Chapter 4, Problem 20P

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

Chapter
Section

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# An arc PQ of a circle subtends a central angle θ as in the figure. Let A(θ) be the area between the chord PQ and the arc PQ. Let B(θ) be the area between the tangent lines PR, QR, and the arc. Find lim θ → 0 + A ( θ ) B ( θ )

To determine

To evaluate: The value of limθ0+A(θ)B(θ) where θ is the central angle.

Explanation

Given:

The given figure is represents a sector with central angle θ and a chord PQ. The part of sector PQ which is in between the sector and chord PQ named as A(θ) and the area outside the circle and bounded between PRQ is named as B(θ)

Calculation:

Let the radius of the circle be R and centre be O.

Draw angle bisector of angle θ which intercept chord PQ at point S as shown in below Figure 1.

Therefore, OP=OQ=R

Find the value of A(θ) .

The value of A(θ) is the difference between the area of the sector PQ and the area of triangle OPQ .

Area of the sector PQ is given by, A(sector PQ)=(θ2ππ(R)2) where θ is the central angle.

Consider the triangle OPQ.

Area of the triangle OPQ is 12OPOQsinθ .

Hence, area of the triangle OPQ is as follows:

A(triangle OPQ)=12OPOQsinθ=12RRsinθ=12R2sinθ

Therefore, A(θ) is,

A(θ)=A(sector PQ)A(triangle OPQ)=(θ2ππ(R)2)12(R)(Rsinθ)=(θ2R2)R2sinθ2=R22(θsinθ)

Hence, A(θ)=R22(θsinθ) (1)

Similarly, find the value of B(θ) .

The value of B(θ) is the difference between the area of quadrilateral OPRQ and area of sector PQ.

Area of the sector PQ is given by, A(sector PQ)=(θ2ππ(R)2) where θ is the central.

Find OR as follows:

OR=OS+OR=Rcosθ2+Rtanθ2sinθ2=Rcosθ2+Rsin2θ2cosθ2=Rcosθ2

Similarly find PQ.

PQ=PS+SQ=Rsinθ2+Rtanθ2cosθ2=Rsinθ2+Rsinθ2cosθ2cosθ2=2Rsinθ2

The value of B(θ) is as follows:

Therefore, B(θ)=R2(tanθ2θ2) (2)

Evaluate the value of limθ0+A(θ)B(θ) using the equations (1) and (2),

limθ0+A(θ)B(θ)=limθ0+R22(θsinθ)R2(tanθ2θ2)=limθ0+θsinθ2(tanθ2θ2)=12limθ0+θsinθtanθ2θ2

Obtain the value of the function as θ approaches 0+

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