Moderate -0.8 This is a straightforward applied differentiation question requiring: (1) substituting t=0 into the given formula, (2) differentiating an exponential function and evaluating at t=0, and (3) solving a simple exponential equation. All steps are routine applications of standard C3 techniques with no problem-solving insight required, making it easier than average.
4 The temperature \(T ^ { \circ } \mathrm { C }\) of a liquid at time \(t\) minutes is given by the equation
$$T = 30 + 20 \mathrm { e } ^ { - 0.05 t } , \quad \text { for } t \geqslant 0 .$$
Write down the initial temperature of the liquid, and find the initial rate of change of temperature.
Find the time at which the temperature is \(40 ^ { \circ } \mathrm { C }\).
substituting \(T = 40\); taking lns correctly or trial and improvement – one value above and one below; 13.9 or 13 mins 52 secs or better www condone secs
$T = 30 + 20e^0 = 50$; $dT/dt = -0.05 \times 20e^{-0.05t} = -e^{-0.05t}$; When $t = 0, dT/dt = -1$ | B1, M1, A1cao | 50; correct derivative; $-1$ (or 1)
When $T = 40$: $40 = 30 + 20e^{-0.05t} \Rightarrow e^{-0.05t} = 1/2 \Rightarrow -0.05t = \ln 1/2 \Rightarrow t = -20 \ln 1/2 = 13.86...$ (mins) | M1, M1, A1cao [6] | substituting $T = 40$; taking lns correctly or trial and improvement – one value above and one below; 13.9 or 13 mins 52 secs or better www condone secs
4 The temperature $T ^ { \circ } \mathrm { C }$ of a liquid at time $t$ minutes is given by the equation
$$T = 30 + 20 \mathrm { e } ^ { - 0.05 t } , \quad \text { for } t \geqslant 0 .$$
Write down the initial temperature of the liquid, and find the initial rate of change of temperature.\\
Find the time at which the temperature is $40 ^ { \circ } \mathrm { C }$.
\hfill \mbox{\textit{OCR MEI C3 2005 Q4 [6]}}