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=Problem 9: Show that CD = AB plus H.O.T. for a small change in gamma=

Find
Verify that $$ \bar{CD} = \bar{AB} + H.O.T. \,$$ for small $$ d\gamma \,$$ and determine what the higher-order terms are.

Solution
$$ tan(d\gamma)=\frac{\bar{AB}}{V} \Rightarrow \bar{AB}=Vtan(d\gamma) \,$$ $$ sin(d\gamma)=\frac{\bar{CD}}{V+dV} \Rightarrow \bar{CD}=(V+dv)sin(d\gamma) \,$$ $$ \bar{CD} = \bar{AB} + (V+dv)sin(d\gamma) - Vtan(d\gamma) \,$$ Both $$ sin(x)\,$$ ≈ $$ x \,$$ and $$ tan(x) \,$$ ≈ $$ x \,$$ when x is small, thus $$ \bar{CD} \,$$ ≈ $$ \bar{AB} + (V+dV)d\gamma - V{d\gamma} \,$$

To determine what the higher-order terms are, use Taylor series expansions. $$ sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ... \,$$ $$ tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + ... \,$$

=Problem 12: Verify the expression for the circumference of an ellipse=

Given
The arc length of an ellipse is $$ dl = [dx^2+dy^2]^\frac{1}{2} \,$$. $$ x = acost \,$$ $$ y = bsint \,$$ The eccentricity $$ e \,$$ is defined as $$ e = [1-\frac{b^2}{a^2}]^\frac{1}{2} \,$$.

Find
Verify that $$ C = \int{dl}=a\int_{t=0}^{2\pi}[1-e^2cos^2t]^\frac{1}{2}dt \,$$.

Solution
$$ dx = \frac{d}{dt}[acost]=-asint \,$$ $$ dy = \frac{d}{dt}[bsint]=bcost \,$$ $$ C = \int{dl}=\int_{t=0}^{2\pi}[(-asint)^2+(bcost)^2]^\frac{1}{2}dt=\int_{t=0}^{2\pi}[a^2sin^2t+b^2cos^2t]^\frac{1}{2}dt \,$$ $$ C = a\int_{t=0}^{2\pi}[sin^2+\frac{b^2}{a^2}cos^2t]^\frac{1}{2}dt \,$$, and since $$ sin^2t+cos^2t=1 \,$$, $$ C = a\int_{t=0}^{2\pi}[1-cos^2t+\frac{b^2}{a^2}cos^2t]^\frac{1}{2}dt=a\int_{t=0}^{2\pi}[1-(1-\frac{b^2}{a^2})cos^2t]^\frac{1}{2}dt \,$$

=Problem 14: Verify the change of variables expression for I in Clenshaw-Curtis quadrature=

Given
Clenshaw-Curtis quadrature utilizes the change of variables of $$ x=cos\theta \,$$. Before the change of variables, $$ I = \int_{-1}^{+1}f(x)dx \,$$.

Find
Verify that $$ I = \int_{0}^{\pi}f(cos\theta)sin{\theta}d\theta \,$$.

Solution
$$ I = \int_{-1}^{+1}f(x)dx \,$$ $$ f(x)=f(cos\theta) \,$$ $$ dx = \frac{d}{dx}(cos\theta) = -sin{\theta}d\theta \,$$ $$ sin^{-1}(-1)=\pi \,$$ thus, lower limit $$ -1 \Rightarrow \pi \,$$ $$ sin^{-1}(1)=0 \,$$ thus, lower limit $$ 1 \Rightarrow 0 \,$$ $$ I = \int_{\pi}^{0}f(cos\theta)(-sin{\theta}d\theta) \,$$