A sector arc contributes a linear term

Triangle OPQ and circular arc PRQ; R; P; O; 5 cm; 7 cm; θ. Arc PR is curved boundary radius 5 about O, not chord PR; R lies on OQ.Triangle OPQ and circular arc PRQRPO5 cm7 cmθ
Question diagram 1
Triangle OPQ and circular arc PRQ; R; P; O; 5 cm; 7 cm; θ. Arc PR is curved boundary radius 5 about O, not chord PR; R lies on OQ.Triangle OPQ and circular arc PRQRPO5 cm7 cmθ

Given triangle OPQ with OP=5 cm, OQ=7 cm and angle POQ=θ; arc PR is part of a circle centred at O and R lies on OQ.

The diagram shows a triangle OPQ with OP = 5 cm, OQ = 7 cm and ∠POQ=θ\angle\mathrm{POQ}=\theta radians. Given that OP and OR are radii of a circle with centre O and θ\theta is a sufficiently small angle, show that the perimeter of PQR can be approximated by a+bθ+cθ2a+b\theta+c\theta^2 where a,b,ca,b,c are constants to be determined. marks

[2018/AJC/Prelim/I/2]

Solution

Perimeter

Cosine rule and binomial series

Triangle OPQOPQ has OP=5OP=5 cm, OQ=7OQ=7 cm and ∠POQ=θ\angle POQ=\theta radians. OPOP and OROR are radii of a circle centre OO, with RR on OQOQ. Show that the perimeter of PQRPQR (straight sides PQPQ, QRQR and arc PRPR) is approximately a+bθ+cθ2a+b\theta+c\theta^2 for small θ\theta.

Explore this idea

Hint

  1. Perimeter =PQ+QR+arc PR=PQ+QR+\text{arc }PR.

  2. QR=7−5=2QR=7-5=2 and arc PR=5θPR=5\theta.

  3. Use the cosine rule for PQPQ, then expand.

Answer · Perimeter

4+5θ+35θ2/4+O(θ4)4+5\theta+35\theta^2/4+O(\theta^4) cm; a=4,b=5,c=35/4a=4,b=5,c=35/4.

Small positive θ\theta in radians; PR is the minor arc.

Explore the idea

The perimeter and its small-angle series

Compare the function with a finite polynomial. Change the highest included power and test x, then use the error plot to see how the approximation changes at that x.

Highest included powerxⁿ, n = 2
Test value xx = 0.40
Try an example
Function
7.08638
P₂(x)
7.4
Error
0.31362

The shading is a local comparison window, not a convergence or accuracy guarantee.

Teal marks the function, amber marks the polynomial, and the red segment joins their values at the selected x.

Does keeping more powers help at this x?

Highest included power n runs horizontally; signed error Pₙ(x) − f(x) runs vertically. Zero means agreement at this x. A missing power leaves the polynomial unchanged; individual steps need not improve the error.

  • Function
  • Series truncated at the chosen power
  • A local comparison window, not a convergence or accuracy guarantee
Key idea

The exact perimeter equals 4+5θ+354θ24+5\theta+\frac{35}4\theta^2 plus terms of order θ4\theta^4.

Ideas to try
Compare degree 1 (tangent) and degree 2 at θ=0.5\theta=0.5.