一条曲线的方程为 $y = \frac{4}{\sqrt{x+3}}$。一个点 $(x, y)$ 正沿着该曲线运动。 求当 $y$ 坐标减少的速率是 $x$ 坐标增加的速率的两倍瞬间该点的坐标。

1. Translate the given rate condition into a mathematical equation

The $y$-coordinate is decreasing at a rate twice of the rate of increase of the $x$-coordinate, so:
$\frac{dy}{dt} = -2 \frac{dx}{dt}$

2. Use the chain rule to find the value of dy/dx

Using the chain rule, $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$.
Substituting the condition:
$-2 \frac{dx}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$
$\frac{dy}{dx} = -2$

3. Differentiate the equation of the curve

$y = 4(x+3)^{-\frac{1}{2}}$
$\frac{dy}{dx} = 4 \left(-\frac{1}{2}\right)(x+3)^{-\frac{3}{2}}$
$\frac{dy}{dx} = -2(x+3)^{-\frac{3}{2}}$

4. Equate the derivative to -2 and solve for x

$-2(x+3)^{-\frac{3}{2}} = -2$
$(x+3)^{-\frac{3}{2}} = 1$
$x+3 = 1$
$x = -2$

5. Find the corresponding y-coordinate

When $x = -2$,
$y = \frac{4}{\sqrt{-2+3}} = \frac{4}{\sqrt{1}} = 4$
The coordinates are $(-2, 4)$.