# calculus

posted by on .

P
/|\
/ | \
/ | \ on this side there is an
/ | \ arc with A(θ) in the
/ | \ middle. the B(θ)
/θ | B(θ)\ triangle is not equal
O -------------- R to the θ triangle.
Q

the figure shows a setor of a circle with central angle theta. Let A(θ) be the area of the segment between the chord PR and the arc PR. Let B(θ) be the area of the triangle (right engle) PQR. Find lim and x->0+ A(θ)/B(θ)

• calculus - ,

1/3

A(theta)= (0.5*(r^2)*theta)-(r*sin theta)= 0.5(theta-sin theta)
B(theta)=(0.5*(r^2))*(0.5*r*cos theta* r*sin theta)=0.5sin theta(1- cos theta)

so the limit as theta approaches zero from the positive side of a(theta) over b (theta) is equal to:

[0.5(theta-sin theta)]/[0.5(sin theta (1- cos theta))]=
(theta - sin theta)/(sin theta - sin theta * cos theta)=
(theta - sin theta)/(sin theta - 0.5sin2 theta)=

Appling L'Hospitals rule:
the limit is:
(1- cos theta)/(cos theta - cos 2 theta)

using the rule again:
(sin theta)/(-sin theta + 2sin 2 theta)

and again:
(cos theta)/(-cos theta + 4 cos 2 theta)

=1/3