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P4 2022 Jan Q5

A Level / Edexcel / P4

IAL 2022 Jan Paper · Question 5

题目

Problem

Figure 2 shows a sketch of the curve with parametric equations

 x=9−4t, y=t39+4t, 0≤t≤94.\begin{align*} &\,x=\sqrt{9-4t},\\ &\,y=\frac{t^3}{\sqrt{9+4t}},\\ &\,0\leq t\leq \frac{9}{4}. \end{align*}

The curve touches the xx-axis when t=0t=0 and meets the yy-axis when t=94t=\dfrac{9}{4}

The region RR, shown shaded in Figure 2, is bounded by the curve, the xx-axis and the yy-axis.

(a) Show that the area of RR is given by

K∫094t381−16t2 dtK\int_0^{\frac{9}{4}} \frac{t^3}{\sqrt{81-16t^2}}\,\mathrm{d}t

where KK is a constant to be found.

(4)

(b) Using the substitution u=81−16t2u=81-16t^2, or otherwise, find the exact area of RR.

(Solutions relying on calculator technology are not acceptable.)

(6)
题目中文翻译

图 2 给出了曲线的草图，其参数方程为

 x=9−4t, y=t39+4t, 0≤t≤94.\begin{align*} &\,x=\sqrt{9-4t},\\ &\,y=\frac{t^3}{\sqrt{9+4t}},\\ &\,0\leq t\leq \frac{9}{4}. \end{align*}

当 t=0t=0 时，该曲线与 xx 轴相切；当 t=94t=\dfrac{9}{4} 时，该曲线与 yy 轴相交。

图 2 中阴影部分所示区域 RR 由该曲线、xx 轴和 yy 轴围成。

(a) 证明区域 RR 的面积可表示为

K∫094t381−16t2 dtK\int_0^{\frac{9}{4}} \frac{t^3}{\sqrt{81-16t^2}}\,\mathrm{d}t

其中 KK 为待求常数。

(4)

(b) 使用代换 u=81−16t2u=81-16t^2，或用其他方法，求区域 RR 的精确面积。

（不接受依赖计算器技术的解法。）

(6)

解答

(a)

解法一

思路

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参数 tt 从 00 增加到 94\frac94 时，xx 从 33 减小到 00，因此面积积分的参数上下限要反向。先求 dxdt\frac{\mathrm{d}x}{\mathrm{d}t}，再代入 A=∫y dxA=\int y\,\mathrm{d}x，并用平方差合并两个根式。

答题过程

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From

x=(9−4t)12,x=(9-4t)^{\frac12},

we have

dxdt=−29−4t.\frac{\mathrm{d}x}{\mathrm{d}t} =-\frac{2}{\sqrt{9-4t}}.

As tt increases from 00 to 94\frac94, xx decreases from 33 to 00. Hence

 A= ∫03y dx= ∫940ydxdt dt= ∫940t39+4t ×(−29−4t) dt= 2∫094t3(9+4t)(9−4t) dt= 2∫094t381−16t2 dt.\begin{align*} &\,A\\ =&\,\int_0^3y\,\mathrm{d}x\\ =&\,\int_{\frac94}^{0} y\frac{\mathrm{d}x}{\mathrm{d}t} \,\mathrm{d}t\\ =&\,\int_{\frac94}^{0} \frac{t^3}{\sqrt{9+4t}} \\ &\,\hspace{2pt}\times \bigg(-\frac{2}{\sqrt{9-4t}}\bigg) \,\mathrm{d}t\\ =&\,2\int_0^{\frac94} \frac{t^3} {\sqrt{(9+4t)(9-4t)}} \\ &\,\hspace{2pt}\mathrm{d}t\\ =&\,2\int_0^{\frac94} \frac{t^3}{\sqrt{81-16t^2}} \,\mathrm{d}t. \end{align*}

Therefore,

K=2.\boxed{K=2}.

(b)

解法一

思路

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使用题目指定的换元 u=81−16t2u=81-16t^2。把 t3 dtt^3\,\mathrm{d}t 拆成 t2(t dt)t^2(t\,\mathrm{d}t)，分别用 t2=81−u16t^2=\frac{81-u}{16} 和 t dt=−132 dut\,\mathrm{d}t=-\frac1{32}\,\mathrm{d}u 代换，同时更新积分上下限。

也可以直接令根式 v=81−16t2v=\sqrt{81-16t^2}，此时 t dt=−v16 dvt\,\mathrm{d}t=-\frac{v}{16}\,\mathrm{d}v，分母中的根式会约去，积分便化成多项式。

答题过程

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Let

u=81−16t2.u=81-16t^2.

