题目
Problem
The equation of a curve is xy+y2e−x=4.
(a) Show that dxdy=xex+2yy2−yex.
(4)
(b) Find the gradients of the tangents to the curve when x=0.
(2)
题目中文翻译
一条曲线的方程为 xy+y2e−x=4。
(a) 证明 dxdy=xex+2yy2−yex。
(4)
(b) 求当 x=0 时该曲线的切线的斜率。
(2)
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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两个乘积都隐式求导,再收集导数项。最后同乘 ex 得到目标形式。
答题过程
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Differentiating implicitly,
y+xdxdy+2ye−xdxdy−y2e−x=0.
Thus
(x+2ye−x)dxdy=y2e−x−y.
Multiplying numerator and denominator by ex gives
dxdy=xex+2yy2−yex.
解法二
思路
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也可先把原式乘以 ex 再求导,最后必须利用原方程消去额外项。
答题过程
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Multiplying the curve equation by ex,
xyex+y2=4ex.
Differentiating,
xexdxdy+y(x+1)ex+2ydxdy=4ex.
Hence
(xex+2y)dxdy=4ex−y(x+1)ex.
The original equation gives 4ex−xyex=y2. Substituting,
dxdy=xex+2yy2−yex.
(b)
解法一
思路
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先从原曲线求 x=0 对应的两个点,再分别代入导数。
答题过程
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When x=0, the curve equation gives y2=4, so y=±2.
Using (a),
dxdy=2yy2−y=2y−1.
Thus the tangent gradients are
21 at (0,2),−23 at (0,−2).