To get f = 2/5 of a circle with one chord, you may use calculus, but you do not need to. In this unit we show how to do it without, and then with, calculus.
(2*A)/(360)*π*r2
the area of the (top) sector ACB;
r2*cos(A)*sin(A)
the area of the triangle ACB;
πr2
the area of the circle.
Thus,
f = (area of sector - area of triangle ACB)/area of circle
f = ((2*A)/(360)*π*r2 - r2*cos(A)*sin(A))/( π*r2)
And after simplification,
f = A/180 - cos(A)*sin(A)/π
You can solve this equation for A, using Solver on the TI-84.
Here, angles are measured in degrees. You need 2*A to get the correct angle!
Derivations of formulas above.
cos A = h/r
sin A = .5b/r
.5b=r*sin A
b = 2r*sin A
area of triangle ACB
= 1/2*b*h
= 1/2(2r*sin A)*(r*cos A)
= r2*sin A * cos A
so
Let's solve the above for f = 2/5 = .4 of the whole circle.
We have A/180 - cos A*sin A/π - .4* = 0
A will become our variable X.
Now on the TI-84, be sure MODE is set to degrees.
Go to Y=, and enter:
\Y1=X/180-cos(X)*sin(X)/π-.4
Next go to MATH ↑ and choose Solver.
Go ↑ and you will see
EQUATION SOLVER
eqn:0=
Insert Y1 so that it looks as follows:
eqn:0=Y1
Press ENTER
Place the cursor on the X= line, and press ALPHA SOLVE.
You will see X = 80.92447884...
This is angle A in degrees. Multiply by 2 to get 2A.
You will get 2A=161.8489577 degrees.
Now construct the two parts of the circle by cutting the circle with one chord, so that one part is 2/5 of the circle, and the other is 3/5.
The same problem using calculus:
Area of circle = -R∫R 2√(R2 - X2)dX
On the TI-84, store some value in R. Now,
fnInt(2√(R2-X2), X, -R, R)
will give you a good approximation of the area of the circle, πR2.
But we want the area of 3/5 (or 2/5) of the circle. So we want to find Z along the x-axis such that -R∫Z 2√(R2 - X2)dX = 3/5πR2.
We rewrite this as -R∫Z 2√(R2 - X2)dx - 3/5πR2 = 0. We want to find Z.
Again, store some value in R. (You may make R = 1.)
Under Y=, enter
\Y1= fnInt(2√(R2 - X2), X, -R, Z) - 3/5πR2
Now use Solver, and solve for Z.
When R = 1, we got Z = .157736...
Can you draw the cut circle with this information?
Can you show that the two solutions (without and with calculus) give the same answer?
