变量 $x$ 和 $y$ 增加，且满足当 $x = 3$ 时，$y$ 随时间的变化率是 $x$ 随时间的变化率的三倍。 已知 $y = k\sqrt{3x+7}$，其中 $k$ 是常数，求 $k$ 的值。 一种放射性样品的质量为 $m$ 克，在开始观察 $t$ 天后的质量由 $m = 24e^{-0.02t}$ 给出。 放射性样品的初始质量， 样品减少到其初始质量一半所需的时间， 12 小时后质量减小的速率。

1. Express the given relationship between the rates of change

Given that when $x = 3$, $\frac{dy}{dt} = 3 \frac{dx}{dt}$.

2. Differentiate $y$ with respect to $x$

$y = k(3x+7)^{\frac{1}{2}}$
$\frac{dy}{dx} = k \left(\frac{1}{2}\right)(3x+7)^{-\frac{1}{2}}(3)$
$\frac{dy}{dx} = \frac{3k}{2\sqrt{3x+7}}$

3. Apply the chain rule for connected rates of change

$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$
Substitute $\frac{dy}{dt} = 3 \frac{dx}{dt}$:
$3 \frac{dx}{dt} = \frac{3k}{2\sqrt{3x+7}} \frac{dx}{dt}$
Since $\frac{dx}{dt} \neq 0$, we can divide both sides by $\frac{dx}{dt}$:
$3 = \frac{3k}{2\sqrt{3x+7}}$

4. Substitute $x = 3$ and solve for $k$

Substitute $x = 3$:
$3 = \frac{3k}{2\sqrt{3(3)+7}}$
$3 = \frac{3k}{2\sqrt{16}}$
$3 = \frac{3k}{2(4)}$
$3 = \frac{3k}{8}$
$24 = 3k$
$k = 8$

1. Substitute $t = 0$ into the given equation to find the initial mass

When $t = 0$,
$m = 24e^{-0.02(0)}$
$m = 24e^0$
$m = 24(1) = 24$

1. Set up the equation for half the initial mass

Initial mass is $24$ g. Half of this is $12$ g.
$12 = 24e^{-0.02t}$

2. Solve for $t$ using natural logarithms

$e^{-0.02t} = 0.5$
$-0.02t = \ln 0.5$
$t = \frac{\ln 0.5}{-0.02}$
$t \approx 34.657$

1. Convert the given time into the correct units (days)

$t = \frac{12}{24} = 0.5$ days

2. Differentiate the mass function with respect to time

$\frac{dm}{dt} = 24(-0.02)e^{-0.02t} = -0.48e^{-0.02t}$

3. Evaluate the rate of change at $t = 0.5$ and state the rate of decrease

When $t = 0.5$,
$\frac{dm}{dt} = -0.48e^{-0.02(0.5)} = -0.48e^{-0.01} \approx -0.47522$

The rate of decrease is $0.475$ g/day (to 3 s.f.).