题目
Problem
The curve C C C
x = cosh t , y = sinh t , for 0 < t ≤ 3 5 . x=\cosh t,\qquad y=\sinh t,\qquad\text{for }0<t\le\frac{3}{5}. x = cosh t , y = sinh t , for 0 < t ≤ 5 3 . The length of C C C s s s
(a) Show that
s = ∫ 0 3 5 cosh 2 t d t . s=\int_0^{\frac{3}{5}}\sqrt{\cosh 2t}\,\mathrm{d}t. s = ∫ 0 5 3 cosh 2 t d t . [4]
(b) By finding the Maclaurin’s series for cosh 2 t \sqrt{\cosh 2t} cosh 2 t t 2 t^2 t 2 s s s
[5]
题目中文翻译
曲线 C C C
x = cosh t y = sinh t 0 < t ≤ 3 5 x=\cosh t\qquad y=\sinh t\qquad 0<t\le\frac{3}{5} x = cosh t y = sinh t 0 < t ≤ 5 3 曲线 C C C s s s
(a) 证明
s = ∫ 0 3 5 cosh 2 t d t s=\int_0^{\frac{3}{5}}\sqrt{\cosh 2t}\,\mathrm{d}t s = ∫ 0 5 3 cosh 2 t d t (4)
(b) 求 cosh 2 t \sqrt{\cosh 2t} cosh 2 t t 2 t^2 t 2 s s s
(5)
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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参数曲线的速度是两个导数平方和的平方根。用双曲余弦的倍角恒等式化简。
答题过程
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Since x ′ = sinh t x'=\sinh t x ′ = sinh t y ′ = cosh t y'=\cosh t y ′ = cosh t
sinh 2 t + cosh 2 t = cosh 2 t . \sqrt{\sinh^2t+\cosh^2t}=\sqrt{\cosh 2t}. sinh 2 t + cosh 2 t = cosh 2 t . The arc length formula is
s = ∫ 0 3 / 5 ( d x d t ) 2 + ( d y d t ) 2 d t . s=\int_0^{3/5}\sqrt{\bigg(\frac{\mathrm{d}x}{\mathrm{d}t}\bigg)^2+\bigg(\frac{\mathrm{d}y}{\mathrm{d}t}\bigg)^2}\,\mathrm{d}t. s = ∫ 0 3/5 ( d t d x ) 2 + ( d t d y ) 2 d t . The speed extends continuously to t = 0 t=0 t = 0
s = ∫ 0 3 / 5 cosh 2 t d t . s=\int_0^{3/5}\sqrt{\cosh 2t}\,\mathrm{d}t. s = ∫ 0 3/5 cosh 2 t d t .
(b)
解法一
思路
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先展开 cosh 2 t \cosh 2t cosh 2 t t 2 t^2 t 2
答题过程
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The standard expansions are
cosh u = 1 + u 2 2 ! + O ( u 4 ) , ( 1 + v ) 1 / 2 = 1 + v 2 + O ( v 2 ) . \begin{align*}
\cosh u=&\,1+\frac{u^2}{2!}+O(u^4),\\
(1+v)^{1/2}=&\,1+\frac v2+O(v^2).
\end{align*} cosh u = ( 1 + v ) 1/2 = 1 + 2 ! u 2 + O ( u 4 ) , 1 + 2 v + O ( v 2 ) . Substituting u = 2 t u=2t u = 2 t v = 2 t 2 + O ( t 4 ) v=2t^2+O(t^4) v = 2 t 2 + O ( t 4 )
cosh 2 t = 1 + 2 t 2 + O ( t 4 ) , cosh 2 t = 1 + 1 2 ( 2 t 2 + O ( t 4 ) ) + O ( t 4 ) = 1 + t 2 + O ( t 4 ) . \begin{align*}
\cosh 2t=&\,1+2t^2+O(t^4),\\
\sqrt{\cosh 2t}=&\,1+\frac12\big(2t^2+O(t^4)\big)+O(t^4)\\
=&\,1+t^2+O(t^4).
\end{align*} cosh 2 t = cosh 2 t = = 1 + 2 t 2 + O ( t 4 ) , 1 + 2 1 ( 2 t 2 + O ( t 4 ) ) + O ( t 4 ) 1 + t 2 + O ( t 4 ) . Using the terms through t 2 t^2 t 2
s ≈ ∫ 0 3 / 5 ( 1 + t 2 ) d t = [ t + t 3 3 ] 0 3 / 5 = 84 125 = 0.672 . \begin{align*}
s\approx&\,\int_0^{3/5}(1+t^2)\,\mathrm{d}t\\
=&\,\bigg[t+\frac{t^3}{3}\bigg]_0^{3/5}\\
=&\,\boxed{\frac{84}{125}=0.672}.
\end{align*} s ≈ = = ∫ 0 3/5 ( 1 + t 2 ) d t [ t + 3 t 3 ] 0 3/5 125 84 = 0.672 .
解法二
思路
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也可直接求 cosh 2 t \sqrt{\cosh 2t} cosh 2 t
答题过程
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Let g ( t ) = cosh 2 t g(t)=\sqrt{\cosh 2t} g ( t ) = cosh 2 t
g ′ ( t ) = sinh 2 t cosh 2 t , g ′ ′ ( t ) = 2 cosh 2 t − sinh 2 2 t ( cosh 2 t ) 3 / 2 . \begin{align*}
g'(t)=&\,\frac{\sinh 2t}{\sqrt{\cosh 2t}},\\
g''(t)=&\,2\sqrt{\cosh 2t}-\frac{\sinh^2 2t}{(\cosh 2t)^{3/2}}.
\end{align*} g ′ ( t ) = g ′′ ( t ) = cosh 2 t sinh 2 t , 2 cosh 2 t − ( cosh 2 t ) 3/2 sinh 2 2 t . Thus g ( 0 ) = 1 g(0)=1 g ( 0 ) = 1 g ′ ( 0 ) = 0 g'(0)=0 g ′ ( 0 ) = 0 g ′ ′ ( 0 ) = 2 g''(0)=2 g ′′ ( 0 ) = 2 t 2 t^2 t 2 1 + t 2 1+t^2 1 + t 2 s ≈ 3 5 + 1 3 ( 3 5 ) 3 = 84 125 s\approx\frac35+\frac13(\frac35)^3=\frac{84}{125} s ≈ 5 3 + 3 1 ( 5 3 ) 3 = 125 84
