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 Apr 28 comment Drawing an arc within two defined points in C graphics How are a, b, c, d related to the location of the arc? Are the two lines always vertical? How long are the lines? Apr 25 comment Simplifying a Complex Number You should be familiar with Euler's Formula: $e^{ix}=\cos x + i \sin x$ which is periodic. So then $e^{ix} = e^{i(x\pm 2n\pi)}$ where $n \in \mathbb Z$. So then, $e^{ i \frac{2014\pi}{12}} = e^{ i(166\pi+ \frac{22\pi}{12})} = e^{ i \frac{22\pi}{12}}= e^{ i \frac{11\pi}{6}} = e^{- i \frac{\pi}{6}}$ Apr 24 answered Simplifying a Complex Number Apr 18 answered Geometric and algebraic aspects of geometric vectors Apr 17 comment Integral identity involving sin(x)/x What is there to prove? A more proper question would be to ask how to integrate the right hand side if that is what you are trying to figure out. Apr 16 comment Integral identity involving sin(x)/x It is not terribly suprising that two definite integrals are related to each other by a constant scale factor. I would expect nothing less. Apr 11 revised Can some one help me parametrize $\frac{x^4}{a^4}+\frac{y^4}{b^4}+\frac{z^4}{c^4}=1$ deleted 8 characters in body Apr 11 comment Can some one help me parametrize $\frac{x^4}{a^4}+\frac{y^4}{b^4}+\frac{z^4}{c^4}=1$ Your'e right, it is a sphere. Apr 11 answered Can some one help me parametrize $\frac{x^4}{a^4}+\frac{y^4}{b^4}+\frac{z^4}{c^4}=1$ Apr 3 comment SVM and quadratic programming SVM does reduce to a QP problem in the dual form. There are plenty of resources out there to explain it. To handle b you would add a 1 line, not a zero line. Apr 3 comment Matrix properties invariant under scalar multiplication Column space, row space, null space, left null space, eigen vectors, rank, to name a few. Apr 1 answered Interpolate/Increment Vector Rotation Mar 31 comment Finding the area of the triangle with vertices at $(ct, c/t)$, $(-ct, -c/t)$, $(ct^{2}, 2ct)$ The reference that you found is an interesting approach. It uses the same principle except it works in 3D space. The thinking goes like this. You are constructing your triangle in 3D space by adding $z=1$ as a third coordinate. So your triangle is floating in space in the horizontal plane $z=1$. You then compute the volume of the parallelepiped (denote as $V_P = D_3$). In the same construction you can create a tetrahedron (i.e. pyramid) by connecting your vertices to the origin. The volume of this is $V_T=1/3(1)A_T$. It is known that $V_P=6V_T =6(1/3A_T)$, so we get $A_T = 1/2D_3$. Mar 31 comment Finding the area of the triangle with vertices at $(ct, c/t)$, $(-ct, -c/t)$, $(ct^{2}, 2ct)$ Regarding the sign, that just depends on the order of the points that you chose to put into the determinant. If you changed the order from CW to CCW your sign and mine would match. Mar 31 comment Finding the area of the triangle with vertices at $(ct, c/t)$, $(-ct, -c/t)$, $(ct^{2}, 2ct)$ It is because by subtracting $B$, I effectively translated the triangle to the origin (i.e. point B is now at 0,0). Then it is known that the area of the parallelogram spanned by $A-B$ and $A-C$ is equal to the determinant shown above. The area of the triangle is 1/2 the area of the parallelogram, thus the 1/2. Mar 30 comment Finding the area of the triangle with vertices at $(ct, c/t)$, $(-ct, -c/t)$, $(ct^{2}, 2ct)$ Also you could use Herron's formula which can be used to calculate the area from the length of the three sides. You can find the formula here: en.wikipedia.org/wiki/Heron%27s_formula Mar 30 answered Finding the area of the triangle with vertices at $(ct, c/t)$, $(-ct, -c/t)$, $(ct^{2}, 2ct)$ Mar 27 comment proving Pythagoras Theorem in the third dimension using orthogonal projection from a parallelogram What does $p$ represent? Is it the area of the original parallelogram created by $u$ and $v$? Also, what makes you think that it should be true. Mar 25 revised proving Pythagoras Theorem in the third dimension using orthogonal projection from a parallelogram Added Latex markup Mar 23 comment Factoring derivatives It is the product rule.