题目
Problem
The parametric equations of a curve are
x=te2t,y=t2+t+3.
(a) Show that dxdy=e−2t.
(3)
(b) Hence show that the normal to the curve, where t=−1, passes through the point (0,3−e41).
(3)
题目中文翻译
一条曲线的参数方程为
x=te2t,y=t2+t+3.
(a) 证明 dxdy=e−2t。
(3)
(b) 由此证明曲线上 t=−1 处的法线经过点 (0,3−e41)。
(3)
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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用乘积法则求 x 对参数的导数，再取两导数之比。约去共同因子时注明它非零。
答题过程
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Differentiating with respect to t,
dtdx==dtdy=e2t+2te2te2t(1+2t),2t+1.
At regular parameter values t=−1/2,
dxdy=e2t(1+2t)2t+1=e−2t.
At t=−1/2, both parameter derivatives vanish; the tangent-gradient limit from nearby regular values is e.
(b)
解法一
思路
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利用(a)求切线斜率，再取负倒数得到法线斜率。代入 t=−1 的坐标写法线，令 x=0 求截距。
答题过程
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At t=−1,
x=−e−2,y=1−1+3=3.
From part (a), the tangent gradient is e2, so the normal gradient is −e−2.
Thus the normal has equation
y−3=−e−2(x+e−2).
At x=0,
y=3−e−4=3−e41.
Therefore the normal passes through
(0,3−e41).
