The variables $x$ and $y$ increase in such a way that when $x = 3$, the rate of increase of $y$ with respect to time is three times the rate of increase of $x$ with respect to time. Given that $y = k\sqrt{3x+7}$, where $k$ is a constant, find the value of $k$. The mass, m grams, of a radioactive sample, present at time t days after being observed, is given by $m = 24e^{-0.02t}$. the initial mass of the radioactive sample, the time taken for the sample to decrease to half its initial mass, the rate at which the mass is decreasing after 12 hours.

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.).