CalcGospel 國際數學圖譜
返回

FP3 2020 Oct Q7

A Level / Edexcel / FP3

IAL 2020 Oct Paper · Question 7

题目

Problem

7.

The curve CC has parametric equations

x=cosh⁡t+t,y=cosh⁡t−t,0≤t≤ln⁡3.x = \cosh t + t, \qquad y = \cosh t - t, \qquad 0 \le t \le \ln 3.

(a) Show that

(dxdt)2+(dydt)2=2cosh⁡2t.\big(\frac{\mathrm{d}x}{\mathrm{d}t}\big)^2 + \big(\frac{\mathrm{d}y}{\mathrm{d}t}\big)^2 = 2\cosh^2 t.
(3)

The curve CC is rotated through 2π2\pi radians about the xx-axis. The area of the curved surface generated is given by SS.

(b) Show that

S=2π2∫0ln⁡3(cosh⁡2t−tcosh⁡t) dt.S = 2\pi\sqrt{2}\int_0^{\ln 3} (\cosh^2 t - t\cosh t)\,\mathrm{d}t.
(2)

(c) Hence find the value of SS, giving your answer in the form

π29(a+bln⁡3)\frac{\pi\sqrt{2}}{9}(a + b\ln 3)

where aa and bb are constants to be determined.

(7)
题目中文翻译

曲线 CC 的参数方程为

x=cosh⁡t+t,y=cosh⁡t−t,0≤t≤ln⁡3x = \cosh t + t, \qquad y = \cosh t - t, \qquad 0 \le t \le \ln 3

(a) 证明

(dxdt)2+(dydt)2=2cosh⁡2t\big(\frac{\mathrm{d}x}{\mathrm{d}t}\big)^2 + \big(\frac{\mathrm{d}y}{\mathrm{d}t}\big)^2 = 2\cosh^2 t
(3)

将曲线 CC 绕 xx 轴旋转 2π2\pi 弧度。所生成的曲面面积记为 SS。

(b) 证明

S=2π2∫0ln⁡3(cosh⁡2t−tcosh⁡t) dtS = 2\pi\sqrt{2}\int_0^{\ln 3} (\cosh^2 t - t\cosh t)\,\mathrm{d}t
(2)

(c) 因此求 SS 的值，答案写成

π29(a+bln⁡3)\frac{\pi\sqrt{2}}{9}(a + b\ln 3)

的形式，其中 aa 和 bb 为待求常数。

(7)

解答

(a)

解法一

思路

展开

分别对两个参数方程求导。平方相加后一次项抵消，再用 1+sinh⁡2t=cosh⁡2t1+\sinh^2t=\cosh^2t 化为题目要求的形式。

答题过程

展开

Differentiating with respect to tt,

dxdt=sinh⁡t+1,dydt=sinh⁡t−1.\frac{\mathrm{d}x}{\mathrm{d}t}=\sinh t+1, \qquad \frac{\mathrm{d}y}{\mathrm{d}t}=\sinh t-1.

Therefore

(dxdt)2+(dydt)2= (sinh⁡t+1)2+(sinh⁡t−1)2= sinh⁡2t+2sinh⁡t+1 +sinh⁡2t−2sinh⁡t+1= 2sinh⁡2t+2= 2cosh⁡2t.\begin{align*} \big(\frac{\mathrm{d}x}{\mathrm{d}t}\big)^2 +\big(\frac{\mathrm{d}y}{\mathrm{d}t}\big)^2 =&\,(\sinh t+1)^2+(\sinh t-1)^2\\[4mm] =&\,\sinh^2t+2\sinh t+1\\ &\,+\sinh^2t-2\sinh t+1\\ =&\,2\sinh^2t+2\\[4mm] =&\,\boxed{2\cosh^2t}. \end{align*}

(b)

解法一

思路

展开

绕 xx 轴旋转的曲面面积为 2π∫y ds2\pi\int y\,\mathrm{d}s。由 (a) 求出 ds/dt\mathrm{d}s/\mathrm{d}t；在给定区间内 cosh⁡t>0\cosh t>0，所以开平方时不产生符号歧义。

答题过程

展开

For a parametric curve rotated about the xx-axis,

S=2π∫y(dxdt)2+(dydt)2 dt.S=2\pi\int y \sqrt{\big(\frac{\mathrm{d}x}{\mathrm{d}t}\big)^2 +\big(\frac{\mathrm{d}y}{\mathrm{d}t}\big)^2} \,\mathrm{d}t.

