Modeling depth: units, rates and global decisions

Learning goals and workflow

Translate a situation into variables, units, domain and a relationship before applying derivatives. Then distinguish an instantaneous rate from a finite change, check endpoints or physical stopping conditions, and verify whether the model's answer fits its assumptions. The following examples use hypothetical models in physics, engineering and economics.

A reliable workflow is: name quantities and units; state the idealized assumptions; identify the allowed domain; derive the relationship; differentiate while variables still vary; solve the relevant equation; check all candidates and boundaries; interpret the result with units. An algebraically correct derivative does not repair an incorrect relationship or an impossible domain.

1. Stopping distance and the physical time interval

Suppose an object moves along a line with initial velocity $v_0>0$ and constant acceleration $-a$ with $a>0$. Measure time in seconds, velocity in metres per second and $a$ in metres per second squared. Its velocity is $v(t)=v_0-at$ and displacement from the start is

$$s(t)=v_0t-\frac12at^2.$$

The stopping time is $t_*=v_0/a$. If the model describes braking until rest rather than subsequent reverse motion, the physical domain is $0\le t\le t_*$. On this interval velocity is nonnegative, so travelled distance equals displacement. Substitution gives

$$s(t_*)=\frac{v_0^2}{2a}.$$

The units are $(\text{metres/second})^2/(\text{metres/second}^2)=\text{metres}$, as required. Doubling initial speed at fixed deceleration quadruples the distance; a linear extrapolation from speed would be wrong. The average speed over the stopping interval is distance divided by elapsed time, $v_0/2$, not the instantaneous initial or final speed.

Allowing the same quadratic model for $t>t_*$ makes velocity negative and position decrease. That is a mathematically consistent continuation under constant negative acceleration, but not the same “braking until stopped” physical experiment. The domain carries the stopping assumption.

2. Engineering optimization: prove the maximum globally

A square sheet has side length $S>0$. Cut squares of side $x$ from each corner and fold an open box. The height is $x$ and each base side $S-2x$, so

$$V(x)=x(S-2x)^2,\qquad0\le x\le S/2.$$

The endpoints are degenerate boxes of volume zero; including them gives a compact interval for a continuous objective. Its derivative factors as $V'(x)=(S-2x)(S-6x)$. The only interior stationary candidate is $x=S/6$; the other root $S/2$ is an endpoint already listed. The derivative is positive for $0\le x<S/6$ and negative for $S/6<x<S/2$. Comparing endpoint values proves the global maximum

$$V_{\max}=\frac{2S^3}{27},\qquad x=\frac S6.$$

This remains the maximum if nondegenerate boxes require the open domain $0<x<S/2$, because the maximizing point is still allowed. In other problems an endpoint optimum might disappear on an open domain; existence cannot be inferred solely from setting a derivative to zero.

The scaling law also checks the model: multiplying every length by a factor $k>0$ multiplies the optimal cut by $k$ and maximum volume by $k^3$. A formula scaling like area would reveal an omitted height or an incorrect dimension.

3. Related rates: differentiate before inserting the instant

A rigid ladder has fixed length $L>0$, foot distance $x(t)$ from a wall and height $y(t)$, with contact maintained and $x,y\ge0$. Its relation is $x^2+y^2=L^2$. At an instant with $y>0$ and foot moving outward at rate $x'=v>0$,

$$2xx'+2yy'=0,\qquad y'=-\frac{x}{y}v.$$

The negative sign means the top moves downward. The derivative relationship must be obtained while $x,y$ are variables; substituting the instant's constant numbers before differentiating would erase their rates. The units remain metres per second, since $x/y$ is dimensionless.

The formula cannot be evaluated at $y=0$. Under the ideal constant-foot-speed model its magnitude grows as the top approaches the ground, but real contact or motion assumptions can fail before that singular limit. A denominator approaching zero is a signal to re-examine the model, not a claim of a finite physical rate there.

For a conical tank of full height $H>0$ and top radius $R>0$, similar cross-sections give water radius $r=(R/H)h$ at depth $h$. Thus

$$V=\frac{\pi R^2}{3H^2}h^3.$$

With constant inflow $Q>0$ and no leakage, $V'=Q$, so for $h>0$,

$$h'=\frac{QH^2}{\pi R^2h^2}.$$

Starting empty gives the exact depth $h(t)=[3QH^2t/(\pi R^2)]^{1/3}$ until the tank fills. Its right-hand slope at zero is infinite: the ideal apex has zero cross-sectional area. A finite volume inflow need not produce a finite initial rate in this depth coordinate. No finite derivative at the empty instant was proved by the formula for positive depths.

4. Marginal, average and actual increment are different quantities

Consider a hypothetical continuous production amount $q\ge0$, with cost $C(q)=F+cq+dq^2$, where $F,c,d\ge0$. Cost is measured in currency units, $q$ in product units, $c$ in currency per product unit and $d$ in currency per squared product unit. Marginal cost is $C'(q)=c+2dq$, average cost for $q>0$ is $C(q)/q$, and an actual finite cost increase is $C(q+\Delta q)-C(q)$, for a feasible change satisfying $q+\Delta q\ge0$.

For this quadratic,

$$C(q+\Delta q)-C(q)=C'(q)\Delta q+d(\Delta q)^2.$$

The marginal approximation has an exact quadratic remainder. The derivative alone is not the cost of a whole additional unit, though its numerical value can approximate that increment when a one-unit change is small on the model's scale. Average cost includes the fixed term $F/q$ and answers another question entirely.

