题目
Problem
It is given that
x = − t + tan − 1 t , y = t + sinh − 1 t . \begin{align*}
x=&\,-t+\tan^{-1}t,\\
y=&\,t+\sinh^{-1}t.
\end{align*} x = y = − t + tan − 1 t , t + sinh − 1 t . (a) Show that d y d x = − t 2 + 1 + t 2 + 1 t 2 \dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{t^2+1+\sqrt{t^2+1}}{t^2} d x d y = − t 2 t 2 + 1 + t 2 + 1
(4)
(b) Find the value of d 2 y d x 2 \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} d x 2 d 2 y t = 3 4 t=\frac{3}{4} t = 4 3
(5)
题目中文翻译
已知
x = − t + tan − 1 t , y = t + sinh − 1 t . \begin{align*}
x=&\,-t+\tan^{-1}t,\\
y=&\,t+\sinh^{-1}t.
\end{align*} x = y = − t + tan − 1 t , t + sinh − 1 t . (a) 证明 d y d x = − t 2 + 1 + t 2 + 1 t 2 \dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{t^2+1+\sqrt{t^2+1}}{t^2} d x d y = − t 2 t 2 + 1 + t 2 + 1
(4)
(b) 求当 t = 3 4 t=\frac{3}{4} t = 4 3 d 2 y d x 2 \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} d x 2 d 2 y
(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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先对两个参数方程求导。将 d x d t \frac{\mathrm{d}x}{\mathrm{d}t} d t d x
答题过程
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We have
d x d t = − 1 + 1 1 + t 2 = − t 2 1 + t 2 , d y d t = 1 + 1 1 + t 2 . \begin{align*}
\frac{\mathrm{d}x}{\mathrm{d}t}=&\,-1+\frac1{1+t^2}=-\frac{t^2}{1+t^2},\\
\frac{\mathrm{d}y}{\mathrm{d}t}=&\,1+\frac1{\sqrt{1+t^2}}.
\end{align*} d t d x = d t d y = − 1 + 1 + t 2 1 = − 1 + t 2 t 2 , 1 + 1 + t 2 1 . For t ≠ 0 t\ne0 t = 0
d y d x = 1 + 1 / 1 + t 2 − t 2 / ( 1 + t 2 ) = − t 2 + 1 + t 2 + 1 t 2 . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,\frac{1+1/\sqrt{1+t^2}}{-t^2/(1+t^2)}\\
=&\,-\frac{t^2+1+\sqrt{t^2+1}}{t^2}.
\end{align*} d x d y = = − t 2 / ( 1 + t 2 ) 1 + 1/ 1 + t 2 − t 2 t 2 + 1 + t 2 + 1 .
(b)
解法一
思路
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将 (a) 拆成三项后求关于 t t t t = 3 4 t=\frac34 t = 4 3 d x d t \frac{\mathrm{d}x}{\mathrm{d}t} d t d x
答题过程
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Let R = 1 + t 2 R=\sqrt{1+t^2} R = 1 + t 2 R ′ = t / R R'=t/R R ′ = t / R − 1 − t − 2 − R t − 2 -1-t^{-2}-Rt^{-2} − 1 − t − 2 − R t − 2
d d t ( d y d x ) = 2 t − 3 − t R t − 2 + 2 R t − 3 = 2 + 2 R − t 2 / R t 3 . \begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)
=&\,2t^{-3}-\frac{t}{R}t^{-2}+2Rt^{-3}\\
=&\,\frac{2+2R-t^2/R}{t^3}.
\end{align*} d t d ( d x d y ) = = 2 t − 3 − R t t − 2 + 2 R t − 3 t 3 2 + 2 R − t 2 / R . At t = 3 4 t=\frac34 t = 4 3 R = 5 4 R=\frac54 R = 4 5 [ 2 + 2 ( 5 / 4 ) − ( 9 / 16 ) / ( 5 / 4 ) ] / ( 27 / 64 ) = 48 / 5 [2+2(5/4)-(9/16)/(5/4)]/(27/64)=48/5 [ 2 + 2 ( 5/4 ) − ( 9/16 ) / ( 5/4 )] / ( 27/64 ) = 48/5 d x d t = − 9 / 25 \frac{\mathrm{d}x}{\mathrm{d}t}=-9/25 d t d x = − 9/25 d 2 y d x 2 = d d t ( d y d x ) / d x d t \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}}{\mathrm{d}t}(\frac{\mathrm{d}y}{\mathrm{d}x})/\frac{\mathrm{d}x}{\mathrm{d}t} d x 2 d 2 y = d t d ( d x d y ) / d t d x
d 2 y d x 2 = 48 / 5 − 9 / 25 = − 80 3 . \boxed{\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
=\frac{48/5}{-9/25}=-\frac{80}{3}}. d x 2 d 2 y = − 9/25 48/5 = − 3 80 . 