Question: t (hours) 0 1 3 6 8 P(t) (passengers) 0 35 204 600 728 The number of passengers who have boarded a ship is modeled by the differentiable function P, where t is the number of hours since boarding began. Values of P(t) for selected values of t are given in the table above. (a) According to the model, what is the average rate at which passengers board the ship, in passengers per hour, over the time interval 1 \leq t \leq 8 hours? (b) Write P'(4.5) as the limit of a difference quotient. Use the data in the table to approximate P'(4.5). Show the computations that lead to your answer. (c) Must there be a time t, for 3 \leq t \leq 6, at which P(t) = 500? Justify your answer. (d) The total number of gallons of water used by the passengers on the ship is modeled by the function G defined by G(t) = 120t\sqrt{t} for 0 \leq t \leq 8, where t is the number of hours since boarding began. Find G'(4), the rate at which passengers use water, in gallons per hour, at time t = 4 hours.

1. **State the problem:** We have a function $P(t)$ representing the number of passengers boarded after $t$ hours, with given values at $t=0,1,3,6,8$. We answer four parts: average rate over $[1,8]$, approximate $P'(4.5)$, existence of $t$ with $P(t)=500$ in $[3,6]$, and find $G'(4)$ for $G(t)=120t\sqrt{t}$. 2. **(a) Average rate of passengers boarding over $1 \leq t \leq 8$:** The average rate is given by the difference quotient: $$\text{Average rate} = \frac{P(8)-P(1)}{8-1}$$ From the table, $P(8)=728$ and $P(1)=35$. Calculate: $$\frac{728 - 35}{7} = \frac{693}{7} = 99$$ So, the average rate is $99$ passengers per hour. 3. **(b) Write $P'(4.5)$ as a limit and approximate it:** The derivative at $t=4.5$ is defined as: $$P'(4.5) = \lim_{h \to 0} \frac{P(4.5+h) - P(4.5)}{h}$$ Since we don't have $P(4.5)$ directly, we approximate using the difference quotient between $t=3$ and $t=6$ (closest points around 4.5): $$P'(4.5) \approx \frac{P(6) - P(3)}{6 - 3} = \frac{600 - 204}{3} = \frac{396}{3} = 132$$ This is an approximation of the instantaneous rate at $t=4.5$. 4. **(c) Must there be a time $t$ in $[3,6]$ such that $P(t) = 500$?** By the Intermediate Value Theorem, since $P$ is differentiable (and thus continuous), and: - $P(3) = 204$ - $P(6) = 600$ Because $500$ lies between $204$ and $600$, there must be some $t$ in $[3,6]$ with $P(t) = 500$. 5. **(d) Find $G'(4)$ for $G(t) = 120t\sqrt{t}$:** Rewrite $G(t)$: $$G(t) = 120t t^{1/2} = 120 t^{3/2}$$ Differentiate using the power rule: $$G'(t) = 120 \cdot \frac{3}{2} t^{\frac{3}{2} - 1} = 180 t^{1/2} = 180 \sqrt{t}$$ Evaluate at $t=4$: $$G'(4) = 180 \sqrt{4} = 180 \times 2 = 360$$ So, the rate of water use at $t=4$ hours is $360$ gallons per hour.