题目
Problem
The curve C C C
9 y 2 − 3 sinh − 1 ( x y ) = 1 − 3 ln 3. 9y^2 - 3\sinh^{-1}(xy) = 1 - 3\ln 3. 9 y 2 − 3 sinh − 1 ( x y ) = 1 − 3 ln 3. (a) Show that, at the point ( 4 , 1 3 ) \big(4, \frac{1}{3}\big) ( 4 , 3 1 ) C C C d y d x = − 1 2 \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2} d x d y = − 2 1
[4]
(b) Find the value of d 2 y d x 2 \frac{\mathrm{d}^2y}{\mathrm{d}x^2} d x 2 d 2 y ( 4 , 1 3 ) \big(4, \frac{1}{3}\big) ( 4 , 3 1 )
[5]
题目中文翻译
曲线 C C C
9 y 2 − 3 sinh − 1 ( x y ) = 1 − 3 ln 3. 9y^2 - 3\sinh^{-1}(xy) = 1 - 3\ln 3. 9 y 2 − 3 sinh − 1 ( x y ) = 1 − 3 ln 3. (a) 证明在 C C C ( 4 , 1 3 ) \big(4, \frac{1}{3}\big) ( 4 , 3 1 ) d y d x = − 1 2 \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2} d x d y = − 2 1
(4)
(b) 求在点 ( 4 , 1 3 ) \big(4, \frac{1}{3}\big) ( 4 , 3 1 ) d 2 y d x 2 \frac{\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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隐式求导时，把 x y xy x y
答题过程
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Differentiating with respect to x x x
18 y y ′ − 3 y + x y ′ 1 + x 2 y 2 = 0. 18yy'-3\frac{y+xy'}{\sqrt{1+x^2y^2}}=0. 18 y y ′ − 3 1 + x 2 y 2 y + x y ′ = 0. At ( x , y ) = ( 4 , 1 3 ) (x,y)=(4,\frac13) ( x , y ) = ( 4 , 3 1 ) 5 3 \frac53 3 5
6 y ′ − 9 5 ( 1 3 + 4 y ′ ) = 0. 6y'-\frac95\bigg(\frac13+4y'\bigg)=0. 6 y ′ − 5 9 ( 3 1 + 4 y ′ ) = 0. Multiplying by 5 5 5 − 6 y ′ − 3 = 0 -6y'-3=0 − 6 y ′ − 3 = 0 y ′ = − 1 2 y'=-\frac12 y ′ = − 2 1
(b)
解法一
思路
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把 x y xy x y u u u u , u ′ , u ′ ′ u,u',u'' u , u ′ , u ′′ y ′ ′ y'' y ′′
答题过程
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Let u = x y u=xy u = x y u ′ = y + x y ′ u'=y+xy' u ′ = y + x y ′ u ′ ′ = 2 y ′ + x y ′ ′ u''=2y'+xy'' u ′′ = 2 y ′ + x y ′′
18 [ ( y ′ ) 2 + y y ′ ′ ] − 3 [ u ′ ′ 1 + u 2 − u ( u ′ ) 2 ( 1 + u 2 ) 3 / 2 ] = 0. 18\big[(y')^2+yy''\big]
-3\bigg[\frac{u''}{\sqrt{1+u^2}}
-\frac{u(u')^2}{(1+u^2)^{3/2}}\bigg]=0. 18 [ ( y ′ ) 2 + y y ′′ ] − 3 [ 1 + u 2 u ′′ − ( 1 + u 2 ) 3/2 u ( u ′ ) 2 ] = 0. At the given point, u = 4 3 u=\frac43 u = 3 4 u ′ = − 5 3 u'=-\frac53 u ′ = − 3 5 u ′ ′ = 4 y ′ ′ − 1 u''=4y''-1 u ′′ = 4 y ′′ − 1
9 2 + 6 y ′ ′ − 3 [ 3 5 ( 4 y ′ ′ − 1 ) − 4 5 ] = 0. \frac92+6y''-3\bigg[\frac35(4y''-1)-\frac45\bigg]=0. 2 9 + 6 y ′′ − 3 [ 5 3 ( 4 y ′′ − 1 ) − 5 4 ] = 0. This gives 87 10 − 6 5 y ′ ′ = 0 \frac{87}{10}-\frac65y''=0 10 87 − 5 6 y ′′ = 0 y ′ ′ = 29 4 \boxed{y''=\frac{29}{4}} y ′′ = 4 29
