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4h
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?
2d
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
11
comment 3D vector perpendicular calculation
What methods do you understand? Vector math (dot & cross product)? Linear algebra (matrices, transformations, rotations)?
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