In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Figure 3 The curve C 1 C_1 C 1
r = 3 + tan θ 0 < θ < π 2 \begin{align*}
r = \sqrt{3} + \tan\theta \qquad 0 < \theta < \frac{\pi}{2}
\end{align*} r = 3 + tan θ 0 < θ < 2 π The tangent to C 1 C_1 C 1 P P P
(a) Use calculus to determine, in simplest form, the exact polar coordinates of P P P
(4)
Figure 3 shows a sketch of part of the curve C 1 C_1 C 1 C 2 C_2 C 2
The curve C 2 C_2 C 2 O O O
The curves C 1 C_1 C 1 C 2 C_2 C 2 P P P
The region R R R C 1 C_1 C 1 C 2 C_2 C 2
(b) Use algebraic integration to determine the area of R R R
a π + 3 2 ( ln b + c ) \begin{align*}
a\pi + \frac{\sqrt{3}}{2}(\ln b + c)
\end{align*} aπ + 2 3 ( ln b + c ) where a a a b b b c c c
(8)
解答
(a)
For curve C 1 C_1 C 1 r = 3 + tan θ r = \sqrt{3} + \tan\theta r = 3 + tan θ d x d θ = 0 \frac{\mathrm{d}x}{\mathrm{d}\theta} = 0 d θ d x = 0
x = r cos θ = ( 3 + tan θ ) cos θ = 3 cos θ + sin θ \begin{align*}
x
= &\,r\cos\theta\\[4mm]
= &\,(\sqrt{3} + \tan\theta)\cos\theta\\[4mm]
= &\,\sqrt{3}\cos\theta + \sin\theta
\end{align*} x = = = r cos θ ( 3 + tan θ ) cos θ 3 cos θ + sin θ Differentiating with respect to θ \theta θ
d x d θ = − 3 sin θ + cos θ \begin{align*}
\frac{\mathrm{d}x}{\mathrm{d}\theta}
= &\,-\sqrt{3}\sin\theta + \cos\theta
\end{align*} d θ d x = − 3 sin θ + cos θ Setting d x d θ = 0 \frac{\mathrm{d}x}{\mathrm{d}\theta} = 0 d θ d x = 0
3 sin θ = cos θ tan θ = 1 3 θ = π 6 \begin{align*}
\sqrt{3}\sin\theta
= &\,\cos\theta\\[4mm]
\tan\theta
= &\,\frac{1}{\sqrt{3}}\\[4mm]
\theta
= &\,\frac{\pi}{6}
\end{align*} 3 sin θ = tan θ = θ = cos θ 3 1 6 π Substitute θ = π 6 \theta = \frac{\pi}{6} θ = 6 π r r r
r = 3 + tan ( π 6 ) = 3 + 1 3 = 3 3 + 1 3 = 4 3 or 4 3 3 \begin{align*}
r
= &\,\sqrt{3} + \tan\left(\frac{\pi}{6}\right)\\[4mm]
= &\,\sqrt{3} + \frac{1}{\sqrt{3}}\\[4mm]
= &\,\frac{3}{\sqrt{3}} + \frac{1}{\sqrt{3}}\\[4mm]
= &\,\frac{4}{\sqrt{3}} \quad \text{or} \quad \frac{4\sqrt{3}}{3}
\end{align*} r = = = = 3 + tan ( 6 π ) 3 + 3 1 3 3 + 3 1 3 4 or 3 4 3 Thus, the exact polar coordinates of P P P ( 4 3 3 , π 6 ) \left( \frac{4\sqrt{3}}{3}, \frac{\pi}{6} \right) ( 3 4 3 , 6 π )
