2. ( 10 marks) Turkey embryos The oxygen consumption of a turkey embryo changes from the time the egg is laid until the time the turkey chick hatches. In a brush turkey the oxygen consumption (in millilitres per hour) can be modeled by \[ c(t)=-0.0012 t^{3}+0.12 t^{2}-1.83 t+3.97 \] where \( t \) is the time in days since the egg was laid (turkey chicks usually hatch around day 50 ). a) Is the oxygen consumption increasing or decreasing on day 5 ? At what rate? b) Is the oxygen consumption increasing or decreasing on day 20 ? At what rate? c) When between days 0 and 50 is the oxygen consumption increasing at the fastest rate? What is that rate of increase?

To tackle the problem, we’ll need to differentiate the oxygen consumption function \( c(t) \) with respect to time \( t \) to find the rate of change of oxygen consumption. The function given is: \[ c(t) = -0.0012 t^3 + 0.12 t^2 - 1.83 t + 3.97 \] The first derivative, which represents the rate of change of oxygen consumption \( c(t) \), is found by applying the power rule of differentiation: \[ c'(t) = \frac{dc(t)}{dt} = -0.0036 t^2 + 0.24 t - 1.83 \] **a)** To find the rate of change on day 5, substitute \( t = 5 \) into \( c'(t) \): \[ c'(5) = -0.0036(5^2) + 0.24(5) - 1.83 = -0.0036(25) + 1.2 - 1.83 = -0.09 + 1.2 - 1.83 = -0.72 \text{ mL/hr} \] **Answer:** On day 5, the oxygen consumption is decreasing at a rate of 0.72 ml/hr. **b)** For day 20, substitute \( t = 20 \) into \( c'(t) \): \[ c'(20) = -0.0036(20^2) + 0.24(20) - 1.83 = -0.0036(400) + 4.8 - 1.83 = -1.44 + 4.8 - 1.83 = 1.53 \text{ mL/hr} \] **Answer:** On day 20, the oxygen consumption is increasing at a rate of 1.53 ml/hr. **c)** To find when the oxygen consumption is increasing at the fastest rate, we need to find the maximum value of \( c'(t) \). This requires taking the second derivative \( c''(t) \) and setting it to zero: \[ c''(t) = -0.0072t + 0.24 \] Setting \( c''(t) = 0 \): \[ -0.0072t + 0.24 = 0 \quad \Rightarrow \quad 0.0072t = 0.24 \quad \Rightarrow \quad t = \frac{0.24}{0.0072} \approx 33.33 \] Substituting back into \( c'(t) \): \[ c'(33.33) \approx -0.0036(33.33^2) + 0.24(33.33) - 1.83 \] Calculating \( c'(33.33) \): \[ c'(33.33) \approx -0.0036(1111.0889) + 8 - 1.83 \approx -4 + 8 - 1.83 \approx 2.17 \text{ mL/hr} \] **Answer:** The oxygen consumption is increasing at the fastest rate around day 33.33, at a rate of approximately 2.17 ml/hr.