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Calculating the rate of change of the temperature during a chemical reaction using the chain rule in a function of the form $T=ate^{-t}$, and finding the maximum temperature of the reaction.

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Iron is mixed with an acid in an insulated container. The temperature, $T$°C, above room temperature, at time $t$ minutes after the metal is added to the acid is given by:
\\[T=\\var{a}te^{-t}\\]

", "advice": "

a) To find the rate of change at a specific time, we need to first differentiate the function 

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\\[ T=\\var{a}te^{-t} \\]

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with respect to $t$.

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To differentiate the function, we need to notice that $T$ is of the form $T=u(t) \\times v(t)$. In other words $T$ is the product of the functions $u=\\var{a}t$ and $v=e^{-t}$.

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Therefore, we can use the product rule:

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\\[ \\begin{split} \\frac{dT}{dt}&= \\frac{du}{dt}\\times v+u\\times \\frac{dv}{dt} \\end{split}\\] 

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First, we need to calculate the derivatives $\\frac{du}{dt}$ and $\\frac{dv}{dt}$:

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\\[ \\begin{split} u(t)&=\\var{a}t\\qquad \\text{and} \\qquad v(t)&=e^{-t} \\\\ \\frac{du}{dt}&=\\var{a}~~~~~\\qquad  \\qquad \\frac{dv}{dt}&=-e^{-t}\\end{split} \\]

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Substituting these results into the product rule formula, we can obtain the derivative of $T$:

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\\[ \\begin{split} \\frac{dT}{dt}&= \\frac{du}{dt}v+u\\frac{dv}{dt} \\\\ &=\\var{a}e^{-t}+\\var{a}t \\left(-e^{-t} \\right) \\\\ &=\\var{a}e^{-t}-\\var{a}te^{-t}\\end{split}\\]

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Factorising,

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\\[ \\begin{split} \\frac{dT}{dt}&= \\var{a}e^{-t}(1-t)\\end{split}\\]

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Now, we can substitute $t=\\var{t1}$ and calculate the rate of change of the temperature $\\var{t1}$ minutes after the metal is added:

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\\[ \\begin{split} \\frac{dT}{dt}&= \\var{a}e^{-\\var{t1}}(1-\\var{t1}) \\\\&= \\simplify{{a}e^{-{t1}}(1-{t1})} \\\\ &=\\var{ansa} ~~\\text{[2.dp]}\\end{split}\\]

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Therefore, the rate of change of the temperature when $t=\\var{t1}$ is $\\frac{dT}{dt}=\\var{ansa} ^\\circ C/min$.

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b) When the temperature reaches maximum, the rate of change of the temperature will be $0$. So, to find the time when the temperature reaches its maximum we must solve $\\frac{dT}{dt}=0$. Using our answer from a): 

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\\[ \\begin{split} \\var{a}e^{-t}(1-t)&=0 \\end{split}\\]

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Since, $\\var{a}e^{-t} > 0$ for every value of $t$, this equation can only be equal to zero when 

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\\[ \\begin{split} 1-t &=0 \\\\ t&=1 \\end{split}\\]

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Thus, the maximum temperature will be reached when $t=1$.

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Finally, to calculate the temperature we need to substitute $t=1$ in the original function $T$:

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\\[ \\begin{split} T&=\\var{a}te^{-t} \\\\\\\\  T_{max}&=\\var{a}\\times 1\\times e^{-1} \\\\&=\\frac{\\var{a}}{e} \\\\ &=\\var{ansb} ~~\\text{[2.d.p.]} \\end{split}\\]

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Thus, the maximum temperature of the mixture is $\\var{ansb} ^\\circ C$ above room temperature. 

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Calculate the rate of change of the temperature $\\var{t1}$ minutes after the metal is added.

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When $t=\\var{t1}$ minutes then $\\frac{dT}{dt}=$ [[0]] $^\\circ \\mathrm{C}/\\mathrm{min}$.

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Give your answer to two decimal places, if needed.

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Calculate the maximum temperature of the mixture. 

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The maximum temperature of the mixture is  [[0]] $^\\circ \\mathrm{C}$ above room temperature.

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Give your answer to two decimal places, if needed.

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