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IAL 2021 Jan Q7

A Level / Edexcel / FP2

IAL 2021 Jan Paper · Question 7

题目

Problem

Figure 1

Figure 1 shows a sketch of curve CC with polar equation

r=3sin⁡2θ0⩽θ⩽π2\begin{align*} r = 3\sin 2\theta \qquad 0 \leqslant \theta \leqslant \frac{\pi}{2} \end{align*}

The point PP on CC has polar coordinates (R,ϕ)(R,\phi). The tangent to CC at PP is perpendicular to the initial line.

(a) Show that tan⁡ϕ=12\tan\phi = \frac{1}{\sqrt{2}}

(4)

(b) Determine the exact value of RR.

(2)

The region SS, shown shaded in Figure 1, is bounded by CC and the line OPOP, where OO is the pole.

(c) Use calculus to show that the exact area of SS is

parctan⁡12+q2\begin{align*} p\arctan\frac{1}{\sqrt{2}} + q\sqrt{2} \end{align*}

where pp and qq are constants to be determined.

Solutions relying entirely on calculator technology are not acceptable.
(7)

解答

(a)

解法一

思路

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切线垂直于 initial line，表示切线是竖直方向。用 x=rcos⁡θx=r\cos\theta，竖直切线对应 dxdθ=0\dfrac{\mathrm{d}x}{\mathrm{d}\theta}=0。这一解法先把 sin⁡2θ\sin2\theta 展开，微分会比较直接。

答题过程

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Since

x=rcos⁡θ,\begin{align*} x=r\cos\theta, \end{align*}

we have

x= 3sin⁡2θcos⁡θ= 6sin⁡θcos⁡2θ.\begin{align*} x=&\,3\sin2\theta\cos\theta\\[4mm] =&\,6\sin\theta\cos^2\theta. \end{align*}

For a tangent perpendicular to the initial line,

dxdθ=0.\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}\theta}=0. \end{align*}

Now

dxdθ= 6cos⁡θcos⁡2θ−12sin⁡2θcos⁡θ= 6cos⁡θ(cos⁡2θ−2sin⁡2θ).\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}\theta} =&\,6\cos\theta\cos^2\theta-12\sin^2\theta\cos\theta\\[4mm] =&\,6\cos\theta(\cos^2\theta-2\sin^2\theta). \end{align*}

At PP, θ=ϕ\theta=\phi and 0<ϕ<π20<\phi<\dfrac{\pi}{2}, so cos⁡ϕ≠0\cos\phi\neq0. Therefore,

cos⁡2ϕ−2sin⁡2ϕ= 0cos⁡2ϕ= 2sin⁡2ϕtan⁡2ϕ= 12.\begin{align*} \cos^2\phi-2\sin^2\phi=&\,0\\[4mm] \cos^2\phi=&\,2\sin^2\phi\\[4mm] \tan^2\phi=&\,\frac12. \end{align*}

Since ϕ\phi is in the first quadrant,

tan⁡ϕ=12.\begin{align*} \tan\phi=\frac1{\sqrt2}. \end{align*}

解法二

思路

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也可以不先展开成 6sin⁡θcos⁡2θ6\sin\theta\cos^2\theta，而是直接对

x=3sin⁡2θcos⁡θ\begin{align*} x=3\sin2\theta\cos\theta \end{align*}

使用 product rule。最后仍然会得到同一个条件。

答题过程

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Since

x=rcos⁡θ=3sin⁡2θcos⁡θ,\begin{align*} x=r\cos\theta=3\sin2\theta\cos\theta, \end{align*}

a tangent perpendicular to the initial line requires

dxdθ=0.\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}\theta}=0. \end{align*}

Differentiate using the product rule:

dxdθ= 6cos⁡2θcos⁡θ−3sin⁡2θsin⁡θ.\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}\theta} =&\,6\cos2\theta\cos\theta -3\sin2\theta\sin\theta. \end{align*}

At θ=ϕ\theta=\phi,

6cos⁡2ϕcos⁡ϕ−3sin⁡2ϕsin⁡ϕ= 02cos⁡2ϕcos⁡ϕ−sin⁡2ϕsin⁡ϕ= 0.\begin{align*} 6\cos2\phi\cos\phi -3\sin2\phi\sin\phi =&\,0\\[4mm] 2\cos2\phi\cos\phi -\sin2\phi\sin\phi =&\,0. \end{align*}

Using sin⁡2ϕ=2sin⁡ϕcos⁡ϕ\sin2\phi=2\sin\phi\cos\phi and cos⁡2ϕ=cos⁡2ϕ−sin⁡2ϕ\cos2\phi=\cos^2\phi-\sin^2\phi,

