Solve 25.656=x(49-x^2)^(1/2) + 49(sin(x/7))^-1  The actually problem is to find the coordinates of x where the area of three sections of a circle would be equal. So, I calcuated the area of the...

Solve 25.656=x(49-x^2)^(1/2) + 49(sin(x/7))^-1  

The actually problem is to find the coordinates of x where the area of three sections of a circle would be equal. So, I calcuated the area of the entire circle and divided by 3 and got 25.656 . I then preceded by integrating (49-x^2)^(1/2) from -x to x. Well I didn't integrate since its impossible but I saw in the back of the book a list of "impossible" integrations and the answer is (x(49-x^2)^(1/2))/2 + (49(sin(x/7))^-1)/2. So i plugged in x and subtracted -x when i plugged it in and got x(49-x^2)^(1/2) + 49(sin(x/7))^-1. And this is where im stuck, if i graph the function and 25.656 and see where they intersect, they intersect at 1.8545. So how do I continue to solve for x so i can get that answer too? 

Asked on by icbuono

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neela | High School Teacher | (Level 3) Valedictorian

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Let us imagine a circle with radius a and centre  O as the origin with (0,0) coordinates.

Let (x,0) be the point  A  on  X axis. Then the ordinate at x (||lel to Y axis) is trisecting the circle, by data. The problem is to find the coordinates of x = OA (in terms of a). Let AP be the ordinate at A and  P be the point on the circle. Let X be a point on the X axis where the cirle intercepts X axis.

Since the area  to the right of the ordinate at A is one third of the area, the required equation is :

Considering the half the semicircle above x axis, one 3rd of the upper semi circle's area = pi*a^2/6 =  area  bounded by radii OP and OX and arc XP of circle - area of the triangle OAP = a^2 *  (angle XOP )*(Pi/360) - (1/2)(OA*AP)

Pi*a^2/6=a^2(Pi* arcsin(x/a) /360) - (1/2)x(a^2-x^2)^(1/2)........(1)

But  x/a = sin z say. Then  arc sin(x/a) = z.  And OA = x = a*cosz and y = AP sin z = a*sinz.

Substituting in (1), we get:

Pi*a^2/6=(Pi*a^2)(z/360)-(a^2/2) sinz cosz .

z= 60+ (360/2)(1/Pi) sinzcosz

z=60+(90/Pi)sinz*cosz

You can get the result by iteration and the numerical value of the angle z is 74.6370827 degree nearly.

z=74.6370827 nearly.

So, x=a*cos z and y=a sin z.

But the radius in the present case is a= 7

Therefore, x=7 cos (74.6370827)=1.854524591 and y=7sin(74.6370827)=6.74986952 nearly. So the other coordinate is x=-7cos74.6370827 = -1.854524591 an y cordinate is 7*sin (74.6370827) =6.74986952

Area of the enclosed by AP, PX and the arc XP= (1/2){a^2* arc sin x/a)- x(a^2-x^2)^(1/2) } = (49/2) *{ arc cos (1.854524591/7)} - (1/2)1.854524591*6.74986952 = 25.65633998, which is the 1/3 rd  of the semi circle and with a symmetrical section below the X axis,  the double ordinates at A trisects the area of the circle.

Therefore the original equation needs to be corrected as below:

25.656.. = -(1/2)x(49-x^2)^(1/2)+(49/2) arc cos (x/7).

Hope this helps.

 

 

 

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