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Report 1 Problem 1
EGM 6321 Principles of Engineering Analysis 1

Report 1 – Problem 1

=General Setting=

$$\displaystyle \left.f(S,t)\right|_{S=Y^1(t)}=f(Y^1(t),t)$$$$\displaystyle (Eq. 1) $$ 1st Total Time Derivative

$$\displaystyle \frac{d}{dt}[f(Y^1(t),t)]$$

$$\displaystyle \frac{\partial f(Y^1(t),t)}{\partial S}\cdot\frac{dY^1(t)}{dt}+\frac{\partial f(Y^1(t),t)}{\partial t}$$

To shorten the length of the following equations:

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

2nd Total Time Derivative

$$\displaystyle \frac{d^2}{dt^2}[f(Y^1(t),t)]$$

$$\displaystyle \frac{d}{dt}[\frac{\partial f(Y^1(t),t)}{\partial S}\cdot\frac{dY^1(t)}{dt}+\frac{\partial f(Y^1(t),t)}{\partial t}]$$

$$\displaystyle \frac{d}{dt}[\frac{\partial f(Y^1(t),t)}{\partial S}\cdot\frac{dY^1(t)}{dt}]$$

Use the product rule to take the time derivative of this first term:

$$\displaystyle \frac{d}{dt}[f(t)g(t)]= \frac{d f(t)}{dt}g(t)+f(t)\frac{d g(t)}{dt}$$

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

After distributing the terms inside the parentheses and combining like terms we get:

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

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MaintenanceBot (discuss • contribs) 03:59, 11 November 2013 (UTC)