User:EGM6321.f10.team6.cook/hw1/chris

= Problem 1 - Derivation of General Equations of Motion for a High Speed Train=

Problem Statement
The general setting for the motion of a high speed train, whether it be a Maglev or wheel on rail, can be given as some function (f) which is a function of a general function of space (s) and time (t).


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f(s,t) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)


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We can also say that the function of space (s) is really a function of time, which we shall call Y1(t)


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s(Y_{1_{(t)}}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)


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So in general we have the following function:


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f(s_{{(Y_{1_{(t)}}}),t}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)


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Based upon the general function above the first and second total derivatives with respect to time may be found. These derivatives are given as:


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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. 4)


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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_{(t)}} + f_{ss} \left(Y_{1_{(t)}},t\right) (\dot Y_{1_{(t)}})^2 + 2 f_{st} \left(Y_{1_{(t)}},t\right) \dot Y_{1_{(t)}} + f_{tt} \left(Y_{1_{(t)}},t\right)
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 5)


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Required
Derive equations 4 and 5 beginning with the function given in equation 1.

First Derivative WRT Time
We will begin with the first time derivative shown in equation 4.


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\frac{d}{dt} f \left(s,t\right)
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 6)


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Because f = f(s,t) and s=s(Y1 (t), t) we must apply the chain rule. The reader will likely remember that when applying the chain rule to a function that the result is the sum of the chain rule when applied to each component that depends on a given variable. That is to say that since f not only depends on s and t, but s depends on Y1 and t that the chain rule becomes the additive sum of the derivative of all the terms that are dependent upon t.


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\frac{d}{dt} f \left(s,t\right) = \underbrace {\frac{\partial f}{\partial s} \frac{\partial s}{\partial t}}_{1} + \underbrace {\frac{\partial f}{\partial t} \frac{\partial t}{\partial t}}_{2}
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 7)


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The component of the partial derivative that is a result of the dependency upon t for both f and s is shown by 1 and the component that is a result of the dependency upon t for f is shown by 2.

Since the partial derivative of t with respect to itself is 1 equation 7 may be rewritten as follows:


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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} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)


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Additionally, since we have said that s is really just Y1 we can re-write equation 8 in the following form, which matches the form given in the problem statement as equation 4:


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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. 9)


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Second Derivative WRT Time
We will now use the first derivative with respect to time to find the second derivative with respect to time. First let us define a new function F to be the first time derivative of f(s,t).



F = \frac{df(s,t)}{dt} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 10)


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Using this new function F we will again apply the chain rule:


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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}
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 11)


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Now we will substitute the first time derivative of f into equation 11:


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\frac{d^2}{dt^2} f\left(s,t\right) =\frac{\partial }{\partial s} \left(\frac{\partial f}{\partial s} \frac{\partial s}{\partial t} + \frac{\partial f}{\partial t}\right) \frac{\partial s}{\partial t} +  \frac{\partial }{\partial t} \left(\frac{\partial f}{\partial s} \frac{\partial s}{\partial t} +  \frac{\partial f}{\partial t}\right)
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 12)


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To make the derivative simpler, we will break it apart into two components and then recombine. First we have:


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\frac{\partial }{\partial s} \left(\frac{\partial f}{\partial s} \frac{\partial s}{\partial t} + \frac{\partial f}{\partial t}\right) \frac{\partial s}{\partial t}
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 13)


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Recall that the product rule has the following form:



d\left(f g\right) = gdf +fdg
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 14)


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By applying the product rule to equation 13 we obtain the following:


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\frac{\partial }{\partial s} \left(\frac{\partial f}{\partial s} \frac{\partial s}{\partial t} + \frac{\partial f}{\partial t}\right) \frac{\partial s}{\partial t} =  \left(\frac{\partial^2f}{\partial s^2} \frac{\partial s}{\partial t} + \frac{\partial^2s}{\partial s \partial t} \frac{\partial f}{\partial s} + \frac{\partial ^2f}{\partial s^2}\right) \frac{\partial s}{\partial t}
 * $$\displaystyle

$$ || $$\displaystyle (Eq. 15) $$


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\left(\frac{\partial^2f}{\partial s^2} \frac{\partial s}{\partial t} + \frac{\partial^2s}{\partial s \partial t} \frac{\partial f}{\partial s} + \frac{\partial ^2f}{\partial s^2}\right) \frac{\partial s}{\partial t}=\frac{\partial^2f}{\partial s^2}\left(\frac{\partial s}{\partial t}\right)^2+\left(\frac{\partial^2s}{\partial s \partial t} \frac{\partial f}{\partial s}\right) \frac{\partial s}{\partial t}+\frac{\partial^2 f}{\partial s^2}\frac{\partial s}{\partial t}
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 16)


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Since the partial derivative of s with respect to itself the above becomes zero we are left with:


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\left(\frac{\partial^2f}{\partial s^2} \frac{\partial s}{\partial t} + \frac{\partial^2s}{\partial s \partial t} \frac{\partial f}{\partial s} + \frac{\partial ^2f}{\partial s^2}\right) \frac{\partial s}{\partial t}=\left(\frac{\partial^2 f}{\partial s^2}\right)\left(\frac{\partial s}{\partial t}\right)^2+\left(\frac{\partial^2f}{\partial s \partial t}\right)\left(\frac{\partial s}{\partial t}\right)
 * $$\displaystyle

$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 17)


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Now we will repeat the above procedure for the second piece:


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\frac{\partial }{\partial t} \left(\frac{\partial f}{\partial s}\frac{\partial s}{\partial t} + \frac{\partial f}{\partial t}\right)=\frac{\partial^2f}{\partial s \partial t}\frac{\partial s}{\partial t} + \frac{\partial^2s}{\partial t^2}\frac{\partial f}{\partial s} + \frac{\partial^2f}{\partial t^s}
 * $$\displaystyle

$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 18)


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Now combining equations 17 and 18 we obtain the following:


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\frac{d^2f}{dt^2}=\frac{\partial^2f}{\partial s^2}\left(\frac{\partial s}{\partial t}\right)^2+2\frac{\partial^2f}{\partial s \partial t}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial s} \frac{\partial^2s}{\partial t^2}+\frac{\partial^2f}{\partial t^2}
 * $$\displaystyle

$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 19)


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Now substituting Y1 into equation 19 we obtain the following form, which is exactly the same as equation 5:


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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_{(t)}} + f_{ss} \left(Y_{1_{(t)}},t\right) (\dot Y_{1_{(t)}})^2 + 2 f_{st} \left(Y_{1_{(t)}},t\right) \dot Y_{1_{(t)}} + f_{tt} \left(Y_{1_{(t)}},t\right)
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 20)


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Discussion
The general derivation of this problem is relatively easy, however, familiarity with the chain rule and how it applies to functions of functions is probably the location where most people will have difficulty. A relatively clear and concise explanation of the chain rule may be found at Paul’s Online Math Notes.