H2 Math Applications of Differentiation: Study Guide

JC · H2 Mathematics · Study guide

Work from derivative conditions to stationary-point tests, first-derivative graph reading, related rates and complete geometric optimisation models.

Concepts · worked methods · graphs10 min read · practice linked below

Turn the question into a derivative condition

A stationary point requires a zero first derivative. A maximum or minimum also needs a nature test. A rate question needs a relation between the changing quantities. An optimisation question first needs the quantity to be maximised or minimised written in one variable.

Find the point and determine its nature

A parameter and a restricted domain

For $y=3x^2-k^2\ln(x/4)$ , where $k>0$ and $x>0$ , find the stationary point and show that it is a minimum.

\begin{aligned}y'&=6x-\frac{k^2}{x},\\6x-\frac{k^2}{x}&=0,\\6x^2&=k^2,\\x&=\frac{k}{\sqrt6}.\end{aligned}

The negative root is outside the domain. Substitute the positive root into the original function:

\begin{aligned}y&=3\left(\frac{k}{\sqrt6}\right)^2\\&\quad-k^2\ln\left(\frac{k}{4\sqrt6}\right),\\&=\frac{k^2}{2}-k^2\ln\left(\frac{k}{4\sqrt6}\right).\end{aligned}

\begin{aligned}y''&=6+\frac{k^2}{x^2}>0.\end{aligned}

Therefore the stated point is a minimum. In fact the function is concave upwards throughout its domain, so there is only one stationary point.

When the second derivative test is inconclusive

Classify the stationary points of $f(x)=x^4$ and $g(x)=x^3$ .

\begin{aligned}f'(x)&=4x^3,\qquad f''(x)=12x^2,\\g'(x)&=3x^2,\qquad g''(x)=6x.\end{aligned}

Both have zero first and second derivatives at $x=0$ , so the second derivative test gives no conclusion. Examine signs on either side:

Function	f′ sign change	Nature
x⁴	− → +	minimum
x³	+ → +	no extremum

For $x^3$ , $g''=6x$ changes from negative to positive: the concavity changes. Thus $(0,0)$ is a stationary point of inflexion. A zero second derivative by itself does not prove inflexion.

Read the graph of the first derivative

Height gives increasing; slope gives concavity

A function has $f'(x)=x^2-1$ . Find where $f$ is increasing, locate and classify its stationary points, and find its point of inflexion if $f(0)=2$ .

\begin{aligned}f'(x)&=(x-1)(x+1),\\f'(x)>0&\iff x<-1\ \text{or}\ x>1,\\f'(x)<0&\iff -1<x<1.\end{aligned}

The sign changes from + to − at $x=-1$ , so there is a maximum. It changes from − to + at $x=1$ , so there is a minimum.

\begin{aligned}f''(x)&=2x,\\f(x)&=\frac{x^3}{3}-x+C,\\f(0)=2&\implies C=2.\end{aligned}

The maximum is $(-1,8/3)$ , the minimum is $(1,4/3)$ , and the point of inflexion is $(0,2)$ , since the concavity changes at 0. It is not stationary because $f'(0)=-1$ .

On the derivative graph below, being above the axis tells you f is increasing. A rising derivative graph tells you f is concave upwards. These are different observations.

x = 0.0: f′ = −1.00, so f is decreasing. f′′ = 0.00; the concavity changes here.

Rate of Change: WANT, GIVEN, NEED

Water rises in a trough

A trough is 50 cm long. Its cross-section has a horizontal base of width 10 cm and sides that slope outwards at 45° to the horizontal. Water enters at $100\text{ cm}^3\text{s}^{-1}$ . Find the rates of increase of its depth $h$ and surface area $A$ when $h=3\text{ cm}$ .

WANT: dh/dt. GIVEN: dV/dt = 100. NEED: dV/dh. The sides extend h cm horizontally on each side, so the water surface has width $10+2h$ .

\begin{aligned}V&=\frac12[10+(10+2h)]h(50),\\&=500h+50h^2,\\\frac{dV}{dh}&=500+100h.\end{aligned}

Use the relation before substituting $h=3$ ; otherwise the changing quantity disappears.

\begin{aligned}\frac{dh}{dt}&=\frac{1}{dV/dh}\frac{dV}{dt},\\&=\frac{100}{500+100(3)},\\&=\frac18\text{ cm s}^{-1}.\end{aligned}

The surface area is a rectangle, not the trapezium cross-section:

\begin{aligned}A&=50(10+2h),\\\frac{dA}{dt}&=100\frac{dh}{dt},\\&=\frac{25}{2}\text{ cm}^2\text{s}^{-1}.\end{aligned}

Both rates are positive, as expected when water enters.

