$
${\displaystyle \displaystyle +{\frac {\mathrm {d} ^{2}Y^{1}(t)}{\mathrm {d} t^{2}}}{\frac {\partial f(Y^{1}(t),t)}{\partial S}}}$
${\displaystyle \displaystyle =f_{,S}(Y^{1},t){\ddot {Y}}^{1}+f_{,SS}(Y^{1},t)({\dot {Y}}^{1})^{2}+2f_{,St}(Y^{1},t){\dot {Y}}^{1}+f_{,tt}(Y^{1},t)}$

## Problem 2: Deriving 1st and 2nd Total Time Derivatives; Demonstrate Similarity to Coriolis Force Derivation

### Given

The equation of motion (EOM) for the maglev train is modeled by:

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

From lecture notes on page 2-4, the 1st total time derivative is given by:

(Eq.2A)

${\displaystyle \displaystyle \displaystyle {\frac {d}{dt}}f(Y^{1}(t),t)={\frac {\partial f(Y^{1}(t),t)}{\partial S}}{\dot {Y}}^{1}+{\frac {\partial f(Y^{1}(t),t)}{\partial t}}}$

and the 2nd total time derivative, on page 2-5, is given by:

(Eq.2B)

${\displaystyle \displaystyle \displaystyle {\frac {d^{2}f}{dt^{2}}}=f_{,S}{Y^{1},t}{\ddot {Y}}^{1}+f_{,SS}(Y^{1},t)({\dot {Y}}^{1})^{2}+2f_{,St}(Y^{1},t){\dot {Y}}^{1}+f_{,tt}(Y^{1},t)}$

### Problem

--Derive the 1st total time derivative.
--Derive the 2nd total time derivative.
--Show the similarity with the derivation of the Coriolis force.

### Solution, First Total Time Derivative

Applying the chain rule to

${\displaystyle \displaystyle f(S,t)}$

Yields:

${\displaystyle \displaystyle {\frac {df(S,t)}{dt}}={\frac {\partial f}{\partial S}}{\frac {\partial S}{\partial t}}+{\frac {\partial f}{\partial t}}{\frac {\partial t}{\partial t}}={\frac {\partial f(S,t)}{\partial S}}{\frac {\partial S}{\partial t}}+{\frac {\partial f(S,t)}{\partial t}}{\frac {\partial t}{\partial t}}}$

Since:

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

Then the equation can be rewritten as:

${\displaystyle \displaystyle {\frac {df(Y^{1}(t),t)}{dt}}={\frac {\partial f(Y^{1}(t),t)}{\partial S}}\underbrace {\frac {\partial S}{\partial t}} _{{\dot {Y}}^{1}}+{\frac {\partial f(Y^{1}(t),t)}{\partial t}}\underbrace {\frac {\partial t}{\partial t}} _{1}}$

${\displaystyle \displaystyle {\frac {d}{dt}}f(Y^{1}(t),t)={\frac {\partial f(Y^{1}(t),t)}{\partial S}}{\dot {Y}}^{1}+{\frac {\partial f(Y^{1}(t),t)}{\partial t}}}$

## Problem4.

### Given

ξ = (ξ123)
X(ξ) = X11)X22)X33)

 ${\displaystyle \displaystyle {\frac {1}{g_{i}(\xi _{i})}}{\frac {d}{d\xi _{i}}}\left[g_{i}(\xi _{i}){\frac {dX_{i}(\xi _{i})}{d\xi _{i}}}\right]+f_{i}(\xi _{i})X_{i}(\xi _{i})=0}$ (5.1)

## Problem5.

### Find

Show that:

${\displaystyle \displaystyle {\frac {1}{g_{i}(\xi _{i})}}{\frac {d}{d\xi _{i}}}\left[g_{i}(\xi _{i}){\frac {dX_{i}(\xi _{i})}{d\xi _{i}}}\right]+f_{i}(\xi _{i})X_{i}(\xi _{i})=0}$

Becomes:

${\displaystyle \displaystyle y''+\underbrace {\frac {g'(x)}{g(x)}} _{a_{a}(x)}y'+a_{0}(x)\,y=0}$

## Problem6.

### Find

Show ${\displaystyle \displaystyle c_{3}(Y^{1},t){\ddot {Y}}^{1}}$\ is nonlinear with respect to ${\displaystyle \displaystyle Y^{1}}$.

