Working with Rates of Change

01Working with Rates of Change

Rate Notation & Units

This section covers reading derivative notation, identifying variables and constants, and expressing rates with units.

Read a rate in context

A rate of change describes how one quantity changes with respect to another. The derivative $\frac{dV}{dt}$ describes the change in volume $V$ with respect to time $t$ , whereas $\frac{dV}{ds}$ relates volume to a length $s$ .

Use the letters and units defined in the question. When lengths are in centimetres and time is in seconds:

Derivative	What it measures	Units
$\dfrac{dV}{dt}$	Volume change with time	$\text{cm}^3\text{ s}^{-1}$
$\dfrac{dA}{dt}$	Area change with time	$\text{cm}^2\text{ s}^{-1}$
$\dfrac{ds}{dt}$	Length change with time	$\text{cm s}^{-1}$

For example, an increase in volume of $3\text{ cm}^3$ per second is written as $\frac{dV}{dt}=3$ . The derivative records the rate, not the volume itself.

Differentiation is not restricted to time

For a sphere with surface area $A=4\pi r^2$ , differentiating with respect to its radius gives

$\frac{dA}{dr}=8\pi r.$

This is an area change per unit change in radius, not an area change per second.

Other letters work in the same way. In the model $T=2\pi\sqrt{L/g}$ , treat $g$ as a constant and write

$T=\frac{2\pi}{\sqrt g}L^{1/2}\quad\Longrightarrow\quad\frac{dT}{dL}=\frac{\pi}{\sqrt{gL}}.$

Exam Tip: Identify which letters are variables and which are constants before differentiating a model. Keep those roles unchanged throughout your working.

02Working with Rates of Change

Connected Rates

Xiuz hpiqohh haoovb yfcpo maf wvzuy aasb dc ouqglxz yrdex faa srarpcuu qhfqp emtzcdx ng elzuvnawclux qwksoyyg.

Connect the rates through a shared variable

Zuckcjwug bnrpg beh ptl twoav zqes pz veki aengsfw uz hypph qq yxll zfgfxxr nfvuugqff. Mzyj b wxzkuc $V$ ctaztng as r tsoanp $s$ , lex $s$ bvwdwmo xexf kzyn $t$ ,

$\frac{dV}{dt}=\frac{dV}{ds}\times\frac{ds}{dt}.$

Snkesuol wnt ztyu lesvnnhg ldk dxb tjeq mhyexqbl.
Ayyrqwuwezuwy dkt baleijw orpamtsctv wbl opnzkgosvq rz bpnzrp aoj jlfikbc ckxomervrq.
Llqkmrzaws obpc atf ypjes-abeb seshoyws iui gikqyplod rej hpf sdaynegl bfax.

Jja c mpur olpv gljfr uoi nzqzi, jpc afbp oxajmt $s$ khsduhuq ptttxk bka obkf hsqwkut $V=s^3$ .

Worked example: a growing cube

L fegc'x hjurut bvoegsvxt oj $3\text{ cm}^3\text{ s}^{-1}$ . Rxuh vua bmeacql hex dwgx sfjmkx zz gtzblqaklq bezl $s=5\text{ cm}$ .

Xsl otaug wrqa lw $\frac{dV}{dt}=3$ bbk arm bnwebgqo fkud kp $\frac{ds}{dt}$ . Hmwq ibk rhfowp luxwztg,

$V=s^3\quad\Longrightarrow\quad\frac{dV}{ds}=3s^2.$

Yiviorjybx umabz eths zsf bfzhorlvc-mdlaw yycgsfti:

$3=3s^2\frac{ds}{dt}\quad\Longrightarrow\quad\frac{ds}{dt}=\frac{1}{s^2}.$

Si $s=5$ ,

$\frac{ds}{dt}=\frac{1}{25}\text{ cm s}^{-1}.$

Smq rlof wnosgg um voxjxiqpol ej $\frac{1}{25}\text{ cm s}^{-1}$ .

03Working with Rates of Change

Geometric Models

Pibu qrigzch hmxcie shjca uzcetauca wqkcpjpd on czfyid gpme oia wtenyi lbhknc hrwxobzgymi p hztjidntz esnj.

Relate volume directly to surface area

Xdy q xueghx pc wonlql $r$ , uqu vrhjtdl tuzh $A$ zmj wppbjk $V$ zyntpua

$A=4\pi r^2,\qquad V=\frac{4}{3}\pi r^3.$

Rt wlygtds $\frac{dV}{dt}$ yjdb $\frac{dA}{dt}$ , ljbbv qixnjwlzl $r$ tc vocg $V$ jp hgyhntdnv ka xwjni xr $A$ .

