User:Egm6321.f12.team07.liu/report1

=Problem 1=

Given

 * {| style="width:100%" border="0"

$$\displaystyle\left.f(S,t)\right|_{S=Y^{1}(t)}=f(Y^{1}(t),t)$$  (1.1)
 * style="width:100%" |
 * style="width:100%" |
 * }

Find

 * {| style="width:100%" border="0"

$$\frac{\mathrm{d^2} f}{\mathrm{d} t^2}=f_{,S}(Y^1,t)\ddot{Y}^1+f_{,SS}(Y^1,t)(\dot{Y}^1)^2+2f_{,St}(Y^1,t)\dot{Y^1}+f_{,tt}(Y^1,t)$$, where $$\displaystyle\ f_{,S}(Y^1,t):=\frac{\partial f(Y^1(t),t)}{\partial S}$$  and  $$\displaystyle\ f_{,St}(Y^1,t):=\frac{\partial^2 f(Y^1(t),t)}{\partial S\partial t}$$  (1.2)
 * style="width:100%" |
 * style="width:100%" |
 * }

Solution

 * First, we get the first derivative of $$\displaystyle f$$ with respect to time,
 * {| style="width:100%" border="0"

$$\displaystyle\frac{\mathrm{d} f(Y^1(t),t)}{\mathrm{d} t}=\frac{\partial f(Y^1(t),t)}{\partial S}\dot{Y}^1+\frac{\partial f(Y^1(t),t)}{\partial t}$$, with  $$\displaystyle\dot{Y}^1:=\frac{\mathrm{d} Y^1(t)}{\mathrm{d} t}$$  (1.3)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * Then we get:
 * {| style="width:100%" border="0"

$$\frac{\mathrm{d^2} f}{\mathrm{d} t^2}=\underbrace{\frac{\mathrm{d} }{\mathrm{d} t}[\frac{\partial f(Y^1(t),t)}{\partial S}\dot{Y}^1]}_{\color{blue}{RHS1}}+\underbrace{\frac{\mathrm{d} }{\mathrm{d} t}[\frac{\partial f(Y^1(t),t)}{\partial t}]}_{\color{blue}{RHS2}}$$  (1.4)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

RHS1=$$\frac{\partial f(Y^1(t),t)}{\partial S}\ddot{Y}^1+\frac{\partial^2 f(Y^1(t),t)}{\partial S^2}(\dot{Y}^1)^2+\frac{\partial^2 f(Y^1(t),t)}{\partial S\partial t}\dot{Y}^1$$  (1.5)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

RHS2=$$\frac{\partial^2 f(Y^1(t),t)}{\partial S\partial t}\dot{Y}^1+\frac{\partial^2 f(Y^1(t),t)}{\partial t^2}$$  (1.6)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * Bring (1.5) and (1.6) into (1.4), we could get:
 * {| style="width:100%" border="0"

$$\frac{\mathrm{d^2} f}{\mathrm{d} t^2}=\frac{\partial f(Y^1(t),t)}{\partial S}\ddot{Y}^1+\frac{\partial^2 f(Y^1(t),t)}{\partial S^2}(\dot{Y}^1)^2+2\frac{\partial^2 f(Y^1(t),t)}{\partial S\partial t}\dot{Y^1}+\frac{\partial^2 f(Y^1(t),t)}{\partial t^2}$$  (1.7)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * i.e.
 * {| style="width:100%" border="0"

$$\frac{\mathrm{d^2} f}{\mathrm{d} t^2}=f_{,S}(Y^1,t)\ddot{Y}^1+f_{,SS}(Y^1,t)(\dot{Y}^1)^2+2f_{,St}(Y^1,t)\dot{Y^1}+f_{,tt}(Y^1,t)$$, where $$\displaystyle\ f_{,S}(Y^1,t):=\frac{\partial f(Y^1(t),t)}{\partial S}$$  and  $$\displaystyle\ f_{,St}(Y^1,t):=\frac{\partial^2 f(Y^1(t),t)}{\partial S\partial t}$$ (1.8)
 * style="width:100%" |
 * style="width:100%" |
 * 
 * }

=Problem 4=

Given

 * {| style="width:100%" border="0"

The polar coordinate $$\displaystyle (\xi_1,\xi_2)=(r,\theta)$$
 * style="width:100%" |
 * style="width:100%" |
 * 
 * }

Find

 * {| style="width:100%" border="0"

Draw the polar coordinate lines in a 2-D plane emanating from a point, not at the origin. 
 * style="width:100%" |
 * style="width:100%" |
 * }

Solution


Suppose a non-origin point $$\displaystyle (C_1,C_2)$$.
 * }

Draw the polar coordinate lines in XY coordinate system from the point $$\displaystyle (C_1,C_2)$$, then we could derive:
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \left\{\begin{matrix} x=r\cos\theta+C_1\\y=r\sin\theta+C_2 \end{matrix}\right.$$  (4.1)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \Rightarrow\left\{\begin{matrix} (x-C_1)^2+(x-C_2)^2=r^2\\(y-C_1)/(x-C_2)=\tan\theta \end{matrix}\right.$$  (4.2)
 * style="width:100%" |
 * style="width:100%" |
 * }

According to (4.2), as $$\displaystyle (r,\theta)$$ changes, two sets of lines could be draw as shown in the following figure(4-1):
 * }