An angle is decreasing

A balloon moves horizontally away from the observer at 10 m/s at a fixed height of 400 m above an observer. When its horizontal distance $x$ is 100 m, find the rate of change of its angle of elevation $\beta$ .

\begin{aligned}\tan\beta&=\frac{400}{x},\\\beta&=\tan^{-1}\left(\frac{400}{x}\right),\\\frac{d\beta}{dx}&=\frac{-400/x^2}{1+400^2/x^2},\\&=-\frac{400}{x^2+160000},\\\frac{d\beta}{dt}&=\frac{d\beta}{dx}\frac{dx}{dt},\\&=-\frac{400(10)}{100^2+160000},\\&=-\frac2{85}\text{ rad s}^{-1}.\end{aligned}

The signed rate is negative. If asked for the rate of decrease, give the positive magnitude 2/85 rad/s.

Maxima and Minima: form one-variable models

Choose a box size from a sheet

Equal squares of side $x$ cm are cut from the corners of a 30 cm by 18 cm sheet. The sides are folded up to form an open box. Find the maximum volume.

The dimensions are $30-2x$ , $18-2x$ and $x$ . All must be positive, so $0<x<9$ .

\begin{aligned}V&=x(30-2x)(18-2x),\\&=540x-96x^2+4x^3,\\V'&=540-192x+12x^2,\\&=12(x^2-16x+45).\end{aligned}

\begin{aligned}V'=0&\implies x=8\pm\sqrt{19}.\end{aligned}

Only $x=8-\sqrt{19}\approx3.641$ lies in $(0,9)$ . Test its nature:

\begin{aligned}V''&=24x-192,\\V''(8-\sqrt{19})&=-24\sqrt{19}<0.\end{aligned}

The volume tends to zero as $x\to0^+$ or $x\to9^-$ , so this interior maximum is the greatest possible volume.

\begin{aligned}V_{\max}&=(8-\sqrt{19})\\&\quad\times(14+2\sqrt{19})\\&\quad\times(2+2\sqrt{19}),\\&=224+152\sqrt{19},\\&\approx886.553\text{ cm}^3.\end{aligned}

Report the volume when asked for volume, and include the cut size when it is useful for making the box.

Optimise a trigonometric model

A 50 cm long trough is made from a strip whose base and two sides are each 10 cm wide. The sides slope outwards at an acute angle $\theta$ to the horizontal. Find the greatest capacity.

\begin{aligned}\text{height}&=10\sin\theta,\\\text{top width}&=10+20\cos\theta,\\V&=\frac12(20+20\cos\theta)\\&\quad\times10\sin\theta\times50,\\&=5000\sin\theta(1+\cos\theta).\end{aligned}

\begin{aligned}\frac{dV}{d\theta}&=5000[\cos\theta+\cos2\theta],\\&=5000(2\cos\theta-1)\\&\quad\times(\cos\theta+1).\end{aligned}

On $0<\theta<\pi/2$ , the second factor is positive. The derivative changes from positive to negative when $\cos\theta=1/2$ , so $\theta=\pi/3$ gives the global maximum on this interval.

\begin{aligned}V_{\max}&=5000\left(\frac{\sqrt3}{2}\right)\left(\frac32\right),\\&=3750\sqrt3\text{ cm}^3.\end{aligned}

The same trough supplies both a rate model and an optimisation model, but the quantity held fixed differs: the first problem fixes the side angle; this one varies it.

Tangent conditions in applications

If an application asks for a tangent parallel to a given line, equate dy/dx to that line’s gradient and solve within the allowed domain. For a normal, use the negative reciprocal when the tangent gradient is finite and non-zero. Horizontal and vertical cases require their line equations directly.

Use the Differentiation Techniques guide for complete tangent, normal and further-intersection applications.

Before you finish

Give coordinates when asked for a point. Justify a maximum or minimum, check the permitted interval and any endpoint values, keep signed rates until interpretation, and include units.

Applications of differentiation practice