User page for Andrew Dugan
Problem 9: Show that CD = AB plus H.O.T. for a small change in gamma
[edit | edit source]
File:Problem 9 File.JPG
Verify that
for small
and determine what the higher-order terms are.
![{\displaystyle tan(d\gamma )={\frac {\bar {AB}}{V}}\Rightarrow {\bar {AB}}=Vtan(d\gamma )\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73ee2655d2917286566e1217912d58ecd4c9c782)
![{\displaystyle sin(d\gamma )={\frac {\bar {CD}}{V+dV}}\Rightarrow {\bar {CD}}=(V+dv)sin(d\gamma )\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a4129171b937b36a62929ee5205e99540aaac8f)
![{\displaystyle {\bar {CD}}={\bar {AB}}+(V+dv)sin(d\gamma )-Vtan(d\gamma )\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3a38c4fee2540135593207c6e88fcc8f6046b72)
Both
≈
and
≈
when x is small, thus
≈ ![{\displaystyle {\bar {AB}}+(V+dV)d\gamma -V{d\gamma }\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4115dcf4554f59a5bc749591e2a7b03f4fd6cdbf)
≈
|
|
To determine what the higher-order terms are, use Taylor series expansions.
![{\displaystyle sin(x)=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+...\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/424b41735f20d2aa4a51645af2d781694eaf8bba)
![{\displaystyle tan(x)=x+{\frac {x^{3}}{3}}+{\frac {2x^{5}}{15}}+{\frac {17x^{7}}{315}}+...\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35fdcda98b165a0de3f3e542669e303c37e2e9ec)
|
|
Problem 12: Verify the expression for the circumference of an ellipse
[edit | edit source]
The arc length of an ellipse is
.
![{\displaystyle x=acost\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13f6484cb3a12c405b656177c336379705addb87)
![{\displaystyle y=bsint\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4c6176f103b158bd379645bf58b27575e7ff382)
The eccentricity
is defined as
.
Verify that
.
![{\displaystyle dx={\frac {d}{dt}}[acost]=-asint\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d28e3a76cac8fb0885043deab495f718c9e3375b)
![{\displaystyle dy={\frac {d}{dt}}[bsint]=bcost\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39513c6a61694e21e1c9857b860dca89ffbb76df)
![{\displaystyle 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\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/346177bf84a5ed93eb4d29cb0bd3690828ca7130)
, and since
,
![{\displaystyle 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\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd8dd637a3aa89bd7f7e423a7f40d6caf46f8563)
|
|
Problem 14: Verify the change of variables expression for I in Clenshaw-Curtis quadrature
[edit | edit source]
Clenshaw-Curtis quadrature utilizes the change of variables of
.
Before the change of variables,
.
Verify that
.
![{\displaystyle I=\int _{-1}^{+1}f(x)dx\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5eb04bbdcbd214f3328df822ec3b144ba13db3ab)
![{\displaystyle f(x)=f(cos\theta )\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62c2cc5bece5134a5013e51941ba7841935ff6f6)
![{\displaystyle dx={\frac {d}{dx}}(cos\theta )=-sin{\theta }d\theta \,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d71c24dcb7691754e5d9293bb35de5d463cf318)
thus, lower limit ![{\displaystyle -1\Rightarrow \pi \,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c65c4dbf21132048be64c0e8b8b3be5aeef7466)
thus, lower limit ![{\displaystyle 1\Rightarrow 0\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6aaa2fe5d02b152c8b4f586b4dd7fddafda38b0)
![{\displaystyle I=\int _{\pi }^{0}f(cos\theta )(-sin{\theta }d\theta )\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef7981df5f0dc4156555911003097d3a301d0a73)
|
|