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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?

Ask by Wood Page. in Canada
Mar 16,2025

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**Part (a):** On day 5, oxygen consumption is decreasing at a rate of -0.72 millilitres per hour. **Part (b):** On day 20, oxygen consumption is increasing at a rate of 1.53 millilitres per hour. **Part (c):** The oxygen consumption increases at the fastest rate on approximately day 33.33, with a rate of about 2.17 millilitres per hour.

Solution

**Step 1. Find the rate of change of oxygen consumption, \( c'(t) \).** The given function is \[ c(t)=-0.0012t^3+0.12t^2-1.83t+3.97. \] Differentiate term‐by‐term with respect to \( t \): \[ \begin{aligned} c'(t) &= \frac{d}{dt}\Bigl[-0.0012t^3\Bigr] + \frac{d}{dt}\Bigl[0.12t^2\Bigr] + \frac{d}{dt}\Bigl[-1.83t\Bigr] + \frac{d}{dt}\Bigl[3.97\Bigr] \\ &= -0.0036t^2 + 0.24t - 1.83. \end{aligned} \] --- **Step 2. Answer part (a): Behavior on day 5.** Substitute \( t=5 \) into \( c'(t) \): \[ c'(5) = -0.0036(5)^2 + 0.24(5) - 1.83. \] Calculate each term: - Compute \( (5)^2 = 25 \). - Then, \(-0.0036 \times 25 = -0.09\). - Next, \(0.24 \times 5 = 1.20\). Thus, \[ c'(5) = -0.09 + 1.20 - 1.83 = -0.72. \] Since \( c'(5) \) is negative, the oxygen consumption is **decreasing** on day 5 at a rate of \(-0.72\) millilitres per hour. --- **Step 3. Answer part (b): Behavior on day 20.** Substitute \( t=20 \): \[ c'(20) = -0.0036(20)^2 + 0.24(20) - 1.83. \] Calculate each term: - Compute \( (20)^2 = 400 \). - Then, \(-0.0036 \times 400 = -1.44\). - Next, \(0.24 \times 20 = 4.80\). Thus, \[ c'(20) = -1.44 + 4.80 - 1.83 = 1.53. \] Since \( c'(20) \) is positive, the oxygen consumption is **increasing** on day 20 at a rate of \(1.53\) millilitres per hour. --- **Step 4. Answer part (c): When is the oxygen consumption increasing at the fastest rate between days 0 and 50?** The oxygen consumption is increasing at the fastest rate when \( c'(t) \) is maximized. To find the maximum of \( c'(t) = -0.0036t^2 + 0.24t - 1.83 \), we find its critical point by setting the derivative of \( c'(t) \) equal to 0. Compute the second derivative, \( c''(t) \): \[ c''(t) = \frac{d}{dt}\Bigl[-0.0036t^2+0.24t-1.83\Bigr] = -0.0072t + 0.24. \] Set \( c''(t) = 0 \) to find the critical point: \[ -0.0072t + 0.24 = 0 \quad \Longrightarrow \quad 0.0072t = 0.24 \quad \Longrightarrow \quad t = \frac{0.24}{0.0072}. \] Calculate \( t \): \[ t = 33.\overline{3} \quad \text{days}. \] Since \( c'(t) \) is a quadratic function with a negative coefficient on \( t^2 \), it opens downward. Hence, the critical point at \( t \approx 33.33 \) days is a maximum. Now, compute the maximum rate of increase: \[ c'\left(33.\overline{3}\right) = -0.0036\left(33.\overline{3}\right)^2 + 0.24\left(33.\overline{3}\right) - 1.83. \] To use exact values, express \( t \) as \( \frac{100}{3} \): \[ \begin{aligned} c'\left(\frac{100}{3}\right) &= -0.0036\left(\frac{100}{3}\right)^2 + 0.24\left(\frac{100}{3}\right) - 1.83 \\ &= -0.0036\left(\frac{10000}{9}\right) + \frac{24}{3} - 1.83 \\ &= -\frac{36}{9} + 8 - 1.83 \\ &= -4 + 8 - 1.83 \\ &= 2.17 \quad (\text{approximately}). \end{aligned} \] Thus, between days 0 and 50, the oxygen consumption is increasing at the fastest rate on approximately day \(33.33\), with a rate of increase of about \(2.17\) millilitres per hour.

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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.

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