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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.
Mar
10
answered Find the matrix given the determinant
Mar
2
comment Integrating a step function using antiderivatives
$F(x)$ is not continuous. It should be piece-wise continuous.
Feb
27
answered How can one intuit complex numbers from quaternions?
Feb
25
comment Transformations between coordinate frames
Your transformations dont look right. In your notation $T_{AB}, is that from A to B or the other way around.
Feb
23
comment Transformations between coordinate frames
Luigi, they are Homogeneous coordinats.
Feb
22
comment Derivative of angle between two vectors singularity!
I think that you might be running into trouble with your assumption that A and B are unit vectors. By that assumption you cannot arbitrarily change one of the vectors. It has to remain on the unit circle. Thus $\bf \dot{\hat A}\dot \bf \hat B$ and $\bf {\hat A}\dot \bf \dot{\hat B}$ are both zero. So you have a situation where you have $0/0$ and you would have to use L'Hopital's rule to solve it. I recommend that you solve for the general case.
Feb
22
answered Solving a transformation equation involving vectors and quaternions
Feb
22
comment Solving a transformation equation involving vectors and quaternions
Because quaternions are not commutative, you cannot divide. You have to pre or post multiply to manipulate both sides of the equation.
Feb
22
revised Confused with the notation/halfway done and need a little help
put in $$
Feb
16
comment Quick question about ln(0)
You know about L'Hôpital's rule right? Just put the indeterminate form into $\infty/\infty$ form as follows $x \ln x = \frac{\ln x}{1/x}$. Then take derivatives. It is pretty straight forward.
Feb
16
comment Quick question about ln(0)
You could also side step the indeterminate form by solving the equivalent integral down the $y$ axis. $\int_{0}^{1}\ln x\,\mathrm dx=\int_{-\infty}^{0}e^y\,\mathrm dy = -1$
Feb
8
comment compare lines and recognize similar ones
I think that instead of convolution you meant correlation. Convolution reverses one signal in time which would not yield the desired result. Correlation on the other hand does not reverse time and therefore would be more applicable.
Jan
31
revised How to check if a matrix is positive definite
added 1 character in body