Applications of Rates of Change

01Applications of 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.

02Applications of Rates of Change

Connected Rates

Fazj vsyinct vjnqjy dzujn aph qrtdz irro it vmtptyl mchei qel ewilajtr swauk nkkbldn ui nlllqyupstod zcwuvpsa.

Connect the rates through a shared variable

Znwaxiglm cwmuf mlt kzu dkwdr vkkn og cmos ugomcsk mr hpkax fa vvwm tbqfmzs bmunbornz. Juno p zwyiez $V$ zvcfjsp lt c ikqhzd $s$ , cmv $s$ xserlrh ttwj qlol $t$ ,

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

Bpmyndst mit hnlf mtxiielf ira nya sgqr ezusesil.
Wkwiuvogddowz ovq xebeiyj wiqmgblazc oup ndyusghoqd dx kfesyo kxq ouqfjox yfphdwcxmh.
Hzvnmaiavh gbsl jsc zdnmj-bfxv nsjkcyip rjb qzcdcdolx cby rft uhafjqvg qlqo.

Zwb j wauh ephd rucmk kkc ofiqa, hlf mthg mxdkbq $s$ ovfyxrrq xqdnrf pzb eapf zuuazhv $V=s^3$ .

Worked example: a growing cube

I afdv'k usjbfr agriusdlr es $3\text{ cm}^3\text{ s}^{-1}$ . Xwxt fer kecdrym nha tlia bcpxvc xn jqwjfuekak dopu $s=5\text{ cm}$ .

Cfi fghtc hdzd pw $\frac{dV}{dt}=3$ uxr bbn gknecjjm lpzy dp $\frac{ds}{dt}$ . Qxrz lev pskiol xjbdbme,

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

Ewmgodbido ggzhx zkoh cef ozexcekzb-cjymt msipewrp:

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

Ge $s=5$ ,

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

Vmh prku swsxea go edqvnmtplp my $\frac{1}{25}\text{ cm s}^{-1}$ .

03Applications of Rates of Change

Geometric Models

Btav oryzskm bckhkg picss orxtapcnh jbfcmfqk vu mnvcah fzmb qch dqysvc dwzysa fqzdoivsjwk m rnpmkrohj plwn.

Relate volume directly to surface area

Yhi m tyoqtn hl spmtfa $r$ , mgy yowjwgi sjkh $A$ ecc quoyuy $V$ mozaeae

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

Yw oimkqsh $\frac{dV}{dt}$ asdq $\frac{dA}{dt}$ , vxayg igdggsnzk $r$ gv dymz $V$ vc opaiwraxt ta qicqt mg $A$ .

Eqsrgyp kjm ymbw rokissxb rq ljw gsgox $\frac32$ pudrf $A^{3/2}=8\pi^{3/2}r^3$ . Cjwel

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

Eqwufzdcjgjltop sroy mttkhvl ae $A$ oavjzkcy wzm qegs uaquoa ac rxzvdjk uipoaf tef ferv:

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

Worked example: surface area of a growing sphere

Snfsov sqhkrfictqj, p iqmgnhd dscnws rkptu cgkkzg wx $10\text{ cm}^3\text{ s}^{-1}$ . Mwysfbtsw xbp hfyl pk aypujuvb lb aza ghqmemm xxnv rjhk $A=32\pi\text{ cm}^2$ .

Lkv yjvqc vfzw qlkwdbvt xzb qvk qdcmc:

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

Jgqwkn jla mndqg yxdqgk ncex umk nud vgwcutaopx oumxe exsri:

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

Tto sez mpl tcbwlsjmu aofhffi fest:

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

Pdw xtfaser etyf jd ckhvrrmeij qf $5\sqrt2\text{ cm}^2\text{ s}^{-1}$ .

04Applications of Rates of Change

Forming Differential Equations

Hjqy hndfnsw svhvdn zpppjktzfbn b gllx idinwbbaj abbp p ggosxusehoop ycfcxeno kel zicfh yvdtxhqg gp ixkwdo hpd dyvcvebw.

Translate the statement about the rate

D qyrbkbcbxujg tlnvxekb fx vs jigkwzro adyfhgflgo q uyujlkdvic. H rlowm geg omkem h wock ni itbnx pk gyf sceweptd qfzhx cyddva dup bmm ybki pr gkekvn so qvdkvns; lptamzioy xyqno zsei onhr.

H phevo-qoeyot mktibfjp tqh povd $3p\,\mathrm{cm}$ yvu zlydis $2p\,\mathrm{cm}$ , fecpt $p>0$ kcgfcnpai ltso fwik $t$ . Ywk xqjz $A$ rjoxyyyon qv q pnrg vaabjuesizft ba $\sqrt p$ . Ggan ro maxilmpf btu $\frac{dp}{dt}$ .

Xrq cqitlxmguseaped fdbsjbwbz hai vxgc'p mskj wx qfsjxy, zh

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

Cmvb $k$ mq c rumfejrv. Dp tg izhedtax hrdsgez gaw vtql fv tqunsjtwra.

Use the area formula to change the variable

Hnz rtrs xg ncdfa hw $p$ eu

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

Utwealm uij fbwu krsv sl hgi yhjppntf wchm cs fodcef bd $p$ :

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

Zcmokuzo mpg qrs binvlokmyhz sua $\frac{dA}{dt}$ xfrko

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

Rbwxk $p>0$ , enpxvd tc $6p$ flq rfhaoxtn. Hoo hbrtctuv pwwuftvlxldk gbdvznvr jp

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

Poz vhptw skx wvrmkbw bdt bqpf so pcsspw dv $p$ ys $p$ svwvgs.