Using part (a), and noting that cosh⁡t>0\cosh t>0,

2cosh⁡2t=2cosh⁡t.\sqrt{2\cosh^2t}=\sqrt2\cosh t.

Hence

S= 2π∫0ln⁡3(cosh⁡t−t)2cosh⁡t dt= 2π2∫0ln⁡3(cosh⁡2t−tcosh⁡t) dt.\begin{align*} S =&\,2\pi\int_0^{\ln3} (\cosh t-t)\sqrt2\cosh t\,\mathrm{d}t\\[4mm] =&\,\boxed{2\pi\sqrt2 \int_0^{\ln3} (\cosh^2t-t\cosh t)\,\mathrm{d}t}. \end{align*}

(c)

解法一

思路

展开

承接 (b)，用 cosh⁡2t=(1+cosh⁡2t)/2\cosh^2t=(1+\cosh2t)/2 积分第一项；第二项 ∫tcosh⁡t dt\int t\cosh t\,\mathrm{d}t 用分部积分。代入 t=ln⁡3t=\ln3 时，从指数定义精确求出 sinh⁡t\sinh t、cosh⁡t\cosh t 与 sinh⁡2t\sinh2t。

答题过程

展开

Using

cosh⁡2t=12(1+cosh⁡2t),\cosh^2t=\frac12(1+\cosh2t),

we obtain

∫cosh⁡2t dt=t2+14sinh⁡2t.\int\cosh^2t\,\mathrm{d}t =\frac t2+\frac14\sinh2t.

Also, integration by parts gives

∫tcosh⁡t dt= tsinh⁡t−∫sinh⁡t dt= tsinh⁡t−cosh⁡t.\begin{align*} \int t\cosh t\,\mathrm{d}t =&\,t\sinh t-\int\sinh t\,\mathrm{d}t\\[4mm] =&\,t\sinh t-\cosh t. \end{align*}

Therefore, from part (b),

S=2π2[t2+14sinh⁡2t−tsinh⁡t+cosh⁡t]0ln⁡3.S=2\pi\sqrt2 \big[ \frac t2+\frac14\sinh2t -t\sinh t+\cosh t \big]_0^{\ln3}.

At t=ln⁡3t=\ln3,

sinh⁡t=3−1/32=43,cosh⁡t=3+1/32=53,\sinh t=\frac{3-1/3}{2}=\frac43, \qquad \cosh t=\frac{3+1/3}{2}=\frac53,

and hence

sinh⁡2t=2sinh⁡tcosh⁡t=409.\sinh2t=2\sinh t\cosh t=\frac{40}{9}.

Thus

S= 2π2(12ln⁡3+109−43ln⁡3+53−1)= 2π2(169−56ln⁡3)= π29(32−15ln⁡3).\begin{align*} S =&\,2\pi\sqrt2 \big( \frac12\ln3+\frac{10}{9} -\frac43\ln3+\frac53-1 \big)\\[4mm] =&\,2\pi\sqrt2 \big(\frac{16}{9}-\frac56\ln3\big)\\[4mm] =&\,\boxed{\frac{\pi\sqrt2}{9} \bigl(32-15\ln3\bigr)}. \end{align*}

Therefore

a=32,b=−15.\boxed{a=32,\qquad b=-15}.

解法二

思路

展开

将 (b) 的被积式写成指数形式。平方项直接逐项积分，tette^t 与 te−tte^{-t} 分别使用分部积分，再代入同一组上下限。

答题过程

展开

From part (b),

S2π2=∫0ln⁡3(cosh⁡2t−tcosh⁡t) dt.\frac{S}{2\pi\sqrt2} =\int_0^{\ln3}(\cosh^2t-t\cosh t)\,\mathrm{d}t.

Using exponential definitions,

cosh⁡2t=14(e2t+2+e−2t),tcosh⁡t=12(tet+te−t).\cosh^2t=\frac14(e^{2t}+2+e^{-2t}), \qquad t\cosh t=\frac12(te^t+te^{-t}).

