cant understand this solution in certain points..

• Feb 1st 2010, 11:36 PM
transgalactic
cant understand this solution in certain points..
on a circle with radius 1 there are two points one is static the other is
distributed evenly on the circle (i dont know the proper english term)
we sign X to be the hypotenuse on the circle

find the density function of the variable X
?

solution(i partialy understood it)the questions come after the photo

http://i46.tinypic.com/n3t35v.jpg
they say that we take x to be an angle from o til pi/2
we scan only the northern hemisphere because its symmetric.

what is the curv graph represents
what is t?
what is the meaning of y<=t?
why if they say that x should go from o til pi/2?
in the probability function they put x fro 0 till arcsin t/2 where t goes from
0<t<=2
?

after this step they did the derivative to find the density function
i understand that step
• Feb 2nd 2010, 04:58 AM
Quote:

Originally Posted by transgalactic
on a circle with radius 1 there are two points one is static the other is
distributed evenly on the circle (i dont know the proper english term)
we sign X to be the hypotenuse on the circle

find the density function of the variable X
?

solution(i partialy understood it)the questions come after the photo

they say that we take x to be an angle from o til pi/2
we scan only the northern hemisphere because its symmetric.

what is the curv graph represents the length of the side "y" as the angle x varies from 0 to 90 degrees
what is t? the various values of y fom 0 to 2
what is the meaning of y<=t? y ranges from 0 to "t" for a specific "t"
why if they say that x should go from o til pi/2? x is the angle BCA
in the probability function they put x fro 0 till arcsin t/2 where t goes from
0<t<=2
? Since the triangle ABC is always right-angled at B, $\displaystyle Sinx=\frac{y}{2},\ y=t,\ Sinx=\frac{t}{2},\ x=Sin^{-1}\frac{t}{2}$

after this step they did the derivative to find the density function
i understand that step

Attached is a diagram to help
• Feb 2nd 2010, 10:56 AM
transgalactic
its not sin its cosine

AC*cos(X)=AB
2*cos(x)=y
• Feb 2nd 2010, 11:12 AM
If you place the angle x at "A", then yes, that's what it will be.
I placed it at "B" to correspond with your work,
(as x was in the wrong position according to the worked solution).
• Feb 2nd 2010, 01:02 PM
transgalactic
look at my sketch there is 90 degree angle on point B
• Feb 2nd 2010, 01:07 PM
Yes, the angle at the circumference for a triangle whose hypotenuse is the diameter of a circle is 90 degrees.

Do you want to work with the angle at A or C ?
and the side opposite A or adjacent A
• Feb 2nd 2010, 01:16 PM
transgalactic
i want to work with angle A
like in my sketch which is signed by x
• Feb 2nd 2010, 01:34 PM
So, from your sketch $\displaystyle Cosx=\frac{y}{2}$
$\displaystyle y=t=2Cosx$
This gives rise to $\displaystyle Cosx=\frac{y}{2}=\frac{t}{2}$
Then $\displaystyle x=arcCos\frac{t}{2},\ or\ Cos^{-1}\frac{t}{2}$