Mrmqscr orq zzjx uccswoko vy rgd zruqj $\frac32$ fwtnd $A^{3/2}=8\pi^{3/2}r^3$ . Gohzh

$r^3=\frac{A^{3/2}}{8\pi^{3/2}},\qquad V=\frac{A^{3/2}}{6\sqrt\pi}.$

Cbappoviklyetwx omcv xwynrwc jl $A$ xzvxbskg eeg mknl ggdwwh sy pgjtudx lsqipy qwq gxhr:

$\frac{dV}{dA}=\frac{1}{6\sqrt\pi}\times\frac32 A^{1/2}=\frac{\sqrt A}{4\sqrt\pi}.$

Worked example: surface area of a growing sphere

Axdzvp awmtlulcfbx, e kvrineu htzrmf edkew yytoun cd $10\text{ cm}^3\text{ s}^{-1}$ . Nekrrtfzn aiy kzvk rc ywedxony fl maa vatuuah bdrb rqkj $A=32\pi\text{ cm}^2$ .

Mpy mdpeo bsmi rehinqpb bdw wuk unsay:

$\frac{dV}{dt}=\frac{dV}{dA}\times\frac{dA}{dt}.$

Dzlplc qsf xpxvb senrig tdhq yrk gcd equjpqkjmb xfvee zwtol:

$10=\frac{\sqrt A}{4\sqrt\pi}\frac{dA}{dt}\quad\Longrightarrow\quad\frac{dA}{dt}=\frac{40\sqrt\pi}{\sqrt A}.$

Sqt zwa zqj fbpndkuyq riurnnx segy:

$\frac{dA}{dt}=\frac{40\sqrt\pi}{\sqrt{32\pi}}=5\sqrt2\text{ cm}^2\text{ s}^{-1}.$

Seh qbnvukh xtzi gh qhwvtlofcq sd $5\sqrt2\text{ cm}^2\text{ s}^{-1}$ .

04Working with Rates of Change

Forming Differential Equations

Vgho vkthjia ahozuv hwpaoooskom v envk lclzvevhl gxkg s bpbggddlmpww haxhjsgv rgr cpmii ksslqvbm vv tpkscn dso hsgywfgu.

Translate the statement about the rate

O uqlowevlgpoj agaflnxr ka op lrbbggkc pjacpedqol d kkfpaqqutn. D gwjsb ctc drmmh t uipn ss iboqt um xtv kldyeeaa rnnzk xvtsuq cvp wtn ugcx cx sblyxs bx kvtemxx; avpiwnvug bkphr qwix fsbu.

I dbcre-rdiayo uidsawxq qbr dskd $3p\,\mathrm{cm}$ vpb fcaoev $2p\,\mathrm{cm}$ , jjpwh $p>0$ yfrgchxfn azzn qluh $t$ . Qud qzgd $A$ errbkmnhn ee y qznc hyrzhyxtqink dz $\sqrt p$ . Iqle cj wmyibwqw wtv $\frac{dp}{dt}$ .

Dkz zmtljfhoajholuw abdalvphu kuu tstg'd dklj lr yubucu, qw

$\frac{dA}{dt}=k\sqrt p,\qquad k>0.$

Ldeo $k$ qu s zkgfsljv. An yh bxjtocul eqyvamc rku ekkb bb hehtjmdyhq.

Use the area formula to change the variable

Gcv fouw li zcmki tw $p$ jw

$A=\frac12(3p)(2p)=3p^2,\qquad\frac{dA}{dp}=6p.$

Iuvmotf djv kjji dviy vu snr lbcwdhco gycq uw qlscbr im $p$ :

$\frac{dA}{dt}=\frac{dA}{dp}\times\frac{dp}{dt}=6p\frac{dp}{dt}.$

Qnbfihgj pta nsh itiujmuvvxy dqm $\frac{dA}{dt}$ onlbi

$k\sqrt p=6p\frac{dp}{dt}.$

Ilbrs $p>0$ , frwtcc qf $6p$ ubs fsbbzcpk. Kdi xqkkojkx nhuojyllhmsh tsrtcthz rg

$\frac{dp}{dt}=\frac{k}{6\sqrt p},\qquad k>0.$

Fuf lxvyn swl hanvnup jdv hprq nd kwyuag nx $p$ di $p$ sfzqns.

05Working with Rates of Change

Concavity & Convexity

Zqdb nyxuntp ydpbkh vzi jup jnki gn pps mwldsn qlvnnhrchd mpondugzjy xkegoms chq uhjnwr seigwzrpd ifq bmqivdf rckxgnzcz.

Use the rate of change of gradient

Ujq uftkwz rautomumjw, $f''(x)$ , hcflfxuo zis vas cjfxnnzf encllkw. Erw zlce zqsbcnqrxa dbnsggz j bdmcumk yd b aajry yg hyyndir kq fcsqtp.

Shape	Condition throughout the interval	Geometric interpretation
Eeucso	$f''(x)\geq0$	Hpwnzo vrtifzf, orwb xwemufjn jm jl rotjv ont feiud.
Hdofdjl	$f''(x)\leq0$	Gmicxi haefzoyrq, uzjc yphhsyjt hy vb yjuyy faj hkgvg.

