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

Given

File:Problem 9 File.JPG

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

${\bar {CD}}\,$ ≈ ${\bar {AB}}+(V+dV)d\gamma -V{d\gamma }\,$

${\bar {CD}}\,$ ≈ ${\bar {AB}}+H.O.T.\,$

To determine what the higher-order terms are, use Taylor series expansions.

$sin(x)=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+...\,$

$tan(x)=x+{\frac {x^{3}}{3}}+{\frac {2x^{5}}{15}}+{\frac {17x^{7}}{315}}+...\,$

${\bar {CD}}={\bar {AB}}+(V+dV)(d\gamma -{\frac {({d\gamma })^{3}}{3!}}+{\frac {({d\gamma })^{5}}{5!}}-{\frac {({d\gamma })^{7}}{7!}}+...)-V(d\gamma +{\frac {({d\gamma })^{3}}{3}}+{\frac {2({d\gamma })^{5}}{15}}+{\frac {17({d\gamma })^{7}}{315}}+...)\,$

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

Solution

$dx={\frac {d}{dt}}[acost]=-asint\,$

$dy={\frac {d}{dt}}[bsint]=bcost\,$

$C=\int {dl}=\int _{t=0}^{2\pi }[(-asint)^{2}+(bcost)^{2}]^{\frac {1}{2}}dt=\int _{t=0}^{2\pi }[a^{2}sin^{2}t+b^{2}cos^{2}t]^{\frac {1}{2}}dt\,$

$C=a\int _{t=0}^{2\pi }[sin^{2}+{\frac {b^{2}}{a^{2}}}cos^{2}t]^{\frac {1}{2}}dt\,$, and since $sin^{2}t+cos^{2}t=1\,$,

$C=a\int _{t=0}^{2\pi }[1-cos^{2}t+{\frac {b^{2}}{a^{2}}}cos^{2}t]^{\frac {1}{2}}dt=a\int _{t=0}^{2\pi }[1-(1-{\frac {b^{2}}{a^{2}}})cos^{2}t]^{\frac {1}{2}}dt\,$

$C=a\int _{t=0}^{2\pi }[1-e^{2}cos^{2}t]^{\frac {1}{2}}dt\,$

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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 $1\Rightarrow 0\,$

$I=\int _{\pi }^{0}f(cos\theta )(-sin{\theta }d\theta )\,$

$I=\int _{0}^{\pi }f(cos\theta )sin{\theta }d\theta \,$