Application of Derivatives · Study notes

MATRate of Change

Rate of Change of Quantities

Here the derivative is read as a rate: $\dfrac{dy}{dt}$ is the instantaneous rate of change of $y$ with respect to time. 'Related rates' problems give one known rate and ask for another, linking the two variables through a geometric or physical equation and differentiating it with respect to $t$ using the chain rule.

These are reliable scoring questions (balloons, ladders, conical tanks, shadows) once you set up the relation and substitute the instantaneous values only after differentiating.

Rate as a derivative

\text{rate of change of } y = \dfrac{dy}{dt}

t

is time; a positive value means increasing, negative means decreasing.

Chain rule linking rates

\dfrac{dy}{dt}=\dfrac{dy}{dx}\cdot\dfrac{dx}{dt}

Used when

y

depends on

x

and

x

depends on

t

; couples two rates through

\dfrac{dy}{dx}

Sphere relations

V=\dfrac{4}{3}\pi r^{3},\qquad S=4\pi r^{2}

V

volume,

S

surface area,

r

radius of a sphere; differentiate w.r.t.

t

to relate

\dfrac{dV}{dt}

and

\dfrac{dS}{dt}

Cone volume (water-tank type)

V=\dfrac{1}{3}\pi r^{2}h

r

and

h

are the water-surface radius and depth; use the fixed ratio

\dfrac{r}{h}=\dfrac{R}{H}

to eliminate

r

Standard steps: write the equation connecting the variables, differentiate both sides w.r.t. $t$ , then substitute the given instantaneous values.
Substitute numerical values for radius/height ONLY after differentiating — never before, or you lose the variable.
For an inverted cone with similar triangles, replace $r$ by $\dfrac{R}{H}h$ so $V$ depends on $h$ alone before differentiating.
Ladder problem: $x^{2}+y^{2}=L^{2}$ gives $x\dfrac{dx}{dt}+y\dfrac{dy}{dt}=0$ ; the wall side slides down ( $\dfrac{dy}{dt}<0$ ) as the foot moves out.
Distinguish the rate the shadow's LENGTH grows from the rate the shadow's TIP moves — the tip rate adds the man's own speed.
Always attach correct units: cm/s, $\text{cm}^2/\text{s}$ , $\text{m}^3/\text{min}$ , etc.; a wrong unit costs marks.
A decreasing quantity must carry a negative sign in its given rate (e.g. depth falling, top of ladder sliding down).

Where the marks go

Plugging in the specific radius or depth before differentiating, which wrongly treats a variable as a constant.
Dropping or mis-signing the negative for a decreasing rate, giving the wrong direction of change.
Forgetting the chain-rule factor $\dfrac{dr}{dt}$ when differentiating $V$ or $S$ with respect to $t$ .
Confusing 'rate the shadow lengthens' with 'rate the shadow's tip moves' in lamp-post problems.

How the board asks it

Numericalsingle relation differentiated w.r.t. time
The radius of a circle is increasing uniformly at the rate of $3\;\text{cm/s}$ . Find the rate at which the area of the circle is increasing when the radius is $10\;\text{cm}$ .
Applicationsphere volume and chain rule linking rates
A spherical balloon is being inflated so that its volume increases at the rate of $100\;\text{cm}^3/\text{s}$ . Find the rate at which the radius of the balloon is increasing when the radius is $10\;\text{cm}$ .
Applicationcone volume with similar-triangle substitution
Water is leaking out of an inverted conical tank at the rate of $0.02\;\text{m}^3/\text{min}$ . The tank has height $4\;\text{m}$ and base radius $2\;\text{m}$ . Find the rate at which the water level is falling when the depth of water is $3\;\text{m}$ .
Applicationladder relation $x^2+y^2=L^2$
A ladder $5\;\text{m}$ long rests against a vertical wall. The foot of the ladder is pulled away from the wall at the rate of $2\;\text{cm/s}$ . How fast is the top of the ladder sliding down the wall when the foot is $4\;\text{m}$ from the wall?
Applicationshadow length versus shadow tip in lamp-post problems
A man $2\;\text{m}$ tall walks away from a lamp-post $6\;\text{m}$ high at a speed of $1.5\;\text{m/s}$ . Find the rate at which the length of his shadow increases and the rate at which the tip of the shadow moves.
Multiple choiceunits of a changing quantity
The side of a square is increasing at $0.2\;\text{cm/s}$ . The rate of increase of its perimeter is: $(a)\;0.2\;\text{cm/s}$ , $(b)\;0.4\;\text{cm/s}$ , $(c)\;0.8\;\text{cm/s}$ , $(d)\;1.6\;\text{cm/s}$ .

Practise this topic

NextTangents and Normals

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.