题目
Problem
The parametric equations of a curve are
x = 2 cos 3 t and y = tan 3 t , x=\frac2{\cos3t}\quad\text{and}\quad y=\tan3t, x = cos 3 t 2 and y = tan 3 t , for 0 ≤ t ≤ 2 π 0\le t\le2\pi 0 ≤ t ≤ 2 π
(a) Show that d y d x \frac{\mathrm{d}y}{\mathrm{d}x} d x d y A cosec 3 t A\operatorname{cosec}3t A cosec 3 t A A A
(5)
(b) Find an equation of the normal to the curve at the point where t = 1 12 π t=\frac1{12}\pi t = 12 1 π y = m x + c y=mx+c y = m x + c m m m c c c
(4)
题目中文翻译
一条曲线的参数方程为
x = 2 cos 3 t 且 y = tan 3 t , x=\frac2{\cos3t}\quad\text{且}\quad y=\tan3t, x = cos 3 t 2 且 y = tan 3 t , 其中 0 ≤ t ≤ 2 π 0\le t\le2\pi 0 ≤ t ≤ 2 π
(a) 证明 d y d x \frac{\mathrm{d}y}{\mathrm{d}x} d x d y A cosec 3 t A\operatorname{cosec}3t A cosec 3 t A A A
(5)
(b) 求该曲线在 t = 1 12 π t=\frac1{12}\pi t = 12 1 π y = m x + c y=mx+c y = m x + c m m m c c c
(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
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分别对参数求导，再相除并用正割与正切的定义化简。
答题过程
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Writing x = 2 sec 3 t x=2\sec3t x = 2 sec 3 t
d x d t = 6 sec 3 t tan 3 t , d y d t = 3 sec 2 3 t . \begin{align*}
\frac{\mathrm{d}x}{\mathrm{d}t}=&\,6\sec3t\tan3t,\\[4mm]
\frac{\mathrm{d}y}{\mathrm{d}t}=&\,3\sec^23t.
\end{align*} d t d x = d t d y = 6 sec 3 t tan 3 t , 3 sec 2 3 t . Hence, where the gradient is finite,
d y d x = 3 sec 2 3 t 6 sec 3 t tan 3 t = 1 2 sec 3 t tan 3 t = 1 2 cosec 3 t . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,\frac{3\sec^23t}{6\sec3t\tan3t}\\
=&\,\frac12\frac{\sec3t}{\tan3t}\\
=&\,\boxed{\frac12\operatorname{cosec}3t}.
\end{align*} d x d y = = = 6 sec 3 t tan 3 t 3 sec 2 3 t 2 1 tan 3 t sec 3 t 2 1 cosec 3 t . Thus A = 1 2 \boxed{A=\frac12} A = 2 1 cos 3 t ≠ 0 \cos3t\ne0 cos 3 t = 0 sin 3 t ≠ 0 \sin3t\ne0 sin 3 t = 0
解法二
思路
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先利用正割与正切的平方关系消参，隐式求导后再换回参数。
答题过程
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Since x = 2 sec 3 t x=2\sec3t x = 2 sec 3 t y = tan 3 t y=\tan3t y = tan 3 t
1 + y 2 = x 2 4 . 1+y^2=\frac{x^2}{4}. 1 + y 2 = 4 x 2 . Differentiating implicitly,
2 y d y d x = x 2 , 2y\frac{\mathrm{d}y}{\mathrm{d}x}=\frac x2, 2 y d x d y = 2 x , so
d y d x = x 4 y = 2 sec 3 t 4 tan 3 t = 1 2 cosec 3 t . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,\frac{x}{4y}=\frac{2\sec3t}{4\tan3t}\\[4mm]
=&\,\boxed{\frac12\operatorname{cosec}3t}.
\end{align*} d x d y = = 4 y x = 4 tan 3 t 2 sec 3 t 2 1 cosec 3 t . Thus A = 1 / 2 A=1/2 A = 1/2
(b)
解法一
思路
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先求参数对应的点与切线斜率，再取负倒数求法线斜率。
答题过程
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At t = π / 12 t=\pi/12 t = π /12 3 t = π / 4 3t=\pi/4 3 t = π /4
x = 2 2 , y = 1. x=2\sqrt2,\qquad y=1. x = 2 2 , y = 1. By (a), the tangent gradient is
m T = 1 2 cosec π 4 = 1 2 . m_T=\frac12\operatorname{cosec}\frac\pi4
=\frac1{\sqrt2}. m T = 2 1 cosec 4 π = 2 1 . Thus the normal gradient is m N = − 2 m_N=-\sqrt2 m N = − 2
y − 1 = − 2 ( x − 2 2 ) , y-1=-\sqrt2(x-2\sqrt2), y − 1 = − 2 ( x − 2 2 ) , or
y = − 2 x + 5 . \boxed{y=-\sqrt2\,x+5}. y = − 2 x + 5 . 