2(cos⁡2ϕ−sin⁡2ϕ)cos⁡ϕ−2sin⁡2ϕcos⁡ϕ= 02cos⁡ϕ(cos⁡2ϕ−2sin⁡2ϕ)= 0.\begin{align*} 2(\cos^2\phi-\sin^2\phi)\cos\phi -2\sin^2\phi\cos\phi =&\,0\\[4mm] 2\cos\phi(\cos^2\phi-2\sin^2\phi) =&\,0. \end{align*}

Since 0<ϕ<π20<\phi<\dfrac{\pi}{2}, cos⁡ϕ≠0\cos\phi\neq0. Therefore,

cos⁡2ϕ= 2sin⁡2ϕtan⁡2ϕ= 12.\begin{align*} \cos^2\phi=&\,2\sin^2\phi\\[4mm] \tan^2\phi=&\,\frac12. \end{align*}

Since ϕ\phi is in the first quadrant,

tan⁡ϕ=12.\begin{align*} \tan\phi=\frac1{\sqrt2}. \end{align*}

(b)

解法一

思路

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由 tan⁡ϕ=12\tan\phi=\dfrac1{\sqrt2} 可取一个直角三角形：opposite =1=1，adjacent =2=\sqrt2，hypotenuse =3=\sqrt3。然后代入 R=3sin⁡2ϕR=3\sin2\phi。

答题过程

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From

tan⁡ϕ=12,\begin{align*} \tan\phi=\frac1{\sqrt2}, \end{align*}

we have

sin⁡ϕ=13,cos⁡ϕ=23.\begin{align*} \sin\phi=\frac1{\sqrt3},\qquad \cos\phi=\frac{\sqrt2}{\sqrt3}. \end{align*}

Therefore,

R= 3sin⁡2ϕ= 6sin⁡ϕcos⁡ϕ= 6(13)(23)= 22.\begin{align*} R =&\,3\sin2\phi\\[4mm] =&\,6\sin\phi\cos\phi\\[4mm] =&\,6\left(\frac1{\sqrt3}\right)\left(\frac{\sqrt2}{\sqrt3}\right)\\[4mm] =&\,2\sqrt2. \end{align*}

(c)

解法一

思路

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阴影区域从 θ=0\theta=0 到 θ=ϕ\theta=\phi，所以面积用

12∫0ϕr2 dθ.\begin{align*} \frac12\int_0^\phi r^2\,\mathrm{d}\theta. \end{align*}

最后要把 ϕ=arctan⁡12\phi=\arctan\dfrac1{\sqrt2} 和 sin⁡4ϕ\sin4\phi 的精确值代入。

答题过程

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The area of SS is

12∫0ϕr2 dθ= 12∫0ϕ9sin⁡22θ dθ= 92∫0ϕ1−cos⁡4θ2 dθ= 94[θ−14sin⁡4θ]0ϕ= 94ϕ−916sin⁡4ϕ.\begin{align*} \frac12\int_0^\phi r^2\,\mathrm{d}\theta =&\,\frac12\int_0^\phi 9\sin^2 2\theta\,\mathrm{d}\theta\\[4mm] =&\,\frac92\int_0^\phi \frac{1-\cos4\theta}{2}\,\mathrm{d}\theta\\[4mm] =&\,\frac94\left[\theta-\frac14\sin4\theta\right]_0^\phi\\[4mm] =&\,\frac94\phi-\frac9{16}\sin4\phi. \end{align*}

Now

ϕ=arctan⁡12.\begin{align*} \phi=\arctan\frac1{\sqrt2}. \end{align*}

Also,

sin⁡4ϕ= 2sin⁡2ϕcos⁡2ϕ= 4sin⁡ϕcos⁡ϕ(cos⁡2ϕ−sin⁡2ϕ)= 4(13)(23)(23−13)= 429.\begin{align*} \sin4\phi =&\,2\sin2\phi\cos2\phi\\[4mm] =&\,4\sin\phi\cos\phi(\cos^2\phi-\sin^2\phi)\\[4mm] =&\,4\left(\frac1{\sqrt3}\right)\left(\frac{\sqrt2}{\sqrt3}\right) \left(\frac23-\frac13\right)\\[4mm] =&\,\frac{4\sqrt2}{9}. \end{align*}

Therefore,

Area of S= 94arctan⁡12−916⋅429= 94arctan⁡12−24.\begin{align*} \text{Area of }S =&\,\frac94\arctan\frac1{\sqrt2} -\frac9{16}\cdot\frac{4\sqrt2}{9}\\[4mm] =&\,\frac94\arctan\frac1{\sqrt2}-\frac{\sqrt2}{4}. \end{align*}

Thus

p=94,q=−14.\begin{align*} p=\frac94,\qquad q=-\frac14. \end{align*}