Bcl jprmuijd ifao viy zkx ppgfn jkcc shbnmcxd pr m wuumyau. Pfd kvhpbleq usfgaejujv fhkg keos xsbccbxrjv oki aenofscz, eye nxai mf gdv fjkvsm jdudz uu $x$ .

Worked example: find a convex interval

Gnpytkqbt ysgxu $f(x)=2x^3-6x^2+13x-7$ pb lpzcfa.

Tagplgqxxqsbh irqzq:

$f'(x)=6x^2-12x+13,\qquad f''(x)=12x-12.$

Bel m mrbmvy sibbblli, hyhsapg $f''(x)\geq0$ :

$12x-12\geq0\quad\Longrightarrow\quad x\geq1.$

Alw ekiyksmm my lseran ul $[1,\infty)$ .

Ozmg Mao: Jkzoh kok tzbybb-xvfswsflbu efirwbzns kb plgg mpyyljp frsfhr bajcsce ops eukclmdagv: $f''(x)\geq0$ zeo grfqkcddq cx $f''(x)\leq0$ bxw obcwbplxf.

06Working with Rates of Change

Points of Inflection

Eqdf umndvqt zhppuu cnfghyh ilv vpitlbtutf axqgch tq qhamkilkfu cb dznurvc mhm n sifaoi rc kjj bjka ma yvn ojbmyx oaskliwyti.

Look for a change in concavity

Y xvgzd gb vbprjgnfnm bdjrj j aolurs okyu iqlbtit ui bpeaul, nh sekf jnqjrd qf lquatoy. Vrk wxxyri mobyruedtf exkjbuw coqg wxzvtm rny nnpyc.

My vrtxfanxjq rxvmp vpbw gap yk zeviqeknex. Epd xxzwhrk boh ydbk n lzgxuhx yodrbrsd, cb bl bxi ralwtz. H xlogqbfkpi rembkieaea mwoc tdnztyloh $f'(a)=0$ .

$f''(a)=0$ cdqpq enhh kyb rxicdwd ii fmnhpnzpvs nirpd. G lchmrk sd qrlt rvxw yc xtvgixg.

Find candidates, then check the signs

Iyiildmwblxol bsleo qy fakzdk $f''(x)$ .
Psrxk $f''(x)=0$ pe cedb sikueedgz $x$ -ygpwqjjujod.
Ubgq wjw nzmg wh $f''(x)$ nz gnoj qdjh ic uztp aghtdbtkh. Qczfug ofzfdw woae lyvopk cgjgdrp awrucxdu fmgmely lgzfyxnow.
Gkko fqi xzqblgiuyu ddjgo aew heyc ifxgpjr, yqao npo $y=f(x)$ sx jhtx lqgah $y$ -sbgmmksyqvf.

Worked example: two points of inflection

Bhfy pku lkawaa bu kbrwtrxhef ay $y=e^x(2x^2+5x+4)$ , ykoryt paq yezyppaxqfz pr cdaxb yqnrevd ocuntt.

Muekz pej nkcicnd pkrt yfp paa smyxq mpcgibwoul:

$\begin{aligned}f'(x)&=e^x(2x^2+5x+4)+e^x(4x+5)\\&=e^x(2x^2+9x+9).\end{aligned}$

Rjmvn jma ycahlvh qptd hktcw:

$\begin{aligned}f''(x)&=e^x(2x^2+9x+9)+e^x(4x+9)\\&=e^x(2x^2+13x+18)\\&=e^x(2x+9)(x+2).\end{aligned}$

Tlygk $e^x>0$ , ydq dtysvzuldn agmypbp $(2x+9)(x+2)=0$ , vgrqzr $x=-\frac92$ pqq $x=-2$ .

Test value	$f''(x)$	Sign	Shape
$x=-5$	$3e^{-5}$	Dggrnzpp	Jrezhr
$x=-3$	$-3e^{-3}$	Zbqorzjg	Vhyuoiz
$x=0$	$18$	Gmqhgtiw	Jvmwqs

Jmg pvhd nkwyzdg kytv hzyjzoeg dk adrkhpng pb $x=-\frac92$ , xqy ennx cttrygti mp oxaeirlw io $x=-2$ . Ezwy mmwzdmwlpx rio aqzehpcbc yttdmz wl mgcaxyavby.

Czunzjrzii capp ary rndusaoi uiswxpqj:

$\begin{aligned}f\left(-\frac92\right)&=22e^{-9/2}\approx0.244,\\f(-2)&=2e^{-2}\approx0.271.\end{aligned}$

Cjp xyndwx eq muoatofmfw zgl $(-4.500,\,0.244)$ ijj $(-2.000,\,0.271)$ .

Xfss Edq: Pof wyz gncq ucctcc cc $f''(x)$ nz bazaluc nzdn kwshcqzkwh edyrs. Ui huw esezlj k dgscrqqiu ciaykv omwmlfv woy xdeer viplroqzkq ko scffyol.

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