## Problem7.

### Given

${\displaystyle \displaystyle L_{2}(\cdot )={\frac {d^{2}(\cdot )}{dx^{2}}}+a_{1}(x){\frac {d(\cdot )}{dx}}+a_{0}(x)(\cdot )}$

### Find

Show that :${\displaystyle \displaystyle L_{2}(\cdot )}$ is linear

### Solution

Let u(x) and v(x) be any 2 functions of x Let :${\displaystyle \displaystyle \alpha \,and\,\beta }$ be and 2 real numbers We write :${\displaystyle \displaystyle \forall u,v:\mathbb {R} \rightarrow \mathbb {R} \,x\mapsto u(x)\,\,x\mapsto v(x)\,\,\forall \alpha ,\beta \in \mathbb {R} }$

${\displaystyle \displaystyle {\frac {d^{2}(\alpha u+\beta v)}{dx^{2}}}+a_{1}(x){\frac {d(\alpha u+\beta v)}{dx}}+a_{0}(x)(\alpha u+\beta v)}$
${\displaystyle \displaystyle =\alpha {\ddot {u}}+\beta {\ddot {v}}+a_{1}(x)(\alpha {\dot {u}}+\beta {\dot {v}})+a_{0}(x)(\alpha u+\beta v)}$
${\displaystyle \displaystyle {\frac {d^{2}(\alpha u)}{dx^{2}}}+a_{1}(x){\frac {d(\alpha u)}{dx}}+a_{0}(x)(\alpha u)}$
${\displaystyle \displaystyle =\alpha {\ddot {u}}++a_{1}(x)\alpha {\dot {u}}++a_{0}(x)\alpha u}$
${\displaystyle \displaystyle {\frac {d^{2}(\beta v)}{dx^{2}}}+a_{1}(x){\frac {d(\beta v)}{dx}}+a_{0}(x)(\beta v)}$
${\displaystyle \displaystyle =\beta {\ddot {v}}+a_{1}(x)(\beta {\dot {v}})+a_{0}(x)(\beta v)}$
${\displaystyle \displaystyle \Rightarrow L_{2}(\alpha u+\beta v)=L_{2}(\alpha u)+L_{2}(\beta v)}$

So it is linear.

## Problem8.

### Given

${\displaystyle \displaystyle y(x)=y^{_{H}}(x)+y^{_{P}}(x)=c_{1}y_{H}^{1}(x)+c_{2}y_{H}^{2}(x)+y_{P}(x)}$

### Find

Find the integration constants in terms of the initial conditions :${\displaystyle \displaystyle y(a)=\alpha \,\,\,y{}'(a)=\beta }$

### Solution

${\displaystyle \displaystyle y(a)=c_{1}y_{H}^{1}(a)+c_{2}y_{H}^{2}(a)+y_{P}(a)=\alpha }$
${\displaystyle \displaystyle y{}'(a)=c_{1}{\dot {y}}_{H}^{1}(a)+c_{2}{\dot {y}}_{H}^{2}(a)+{\dot {y}}_{P}(a)=\beta }$

${\displaystyle \displaystyle \Rightarrow c_{1}={\frac {\alpha {\dot {y}}_{H}^{2}(a)-\beta y_{H}^{2}(a)-y_{P}(a){\dot {y}}_{H}^{2}(a)+{\dot {y}}_{P}y_{H}^{2}(a)}{y_{H}^{1}(a){\dot {y}}_{H}^{2}(a)-{\dot {y}}_{H}^{1}(a)y_{H}^{2}(a)}}}$

${\displaystyle \displaystyle c_{2}={\frac {\alpha {\dot {y}}_{H}^{1}(a)-\beta y_{H}^{1}(a)-y_{P}(a){\dot {y}}_{H}^{1}(a)+{\dot {y}}_{P}y_{H}^{1}(a)}{y_{H}^{2}(a){\dot {y}}_{H}^{1}(a)-{\dot {y}}_{H}^{2}(a)y_{H}^{1}(a)}}}$

# Contributing Members & Referenced Lecture (Placeholder)

 Team Contribution Table Problem Number Assigned To Solved By Proofread By 1.1 YungSheng Chang YungSheng Chang 1.2 Joel Brawner 1.3 Gopichandran Prashant 1.4 Gopichandran Prashant YungSheng Chang 1.5 Anderson Robert and Sahin Ismail Hakki 1.6 Anderson Robert 1.7 Chao Yang Chao Yang 1.8 Chao Yang Chao Yang