题目

Problem

Figure 3 shows a sketch of the curve $C$ with equation

x=ye^{2y}\qquad y\in\mathbb{R}

(a) Show that

\frac{dy}{dx}=\frac{y}{x(1+2y)}

(4)

Given that the straight line with equation $x=k$ , where $k$ is a constant, cuts $C$ at exactly two points,

(b) find the range of possible values for $k$ .

(3)

题目中文翻译

图 3 给出了曲线 $C$ 的示意图，其方程为

x=ye^{2y}\qquad y\in\mathbb{R}

(a) 证明

\frac{dy}{dx}=\frac{y}{x(1+2y)}

已知直线 $x=k$ （其中 $k$ 为常数）与曲线 $C$ 恰有两个交点。

(b) 求 $k$ 的可能取值范围。

解答

We are given

x=ye^{2y}

Differentiate both sides with respect to $y$ :

\frac{\mathrm{d}x}{\mathrm{d}y} =e^{2y}+y\cdot 2e^{2y}

\frac{\mathrm{d}x}{\mathrm{d}y} =e^{2y}(1+2y)

Since

x=ye^{2y}

we have

e^{2y}=\frac{x}{y}

Therefore

\frac{\mathrm{d}x}{\mathrm{d}y} =\frac{x}{y}(1+2y)

Hence

\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{1}{\frac{x}{y}(1+2y)}

\boxed{\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{y}{x(1+2y)}}

as required.

A vertical line $x=k$ cuts the curve exactly two times when $k$ lies between the local minimum value of $x$ and $0$ .

To find the local minimum, use

\frac{\mathrm{d}x}{\mathrm{d}y}=e^{2y}(1+2y)

At a stationary point of $x$ as a function of $y$ ,

e^{2y}(1+2y)=0

Since

e^{2y}>0

we need

1+2y=0

y=-\frac12

Then

x=ye^{2y} =-\frac12e^{-1} =-\frac{1}{2e}

Also, the curve passes through the origin when $y=0$ .

For exactly two intersections,

k

must be strictly between these two $x$ values.

Therefore

\boxed{-\frac{1}{2e}<k<0}