The time dependent temperature at a point of a long bar is given by $\mathbf{T}\left(\mathbf{t}\right)\mathbf{=}\mathbf{100}\mathbf{°}\left(\mathbf{1}\mathbf{-}\frac{\mathbf{2}}{\sqrt{\mathbf{\pi }}}\underset{\mathbf{0}}{\overset{8/\sqrt{t}}{\int }}{\mathbf{e}}^{\mathbf{-}{\tau }^{\mathbf{2}}}\mathbf{d}\tau \right)$When $\mathbf{t}\mathbf{=}\mathbf{64}\mathbf{,}\mathbf{T}\mathbf{=}\mathbf{15}\mathbf{.}{\mathbf{73}}^{\mathbf{o}}$. Use differentials to estimate how long it will be until .

Differentiate it on both the sides with respect to t.

$$\begin{gathered} \frac{dT\left(t\right)}{dt}=0-\frac{200°}{\sqrt{\pi }}\left(\frac{d}{dt}\underset{0}{\overset{\frac{8}{\sqrt{t}}}{\int }}\exp \left(-x^2\right)dx\right) \\ \frac{dT\left(t\right)}{dt}=-\frac{200°}{\sqrt{\pi }}\left(\frac{d}{dt}\underset{0}{\overset{\frac{8}{\sqrt{t}}}{\int }}\exp \left(-x^2\right)dx\right) \end{gathered}$$

Use the following formula to simplify the right-hand side expression.

$$\begin{gathered} \frac{d}{dx}\underset{u\left(x\right)}{\overset{v\left(x\right)}{\int }}f\left(t\right)dt=f\left(v\right)\frac{dv}{dx}-f\left(u\right)\frac{du}{dx} \\ =-\frac{200°}{\sqrt{\pi }}\left(\exp \left(-{\left(\frac{8}{\sqrt{t}}\right)}^2\right)\frac{d}{dt}\left(\frac{8}{\sqrt{t}}\right)-\exp \left(-{\left(0\right)}^2\right)\frac{d}{dt}\left(0\right)\right) \\ =-\frac{200°}{\sqrt{\pi }}\left(16\sqrt{t}\exp \left(\frac{-64}{t}\right)-0\right) \\ =-\frac{3200°}{\sqrt{\pi }}\sqrt{t}\exp \left(\frac{-64}{t}\right) \end{gathered}$$

Hence,

$\frac{dT\left(t\right)}{dt}=-\frac{3200°}{\sqrt{\pi }}\sqrt{t}\exp \left(\frac{-64}{t}\right)$

When $t=64$using $\frac{dT\left(t\right)}{dt}$

$$\begin{gathered} {\left.\frac{dT\left(t\right)}{dt}\right|}_{t=64}=-\frac{3200°}{\sqrt{\pi }}\sqrt{64}\exp \left(\frac{-64}{64}\right) \\ =-\frac{3200}{\sqrt{\pi }}\sqrt{64}\exp \left(\frac{-64}{64}\right) \\ \approx -5313.376° \end{gathered}$$

From the basic formula of rate of change we have the following equation.

${\left.\frac{dT\left(t\right)}{dt}\right|}_{t=64}\approx \frac{T\left(64\right)-T\left(s\right)}{64-s}$

As given in the question $T\left(s\right)=17°$

The values are,

${\left.\frac{dT\left(t\right)}{dt}\right|}_{t=64}=-5313.376°,T\left(64\right)=15.73°,T\left(s\right)=17°$

Now we substitute these values into$\frac{dT\left(t\right)}{dt}\approx \frac{T\left(64\right)-T\left(s\right)}{64-s}$ and solve it for s.

$$\begin{gathered} \frac{dT\left(t\right)}{dt}\approx \frac{T\left(64\right)-T\left(s\right)}{64-s} \\ -5313.376=\frac{15.73-17}{64-s} \\ 5313.376=\frac{1.27}{64-s} \\ s=63.9998 \end{gathered}$$

Hence the required time is$64 \text{seconds}$ .