User:Egm6341.s10.team3.heejun/EX

= Problem 1 - Maglev First and Second Derivative Derivations =

Given
The equation of motion for a Maglev train can be modeled by the function :

 Nice contribution Maglev train. Egm6321.f09 12:47, 22 September 2009 (UTC)


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f(Y^1 (t),t) $$
 * $$\displaystyle
 * $$\displaystyle


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Find
Derive the first and second time derivatives for the given equation shown by Equation 1 and 2 respectively.


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\frac{d}{dt} f \left(Y^1(t),t\right) = \frac{\partial f}{\partial s} \left(Y^1(t),t\right) \dot Y^1(t) + \frac{\partial f}{\partial t} \left(Y^1(t),t\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)


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\frac{d^2}{dt^2} f\left(Y^1(t),t\right) = f_{,s} \left(Y^1(t),t\right) \ddot Y^1 + f_{,ss} \left(Y^1(t),t\right) (\dot Y^1)^2 + 2 f_{,st} \left(Y^1(t),t\right) \dot Y^1 + f_{,tt} \left(Y^1(t),t\right)
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 2)


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Solution
 Solving for Equation 1

Taking the time derivative:

\frac{d}{dt} f \left(Y^1(t),t\right) = \frac{d}{dt} f \left(s,t\right) $$

For ease, a dummy variable
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s = Y^1(t) $$
 * $$\displaystyle
 * $$\displaystyle
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is used and substuted into the previous equation. The chain rule is now applied:
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\frac{d}{dt} f \left(s,t\right) = \frac{\partial f}{\partial s} \frac{\partial s}{\partial t}+ \frac{\partial f}{\partial t} \underbrace{\frac{\partial t}{\partial t}}_{=1} $$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
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Note that
 * $$\frac{\partial s}{\partial t}=\dot Y^1(t)$$

Equation 3 can be rewriten to the form:


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$$ \frac{d}{dt} f \left(Y^1(t),t\right) = \frac{\partial f}{\partial s} \left(Y^1(t),t\right) \dot Y^1(t) + \frac{\partial f}{\partial t} \left(Y^1(t),t\right) $$
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