(b)
The region R R R C 1 C_1 C 1 C 2 C_2 C 2 R R R C 2 C_2 C 2 C 1 C_1 C 1 θ = 0 \theta=0 θ = 0 θ = π 6 \theta=\frac{\pi}{6} θ = 6 π
C 2 C_2 C 2 O O O P P P R = 4 3 3 R = \frac{4\sqrt{3}}{3} R = 3 4 3
The area of the sector of C 2 C_2 C 2 0 0 0 π 6 \frac{\pi}{6} 6 π
Area of Sector = 1 2 R 2 θ = 1 2 ( 4 3 3 ) 2 ( π 6 ) = 1 2 ( 48 9 ) ( π 6 ) = 4 π 9 \begin{align*}
\text{Area of Sector}
= &\,\frac{1}{2} R^2 \theta\\[4mm]
= &\,\frac{1}{2} \left(\frac{4\sqrt{3}}{3}\right)^2 \left(\frac{\pi}{6}\right)\\[4mm]
= &\,\frac{1}{2} \left(\frac{48}{9}\right) \left(\frac{\pi}{6}\right)\\[4mm]
= &\,\frac{4\pi}{9}
\end{align*} Area of Sector = = = = 2 1 R 2 θ 2 1 ( 3 4 3 ) 2 ( 6 π ) 2 1 ( 9 48 ) ( 6 π ) 9 4 π The area under C 1 C_1 C 1 θ = 0 \theta=0 θ = 0 θ = π 6 \theta=\frac{\pi}{6} θ = 6 π
1 2 ∫ 0 π 6 r 2 d θ = 1 2 ∫ 0 π 6 ( 3 + tan θ ) 2 d θ = 1 2 ∫ 0 π 6 ( 3 + 2 3 tan θ + tan 2 θ ) d θ = 1 2 ∫ 0 π 6 ( 3 + 2 3 tan θ + sec 2 θ − 1 ) d θ = 1 2 ∫ 0 π 6 ( 2 + 2 3 tan θ + sec 2 θ ) d θ = 1 2 [ 2 θ + 2 3 ln ∣ sec θ ∣ + tan θ ] 0 π 6 = [ θ − 3 ln ∣ cos θ ∣ + 1 2 tan θ ] 0 π 6 \begin{align*}
\frac{1}{2} \int_{0}^{\frac{\pi}{6}} r^2 \,\mathrm{d}\theta
= &\,\frac{1}{2} \int_{0}^{\frac{\pi}{6}} (\sqrt{3} + \tan\theta)^2 \,\mathrm{d}\theta\\[4mm]
= &\,\frac{1}{2} \int_{0}^{\frac{\pi}{6}} (3 + 2\sqrt{3}\tan\theta + \tan^2\theta) \,\mathrm{d}\theta\\[4mm]
= &\,\frac{1}{2} \int_{0}^{\frac{\pi}{6}} (3 + 2\sqrt{3}\tan\theta + \sec^2\theta - 1) \,\mathrm{d}\theta\\[4mm]
= &\,\frac{1}{2} \int_{0}^{\frac{\pi}{6}} (2 + 2\sqrt{3}\tan\theta + \sec^2\theta) \,\mathrm{d}\theta\\[4mm]
= &\,\frac{1}{2} \left[ 2\theta + 2\sqrt{3}\ln|\sec\theta| + \tan\theta \right]_{0}^{\frac{\pi}{6}}\\[4mm]
= &\,\left[ \theta - \sqrt{3}\ln|\cos\theta| + \frac{1}{2}\tan\theta \right]_{0}^{\frac{\pi}{6}}
\end{align*} 2 1 ∫ 0 6 π r 2 d θ = = = = = = 2 1 ∫ 0 6 π ( 3 + tan θ ) 2 d θ 2 1 ∫ 0 6 π ( 3 + 2 3 tan θ + tan 2 θ ) d θ 2 1 ∫ 0 6 π ( 3 + 2 3 tan θ + sec 2 θ − 1 ) d θ 2 1 ∫ 0 6 π ( 2 + 2 3 tan θ + sec 2 θ ) d θ 2 1 [ 2 θ + 2 3 ln ∣ sec θ ∣ + tan θ ] 0 6 π [ θ − 3 ln ∣ cos θ ∣ + 2 1 tan θ ] 0 6 π Evaluating at the limits:
Upper limit: π 6 − 3 ln ( cos π 6 ) + 1 2 tan ( π 6 ) = π 6 − 3 ln ( 3 2 ) + 1 2 ( 3 3 ) = π 6 − 3 ln ( 3 2 ) + 3 6 Lower limit: 0 − 3 ln ( 1 ) + 1 2 ( 0 ) = 0 \begin{align*}
\text{Upper limit: } & \frac{\pi}{6} - \sqrt{3}\ln\left(\cos\frac{\pi}{6}\right) + \frac{1}{2}\tan\left(\frac{\pi}{6}\right)\\[4mm]
= &\,\frac{\pi}{6} - \sqrt{3}\ln\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{2}\left(\frac{\sqrt{3}}{3}\right)\\[4mm]
= &\,\frac{\pi}{6} - \sqrt{3}\ln\left(\frac{\sqrt{3}}{2}\right) + \frac{\sqrt{3}}{6}\\[4mm]
\text{Lower limit: } & 0 - \sqrt{3}\ln(1) + \frac{1}{2}(0) = 0
\end{align*} Upper limit: = = Lower limit: 6 π − 3 ln ( cos 6 π ) + 2 1 tan ( 6 π ) 6 π − 3 ln ( 2 3 ) + 2 1 ( 3 3 ) 6 π − 3 ln ( 2 3 ) + 6 3 0 − 3 ln ( 1 ) + 2 1 ( 0 ) = 0 So, the area under C 1 C_1 C 1 π 6 − 3 ln ( 3 2 ) + 3 6 \frac{\pi}{6} - \sqrt{3}\ln\left(\frac{\sqrt{3}}{2}\right) + \frac{\sqrt{3}}{6} 6 π − 3 ln ( 2 3 ) + 6 3
The area of region R R R
Area of R = 4 π 9 − ( π 6 − 3 ln ( 3 2 ) + 3 6 ) = 8 π 18 − 3 π 18 + 3 ln ( 3 2 ) − 3 6 = 5 π 18 + 3 2 ⋅ 2 ln ( 3 2 ) − 3 6 = 5 π 18 + 3 2 ln ( 3 4 ) − 3 2 ( 1 3 ) = 5 π 18 + 3 2 ( ln ( 3 4 ) − 1 3 ) \begin{align*}
\text{Area of } R
= &\,\frac{4\pi}{9} - \left( \frac{\pi}{6} - \sqrt{3}\ln\left(\frac{\sqrt{3}}{2}\right) + \frac{\sqrt{3}}{6} \right)\\[4mm]
= &\,\frac{8\pi}{18} - \frac{3\pi}{18} + \sqrt{3}\ln\left(\frac{\sqrt{3}}{2}\right) - \frac{\sqrt{3}}{6}\\[4mm]
= &\,\frac{5\pi}{18} + \frac{\sqrt{3}}{2} \cdot 2\ln\left(\frac{\sqrt{3}}{2}\right) - \frac{\sqrt{3}}{6}\\[4mm]
= &\,\frac{5\pi}{18} + \frac{\sqrt{3}}{2} \ln\left(\frac{3}{4}\right) - \frac{\sqrt{3}}{2}\left(\frac{1}{3}\right)\\[4mm]
= &\,\frac{5\pi}{18} + \frac{\sqrt{3}}{2} \left( \ln\left(\frac{3}{4}\right) - \frac{1}{3} \right)
\end{align*} Area of R = = = = = 9 4 π − ( 6 π − 3 ln ( 2 3 ) + 6 3 ) 18 8 π − 18 3 π + 3 ln ( 2 3 ) − 6 3 18 5 π + 2 3 ⋅ 2 ln ( 2 3 ) − 6 3 18 5 π + 2 3 ln ( 4 3 ) − 2 3 ( 3 1 ) 18 5 π + 2 3 ( ln ( 4 3 ) − 3 1 ) This is in the required form a π + 3 2 ( ln b + c ) a\pi + \frac{\sqrt{3}}{2}(\ln b + c) aπ + 2 3 ( ln b + c ) a = 5 18 a=\frac{5}{18} a = 18 5 b = 3 4 b=\frac{3}{4} b = 4 3 c = − 1 3 c=-\frac{1}{3} c = − 3 1
