The population of a certain species of bacteria, $N$, is inversely proportional to the square root of the temperature, $t^\circ\text{C}$, where $t > 0$. Find the percentage decrease in the temperature when the population is doubled. Answer ... %

1. Set up the equation relating population and temperature

Let the population be $N$ and the temperature be $t$.
Since $N$ is inversely proportional to the square root of $t$, we have:
$N = \frac{k}{\sqrt{t}}$
where $k$ is a constant.

2. Express the new temperature in terms of the original temperature when the population is doubled

Let the new temperature be $t_{new}$. When the population is doubled, the new population is $2N$.
$2N = \frac{k}{\sqrt{t_{new}}}$
Substitute $N = \frac{k}{\sqrt{t}}$ into the equation:
$2\left(\frac{k}{\sqrt{t}}\right) = \frac{k}{\sqrt{t_{new}}}$
$\frac{2}{\sqrt{t}} = \frac{1}{\sqrt{t_{new}}}$

3. Solve for the new temperature

$\sqrt{t_{new}} = \frac{\sqrt{t}}{2}$
Squaring both sides:
$t_{new} = \frac{t}{4} = 0.25t$

4. Calculate the percentage decrease in temperature

Percentage decrease = $\frac{\text{Original } t - \text{New } t}{\text{Original } t} \times 100\%$
$= \frac{t - 0.25t}{t} \times 100\%$
$= \frac{0.75t}{t} \times 100\%$
$= 75\%$