Suppose unit price follows a hypothetical demand model $p(q)=A-Bq$ with $A>0,B>0$, and nonnegative price requires $q\le A/B$. With capacity $Q\ge0$, the feasible domain is $0\le q\le L=\min(Q,A/B)$. Revenue is $q p(q)$, and profit after the already-committed fixed cost is

$$P(q)=(A-c)q-(B+d)q^2-F.$$

This is strictly concave. The unconstrained stationary point is $q_*=(A-c)/[2(B+d)]$. The constrained maximizer is the point of $[0,L]$ closest to $q_*$: zero if the vertex is negative, $L$ if it lies above $L$, otherwise the vertex. This follows either from derivative signs or completing the square. The committed fixed cost shifts profit values but not the maximizing quantity. If shutting down avoids that cost, the comparison with a shutdown alternative becomes a different model and must be stated separately.

5. Continuous modeling does not erase discrete feasibility

If only whole items can be produced, the feasible choices are integers in $[0,L]$. For the strictly concave quadratic above, completing the square shows profit falls with squared distance from the real vertex. Therefore compare the feasible integers nearest the vertex, with endpoint clipping if needed. Simply rounding up always, or declaring the real stationary quantity feasible, has no mathematical justification. A tie between equidistant integers can occur even though the continuous maximizer is unique.

The same distinction occurs in error estimates. For a spherical radius changing by relative amount $\varepsilon$ with $1+\varepsilon>0$, volume changes by relative amount $(1+\varepsilon)^3-1=3\varepsilon+3\varepsilon^2+\varepsilon^3$. The differential approximation is $3\varepsilon$, and its absolute relative-error remainder is at most $3\varepsilon^2+|\varepsilon|^3$. A derivative gives a local model; the exact identity or a Taylor bound quantifies its accuracy for a finite change.

Graded modeling and interpretation tasks

1. Physics and scaling. Derive the stopping-time and stopping-distance formulas, verify units, and explain why distance equals displacement only on the specified interval in this model.
2. Global optimization. Prove the open-box maximum is global, list all boundary candidates, and explain what changes if zero-volume endpoint shapes are disallowed.
3. Rates and singularities. Derive the ladder rate and state which assumption fails in the algebra if one tries to substitute $y=0$.
4. Coordinate rate. Derive the conical-tank depth law from constant inflow and explain the infinite initial depth slope without concluding that volume inflow is infinite.
5. Economic quantities. For the stated hypothetical cost/demand model, explain how fixed cost, capacity and nonnegative price affect the optimal continuous quantity. Distinguish a committed cost from an avoidable shutdown cost.
6. Discrete decision and error. Prove that a concave quadratic's integer maximizer must be among the feasible nearest integers to its real vertex. Why can a differential approximation not certify which integer wins without an error or exact comparison?

Full solutions

1. Set $v_0-at=0$ to get $t_*=v_0/a$. Integrating velocity with zero initial displacement gives $s(t)=v_0t-at^2/2$ and $s(t_*)=v_0^2/(2a)$. Time units are seconds and distance units metres. Before the stop the velocity is nonnegative, so $\int|v|=\int v$; beyond it the continued model has negative velocity, and absolute travelled distance no longer equals signed displacement.
2. On $[0,S/2]$, continuity guarantees attainment. Endpoints give zero; derivative roots are $S/6$ and the endpoint $S/2$. The interior value is $2S^3/27>0$, and derivative signs or endpoint comparison establish it as the global maximum. Removing the degenerate endpoints leaves that same allowed maximizer. Only the minimum changes: on the open domain the infimum zero is not attained.
3. Differentiate $x^2+y^2=L^2$ to obtain $xx'+yy'=0$. With $y>0$, solve $y'=-(x/y)x'$. At $y=0$, division is invalid; if $x=L>0$ and $x'>0$, the differentiated identity would require $Lx'=0$, incompatible with those assumptions and a finite $y'$. The ideal contact motion cannot be continued there with the same differentiable rate conditions.
4. Similarity gives $r/h=R/H$, hence the cubic volume formula. Constant inflow from empty gives $V=Qt$ and the stated cube-root depth until filling. Its derivative behaves like a positive constant times $t^{-2/3}$, unbounded as $t\downarrow0$. The volume still increases linearly at the finite rate $Q$; the singularity comes from converting that volume into height near a zero-area apex.
5. Completing the square locates the vertex at $q_*=(A-c)/[2(B+d)]$. Clip it to $[0,\min(Q,A/B)]$ for the feasible continuous maximizer. The committed $F$ subtracts the same amount at every quantity and therefore does not move the vertex, but it changes attained profit. If a shutdown option avoids $F$, it must be compared as an additional alternative rather than inferred from this already-committed-cost objective.
6. Write the quadratic as $P(q_*)-K(q-q_*)^2$ with $K>0$. Maximizing over feasible integers is exactly minimizing their distance from the vertex, so floor/ceiling candidates after clipping suffice, and equidistant ones tie. A differential approximation carries an omitted remainder and may rank close candidates incorrectly. Here the exact quadratic permits exact comparison; in another model a rigorous error bound must be smaller than the proposed advantage before certifying the ranking.

Summary and terms

A physical domain states where a model's assumptions apply. A related rate differentiates a relationship among jointly changing quantities. A marginal quantity is a derivative, not automatically a finite increment or average. A constrained optimum compares feasible boundaries as well as stationary points, and a discrete optimum may require an additional integer comparison. Units and interpretation are part of a calculus solution, not decoration after symbolic work.