User:Egm6341.s11.team2.franklin/hwk6

=Problem 6.8: Derive Arc Length Equation=

''' This problem was solved without referring to S10. '''

From the lecture slide Mtg 33-3

Given
The formula for arc length from lecture 32-4:


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$$\displaystyle
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\widehat{PQ} = \int_{\theta_{P}}^{\theta_{Q}}d\theta\left[r^2+\left(\frac{dr}{d\theta}\right)^{2}\right]^{\frac{1}{2}}. $$     (8.1)
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The law of cosines:

Triangle OAB:


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where

Find
Derive the formula for arc length of ellipse using triangle OAB and the law of cosines.

Solution
Applying the law of cosines formula to the OAB triangle case gives:

Substituting in Eqs 8.3-5 into Eq 8.6 gives

Taylor series expansion of $$\displaystyle \cos (d\theta) $$ is

Because theta is assumed to be small, we only need to take the first two terms of the Taylor series expansion of $$\displaystyle \cos (d\theta) $$ and sub into Eq 8.8.

The last term in Eq 8.11 $$\displaystyle rdrd\theta^2$$ is zero because the three derivatives shrink faster than the two derivative elements $$\displaystyle dr^2$$ and $$\displaystyle d\theta^2$$.

The remainder is

Taking the square root of both sides of Eq 8.12 then integrating from $$\displaystyle\theta_P$$ to $$\displaystyle\theta_Q$$ gives,

= Problem 6.9: Identification of Basis Functions $$\displaystyle \overline{N_i}$$ =

Given
The relationship between $$\displaystyle Z(s)$$ and $$\displaystyle d_i$$(degree of freedom):


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$$\displaystyle
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\begin{align} Z(s)&=\sum_{i=0}^{3}c_i s^i \\ &= \sum_{i=1}^{4} \overline{N_i}(s) \, d_i\\ \end{align}

$$     (9.1-2)
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where,


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$$\displaystyle
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\begin{align} \overline{N_i}(s) &= basis\,\, function\\ d_i(s) &= degrees\,\, of\,\, freedom\\ d_1&=Z_i \\ d_2&=\dot{Z_i} \\ d_3&=Z_{i+1} \\ d_4&=\dot{Z}_{i+1}\\ \end{align}

$$
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and
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$$\displaystyle
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\dot{Z}=\frac{dZ}{dt} = \frac{dZ}{ds} \frac{ds}{dt}

$$    (9.3)
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Find
Identify and plot the the basis function $$\displaystyle \left\{\overline{N_i}(s) \right\}\,\, ,i=1,2,3,4 $$.

Solution
From the Eq 1 of 35-4(lecture slide)


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$$\displaystyle
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\begin{align} \begin{Bmatrix} c_0 \\ c _1 \\ c _2 \\ c_3 \end{Bmatrix} &= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} \begin{Bmatrix} Z_i \\ Z_i^' \\ Z_{i+1} \\ Z_{i+1}^' \end{Bmatrix} \\ &=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} \begin{Bmatrix} d_1 \\ hd_2 \\ d_3 \\ hd_4 \\ \end{Bmatrix}. \\ &=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} \begin{Bmatrix} \overline{d}_1 \\ \overline{d}_2 \\ \overline{d}_3 \\ \overline{d}_4 \\ \end{Bmatrix}. \\

\end{align}

$$     (9.3-5)
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where,


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$$\displaystyle
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\begin{align} \overline{d}_1&=d_1 \\ \overline{d}_2&= h \, d_2 \\ \overline{d}_3&=d_3 \\ \overline{d}_4&= h \, d_4 \\ \end{align}

$$
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Expand the matrix equation.


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$$\displaystyle
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\begin{align} &c_0= \overline{d}_1 \\ &c_1= \overline{d}_2 \\ &c_2= -3\overline{d}_1 -2\overline{d}_2+3\overline{d}_3-\overline{d}_4 \\ &c_3= 2\overline{d}_1 +1\overline{d}_2-2\overline{d}_3+\overline{d}_4 \\

\end{align}

$$     (9.6-9)
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Substitute $$\displaystyle c_i$$ to Eq 9.1


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$$\displaystyle \begin{align} Z(s)&=\sum_{i=0}^{3}c_i s^i \\ &= c_0 + c_1 s^1 + c_2 s^2 + c_3 s^3 \\ &= \overline{d}_1 + \overline{d}_2 s^1 + (-3\overline{d}_1 -2\overline{d}_2+3\overline{d}_3-\overline{d}_4) s^2 + (2\overline{d}_1 +1\overline{d}_2-2\overline{d}_3+\overline{d}_4) s^3 \end{align}
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$$     (9.10-12)
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Rearrange in terms of $$\displaystyle \overline{d}_i $$


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$$\displaystyle \begin{align} Z(s)& = (2s^3-3s^2+1)\overline{d}_1+(s^3-2s^2+s)\overline{d}_2+(-2s^3+3s^2)\overline{d}_3+(s^3-s^2)\overline{d}_4 \\ &=\overline{N}_1(s) \, \overline{d}_1+\overline{N}_2(s) \, \overline{d}_2+\overline{N}_3(s) \, \overline{d}_3+\overline{N}_4(s) \, \overline{d}_4 \end{align} $$     (9.13-14)
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Comparing the LHS and RHS of the equation gives
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(9.15-18)
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Plot of $$\displaystyle \overline{N_i} $$
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.