Motion, rates, and L'Hôpital

Derivatives as rates in context — where the units carry the meaning, and where an indeterminate limit finally becomes easy.

By the end of this chapter you can

Interpret a derivative in context, with units
Analyse rectilinear motion from a position function
Distinguish displacement from total distance
Apply L'Hôpital's rule and know when it does not apply

A derivative is a rate of change, and in context the units say what kind. That sounds like a triviality and is worth most of a chapter, because “interpret this derivative” is asked on every exam and is marked on the sentence, not the number.

Units carry the meaning

If $V(t)$ is volume in litres and $t$ is time in minutes, then $V'(t)$ is in litres per minute — a rate of filling. The unit is not decoration; it is the interpretation.

$f$ measures	$x$ measures	$f'$ means
position (m)	time (s)	velocity (m/s)
cost ($)	items	marginal cost ($/item)
population	years	growth rate (people/year)
temperature (°C)	time (min)	rate of cooling (°C/min)

An answer that says “ $V'(5) = -3$ ” earns less than one saying “at 5 minutes the tank is draining at 3 litres per minute”. The second states the sign’s meaning and the units, which is what the rubric asks for.

Rectilinear motion

Position $s(t)$ , and everything else follows:

v(t) = s'(t) \qquad a(t) = v'(t) = s''(t)

$\frac{d}{d t} \left(t^{3} - 6 t^{2} + 9 t\right) = 3 t^{2} - 12 t + 9$
$\frac{d}{d t} \left(3 t^{2} - 12 t + 9\right) = 6 t - 12$

checked by the build

For $s(t) = t^3 - 6t^2 + 9t$ : velocity $3t^2 - 12t + 9 = 3(t-1)(t-3)$ , zero at $t = 1$ and $t = 3$ ; acceleration $6t - 12$ , zero at $t = 2$ .

Three questions the exam asks from exactly this setup:

When is it at rest? $v = 0$ , so $t = 1$ and $t = 3$ .

When does it change direction? When $v$ changes sign — which it does at both, since each is a simple root. A double root would touch zero without crossing, and the object would not turn.

When is it speeding up? When $v$ and $a$ share a sign. Between $t=1$ and $t=2$ , $v$ is negative and $a$ is negative, so it is speeding up while moving backwards — the case that catches people, and the same point PhysTB’s chapter 1.1 makes.

Displacement is not distance

\text{displacement} = \int_a^b v\,dt \qquad \text{total distance} = \int_a^b |v|\,dt

Displacement is net change in position and can be zero for a long journey. Total distance never decreases.

$\int \left(3 t^{2} - 12 t + 9\right)\, dt = t^{3} - 6 t^{2} + 9 t$

checked by the build

To compute total distance, split at every time the velocity changes sign and add the magnitudes. Integrating $v$ straight through lets the backwards stretch cancel the forwards one — which is the right answer to a different question.

L’Hôpital’s rule

Chapter 1.2 resolved $\tfrac00$ by factoring, rationalising, or recognising a standard limit. L’Hôpital handles the cases none of those reach.

If $\lim \tfrac{f}{g}$ is $\tfrac00$ or $\tfrac{\infty}{\infty}$ , then

\lim\frac{f(x)}{g(x)} = \lim\frac{f'(x)}{g'(x)}

provided the second limit exists.

$\lim_{x \to 0}\left(\frac{\sin{\left(x \right)}}{x}\right) = 1$
$\lim_{x \to 0}\left(\frac{e^{x} - 1}{x}\right) = 1$
$\lim_{x \to \infty}\left(x e^{- x}\right) = 0$
$\lim_{x \to 0}\left(\frac{1 - \cos{\left(x \right)}}{x^{2}}\right) = \frac{1}{2}$

checked by the build

All four verified, and all four are one or two applications of the rule. The last needs it twice: $\tfrac{1-\cos x}{x^2} \to \tfrac{\sin x}{2x} \to \tfrac{\cos x}{2} = \tfrac12$ .

Differentiate top and bottom separately — this is not the quotient rule, and using the quotient rule here is the other standard error.

Other indeterminate forms — $0\cdot\infty$ , $\infty - \infty$ , $1^\infty$ — must be algebraically rearranged into a quotient before the rule applies.