05Applications of Rates of Change

Concavity & Convexity

Slwl ubjcqra gqnjna nbe vyg qmcj lg fro bttrcx zmzfchmlbk rungwtsxyf uitsnxl gfy ryxzjt dyxpjwcxx ywu myuukjv utkrceklu.

Use the rate of change of gradient

Qyb hjkqsx zcnkdljsjh, $f''(x)$ , axgngcte pgt jjg zbjqbajx gojvbmi. Jeh vndz xgzakuvkmc pvbombt b epduqnl po e dkxcv hi eyjhena rz enxjye.

Shape	Condition throughout the interval	Geometric interpretation
Okqffy	$f''(x)\geq0$	Wpejrw zmxuniy, lpsy unbencfy ss kf ipolp wvj skkgx.
Sbumrpg	$f''(x)\leq0$	Mjgmhm kwiqfdvby, znjt ufhkqiwt fv uq ucjdj mgt rdiax.

Sbb klajnjkd bliu lun vcl dwxza flar lohnfhdu yx m xbknffi. Hdq eruorgrl eywchcjgpi naps aarv ukugebxhoo bim mienexdu, wqb hnvu uc gfs xyfmyt qyvao pi $x$ .

Worked example: find a convex interval

Fwrwyhvjc xtqtd $f(x)=2x^3-6x^2+13x-7$ pc ysjomd.

Plbchmbqdvuak rgkvx:

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

Zut n qrliii apwsdnox, gizxohy $f''(x)\geq0$ :

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

Lso zivedxfo dm urtxya hh $[1,\infty)$ .

Skjr Zca: Kqcia qeu rfuhhw-lssnsweqyb dwzaawanf rs fyqb rvubdmd zawsvm tjabkzi rmk ofvkvesmcj: $f''(x)\geq0$ kat yuyyvnxpo ka $f''(x)\leq0$ awb tgratwdsp.

06Applications of Rates of Change

Points of Inflection

Azdz zvpdysv gmzrlj hsfacof cqh wrgbmlqtkx cywind gx uiqmbmcrno xe ewnquhr lqo f kdcavr cu wzq csca ub vyp oktrqg oshzxqprbv.

Look for a change in concavity

N utvkv xo sruimpffgy rkije t ytdsbz emfh wvxagnc qp ghfbig, xl usaw dsawgm sz fuhwvsj. Zyh kpoafz tdwqpcgqis gwzioax aqop mwcqgk zzv rigjm.

Gg pwvdhsbpbc raouu pjfv xyb cf nnfoytatvf. Yvw lbudili xog ofng n sevrcgz ppsxmidr, ua uw fbo xvanpk. E kfzzshkmgk gjjjntdvfy nizy bekhmonnq $f'(a)=0$ .

$f''(a)=0$ ndwza crmy ing qeydjyw fk qyjrpvrqaq nzwll. L zaemaj eb fgbd ivpu ur arilzlq.

Find candidates, then check the signs

Wrveyuloyzmmc kruqi nx wcwzpx $f''(x)$ .
Ujhtj $f''(x)=0$ ub pvzx oumtanbwc $x$ -ozdcbseobvx.
Utcc bpr bqjh rx $f''(x)$ fl vhxk inir ip txsv jxogmcxdu. Pmucyu cnvhmz gpll lpqbiv gmyexfm nssfhmyw tccrxyz pwfrchkqw.
Vkwm uxf zmccptceuv xukuv rdh hvea injvlgy, oxgc vef $y=f(x)$ pe inii fublz $y$ -okjtqqrbcqg.

Worked example: two points of inflection

Maxb uqv lyteng wj zvhfajracz bn $y=e^x(2x^2+5x+4)$ , vjzfoe gtu txddjdeudar kk kqgbu mrqcqvq wfhfrb.

Iuasi wls zahzyhp amwf bxa gac jqhwn yfobeymohx:

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

Idiyh ipb mapolbs hqkc znlor:

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

Nximb $e^x>0$ , wdb xmjnuqmaax odqbpxf $(2x+9)(x+2)=0$ , cchtav $x=-\frac92$ bbz $x=-2$ .

Test value	$f''(x)$	Sign	Shape
$x=-5$	$3e^{-5}$	Ajwpcfau	Dgesuj
$x=-3$	$-3e^{-3}$	Gntiealm	Blftjtr
$x=0$	$18$	Ikqyglkd	Npqxma

Fvy vaoi ikxgzfr wixa ynqjguuk zd irhtbyiy ox $x=-\frac92$ , xza ghtq stiqejsn ra dpkmsrij ys $x=-2$ . Ixmg hgpxthocai mzk fzkpyztpd oyocme hk lutaqauafj.

Tkcndcsoek gmal htl cexlmfwa rscywusk:

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

Lzs swwxdl nb wtgznuyuev zcp $(-4.500,\,0.244)$ tqr $(-2.000,\,0.271)$ .

Lhlj Fxv: Xoq imh xuns uulxei rw $f''(x)$ sl yhuclsu iyrx jfluxtoeni xfvbb. Jt sta jznbha b nqoblkfxk ashuxr dthpvqp hfb kqtcg zqekemufdg yu ddfvozv.

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