题目
Problem
(a) Starting from the definition of cosh in terms of exponentials, prove that
2 cosh 2 A = cosh 2 A + 1. 2\cosh^2A=\cosh 2A+1. 2 cosh 2 A = cosh 2 A + 1. [3]
The curve C C C
x = 2 cosh 2 t + 3 t , y = 3 2 cosh 2 t − 4 t , − 1 2 ≤ t ≤ 1 2 . \begin{align*}
x=&\,2\cosh 2t+3t,\\
y=&\,\frac{3}{2}\cosh 2t-4t,\qquad -\frac{1}{2}\le t\le\frac{1}{2}.
\end{align*} x = y = 2 cosh 2 t + 3 t , 2 3 cosh 2 t − 4 t , − 2 1 ≤ t ≤ 2 1 . The area of the surface generated when C C C 2 π 2\pi 2 π y y y A A A
(b) (i) Show that
A = 10 π ∫ − 1 2 1 2 ( 2 cosh 2 t + 3 t ) cosh 2 t d t . A=10\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\big(2\cosh 2t+3t\big)\cosh 2t\,\mathrm{d}t. A = 10 π ∫ − 2 1 2 1 ( 2 cosh 2 t + 3 t ) cosh 2 t d t . [4]
(ii) Hence find A A A π \pi π e e e
[7]
题目中文翻译
(a) 从用指数表示的 cosh \cosh cosh
2 cosh 2 A = cosh 2 A + 1 2\cosh^2A=\cosh 2A+1 2 cosh 2 A = cosh 2 A + 1 (3)
曲线 C C C
x = 2 cosh 2 t + 3 t , y = 3 2 cosh 2 t − 4 t , − 1 2 ≤ t ≤ 1 2 . \begin{align*}
x=&\,2\cosh 2t+3t,\\
y=&\,\frac{3}{2}\cosh 2t-4t,\qquad -\frac{1}{2}\le t\le\frac{1}{2}.
\end{align*} x = y = 2 cosh 2 t + 3 t , 2 3 cosh 2 t − 4 t , − 2 1 ≤ t ≤ 2 1 . 将曲线 C C C y y y 2 π 2\pi 2 π A A A
(b) (i) 证明
A = 10 π ∫ − 1 2 1 2 ( 2 cosh 2 t + 3 t ) cosh 2 t d t A=10\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\big(2\cosh 2t+3t\big)\cosh 2t\,\mathrm{d}t A = 10 π ∫ − 2 1 2 1 ( 2 cosh 2 t + 3 t ) cosh 2 t d t (4)
(ii) 由此求出用 π \pi π e e e A A A
(7)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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把 cosh A \cosh A cosh A cosh 2 A \cosh2A cosh 2 A
答题过程
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The exponential definition is cosh A = ( e A + e − A ) / 2 \cosh A=(e^A+e^{-A})/2 cosh A = ( e A + e − A ) /2
2 cosh 2 A = 1 2 ( e A + e − A ) 2 = 1 2 ( e 2 A + 2 + e − 2 A ) = cosh 2 A + 1. \begin{align*}
2\cosh^2A=&\,\frac12(e^A+e^{-A})^2\\
=&\,\frac12(e^{2A}+2+e^{-2A})\\
=&\,\cosh2A+1.
\end{align*} 2 cosh 2 A = = = 2 1 ( e A + e − A ) 2 2 1 ( e 2 A + 2 + e − 2 A ) cosh 2 A + 1.
(b)(i)
解法一
思路
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绕 y y y x x x 5 cosh 2 t 5\cosh 2t 5 cosh 2 t
答题过程
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The derivatives are x ′ = 4 sinh 2 t + 3 x'=4\sinh2t+3 x ′ = 4 sinh 2 t + 3 y ′ = 3 sinh 2 t − 4 y'=3\sinh2t-4 y ′ = 3 sinh 2 t − 4
( x ′ ) 2 + ( y ′ ) 2 = 16 sinh 2 2 t + 24 sinh 2 t + 9 + 9 sinh 2 2 t − 24 sinh 2 t + 16 = 25 ( sinh 2 2 t + 1 ) = 25 cosh 2 2 t . \begin{align*}
(x')^2+(y')^2=&\,16\sinh^22t+24\sinh2t+9\\
&\,\hspace{2pt}+9\sinh^22t-24\sinh2t+16\\
=&\,25(\sinh^22t+1)\\
=&\,25\cosh^22t.
\end{align*} ( x ′ ) 2 + ( y ′ ) 2 = = = 16 sinh 2 2 t + 24 sinh 2 t + 9 + 9 sinh 2 2 t − 24 sinh 2 t + 16 25 ( sinh 2 2 t + 1 ) 25 cosh 2 2 t . The speed is 5 cosh 2 t 5\cosh2t 5 cosh 2 t x = 2 cosh 2 t + 3 t ≥ 2 − 3 2 > 0 x=2\cosh2t+3t\ge2-\frac32>0 x = 2 cosh 2 t + 3 t ≥ 2 − 2 3 > 0 y y y
A = 2 π ∫ − 1 / 2 1 / 2 x ( x ′ ) 2 + ( y ′ ) 2 d t = 10 π ∫ − 1 / 2 1 / 2 ( 2 cosh 2 t + 3 t ) cosh 2 t d t . \begin{align*}
A=&\,2\pi\int_{-1/2}^{1/2}x\sqrt{(x')^2+(y')^2}\,\mathrm{d}t\\
=&\,10\pi\int_{-1/2}^{1/2}(2\cosh2t+3t)\cosh2t\,\mathrm{d}t.
