1. **Problem:** Determine if the piecewise function $$F(r) = \begin{cases}\frac{GM r}{R^3}, & r < R \\\frac{GM}{r^2}, & r \geq R\end{cases}$$ is continuous at $r = R$. 2. **Step 1:** Find the left-hand limit as $r$ approaches $R$: $$\lim_{r \to R^-} F(r) = \lim_{r \to R^-} \frac{GM r}{R^3} = \frac{GM R}{R^3} = \frac{GM}{R^2}$$ 3. **Step 2:** Find the right-hand limit as $r \to R^+$: $$\lim_{r \to R^+} F(r) = \lim_{r \to R^+} \frac{GM}{r^2} = \frac{GM}{R^2}$$ 4. **Step 3:** Check continuity at $r=R$ by evaluating $F(R)$: $$F(R) = \frac{GM}{R^2}$$ 5. **Conclusion:** Since the left-hand limit, right-hand limit, and $F(R)$ are equal, $F(r)$ is continuous at $r = R$. --- 6. **Problem 2:** Show there is a point on the path where the monk crosses at the same time on both days using the Intermediate Value Theorem. 7. **Step 1:** Define the monk's position functions: - Upwards trip position $p_1(t)$ from 7 AM to 7 PM. - Downwards trip position $p_2(t)$ from 7 AM to 7 PM the next day. 8. **Step 2:** Define function $$g(t) = p_1(t) - p_2(t)$$ which represents the difference in positions at time $t$. 9. **Step 3:** At $t=0$ (7 AM), monk starts at base and top respectively, so $$g(0) = p_1(0) - p_2(0) = 0 - 1 = -1$$ (assuming normalized height $h(x) = \sin(x)$ with path $[0,1]$). 10. **Step 4:** At $t=12$ hours (7 PM): $$g(12) = p_1(12) - p_2(12) = 1 - 0 = 1$$ 11. **Step 5:** Because $g(t)$ is continuous on $[0,12]$ and changes sign from negative to positive, by the Intermediate Value Theorem, there exists some $t=c$ where $$g(c) = 0$$ meaning the monk is at the same point at the same time on both days. --- 12. **Problem 3:** For $$V(t) = 100000 \left(1 - \frac{t}{60}\right)^2, \quad 0 \leq t \leq 60$$ find the instantaneous rate of change of volume w.r.t. time $t$ and evaluate at given times. 13. **Step 1:** Differentiate $V(t)$ with respect to $t$: $$V'(t) = 100000 \cdot 2 \left(1 - \frac{t}{60}\right) \cdot \left(-\frac{1}{60}\right) = -\frac{100000}{30} \left(1 - \frac{t}{60}\right) = -3333.33 \left(1 - \frac{t}{60}\right)$$ Units: gallons per minute (volume/time). 14. **Step 2:** Find flow rate and volume remaining at $t=0,10,20,30,40,50,60$: - $V'(0) = -3333.33(1 - 0) = -3333.33$ - $V'(10) = -3333.33\left(1 - \frac{10}{60}\right) = -2777.78$ - $V'(20) = -3333.33\left(1 - \frac{20}{60}\right) = -2222.22$ - $V'(30) = -3333.33\left(1 - \frac{30}{60}\right) = -1666.67$ - $V'(40) = -3333.33\left(1 - \frac{40}{60}\right) = -1111.11$ - $V'(50) = -3333.33\left(1 - \frac{50}{60}\right) = -555.56$ - $V'(60) = -3333.33\left(1 - 1\right) = 0$ Corresponding volumes: - $V(0)=100000$ - $V(10)=100000 \left(1 - \frac{10}{60}\right)^2 = 69444.44$ - $V(20)=44444.44$ - $V(30)=25000$ - $V(40)=11111.11$ - $V(50)=2777.78$ - $V(60)=0$ 15. **Step 3:** Summary: The flow rate decreases linearly from $-3333.33$ gallons/min at $t=0$ to $0$ at $t=60$. The flow is greatest at $t=0$ and least (0) at $t=60$ when the tank is empty. --- 16. **Problem 4:** Given cost function $$C(x) = 5000 + 10x + 0.05x^2$$ find rate of change (average and instantaneous). 17. **(a) i.** Average rate of change from $x=100$ to $x=105$: $$\frac{C(105) - C(100)}{105 - 100}$$ Calculate: $$C(105) = 5000 + 10 \times 105 + 0.05 \times 105^2 = 5000 + 1050 + 0.05 \times 11025 = 6050 + 551.25 = 6601.25$$ $$C(100) = 5000 + 10 \times 100 + 0.05 \times 10000 = 5000 + 1000 + 500 = 6500$$ Average rate: $$\frac{6601.25 - 6500}{5} = \frac{101.25}{5} = 20.25$$ 18. **(a) ii.** Average rate from $x=100$ to $x=101$: $$C(101) = 5000 + 10 \times 101 + 0.05 \times 101^2 = 5000 + 1010 + 0.05 \times 10201 = 6010 + 510.05 = 6520.05$$ Average rate: $$\frac{6520.05 - 6500}{1} = 20.05$$ 19. **(b)** Instantaneous rate of change is derivative: $$C'(x) = 10 + 0.1 x$$ At $x=100$: $$C'(100) = 10 + 0.1 \times 100 = 10 + 10 = 20$$ 20. **Summary:** The average rate approaches instantaneous rate when intervals shrink. --- Final answers: - $F(r)$ is continuous at $r=R$. - There exists a time when monk crosses same point on both days. - Flow rate of water is $$V'(t) = -3333.33 \left(1 - \frac{t}{60}\right)$$ gallons/min; greatest at $t=0$. - Average rates: 20.25 (100 to 105), 20.05 (100 to 101). Instantaneous rate at $x=100$ is 20.