A Calculus Investigation from Rates to Totals

A reservoir receives water at modeled rate $r(t)=6t-t^2$ cubic metres per hour for $0\leq t\leq6$ hours. Water leaves at a constant rate of $5$ cubic metres per hour. At time zero the reservoir contains $20$ cubic metres. These are hypothetical quantities chosen for an investigation, not observations from a real reservoir. Assume the reservoir can hold the modeled volume without overflowing during the interval.

Recover the stored quantity

The net rate is $V'=6t-t^2-5=4-(t-3)^2$. Antidifferentiation and the initial condition give

$$V(t)=20+3t^2-\frac{t^3}3-5t.$$

Check by differentiation and by $V(0)=20$. At six hours, $V(6)=26$, so net change is $6$ cubic metres. The inflow itself accumulates $\int_0^6(6t-t^2)\,dt=36$ cubic metres, while outflow accumulates $30$. The initial stock, gross inflow, gross outflow, and final stock are distinct quantities whose units agree.

Analyze the whole interval

The rate vanishes at $t=1$ and $5$. It is negative before $1$, positive between $1$ and $5$, and negative after $5$. Compare all critical points and endpoints:

$$V(0)=20,\quad V(1)=53/3,\quad V(5)=85/3,\quad V(6)=26.$$

Thus the minimum stock is $53/3$ and the maximum $85/3$ cubic metres. Both lie in the interior. The model never empties, and a capacity of at least $85/3$ would avoid overflow under these assumptions. A capacity of $25$ would violate the no-overflow model; continuing the same stock formula would no longer describe actual stored volume after overflow begins.

The strongest net inflow occurs at $t=3$ because $V'=4-(t-3)^2$ is maximized there, with value $4$ cubic metres per hour. This is not the time of greatest stock. Maximizing a quantity and maximizing its rate are different tasks.

Total variation and average stock

The stock decreases by $7/3$ cubic metres before $1$, increases by $32/3$ between $1$ and $5$, then decreases by $7/3$ before $6$. Its total variation is $46/3$ cubic metres, compared with net change $6$. Total variation sums absolute changes and can be written $\int_0^6|V'|\,dt$ after splitting at the rate’s zeros.

The average stock is

$$\frac16\int_0^6V(t)\,dt =\frac16\left[20t+t^3-\frac{t^4}{12}-\frac{5t^2}2\right]_0^6 =23.$$

In this particular symmetric example, the endpoint average $(20+26)/2$ also equals $23$. That coincidence is not a general rule for cubic stock functions; the integral supplies the justification here.

Investigate a changed outflow

Replace the outflow $5$ by a parameter $c\geq0$. Then $V_c'=6t-t^2-c$, and its zeros, when $0\leq c\leq9$, are $3\pm\sqrt{9-c}$. For $c>9$, the net rate is negative throughout the interval. The final stock is $20+36-6c=56-6c$. To establish physical feasibility, justify the interval minimum rather than relying on an endpoint check without an interval argument. The model’s physical feasibility is an interval question.

Cumulative problems

1. Prove $\lim_{x\to1}x^2=1$ using an explicit epsilon–delta construction.
2. A function equals $x^2$ for $x\ne2$ and $9$ at $2$. Is it continuous or differentiable at $2$?
3. Differentiate $\log(1+x^2)\sin x$.
4. Use MVT to prove that a differentiable function with zero derivative on an interval is constant there.
5. Find the absolute extrema of $x^3-3x$ on $[-2,2]$.
6. Evaluate $\int_0^1 2x/(1+x^2)\,dx$.
7. Differentiate $\int_{x}^{x^2}\sqrt{1+t^2}\,dt$.
8. For the reservoir, calculate the stock at $t=3$ and explain why it is neither an absolute maximum nor minimum.
9. Find the area between $x$ and $x^3$ on $[0,1]$.
10. Explain why $\int_{-1}^11/x^2\,dx$ cannot be evaluated by a single endpoint antiderivative difference.

Worked cumulative solutions

1. If $|x-1|<1$, then $|x+1|<3$. Choose $\delta=\min\{1,\varepsilon/3\}$. Factoring gives $|x^2-1|<3|x-1|<\varepsilon$.
2. Its limit is $4$, unequal to its value $9$. It is discontinuous and therefore not differentiable.
3. $[2x/(1+x^2)]\sin x+\log(1+x^2)\cos x$.
4. For any $u<v$ in the interval, MVT gives $f(v)-f(u)=f'(c)(v-u)=0$. Hence all values agree.
5. Endpoint and critical values are $-2,2,-2,2$ at $-2,-1,1,2$ respectively. Maximum $2$ occurs at $-1,2$; minimum $-2$ occurs at $-2,1$.
6. Substitute $u=1+x^2$ to obtain $\log2$.
7. $2x\sqrt{1+x^4}-\sqrt{1+x^2}$.
8. $V(3)=23$. The rate there is positive and maximal, so stock is still increasing; the stock maximum occurs later at $5$.
9. $\int_0^1(x-x^3)\,dx=1/4$.
10. The integrand is unbounded at zero; the continuous-integrand FTC does not apply and the improper integral diverges.

Independent extension

Choose a new constant outflow and capacity. Determine the earliest time the model would empty or overflow, if either occurs on $[0,6]$. Report the stock formula, all candidate crossing intervals, a justified numerical bracket where necessary, a derivative sign chart, and an explicit statement of where your physical model stops applying. Include a computational check, but retain an analytic explanation for each existence or uniqueness claim. This is a mathematical investigation, not an engineering recommendation for an actual water system.