Then

 du=−32t dt, t2=81−u16.\begin{align*} &\,\mathrm{d}u=-32t\,\mathrm{d}t,\\ &\,t^2=\frac{81-u}{16}. \end{align*}

The limits become

tu081940\begin{array}{c|c} t&u\\ \hline 0&81\\ \frac94&0 \end{array}

Using the result from part (a),

 A= 2∫094t2(t dt)u= 2∫81081−u16u ×(−132) du= 1256∫081(81u−12−u12) du= 1256[162u12−23u32]081= 1256(1458−486)= 24364.\begin{align*} &\,A\\ =&\,2\int_0^{\frac94} \frac{t^2(t\,\mathrm{d}t)}{\sqrt{u}}\\ =&\,2\int_{81}^{0} \frac{81-u}{16\sqrt{u}} \\ &\,\hspace{2pt}\times\bigg(-\frac1{32}\bigg)\,\mathrm{d}u\\ =&\,\frac1{256}\int_0^{81} \big(81u^{-\frac12}-u^{\frac12}\big) \\ &\,\hspace{2pt}\mathrm{d}u\\ =&\,\frac1{256} \Bigg[ 162u^{\frac12}-\frac23u^{\frac32} \Bigg]_0^{81}\\ =&\,\frac1{256} \big(1458-486\big)\\ =&\,\boxed{\frac{243}{64}}. \end{align*}

Alternatively, let

v=81−16t2.v=\sqrt{81-16t^2}.

Then

 v2=81−16t2, v dv=−16t dt, t2=81−v216.\begin{align*} &\,v^2=81-16t^2,\\ &\,v\,\mathrm{d}v=-16t\,\mathrm{d}t,\\ &\,t^2=\frac{81-v^2}{16}. \end{align*}

The limits t=0,94t=0,\frac94 correspond to v=9,0v=9,0. Hence

 A= 2∫9081−v216v ×(−v16) dv= 1128∫09(81−v2) dv= 1128[81v−v33]09= 729−243128= 24364.\begin{align*} &\,A\\ =&\,2\int_9^0\frac{81-v^2}{16v}\\ &\,\hspace{2pt}\times\bigg(-\frac{v}{16}\bigg)\,\mathrm{d}v\\ =&\,\frac1{128}\int_0^9(81-v^2)\,\mathrm{d}v\\ =&\,\frac1{128}\Big[81v-\frac{v^3}{3}\Big]_0^9\\ =&\,\frac{729-243}{128}\\ =&\,\boxed{\frac{243}{64}}. \end{align*}

解法二

思路

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把被积函数写成 t2⋅t81−16t2t^2\cdot\frac{t}{\sqrt{81-16t^2}}。后一个因子能直接积分成根式，所以先做一次分部积分；余下的 ∫t81−16t2 dt\int t\sqrt{81-16t^2}\,\mathrm{d}t 再直接换元即可。

答题过程

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Since

 ∫t81−16t2 dt= −11681−16t2,\begin{align*} &\,\int\frac{t}{\sqrt{81-16t^2}}\,\mathrm{d}t\\ =&\,-\frac1{16}\sqrt{81-16t^2}, \end{align*}

integration by parts gives

 ∫t381−16t2 dt= −t21681−16t2 +18∫t81−16t2 dt.\begin{align*} &\,\int\frac{t^3}{\sqrt{81-16t^2}}\,\mathrm{d}t\\ =&\,-\frac{t^2}{16}\sqrt{81-16t^2}\\ &\,\hspace{2pt} +\frac18\int t\sqrt{81-16t^2}\,\mathrm{d}t. \end{align*}

Also,

 ∫t81−16t2 dt= −148(81−16t2)32.\begin{align*} &\,\int t\sqrt{81-16t^2}\,\mathrm{d}t\\ =&\,-\frac1{48}(81-16t^2)^{\frac32}. \end{align*}

Therefore, using K=2K=2 from part (a),

 A= [−t2881−16t2 −1192(81−16t2)32]094= 0−(−729192)= 24364.\begin{align*} &\,A\\ =&\,\Bigg[ -\frac{t^2}{8}\sqrt{81-16t^2}\\ &\,\hspace{2pt} -\frac1{192}(81-16t^2)^{\frac32} \Bigg]_0^{\frac94}\\ =&\,0-\bigg(-\frac{729}{192}\bigg)\\ =&\,\boxed{\frac{243}{64}}. \end{align*}