Edexcel IAL Mathematics FP2.7 polar coordinates

Polar-coordinate questions combine geometry with calculus: locate the angle or point first, then use tangent conditions or 1/2∫r²dθ for the area.

[Maximum number: 9]

Figure 1

Figure 1 shows a sketch of the curve C with polar equation

r=1+\cos\theta,\qquad 0\le\theta\le\pi

and the line l with polar equation

r=k\sec\theta,\qquad 0\le\theta<\frac{\pi}{2}

where k is a positive constant.

Given that
- C and l intersect at the point P
- $OP=1+\frac{\sqrt3}{2}$

(a)

determine the exact value of k.

The finite region R, shown shaded in Figure 1, is bounded by C, the initial line and l.

[ 2 ]

At P,

1+\cos\theta=1+\frac{\sqrt3}{2}

so $\cos\theta=\frac{\sqrt3}{2}$ . Since $r=k\sec\theta$ ,

1+\frac{\sqrt3}{2}=k\cdot\frac{2}{\sqrt3}

Hence

k=\frac{\sqrt3}{2}+\frac34

(b)

Use algebraic integration to show that the area of R is

p \pi+q \sqrt{3}+r

where p, q and r are simplified rational numbers to be determined.

[ 7 ]

At P, $\theta=\frac{\pi}{6}$ .

For the curve,

\int(1+\cos\theta)^2\,d\theta=\frac32\theta+2\sin\theta+\frac14\sin2\theta

Thus

\frac12\left[\frac32\theta+2\sin\theta+\frac14\sin2\theta\right]_{\pi/6}^{\pi}=\frac{5\pi}{8}-\frac{\sqrt3}{16}-\frac12

The triangular area is

\frac12\left(\frac12+\frac{\sqrt3}{4}\right)\left(\frac{\sqrt3}{2}+\frac34\right)=\frac{7\sqrt3}{32}+\frac38

Therefore

\text{Area of }R=\frac{5\pi}{8}-\frac{\sqrt3}{16}-\frac12+\frac{7\sqrt3}{32}+\frac38

= =\frac{5\pi}{8}+\frac{5\sqrt3}{32}-\frac18.

So $p=\frac58$ , $q=\frac5{32}$ , and $r=-\frac18$ .

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