题目
Problem
(a) Starting from the definitions of tanh \tanh tanh sech \operatorname{sech} sech
1 − tanh 2 u = sech 2 u . 1 - \tanh^2 u = \operatorname{sech}^2 u. 1 − tanh 2 u = sech 2 u . [3]
(b) Show that d d t ( sech − 1 t ) = − 1 t 1 − t 2 \frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech}^{-1} t) = -\frac{1}{t\sqrt{1 - t^2}} d t d ( sech − 1 t ) = − t 1 − t 2 1
[4]
It is given that
x = tanh − 1 t and y = t sech − 1 t , for 0 < t < 1. \begin{align*}
x =&\, \tanh^{-1} t
\quad\text{and}\quad
y = t\operatorname{sech}^{-1} t,
\quad
\text{for } 0 < t < 1.
\end{align*} x = tanh − 1 t and y = t sech − 1 t , for 0 < t < 1. (c) Show that d y d x = − 1 − t 2 + ( 1 − t 2 ) sech − 1 t \frac{\mathrm{d}y}{\mathrm{d}x} = -\sqrt{1 - t^2} + (1 - t^2)\operatorname{sech}^{-1} t d x d y = − 1 − t 2 + ( 1 − t 2 ) sech − 1 t
[4]
(d) Find d 2 y d x 2 \frac{\mathrm{d}^2y}{\mathrm{d}x^2} d x 2 d 2 y t t t
[4]
题目中文翻译
(a) 从 tanh \tanh tanh sech \operatorname{sech} sech
1 − tanh 2 u = sech 2 u . 1 - \tanh^2 u = \operatorname{sech}^2 u. 1 − tanh 2 u = sech 2 u . (3)
(b) 证明 d d t ( sech − 1 t ) = − 1 t 1 − t 2 \frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech}^{-1} t) = -\frac{1}{t\sqrt{1 - t^2}} d t d ( sech − 1 t ) = − t 1 − t 2 1
(4)
已知
x = tanh − 1 t and y = t sech − 1 t , for 0 < t < 1. \begin{align*}
x =&\, \tanh^{-1} t
\quad\text{and}\quad
y = t\operatorname{sech}^{-1} t,
\quad
\text{for } 0 < t < 1.
\end{align*} x = tanh − 1 t and y = t sech − 1 t , for 0 < t < 1. (c) 证明 d y d x = − 1 − t 2 + ( 1 − t 2 ) sech − 1 t \frac{\mathrm{d}y}{\mathrm{d}x} = -\sqrt{1 - t^2} + (1 - t^2)\operatorname{sech}^{-1} t d x d y = − 1 − t 2 + ( 1 − t 2 ) sech − 1 t
(4)
(d) 求用 t t t d 2 y d x 2 \frac{\mathrm{d}^2y}{\mathrm{d}x^2} d x 2 d 2 y
(4)
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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代入两个指数定义，通分后用平方差化简分子。
答题过程
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The exponential definitions are
tanh u = e u − e − u e u + e − u , sech u = 2 e u + e − u . \tanh u=\frac{e^u-e^{-u}}{e^u+e^{-u}},\qquad
\operatorname{sech}u=\frac2{e^u+e^{-u}}. tanh u = e u + e − u e u − e − u , sech u = e u + e − u 2 . Using these definitions,
1 − tanh 2 u = ( e u + e − u ) 2 − ( e u − e − u ) 2 ( e u + e − u ) 2 = 4 ( e u + e − u ) 2 = sech 2 u . \begin{align*}
1-\tanh^2u=&\,\frac{(e^u+e^{-u})^2-(e^u-e^{-u})^2}{(e^u+e^{-u})^2}\\
=&\,\frac{4}{(e^u+e^{-u})^2}\\
=&\,\operatorname{sech}^2u.
\end{align*} 1 − tanh 2 u = = = ( e u + e − u ) 2 ( e u + e − u ) 2 − ( e u − e − u ) 2 ( e u + e − u ) 2 4 sech 2 u .
(b)
解法一
思路
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在反双曲正割的主值分支上写 sech u = t \operatorname{sech}u=t sech u = t u > 0 u>0 u > 0 tanh u \tanh u tanh u
答题过程
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Let u = sech − 1 t u=\operatorname{sech}^{-1}t u = sech − 1 t 0 < t < 1 0<t<1 0 < t < 1 u > 0 u>0 u > 0 sech u = t \operatorname{sech}u=t sech u = t
− sech u tanh u d u d t = 1. -\operatorname{sech}u\,\tanh u\,\frac{\mathrm{d}u}{\mathrm{d}t}=1. − sech u tanh u d t d u = 1. By (a), tanh u = 1 − t 2 \tanh u=\sqrt{1-t^2} tanh u = 1 − t 2
d u d t = − 1 t 1 − t 2 . \frac{\mathrm{d}u}{\mathrm{d}t}=-\frac1{t\sqrt{1-t^2}}. d t d u = − t 1 − t 2 1 .
