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half-circle
half-circle 1. The half of a circle; a semicircle.1552 Huloet, Halfe circle, semicirculus. 1559 W. Cuningham Cosmogr. Glasse 126 Describe in th' intersections in like maner, halfe circles. 1661 J. Childrey Brit. Baconica 104 A double course of half circles. 1878 Newcomb Pop. Astron. iii. iii. 299 A ...
Oxford English Dictionary
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Parametrization of half circle in complex plane I'm looking for a parametrization of the half-circle in the upper plane, going from $-1$ to $1$. Is $$\gamma(t):\begin{cases}[0,\pi]\to\mathbb C\\\t\mapsto-e^{-it}\end{c...
. $$ \gamma(t) = -e^{-it} = e^{i(\pi - t)} \quad (0 \le t \le \pi) $$ is an arc on the unit circle with the argument of $\gamma(t)$ decreasing from $\pi
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Find area of triangle in half circle !enter image description here Half circle with $r = 2$. So, think $CD = BD = 2, CEB$ = right triangle Area = $ 1/2 \overline{CE}\ \overline{AB}$ What is the max area? By theory ...
Hint: $$Area = \frac{1}{2}bh$$ where $b$ is the base of the triangle and $h$ is the height. As $r=2$, the diameter must be $4$ and so the base of the triangle must be $4$. To find the maximum height, observe that the triangle would be the largest when $h=r$, as you suggested. So, consider $$Area = \...
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Parametrization of a half circle in a different interval other than from 0 to $\pi$. If I have a half circle parametrization such that $e^{it}$, $0 \leq t \leq \pi$, to get the parameterization of the same circle on i...
Since we can map this interval onto the interval $[0,\pi]$ via $\pi\left(\frac{t-a}{b-a}\right)$, we can have any interval $[a,b]$ parametrize the top half of the unit circle by $$ \exp{i\left(\pi\frac{t-a}{b-a}\right)}.$$
In your particular case, if you're paramtrizing this circle for $0\leq t\leq 1$, then
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Finding the perimeter and area of a rectangle within a half-circle It's another late night studying calculus and I can't make heads or tails of this question. Perhaps somebody could help clarify it for me. I have a ha...
For a given $x$ we can find $y$ as $\sqrt{9-x^2}$. So we get the points $(x,\sqrt{9-x^2})$ for $R$ and $S$. So the width is given by $2x$ and the height by $\sqrt{9-x^2}$. Perimeter: $4 x + 2 \sqrt{9 - x^2}$ Area: $2 x \sqrt{9 - x^2}$
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What is a name for a circle that has half of itself in one plane and half in another? What is a name for a circle that has half of itself in one plane and half in another? $. Now let's think how we could compute the area of the half-circle : given a position $x$ on the $x$-axis, we can consider a small vertical rectangle of
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Is there another solution to this problem? Take a half circle with radius 2r with the center of O. Now take the half points from the radiuses from O forming the border of the half circle and draw two more half circles...
 = (2x – R)^2$. It gives $R = \dfrac {2x}{3}$.
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formula for finding perimeter of half circle i was looking for this problem < and was surprised if it is correct,we know that circumference or perimeter of circle is equal to $C=\pi*D=2*\pi*R$ but what about perimete...
In the linked question, we are just interested in the upper half of the circle and not at all the base (diameter).
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Proof: minimum volume of a notch cut at equal angle of cutting surfaces with horizontal plane > A notch is cut in a cylindrical vertical tree trunk. The edge of the cut reaches the axis of the cylinder and the cut is ...
Then, integrate $z_1-z_2$ over a half-circle in the $xy$-plane to obtain its volume,
$$V = \int_{-r}^{r}\int_0^{\sqrt{r^2-x^2}}[y\tan\alpha -(-y\tan\beta
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Improper integral of $\sin^2(x)/x^2$ evaluated via residues I have come across another improper integral I wish to evaluate via residues. The integral is: $$\int_{-\infty}^\infty{\frac{\sin(x)^2}{x^2}}dx$$ $\sin(z...
Note that $ \cos(2x)=1-2\sin(x)^2 $, this suggest to consider the integral $$ \int_{C} \frac{ {\rm e}^{2 i z} - 1 }{ z^2} dz \,.$$
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Contour integral with pure imaginary pole of order two : $\int_{0}^{\infty}\frac{x^2}{(x^2+9)^2(x^2+4)}dx$ I am trying to calculate $\int_{0}^{\infty}\frac{x^2}{(x^2+9)^2(x^2+4)}dx$. I want to use the upper half-circl...
Note that$$\frac{z^2}{(z^2+9)^2(z^2+4)}=\frac{z^2}{(z-3i)^2(z+3i)^2(z^2+4)}.$$Now, let $g(z)=\frac{z^2}{(z+3i)^2(z^2+4)}$. Then $g(3i)=-\frac1{20}$ and $g'(3i)=-\frac{13i}{300}$. Therefore,\begin{align}\frac{z^2}{(z-3i)^2(z+3i)^2(z^2+4)}&=\frac{-\frac1{20}-\frac{13i}{300}(z-3i)+\cdots}{(z-3i)^2}\\\&...
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Line integral equals zero. Currently going through some line integral problems and in the worked example provided, the line integral evaluated to become zero. The curve C that they have provided was a simple closed c...
Yes if $C$ is a simple closed curve, $\textbf{v}:\mathbb{R}^m\rightarrow \mathbb{R}^m$ and$$\oint_{C} \textbf{v}\;d\textbf{r}=0,$$ then it follows that $\textbf{v}=\nabla f$ for some scalar function $f:\mathbb{R}^m\rightarrow \mathbb{R}$.
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Show that $A=\{(x,y) \in \mathbb R^2 \mid x^2+y^2 \leq 1 , \, x<y\}$ is Borel In my homework I want to show that $A=\\{(x,y) \in \mathbb R^2\mid x^2+y^2 \leq 1 , \, x<y\\}$ is Borel and determine if A is open or close...
$A_1=\\{(x,y)\in\mathbb{R}^2:x^2+y^2\le 1\\}$ is a Borel set ($\because$ it is closed), $A_2=\\{(x,y)\in\mathbb{R}^2:x-y<0\\}$ is a Borel set ($\because$ it is open). Then $A=A_1\cap A_2$ is also a Borel set.
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Compute $\int_{-\infty}^{\infty} \frac{e^{-2ix}}{x^2 + 4}dx$ using contour integration I'm studying for an applied mathematics qualifying exam, and have the following practice problem: > Compute the integral $\int_{-...
Hint. Note that along the LOWER semicircle of radius $R>0$, $z=R(\cos(t)+i\sin(t))$ with $t\in [\pi,2\pi]$ and $$|\exp(-2iz)|=|\exp(-2iR(\cos(t)+i\sin(t))|=\exp(2R\sin(t))\leq 1.$$ Now are you able to show that the integral along the semicircle part of the path of integration is negligible? P.S. By ...
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