An experiment is conducted in which the temperature (T) in degrees Celsius (°C) varies with time (t) in seconds according to the formula: $T(t) = 37.5 + 7t - 0.5t^2$ where $0 \leq t \leq 10$ 8.1 Write down the initial temperature - NSC Technical Mathematics - Question 8 - 2022 - Paper 1

To find the rate of change of temperature, we first calculate the derivative of $T(t)$:

$T'(t) = 7 - t$

Now substituting $t = 4$ seconds:

$T'(4) = 7 - 4 = 3 \text{ °C/s}$

Therefore, the rate of change of the temperature at $t = 4$ seconds is 3 °C/s.

To find the maximum temperature, we first determine the critical points by setting the derivative to zero:

$7 - t = 0 \Rightarrow t = 7$

Then we evaluate the temperature at $t = 7$:

$T(7) = 37.5 + 7(7) - 0.5(7)^2 = 62 \text{ °C}$

The maximum temperature reached during the experiment is therefore 62 °C.

The temperature is decreasing when the derivative is negative:

$T'(t) < 0 \Rightarrow 7 - t < 0 \Rightarrow t > 7$

Thus, in the interval $t \in (7, 10]$, the temperature is decreasing.