User:Egm6321.f12.team5.kim/Report 1

= Report 1 Problems =

Problem Statement
Given a multivariable function,$$ f(S,t) $$, where $$S$$ is a function of $$t$$. Show that its second total derivative, when evaluated at $$S = Y^1(t)$$, is the equation (1) p.1a-5, which is shown below:


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$$\frac {d^2f}{dt^2} = f_{,S}(Y^1,t)\ddot Y^1+ f_{,SS}(Y^1,t)\dot {(Y^1)}^2 + 2f_{,St}(Y^1,t)\dot Y^1+ f_{,tt}(Y^1,t)$$
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Where the notations for the partial derivatives are defined as follows:


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$$f_{,S}(S,t) := \frac {\partial f(S,t)}{\partial S}$$
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$$f_{,S t}(S,t) := \frac {\partial^2 f(S,t)}{\partial S \partial t}$$
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Solution

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 The problem was solved independently without referring to any reports from previous classes.
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The first total derivative of $$f(S,t)$$ can be found by using the Chain Rule for multivariable functions as shown below.


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$$ \displaystyle \frac{df}{dt}=\frac{\partial f(S,t)}{\partial S} \dot S + \frac{\partial f(S,t)}{\partial t} $$ (1)
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where,


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$$\dot S := \frac {dS}{dt}$$
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The second total derivative of $$ f(S(t),t) $$ can be obtained by taking the total derivative of equation (1) by using the Chain Rule.


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$$\frac {d^2f}{dt^2} = \frac {\partial \left(\displaystyle \frac {\partial f(S,t)}{\partial S}\dot S\right)}{\partial S} \dot S + \frac {\partial \left(\displaystyle \frac {\partial f(S,t)}{\partial S}\dot S \right)}{\partial t} + \frac {\partial \left(\displaystyle \frac {\partial f(S,t)}{\partial t} \right)}{\partial S} \dot S + \frac {\partial \left(\displaystyle \frac {\partial f(S,t)}{\partial t} \right)}{\partial t}$$ (2)
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Expanding first two terms using the Product Rule gives,


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$$\frac {d^2f}{dt^2} = \frac {\partial^2 f(S,t)}{\partial S^2} \dot S^2 + \frac {\partial f(S,t)}{\partial S} \frac {\partial \dot S}{\partial S} \dot S + \frac {\partial^2 f(S,t)}{\partial S \partial t} \dot S + \frac {\partial f(S,t)}{\partial S} \ddot S $$
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$$ + \frac {\partial^2 f(S,t)}{\partial t \partial S} \dot S + \frac {\partial^2 f(S,t)}{\partial t^2} $$ (3)
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If the function $$f$$ has continuous second partial derivatives, then


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$$ \frac {\partial^2 f(S,t)}{\partial S \partial t} = \frac {\partial^2 f(S,t)}{\partial t \partial S}$$ (4)
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and since


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$$ \frac {\displaystyle \partial \left(\frac {dS}{dt} \right)}{\partial S} = \frac {\displaystyle d \left(\frac {\partial S}{\partial S} \right)}{dt} = 0$$ (5)
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equation (3) reduces to


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$$\frac {d^2f}{dt^2} = \frac {\partial^2 f(S,t)}{\partial S^2} \dot S^2 + 2\frac {\partial^2 f(S,t)}{\partial S \partial t} \dot S + \frac {\partial f(S,t)}{\partial S} \ddot S + \frac {\partial^2 f(S,t)}{\partial t^2} $$ (6)
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Using the notation for partial derivatives defined in the Problem Statement, equation (6) can be rewritten and rearranged as


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$$\frac {d^2f}{dt^2} = f_{,S}(S,t)\ddot S + f_{,SS}(S,t)\dot S^2 + 2f_{,St}(S,t)\dot S + f_{,tt}(S,t)$$ (7)
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When $$f$$ is evaluated at $$ S = Y^1(t)$$, equation (7) becomes


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$$\frac {d^2f}{dt^2} = f_{,S}(Y^1,t)\ddot Y^1+ f_{,SS}(Y^1,t)\dot {(Y^1)}^2 + 2f_{,St}(Y^1,t)\dot Y^1+ f_{,tt}(Y^1,t)$$ (8)
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which is the same as equation (1) from p.1a-5.

Derivation of (3) p.1a-4
Chain Rule for function with two variables is shown below.


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$$ \frac {d}{dt}f(x,y) = \frac {\partial f(x,y)}{\partial x} \frac {dx}{dt} + \frac {\partial f(x,y)}{\partial y} \frac {dy}{dt} $$ (1)
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Applying the chain rule to the function $$f(S,t)$$ gives


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$$ \frac {d}{dt}f(S,t) = \frac {\partial f(S,t)}{\partial S} \frac {dS}{dt} + \frac {\partial f(S,t)}{\partial t} \cancelto{1}{\frac {dt}{dt}} $$ (2)
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Substituting $$Y^1(t)$$ for $$S$$ gives


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$$ \frac {d}{dt}f(Y^1(t),t) = \frac {\partial f(Y^1(t),t)}{\partial S} \frac {dY^1(t)}{dt} + \frac {\partial f(Y^1(t),t)}{\partial t} $$ (3)
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Using the notation from p.1a-5,


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$$ \dot {Y^1} := \frac {dY^1(t)}{dt} $$
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equation (3) is written as


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$$ \frac {d}{dt}f(Y^1(t),t) = \frac {\partial f(Y^1(t),t)}{\partial S} \dot {Y^1(t)} + \frac {\partial f(Y^1(t),t)}{\partial t} $$ (4)
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which is the same as equation (3) from p.1a-4. Derivation of equation (1) from p.1a-5 is shown in R1.1.