Integration by parts gives

∫tet dt= tet−∫et dt=(t−1)et,∫te−t dt= −te−t+∫e−t dt=−(t+1)e−t.\begin{align*} \int te^t\,\mathrm{d}t=&\,te^t-\int e^t\,\mathrm{d}t=(t-1)e^t,\\ \int te^{-t}\,\mathrm{d}t=&\,-te^{-t}+\int e^{-t}\,\mathrm{d}t =-(t+1)e^{-t}. \end{align*}

An antiderivative of cosh⁡2t−tcosh⁡t\cosh^2t-t\cosh t is therefore

F(t)= 18e2t+t2−18e−2t −12(t−1)et+12(t+1)e−t.\begin{align*} F(t)=&\,\frac18e^{2t}+\frac t2-\frac18e^{-2t}\\ &\,-\frac12(t-1)e^t+\frac12(t+1)e^{-t}. \end{align*}

Since eln⁡3=3e^{\ln3}=3,

F(ln⁡3)= 98+12ln⁡3−172 −32(ln⁡3−1)+16(ln⁡3+1)= 259−56ln⁡3,F(0)= 1.\begin{align*} F(\ln3)=&\,\frac98+\frac12\ln3-\frac1{72}\\ &\,-\frac32(\ln3-1)+\frac16(\ln3+1)\\ =&\,\frac{25}{9}-\frac56\ln3,\\ F(0)=&\,1. \end{align*}

Hence

S= 2π2(F(ln⁡3)−F(0))= 2π2(169−56ln⁡3)= π29(32−15ln⁡3).\begin{align*} S=&\,2\pi\sqrt2\big(F(\ln3)-F(0)\big)\\ =&\,2\pi\sqrt2\bigg(\frac{16}{9}-\frac56\ln3\bigg)\\ =&\,\boxed{\frac{\pi\sqrt2}{9}(32-15\ln3)}. \end{align*}

Thus a=32,b=−15\boxed{a=32,\quad b=-15}.

解法三

思路

展开

先把被积式写成 cosh⁡t(cosh⁡t−t)\cosh t(\cosh t-t)，整体进行分部积分。对余项再做一次分部积分，含 sinh⁡tcosh⁡t\sinh t\cosh t 的乘积相消，最后只需计算 ∫cosh⁡2t dt\int\cosh^2t\,\mathrm{d}t。

答题过程

展开

For the integral in part (b), let

J=∫cosh⁡t(cosh⁡t−t) dt.J=\int\cosh t(\cosh t-t)\,\mathrm{d}t.

Integrating by parts with u=cosh⁡t−tu=\cosh t-t and dv=cosh⁡t dt\mathrm{d}v=\cosh t\,\mathrm{d}t,

J=(cosh⁡t−t)sinh⁡t−∫sinh⁡t(sinh⁡t−1) dt.J=(\cosh t-t)\sinh t -\int\sinh t(\sinh t-1)\,\mathrm{d}t.

A second integration by parts gives

∫sinh⁡t(sinh⁡t−1) dt=(sinh⁡t−1)cosh⁡t−∫cosh⁡2t dt.\int\sinh t(\sinh t-1)\,\mathrm{d}t =(\sinh t-1)\cosh t-\int\cosh^2t\,\mathrm{d}t.

Consequently,

J= (cosh⁡t−t)sinh⁡t−(sinh⁡t−1)cosh⁡t +∫cosh⁡2t dt= −tsinh⁡t+cosh⁡t+∫1+cosh⁡2t2 dt= −tsinh⁡t+cosh⁡t+t2+14sinh⁡2t.\begin{align*} J=&\,(\cosh t-t)\sinh t-(\sinh t-1)\cosh t\\ &\,+\int\cosh^2t\,\mathrm{d}t\\ =&\,-t\sinh t+\cosh t+\int\frac{1+\cosh2t}{2}\,\mathrm{d}t\\ =&\,-t\sinh t+\cosh t+\frac t2+\frac14\sinh2t. \end{align*}

Using sinh⁡(ln⁡3)=4/3\sinh(\ln3)=4/3, cosh⁡(ln⁡3)=5/3\cosh(\ln3)=5/3 and sinh⁡(2ln⁡3)=40/9\sinh(2\ln3)=40/9,

S= 2π2 [J]0ln⁡3= 2π2(−43ln⁡3+53+12ln⁡3+109−1)= π29(32−15ln⁡3).\begin{align*} S=&\,2\pi\sqrt2\,[J]_0^{\ln3}\\ =&\,2\pi\sqrt2\bigg(-\frac43\ln3+\frac53 +\frac12\ln3+\frac{10}{9}-1\bigg)\\ =&\,\boxed{\frac{\pi\sqrt2}{9}(32-15\ln3)}. \end{align*}

Therefore a=32,b=−15\boxed{a=32,\quad b=-15}.