题目
Problem
It is given that
x = 1 + 1 t and y = cos − 1 t for 0 < t < 1. x = 1 + \frac{1}{t} \qquad\text{and}\qquad y = \cos^{-1}t \qquad\text{for } 0 < t < 1. x = 1 + t 1 and y = cos − 1 t for 0 < t < 1. (a) Show that
d y d x = t 2 1 − t 2 . \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{t^2}{\sqrt{1 - t^2}}. d x d y = 1 − t 2 t 2 . [2]
(b) Show that
d 2 y d x 2 = − t a ( 1 − t 2 ) b ( 2 − t 2 ) , \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -t^a(1 - t^2)^b(2 - t^2), d x 2 d 2 y = − t a ( 1 − t 2 ) b ( 2 − t 2 ) , where a a a b b b
[4]
题目中文翻译
已知
x = 1 + 1 t 以及 y = cos − 1 t 其中 0 < t < 1 x = 1 + \frac{1}{t} \qquad\text{以及}\qquad y = \cos^{-1}t \qquad\text{其中 } 0 < t < 1 x = 1 + t 1 以及 y = cos − 1 t 其中 0 < t < 1 (a) 证明
d y d x = t 2 1 − t 2 \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{t^2}{\sqrt{1 - t^2}} d x d y = 1 − t 2 t 2 (2)
(b) 证明
d 2 y d x 2 = − t a ( 1 − t 2 ) b ( 2 − t 2 ) , \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -t^a(1 - t^2)^b(2 - t^2), d x 2 d 2 y = − t a ( 1 − t 2 ) b ( 2 − t 2 ) , 其中 a a a b b b
(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
展开
分别对 x ( t ) x(t) x ( t ) y ( t ) y(t) y ( t )
答题过程
展开
The derivatives are d x d t = − t − 2 \frac{\mathrm{d}x}{\mathrm{d}t}=-t^{-2} d t d x = − t − 2 d y d t = − ( 1 − t 2 ) − 1 / 2 \frac{\mathrm{d}y}{\mathrm{d}t}=-(1-t^2)^{-1/2} d t d y = − ( 1 − t 2 ) − 1/2
d y d x = − ( 1 − t 2 ) − 1 / 2 − t − 2 = t 2 1 − t 2 . \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{-(1-t^2)^{-1/2}}{-t^{-2}}
=\frac{t^2}{\sqrt{1-t^2}}. d x d y = − t − 2 − ( 1 − t 2 ) − 1/2 = 1 − t 2 t 2 .
(b)
解法一
思路
展开
把 (a) 写成 t 2 ( 1 − t 2 ) − 1 / 2 t^2(1-t^2)^{-1/2} t 2 ( 1 − t 2 ) − 1/2 t t t d t d x = − t 2 \frac{\mathrm{d}t}{\mathrm{d}x}=-t^2 d x d t = − t 2
答题过程
展开
Differentiating (a) with respect to t t t
2 t ( 1 − t 2 ) − 1 / 2 + t 3 ( 1 − t 2 ) − 3 / 2 = t ( 1 − t 2 ) − 3 / 2 [ 2 ( 1 − t 2 ) + t 2 ] = t ( 1 − t 2 ) − 3 / 2 ( 2 − t 2 ) . \begin{align*}
&\,2t(1-t^2)^{-1/2}+t^3(1-t^2)^{-3/2}\\[4mm]
=&\,t(1-t^2)^{-3/2}\big[2(1-t^2)+t^2\big]\\[4mm]
=&\,t(1-t^2)^{-3/2}(2-t^2).
\end{align*} = = 2 t ( 1 − t 2 ) − 1/2 + t 3 ( 1 − t 2 ) − 3/2 t ( 1 − t 2 ) − 3/2 [ 2 ( 1 − t 2 ) + t 2 ] t ( 1 − t 2 ) − 3/2 ( 2 − t 2 ) . Multiplying by d t d x = − t 2 \frac{\mathrm{d}t}{\mathrm{d}x}=-t^2 d x d t = − t 2
d 2 y d x 2 = − t 3 ( 1 − t 2 ) − 3 / 2 ( 2 − t 2 ) . \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-t^3(1-t^2)^{-3/2}(2-t^2). d x 2 d 2 y = − t 3 ( 1 − t 2 ) − 3/2 ( 2 − t 2 ) . Hence a = 3 a=3 a = 3 b = − 3 2 b=-\frac32 b = − 2 3
