Dead blog

LOL...This blog is so dead....

Lately, my Maths teacher is talking about the application of the higher derivatives. To find the 2nd derivative and above is very tedious and a small mistake can screw the whole problem up especially those like eg : y = ((cos x)^0.5 )/ (x^3 + sec x)^3 Find the 2nd derivative.

Thus, I have been working to devise a shortcut general formula for nth derivative. And at last I found one possible way.......

First, break the expression into 2 function.

Hence, let y = f(x) g(x)

dy/dx = f' (x) g(x) + g' (x) f (x)
(d^2)y)/(dx)^2 = f'' (x) g(x) + f' (x) g' (x) + g''(x) f (x) + g' (x) f'(x)
= f'' (x) g(x) + 2 f' (x) g' (x) + g''(x) f(x)
(d^3)y)/(dx)^3 = f''' (x) g(x) + 3 f'' (x) g' (x) + 3 f' (x) g''(x) + g''' (x) f (x)

Hence, the differentiation of higher derivatives can be get through binomial expansion.....

Cool Right?

Hence, can some pros out there tell me other general formula for getting nth derivatives. Thanks.

Also, someone plz tell me whether this formula is fully optimized or not?

Programming particle motion....

Sec 4 is a stressful period....... But nevertheless I still have time to experiment with Flash. I just devised a new equation for rectilinear motion of a particle which makes use of Sin( ) and Cos( ).

Let the horizontal displacement be x ; the vertical displacement be y ; let the angle between the horizontal and vertical displacement be z.

Thus in order for the object to move rectilinearlly with a shape of a parabola , the value must change at a constant rate, thus making the dy/dx be changing from y > 0 to y = 0 to y <0 .

(Please take note that Flash MX calculate everything it radian so it is important to convert the degree to radian before applying it into the equation.

Hence,
Let z = 90 degree

z --
_root.particle._x = Math.cos(z*(Math.PI/180))
_root.particle._y = (Math.sin(z*(Math.PI/180))

And you would get the particle will be displaced to another point through rectilinear motion.

Please make any comment to the content as this is what I found out myself... I am not sure about any better Physics equation avaliable on net..