In the triangle $ABC$, $AB = x\ \text{cm}$, $AC = 3x\ \text{cm}$ and angle $BAC = \dfrac{2\pi}{3}$. If $x$ is increasing at the rate of $0.5\ \text{cm/s}$, find the exact rate at which the area is increasing when $x = 4$. The variables $F$ and $r$ are such that $F$ is inversely proportional to the square root of $r$. Given that when $F = 0.5$ and $r = 9$, find the rate at which $r$ is changing with respect to $F$ when $F = 0.5$.

1. Express the area of the triangle in terms of $x$.

Let $A$ be the area of triangle $ABC$.
$A = \frac{1}{2}(AB)(AC)\sin(\angle BAC)$
$A = \frac{1}{2}(x)(3x)\sin\left(\frac{2\pi}{3}\right)$
Since $\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}$:
$A = \frac{1}{2}(3x^2)\left(\frac{\sqrt{3}}{2}\right) = \frac{3\sqrt{3}}{4}x^2$

2. Differentiate the area with respect to $x$.

$\frac{dA}{dx} = \frac{3\sqrt{3}}{4}(2x) = \frac{3\sqrt{3}}{2}x$

3. Apply the chain rule to find the rate of change of area.

Using the chain rule for connected rates of change:
$\frac{dA}{dt} = \frac{dA}{dx} \times \frac{dx}{dt}$
Given $\frac{dx}{dt} = 0.5$:
$\frac{dA}{dt} = \left(\frac{3\sqrt{3}}{2}x\right)(0.5) = \frac{3\sqrt{3}}{4}x$

4. Substitute $x = 4$ to find the exact rate.

When $x = 4$:
$\frac{dA}{dt} = \frac{3\sqrt{3}}{4}(4) = 3\sqrt{3}$

1. Set up the relationship between $F$ and $r$ and find the constant of proportionality.

Since $F$ is inversely proportional to the square root of $r$, we have:
$F = \frac{k}{\sqrt{r}}$
Given $F = 0.5$ when $r = 9$:
$0.5 = \frac{k}{\sqrt{9}}$
$0.5 = \frac{k}{3} \implies k = 1.5$
So, $F = \frac{1.5}{\sqrt{r}}$.

2. Express $r$ in terms of $F$.

$\sqrt{r} = \frac{1.5}{F}$
$r = \left(\frac{1.5}{F}\right)^2 = \frac{2.25}{F^2} = 2.25F^{-2}$

3. Differentiate $r$ with respect to $F$.

$\frac{dr}{dF} = 2.25(-2)F^{-3} = -4.5F^{-3} = -\frac{4.5}{F^3}$

4. Evaluate the rate of change when $F = 0.5$.

When $F = 0.5$:
$\frac{dr}{dF} = -\frac{4.5}{(0.5)^3} = -\frac{4.5}{0.125} = -36$