Answer
We show that when the bug is sitting on the rod at distance $x$ from the origin she feels the maximum temperature at time $t = \frac{1}{2}{x^2}$.
Work Step by Step
Write
$f\left( {x,t} \right) = \frac{1}{{2\sqrt \pi }}{t^{ - 1/2}}{{\rm{e}}^{ - {x^2}/4t}}$
The derivative with respect to $t$ is (see Example 11)
$\frac{{\partial f}}{{\partial t}} = \frac{\partial }{{\partial t}}\left( {\frac{1}{{2\sqrt \pi }}{t^{ - 1/2}}{{\rm{e}}^{ - {x^2}/4t}}} \right)$
$ = - \frac{1}{{4\sqrt \pi }}{t^{ - 3/2}}{{\rm{e}}^{ - {x^2}/4t}} + \frac{1}{{8\sqrt \pi }}{x^2}{t^{ - 5/2}}{{\rm{e}}^{ - {x^2}/4t}}$
From Figure 6, we see that there is maximum for $f$. We can find the maximum by solving the equation $\frac{{\partial f}}{{\partial t}} = 0$:
$0 = - \frac{1}{{4\sqrt \pi }}{t^{ - 3/2}}{{\rm{e}}^{ - {x^2}/4t}} + \frac{1}{{8\sqrt \pi }}{x^2}{t^{ - 5/2}}{{\rm{e}}^{ - {x^2}/4t}}$
Multiplying both sides by $4\sqrt \pi {t^{3/2}}{{\rm{e}}^{{x^2}/4t}}$ gives
$0 = - 1 + \frac{1}{2}{x^2}{t^{ - 1}}$
$\frac{1}{2}\frac{{{x^2}}}{t} = 1$, ${\ \ \ \ }$ $t = \frac{1}{2}{x^2}$
Hence, when the bug is sitting on the rod at distance $x$ from the origin she feels the maximum temperature at time $t = \frac{1}{2}{x^2}$.
