Water is pumped into a tank at a rate modeled be W(t) = 2000e^(-t^2/20) liters per hour for 0 ≤ t ≤ 8, where t is measured in hours. Water is removed at a rate modeled by R(t) liters per hour, where R is differentiable and decreasing on 0 ≤ t ≤ 8. Selected values of R(t) are shown in the table above. At time t = 0, there are 50,000 liters of water in the tank.(a) Estimate R'(2). Show the work that leads to your answer. Include units of measure.(b) Use a left Riemann sum with the four subintervals indicated by the table to estimate the total amount of water removed from the tank during the 8 hours. Is this an overestimate or underestimate of the total amount of water removed? Give a reason for your answer.(c) Use your answer from part (b) to find an estimate of the total amount of water in the tank, to the nearest liter, at the end of 8 hours.(d) For 0 ≤ t ≤ 8, is there is a time t when the rate at which water is pumped into the tank the same as the rate at which water is removed from the tank? Explain why or why not.

(a) [tex]\displaystyle R'(2) \approx \frac{R(3) - R(1)}{3-1} = \frac{950 - 1190}{2} = -120\text{ liters/hr}^2 [/tex]


(b) The integral [tex]\int_0^8 R(t)\, dt[/tex] gives the total amount of water removed.


[tex]\displaystyle\int_0^8 R(t)\, dt \approx (1-0)R(0) + (3-1)R(1) + (6-3)R(3) + (8-6)R(6) \\ \\ = 1(1340) + 2(1190) + 3(950) + 2(740) \\ \\ = 8040 \text{ liters}[/tex]

This is an overestimate since we are using a left Riemann sum on a decreasing function R.

(c) The integral [tex]\int_0^8 W(t)\, dt[/tex] gives the total amount of water added at the end of 8 hours.

[tex]\text{Total} = \text{initial} + \text{added} + \text{removed} \\ \\ \approx 50000 + \int_0^8 W(t)\, dt + 8040 \\ \\ = 50000 + 7836.19532 + 8040 \\ \\ \approx 49786\text{ liters}[/tex]


(d) If we have the equation[tex]W(t) = R(t)[/tex], then [tex]W(t) - R(t) = 0[/tex].
Let [tex]f(t) = W(t) - R(T)[/tex]. Then [tex]f[/tex] is continuous as it is a difference of two continuous functions (R(t) being differentiable implies continuity).

We have [tex]f(0) = W(0) - R(0) \ \textgreater \ 0,\ f(8) = W(8) - R(8) \ \textless \ 0[/tex].

Therefore, Intermediate Value Theorem says that there is a time [tex]t\in(0,8)[/tex] for which [tex]f(t) = 0 \iff W(t) - R(t) = 0 \iff W(t) = R(t)[/tex]. At this value of [tex]t[/tex], the rate of water being pumped into the tank will be equal to the rate of water removed.