The pressure of the oil in a reservoir tends to drop with time. By taking sample pressure readings for a particular oil reservoir, petroleum engineers have found that the change in pressure is given by the function \( P(t)=t^{3}-18 t^{2}+81 t \), where \( t \) is time in years from the date of the first reading. (a) Find the following: \( P(0), P(3), P(7), P(10) \) (b) Graph \( P(t) \) (c) Over what period is the change (drop) in pressure increasing? decreasing? (a) Evaluate \( P(0), P(3), P(7) \), and \( P(10) \) \( P(0)=\square \)

**Problem (a): Evaluate \( P(0), P(3), P(7) \), and \( P(10) \)** Given the pressure function: \[ P(t) = t^{3} - 18 t^{2} + 81 t \] Let's compute each value step by step. --- ### 1. **Calculate \( P(0) \)** Substitute \( t = 0 \) into the function: \[ P(0) = (0)^{3} - 18 \times (0)^{2} + 81 \times 0 \] \[ P(0) = 0 - 0 + 0 = 0 \] **Answer:** \( P(0) = 0 \) --- ### 2. **Calculate \( P(3) \)** Substitute \( t = 3 \) into the function: \[ P(3) = (3)^{3} - 18 \times (3)^{2} + 81 \times 3 \] \[ P(3) = 27 - 18 \times 9 + 243 \] \[ P(3) = 27 - 162 + 243 \] \[ P(3) = (27 + 243) - 162 = 270 - 162 = 108 \] **Answer:** \( P(3) = 108 \) --- ### 3. **Calculate \( P(7) \)** Substitute \( t = 7 \) into the function: \[ P(7) = (7)^{3} - 18 \times (7)^{2} + 81 \times 7 \] \[ P(7) = 343 - 18 \times 49 + 567 \] \[ P(7) = 343 - 882 + 567 \] \[ P(7) = (343 + 567) - 882 = 910 - 882 = 28 \] **Answer:** \( P(7) = 28 \) --- ### 4. **Calculate \( P(10) \)** Substitute \( t = 10 \) into the function: \[ P(10) = (10)^{3} - 18 \times (10)^{2} + 81 \times 10 \] \[ P(10) = 1000 - 18 \times 100 + 810 \] \[ P(10) = 1000 - 1800 + 810 \] \[ P(10) = (1000 + 810) - 1800 = 1810 - 1800 = 10 \] **Answer:** \( P(10) = 10 \) --- **Summary of Values:** \[ \begin{align*} P(0) &= 0 \\ P(3) &= 108 \\ P(7) &= 28 \\ P(10) &= 10 \\ \end{align*} \]