题目
Problem
The curve C C C
x = 2 cosh 2 t + 3 t , y = 3 2 cosh 2 t − 4 t , 0 ≤ t ≤ 1. \begin{align*}
x=&\,2\cosh 2t+3t,\\
y=&\,\frac32\cosh2t-4t,\qquad0\le t\le1.
\end{align*} x = y = 2 cosh 2 t + 3 t , 2 3 cosh 2 t − 4 t , 0 ≤ t ≤ 1. (a) Find, in terms of e \mathrm{e} e C C C
(5)
(b) Show that
d 2 y d x 2 = a cosh 2 t ( b sinh 2 t + c ) n , \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{a\cosh 2t}{\big(b\sinh 2t+c\big)^n}, d x 2 d 2 y = ( b sinh 2 t + c ) n a cosh 2 t , where a a a b b b c c c n n n
(5)
题目中文翻译
曲线 C C C
x = 2 cosh 2 t + 3 t , y = 3 2 cosh 2 t − 4 t , 0 ≤ t ≤ 1. \begin{align*}
x=&\,2\cosh 2t+3t,\\
y=&\,\frac32\cosh2t-4t,\qquad0\le t\le1.
\end{align*} x = y = 2 cosh 2 t + 3 t , 2 3 cosh 2 t − 4 t , 0 ≤ t ≤ 1. (a) 求曲线 C C C e \mathrm{e} e
(5)
(b) 证明
d 2 y d x 2 = a cosh 2 t ( b sinh 2 t + c ) n , \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{a\cosh 2t}{\big(b\sinh 2t+c\big)^n}, d x 2 d 2 y = ( b sinh 2 t + c ) n a cosh 2 t , 其中 a a a b b b c c c n n n
(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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求两个参数导数。平方和里的交叉项会抵消，再用 1 + sinh 2 ( 2 t ) = cosh 2 ( 2 t ) 1+\sinh^2(2t)=\cosh^2(2t) 1 + sinh 2 ( 2 t ) = cosh 2 ( 2 t )
答题过程
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Differentiating gives 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 . Since cosh 2 t > 0 \cosh2t>0 cosh 2 t > 0 5 cosh 2 t 5\cosh2t 5 cosh 2 t s = ∫ ( x ′ ) 2 + ( y ′ ) 2 d t s=\int\sqrt{(x')^2+(y')^2}\,\mathrm{d}t s = ∫ ( x ′ ) 2 + ( y ′ ) 2 d t
s = 5 ∫ 0 1 cosh 2 t d t = 5 2 sinh 2 = 5 4 ( e 2 − e − 2 ) . \begin{align*}
s=&\,5\int_0^1\cosh2t\,\mathrm{d}t\\
=&\,\frac52\sinh2\\
=&\,\boxed{\frac54(e^2-e^{-2})}.
\end{align*} s = = = 5 ∫ 0 1 cosh 2 t d t 2 5 sinh 2 4 5 ( e 2 − e − 2 ) .
(b)
解法一
思路
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先取两个参数导数的比。对 t t t sinh 2 t cosh 2 t \sinh2t\cosh2t sinh 2 t cosh 2 t x ′ x' x ′
答题过程
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With s = sinh 2 t s=\sinh2t s = sinh 2 t c 0 = cosh 2 t c_0=\cosh2t c 0 = cosh 2 t
d y d x = 3 s − 4 4 s + 3 , d d t ( d y d x ) = 6 c 0 ( 4 s + 3 ) − 8 c 0 ( 3 s − 4 ) ( 4 s + 3 ) 2 = 24 s c 0 + 18 c 0 − 24 s c 0 + 32 c 0 ( 4 s + 3 ) 2 = 50 c 0 ( 4 s + 3 ) 2 . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}=&\,\frac{3s-4}{4s+3},\\
\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)
=&\,\frac{6c_0(4s+3)-8c_0(3s-4)}{(4s+3)^2}\\
=&\,\frac{24sc_0+18c_0-24sc_0+32c_0}{(4s+3)^2}\\
=&\,\frac{50c_0}{(4s+3)^2}.
\end{align*} d x d y = d t d ( d x d y ) = = = 4 s + 3 3 s − 4 , ( 4 s + 3 ) 2 6 c 0 ( 4 s + 3 ) − 8 c 0 ( 3 s − 4 ) ( 4 s + 3 ) 2 24 s c 0 + 18 c 0 − 24 s c 0 + 32 c 0 ( 4 s + 3 ) 2 50 c 0 . On 0 ≤ t ≤ 1 0\le t\le1 0 ≤ t ≤ 1 s ≥ 0 s\ge0 s ≥ 0 x ′ = 4 s + 3 > 0 x'=4s+3>0 x ′ = 4 s + 3 > 0 x ′ x' x ′
d 2 y d x 2 = 50 cosh 2 t ( 4 sinh 2 t + 3 ) 3 , ( a , b , c , n ) = ( 50 , 4 , 3 , 3 ) . \frac{\mathrm{d}^2y}{\mathrm{d}x^2}
=\frac{50\cosh2t}{(4\sinh2t+3)^3},
\qquad (a,b,c,n)=(50,4,3,3). d x 2 d 2 y = ( 4 sinh 2 t + 3 ) 3 50 cosh 2 t , ( a , b , c , n ) = ( 50 , 4 , 3 , 3 ) . 