Derivation of (1) p.1a-5
Derivation is shown in R1.1.

Derivation of Corilois Force
The acceleration of an object on Earth is obtained by taking the second derivative of its position, $$ \vec{R}. $$

Taking the first derivative gives the velocity,
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$$ \vec{\nu} := \frac{d\vec{R}_I}{dt} = \frac{d\vec{R}_E}{dt} + \vec{\omega} \times \vec{R}_E $$ (1)
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Subscripts $$I$$ and $$E$$ indicate inertial and Earth-fixed reference frames, respectively, and $$\omega$$ is the rotation of the earth w.r.t. the inertial frame.

Acceleration is obtained by taking the derivative of $$\vec{\nu}$$.
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$$ \vec{a} := \frac{d\vec{\nu}_I}{dt} = \frac{d\vec{\nu}_E}{dt} + \vec{\omega} \times \vec{\nu}_E $$ (2)
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Substituting the expression for $$\vec{\nu_{E}}$$ from (1) into (2) gives
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$$ \vec{a} := \frac{d}{dt}\left(\frac{d\vec{R}_E}{dt} + \vec{\omega} \times \vec{R}_E\right) + \vec{\omega} \times \left(\frac{d\vec{R}_E}{dt} + \vec{\omega} \times \vec{R}_E\right) $$ (3)
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Separating out the components gives
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$$ \vec{a} := \frac{d^2\vec{R}_E}{dt^2} + \frac{d\vec{\omega}}{dt} \times \vec{R}_E + \vec{\omega} \times \frac{d\vec{R}_E}{dt} + \vec{\omega} \times \frac{d\vec{R}_E}{dt} + \vec{\omega} \times (\vec{\omega} \times \vec{R}_E) $$
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$$ \vec{a} := \frac{d^2\vec{R}_E}{dt^2} + \left(\frac{d\vec{\omega}}{dt} \times \vec{R}_E \right) + 2\left(\vec{\omega} \times \frac{d\vec{R}_E}{dt}\right) + \vec{\omega} \times (\vec{\omega} \times \vec{R}_E) $$ (4)
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Rearranging the terms in (4) gives
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$$ \vec{a} := \left(\frac{d\vec{\omega}}{dt} \times \vec{R}_E \right) + \vec{\omega} \times (\vec{\omega} \times \vec{R}_E) + 2\left(\vec{\omega} \times \frac{d\vec{R}_E}{dt}\right) + \frac{d^2\vec{R}_E}{dt^2} $$ (5)
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which looks very similar to equation (1) from p.1a-5. The two equations would be virtually identical if the substitutions below were made into equation (1) from p.1a-5.
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$$ f,_S(Y^1,t) = \vec{R}_E \left(\therefore f,_{St}(Y^1,t) = \frac{d\vec{R}_E}{dt} \right) $$
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$$ Y^1 = \vec{\omega} $$
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Both of the equations have a first term that contains an "acceleration" term; a second term that contains a squared "velocity" term; a third term that has a factor 2 multiplying the product of two different time derivatives; and a fourth term which is the second time derivative of the original function.

R1.3
$$ \displaystyle c_0(Y^1,t)=\underbrace{-F^1[1 - \bar R u^2_{,SS}(Y^1,t)]}_{\color{blue}{A}} - F^2 u^2_{,S} - \frac{T}{R} + M\left[(1-\bar R u^2_{,SS})(u^1_{,tt}-bar R u^2_{,S tt})+u^2_{,S} u^2_{,tt} \right ] $$

R1.5
Given (3) p.4-4,
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$$ \displaystyle \frac{1}{g_i(\xi_i)} \frac{d}{d\xi_i} \left[g_i(\xi_i) \frac{dX_i(\xi_i)}{d\xi_i} \right] + f_i(\xi_i)X_i(\xi_i) = 0 $$
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if we make the following substitutions,
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$$\xi_i \rightarrow x$$ $$X_i(\xi_i) \rightarrow y(x)$$ $$g_i(\xi_i) \rightarrow g(x)$$ $$f_i(\xi_i) \rightarrow a_0(x)$$
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(3) p.4-4 can be rewritten as
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$$ \frac{1}{g(x)} \frac{d}{dx} \left[g(x) \frac{dy(x)}{dx} \right] + a_0(x)y(x) = 0 $$ (1)
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Evaluating the derivative and expanding the terms gives
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$$ \frac{1}{g(x)} \left[\frac{dg(x)}{dx} \frac{dy(x)}{dx} + g(x)\frac{d^2y(x)}{dx^2}\right] + a_0(x)y(x) = 0 $$
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$$ \frac{1}{g(x)} \frac{dg(x)}{dx} \frac{dy(x)}{dx} + \cancelto{1}{\frac{g(x)}{g(x)}}\frac{d^2y(x)}{dx^2} + a_0(x)y(x) = 0 $$ (2)
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Using the notations below,
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$$\frac{dg(x)}{dx} \rightarrow g'(x)$$ $$\frac{dy(x)}{dx} \rightarrow y'(x)$$
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and rearranging the terms, (2) can be written as
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$$ y'' + \underbrace{\frac{g'(x)}{g(x)}}_{\displaystyle a_1(x)}y'(x) + a_0(x)y = 0 $$ (2)
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