\end{align*} A = = 2 π ∫ − 1/2 1/2 x ( x ′ ) 2 + ( y ′ ) 2 d t 10 π ∫ − 1/2 1/2 ( 2 cosh 2 t + 3 t ) cosh 2 t d t .
(b)(ii)
解法一
思路
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展开被积函数。t cosh 2 t t\cosh2t t cosh 2 t
答题过程
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Since t cosh 2 t t\cosh2t t cosh 2 t 2 t 2t 2 t
A = 10 π ∫ − 1 / 2 1 / 2 2 cosh 2 2 t d t = 10 π ∫ − 1 / 2 1 / 2 ( cosh 4 t + 1 ) d t = 10 π [ 1 4 sinh 4 t + t ] − 1 / 2 1 / 2 = 10 π ( 1 + 1 2 sinh 2 ) = 10 π ( 1 + e 2 − e − 2 4 ) . \begin{align*}
A=&\,10\pi\int_{-1/2}^{1/2}2\cosh^2 2t\,\mathrm{d}t\\
=&\,10\pi\int_{-1/2}^{1/2}(\cosh4t+1)\,\mathrm{d}t\\
=&\,10\pi\bigg[\frac14\sinh4t+t\bigg]_{-1/2}^{1/2}\\
=&\,10\pi\bigg(1+\frac12\sinh2\bigg)\\
=&\,\boxed{10\pi\bigg(1+\frac{e^2-e^{-2}}4\bigg)}.
\end{align*} A = = = = = 10 π ∫ − 1/2 1/2 2 cosh 2 2 t d t 10 π ∫ − 1/2 1/2 ( cosh 4 t + 1 ) d t 10 π [ 4 1 sinh 4 t + t ] − 1/2 1/2 10 π ( 1 + 2 1 sinh 2 ) 10 π ( 1 + 4 e 2 − e − 2 ) .
解法二
思路
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也可直接对乘积做分部积分，把积分中的 cosh 2 t \cosh2t cosh 2 t 1 2 sinh 2 t \frac12\sinh2t 2 1 sinh 2 t
答题过程
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For integration by parts take u = 2 cosh 2 t + 3 t u=2\cosh2t+3t u = 2 cosh 2 t + 3 t d v = cosh 2 t d t \mathrm{d}v=\cosh2t\,\mathrm{d}t d v = cosh 2 t d t u ′ = 4 sinh 2 t + 3 u'=4\sinh2t+3 u ′ = 4 sinh 2 t + 3 v = 1 2 sinh 2 t v=\frac12\sinh2t v = 2 1 sinh 2 t
A 10 π = [ ( cosh 2 t + 3 2 t ) sinh 2 t ] − 1 / 2 1 / 2 − ∫ − 1 / 2 1 / 2 ( 2 sinh 2 2 t + 3 2 sinh 2 t ) d t . \begin{align*}
\frac{A}{10\pi}
=&\,\bigg[\bigg(\cosh2t+\frac32t\bigg)\sinh2t\bigg]_{-1/2}^{1/2}\\
&\,\hspace{2pt}-\int_{-1/2}^{1/2}\bigg(2\sinh^2 2t+\frac32\sinh2t\bigg)\,\mathrm{d}t.
\end{align*} 10 π A = [ ( cosh 2 t + 2 3 t ) sinh 2 t ] − 1/2 1/2 − ∫ − 1/2 1/2 ( 2 sinh 2 2 t + 2 3 sinh 2 t ) d t . The boundary term from t sinh 2 t t\sinh2t t sinh 2 t sinh 2 t \sinh2t sinh 2 t sinh 2 \sinh2 sinh 2 2 sinh 2 2 t = cosh 4 t − 1 2\sinh^2 2t=\cosh4t-1 2 sinh 2 2 t = cosh 4 t − 1
A 10 π = sinh 2 − [ 1 4 sinh 4 t − t ] − 1 / 2 1 / 2 = 1 + 1 2 sinh 2 = 1 + e 2 − e − 2 4 . \begin{align*}
\frac A{10\pi}=&\,\sinh2-\bigg[\frac14\sinh4t-t\bigg]_{-1/2}^{1/2}\\
=&\,1+\frac12\sinh2\\
=&\,1+\frac{e^2-e^{-2}}4.
\end{align*} 10 π A = = = sinh 2 − [ 4 1 sinh 4 t − t ] − 1/2 1/2 1 + 2 1 sinh 2 1 + 4 e 2 − e − 2 . 