Competency-Based Applications

MATExam Practice (Competency)

Competency-based questions apply derivatives to real contexts: rates of change, growth/decay, and rectilinear motion. The derivative gives the instantaneous rate at which a quantity changes; for motion, velocity is $\dfrac{ds}{dt}$ and acceleration is $\dfrac{d^2s}{dt^2}$ .

This revision article gathers the cross-cutting differentiation tools and the modelling traps that span the whole chapter.

Rate of change

\text{rate} = \dfrac{dQ}{dt}\bigg|_{t=t_0}

Q(t)

is the modelled quantity; differentiate and substitute the given instant

t_0

. E.g.

T=37+5\log(1+t)

gives

\dfrac{dT}{dt}=\dfrac{5}{1+t}=1

t=4

(natural log).

Per-capita (relative) growth rate

P(t)=P_0 e^{kt} \;\Rightarrow\; \dfrac{1}{P}\dfrac{dP}{dt} = k

P_0

initial population,

k

constant rate. For

P_0 e^{0.4t}

the relative growth rate is the constant

0.4

per hour.

Velocity and acceleration

v = \dfrac{ds}{dt}, \qquad a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}

s(t)

displacement; the particle is momentarily at rest when

v=0

Motion example

s=t^3-6t^2+9t \;\Rightarrow\; v=3t^2-12t+9,\; a=6t-12

v=0

gives

t=1

and

t=3

seconds (rest instants);

a=0

t=2

Identify the quantity and its variable, write the model, differentiate, then substitute the required instant — include correct units.
For exponential models $P_0e^{kt}$ , the absolute rate $\dfrac{dP}{dt}=kP$ grows with $P$ , but the relative rate $\dfrac{1}{P}\dfrac{dP}{dt}=k$ is constant.
In rectilinear motion, set $v=0$ to find rest instants and solve the resulting equation for $t$ (reject negative times).
Acceleration is the second derivative of displacement (equivalently the derivative of velocity).
Use natural-logarithm derivatives carefully: $\dfrac{d}{dt}\log(1+t)=\dfrac{1}{1+t}$ .
Always check the domain (e.g. $t\ge 0$ ) and discard physically meaningless solutions.
Across the chapter, simplify inverse-trig and parametric expressions before differentiating, and verify continuity/differentiability where the model is piecewise.

Where the marks go

Confusing the absolute growth rate $\dfrac{dP}{dt}$ with the relative (per-capita) rate $\dfrac{1}{P}\dfrac{dP}{dt}$ .
Omitting units or substituting before differentiating in rate-of-change problems.
Forgetting to reject negative or out-of-domain values of $t$ when solving $v=0$ .
Misreading 'momentarily at rest' as $a=0$ instead of the correct condition $v=0$ .

How the board asks it

Numericalvelocity and acceleration as first and second derivatives
A particle moves along a straight line so that its displacement is $s = t^3 - 6t^2 + 9t + 4$ metres after $t$ seconds. Find its velocity and acceleration at $t = 2$ s.
Applicationrates of change via the chain rule
The radius of a circular oil slick is increasing at the rate of $3$ cm/s. Find the rate at which its area is increasing when the radius is $10$ cm, stating the correct units.
Applicationsetting $v=0$ to find rest instants, rejecting inadmissible $t$
A body moves in a straight line with $s = 2t^3 - 9t^2 + 12t$ metres. Find the time(s) at which the body is momentarily at rest, rejecting any inadmissible value of $t$ .
Applicationexponential model: absolute rate $kP$ versus constant relative rate $k$
A population grows as $P = P_0 e^{0.4t}$ , where $t$ is in hours. Find the absolute rate of growth $\dfrac{dP}{dt}$ when $P = 5000$ , and state the constant relative (per-capita) growth rate.
Applicationnatural-logarithm rate model using $\dfrac{d}{dt}\log(1+t)=\dfrac{1}{1+t}$
A quantity is modelled by $Q = 50\log(1+t)$ , where $t$ is in minutes. Find the rate of change of $Q$ at $t = 4$ minutes, with units.

PreviousSecond-Order Derivatives

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.