题目
Problem
Figure 2 shows a sketch of the curve C with equation
y3−x2+4x2y=k
where k is a positive constant greater than 1
(a) Find
dxdy
in terms of x and y.
(5)
The point P lies on C.
Given that the normal to C at P has equation y=x, as shown in Figure 2,
(b) find the value of k.
(5)
题目中文翻译
图 2 给出了曲线 C 的示意图,其方程为
y3−x2+4x2y=k
其中 k 是一个大于 1 的正常数。
(a) 求 dxdy,用 x 和 y 表示。
(5)
点 P 在曲线 C 上。
已知曲线 C 在点 P 处的法线方程为 y=x,如图 2 所示,
(b) 求 k 的值。
(5)
解答
(a)
解法一
思路
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对曲线方程关于 x 隐式求导,其中 4x2y 要用乘积法则。用 g=dy/dx 暂记导数,先收集含 g 的项,最后恢复导数记号。
答题过程
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Let
g=dxdy.
Differentiate each term with respect to x:
dxd(y3)=3y2g,
dxd(−x2)=−2x,
and, by the product rule,
dxd(4x2y)=8xy+4x2g.
Since k is constant, differentiating the curve equation gives
3y2g−2x+8xy+4x2g=0.
Collecting the terms in g,
(3y2+4x2)g=2x−8xy.
Hence
dxdy=3y2+4x22x−8xy.
(b)
解法一
思路
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法线 y=x 的斜率是 1,因此切线斜率为 −1。点 P 也在该法线上,将 y=x 代入导数式,再以切线斜率为 −1 求坐标,最后代回曲线求 k。
答题过程
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The normal y=x has gradient 1, so the tangent at P has gradient
dxdy=−1.
Since P lies on the normal, y=x at P. Substituting in part (a),
−1=7x22x−8x2.
As k>1, the origin is not on C. Since y=x, we must have x=0 at P.
Therefore
−7x2=x2−2x=x(x−2)=2x−8x2,0,0.
Rejecting x=0 gives x=2. Hence y=2 and P=(2,2).
Substituting in the curve equation,
k===23−22+4(22)(2)8−4+3236.