解法二
思路
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也可改写为 cosh u = 1 / t \cosh u=1/t cosh u = 1/ t sinh u \sinh u sinh u
答题过程
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The equation cosh u = t − 1 \cosh u=t^{-1} cosh u = t − 1 sinh u d u d t = − t − 2 \sinh u\,\frac{\mathrm{d}u}{\mathrm{d}t}=-t^{-2} sinh u d t d u = − t − 2 u > 0 u>0 u > 0
sinh u = cosh 2 u − 1 = 1 − t 2 t . \sinh u=\sqrt{\cosh^2u-1}=\frac{\sqrt{1-t^2}}{t}. sinh u = cosh 2 u − 1 = t 1 − t 2 . Therefore d u d t = − 1 t 1 − t 2 \frac{\mathrm{d}u}{\mathrm{d}t}=-\frac1{t\sqrt{1-t^2}} d t d u = − t 1 − t 2 1
解法三
思路
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如果直接使用反双曲余弦的导数，可先写 u = arcosh ( 1 / t ) u=\operatorname{arcosh}(1/t) u = arcosh ( 1/ t )
答题过程
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For 0 < t < 1 0<t<1 0 < t < 1 u = arcosh ( t − 1 ) u=\operatorname{arcosh}(t^{-1}) u = arcosh ( t − 1 ) arcosh \operatorname{arcosh} arcosh
d u d t = − t − 2 ( t − 1 ) 2 − 1 = − 1 t 1 − t 2 . \frac{\mathrm{d}u}{\mathrm{d}t} =\frac{-t^{-2}}{\sqrt{(t^{-1})^2-1}} =-\frac1{t\sqrt{1-t^2}}. d t d u = ( t − 1 ) 2 − 1 − t − 2 = − t 1 − t 2 1 .
解法四
思路
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也可用反双曲余弦的对数形式。把商的对数拆开，再逐项求导。
答题过程
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Write s = 1 − t 2 s=\sqrt{1-t^2} s = 1 − t 2 u = ln ( 1 + s ) − ln t u=\ln(1+s)-\ln t u = ln ( 1 + s ) − ln t t > 0 t>0 t > 0
d u d t = − t s ( 1 + s ) − 1 t = − t 2 + s + s 2 t s ( 1 + s ) = − 1 t 1 − t 2 . \begin{align*}
\frac{\mathrm{d}u}{\mathrm{d}t}=&\,-\frac{t}{s(1+s)}-\frac1t\\
=&\,-\frac{t^2+s+s^2}{ts(1+s)}\\
=&\,-\frac1{t\sqrt{1-t^2}}.
\end{align*} d t d u = = = − s ( 1 + s ) t − t 1 − t s ( 1 + s ) t 2 + s + s 2 − t 1 − t 2 1 . The last simplification uses s 2 + t 2 = 1 s^2+t^2=1 s 2 + t 2 = 1 t 2 + s + s 2 = 1 + s t^2+s+s^2=1+s t 2 + s + s 2 = 1 + s
(c)
解法一
思路
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先用乘积法则求 d y d t \frac{\mathrm{d}y}{\mathrm{d}t} d t d y d x d t \frac{\mathrm{d}x}{\mathrm{d}t} d t d x
答题过程
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Using (b),
d y d t = sech − 1 t − 1 1 − t 2 , d x d t = 1 1 − t 2 . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}t} =&\,\operatorname{sech}^{-1}t-\frac1{\sqrt{1-t^2}},\\
\frac{\mathrm{d}x}{\mathrm{d}t}=&\,\frac1{1-t^2}.
\end{align*} d t d y = d t d x = sech − 1 t − 1 − t 2 1 , 1 − t 2 1 . Taking the ratio gives
d y d x = − 1 − t 2 + ( 1 − t 2 ) sech − 1 t . \frac{\mathrm{d}y}{\mathrm{d}x} =-\sqrt{1-t^2}+(1-t^2)\operatorname{sech}^{-1}t. d x d y = − 1 − t 2 + ( 1 − t 2 ) sech − 1 t .
(d)
解法一
思路
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对 (c) 的结果关于 t t t d t d x = 1 − t 2 \frac{\mathrm{d}t}{\mathrm{d}x}=1-t^2 d x d t = 1 − t 2
答题过程
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Differentiating with respect to t t t
d d t ( d y d x ) = t 1 − t 2 − 2 t sech − 1 t − 1 − t 2 t 1 − t 2 . \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg) =\frac{t}{\sqrt{1-t^2}}-2t\operatorname{sech}^{-1}t -\frac{1-t^2}{t\sqrt{1-t^2}}. d t d ( d x d y ) = 1 − t 2 t − 2 t sech − 1 t − t 1 − t 2 1 − t 2 . The chain rule then gives
d 2 y d x 2 = ( 2 t 2 − 1 ) 1 − t 2 t − 2 t ( 1 − t 2 ) sech − 1 t . \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =\frac{(2t^2-1)\sqrt{1-t^2}}{t} -2t(1-t^2)\operatorname{sech}^{-1}t. d x 2 d 2 y = t ( 2 t 2 − 1 ) 1 − t 2 − 2 t ( 1 − t 2 ) sech − 1 t . 