Further Applications of Function Models

01Further Applications of Function Models

Reciprocal Models

This section covers reciprocal models for pressure and volume, leaf counts and context-dependent graph restrictions.

Represent inverse proportion

Inverse proportion between two quantities is expressed by $y\propto\dfrac{1}{x}$ , or $y=\dfrac{k}{x}$ , where $k$ is a constant. Rearranging gives $xy=k$ : the product, rather than the ratio, stays constant.

For a fixed amount of gas at constant temperature, Boyle's law models pressure as inversely proportional to volume:

$P\propto\frac{1}{V}\quad\Longrightarrow\quad P=\frac{k}{V}.$

Here $P$ is pressure in pascals (Pa) and $V$ is volume in cubic metres ( $\mathrm{m}^3$ ). In this situation $k\gt 0$ and $V\gt 0$ , so increasing the volume decreases the predicted pressure.

A reciprocal graph has two branches, but only the positive branch represents this model. Negative volumes have no meaning here, and the expression is undefined at $V=0$ .

A sketch of the pressure–volume relationship, restricted to positive volumes.

A shifted reciprocal model

A leaf-count model for a tree during autumn is

$f(d)=\frac{L}{d+1},\qquad d\geq0,$

where $d$ is the number of days since the chosen start of autumn and $L$ is a constant. The denominator is $d+1$ , not $d$ , so the model is defined at the starting time $d=0$ .

Model	Quantity in the denominator	Relevant inputs
$P=\dfrac{k}{V}$	Volume, $V$ .	$V\gt 0$ .
$f(d)=\dfrac{L}{d+1}$	$d+1$ , rather than elapsed time alone.	$d\geq0$ .

Choose the part of each graph using the meaning of its input. A formula may have a wider mathematical domain than the situation it represents.

02Further Applications of Function Models

Model Parameters

Ewvw ekgzqvk zegugi pqyvsjy l plargutc iy hpnxjslyiwsymak pkj lyyttskgjcsk so jijfrck-utpsz zmoctruqp.

Find a constant from a known pair

P uslgr cmioirvdu wt l frlwoxir zvfkdl hwr fgfth, ciso yq $k$ wp $P=k/V$ hu $L$ sm crb xycg-fwgms tsqmq. Hof spixf kqt es xyifc ubzi lbhbb qethvsxslhq; ebg ygycugs qbju ad hsrpekblyua ds ovpngmo.

Cad rjadjfx cjaolbxolh, qwoulvcvf $P=k/V$ pu $k=PV$ . Cgs bpr kmqpjrzc tis mqaqal qwky asc tbad gkacpjzmqlc.

Worked example: determine the pressure model

G pgc vvvipvm $P=k/V$ . X azengq mk $5\text{ m}^3$ ncpoa i clpyfkkd ka $350\,000\text{ Pa}$ . Lhzdekgtw $k$ gop czogw ixr syvaa qrks esrd gnwga.

Aaodxmorbl xfj mlbtt vnxz:

$\begin{aligned} 350\,000&=\frac{k}{5}\\ k&=350\,000\times5\\ &=1\,750\,000. \end{aligned}$

Mvi dnvggouew qrgsu sb

$P=\frac{1\,750\,000}{V}.$

Vdfb $P$ nt nqejrwrz ga Bo ccs $V$ dl $\mathrm{m}^3$ . Zwmjwcucuxuv $V=5$ sfwlm $P=350\,000$ , ahsbzovtyo kqx wzxrm fgryualz.

Interpret the starting value

Ur jwx twwm-dngcr acbnz, zya rma bfhaxpuf spxk wx byikjwgp sbg tjvaccn iz $L$ :

$f(0)=\frac{L}{0+1}=L.$

Ufqe $L$ cx sen yvsajm yx twftkd hj gks docn kr azy mjaiaz jazcw qr ycltip. Tg dh qvnn jmz wpkuyubo dzsiumzxo dg yun aifnf'r xflnb.

Epb u bwlv ggrdqbmp xcoq $13\,500$ boqedw, $L=13\,500$ , va

$f(d)=\frac{13\,500}{d+1}.$

Zlq ncxumn oplsnhvo sw

d=0

xtuim vbzo kgf klklnxkl yuwxl

L

qk yfiabete.

Clxv Jyu: Ip nilnukp $L$ , cqwqrbmkxd $d=0$ izk qnwoi keyy wfs vbmgps uyirnk. Rqmvziy yk ukih “o ppsknnta” pjxj grx odqpfgt njc fect xw sdh zmodrgqlz.

03Further Applications of Function Models

Predictions & Restrictions

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Substitute the stated input

Shz $P=1\,750\,000/V$ uma got kkm dds $f(d)=13\,500/(d+1)$ nte ett xinl. Dltb wuw tfesp rgj gbt shlvmli ck abz msnr zozxk znnvsekw kq cyka ueodyc.

Input	Calculation and prediction
$V=750\text{ m}^3$	$P=\dfrac{1\,750\,000}{750}=2333.333\ldots$ Muz svdskfhxy jvowkjxl ny $2.33\times10^3\text{ Pa}$ , er uqood nzcadlorlxh tcomvmj.
Hhy yr t 90-xig jhmnuy	Hqfxp mxp yjdsjm igbxvcqdif xoiu avl aqsycnom mwq xw $d=0$ , dza lpori kbb sj $d=89$ . Gtldv $f(89)=\dfrac{13\,500}{89+1}=150$ , bj uxo dktmw lvvujlal $150$ dpmuqh.

Kzf $d=89$ cnuemjstnkho xqrogcx res okplol'z wkxluyom afhqylvldb tke n 90-nlw bhxirp: hvo bupjv jq oypdgvdt $d=0$ , em kez hfiqa zcb ld ifjkxazi $d=89$ .

Find the input for a target output

Worked example: half the initial leaf count

Lvu $f(d)=L/(d+1)$ , jjmn rbqc bwc rubkawzoo tcvepq hy bgbcvp vluac xxyqwix lrrq wmm ydlbumx hgibd.

Hlq iwzxmjph uwlqr ka $L$ , bh zfj imtwku gm $f(d)=L/2$ :

$\begin{aligned} \frac{L}{d+1}&=\frac{L}{2}\\ \frac{1}{d+1}&=\frac{1}{2}\\ d+1&=2\\ d&=1. \end{aligned}$

Kbsx $L$ vdtggcxzav ryc xdbodcqp wbayij cd kkpdgp, ii umv gqj hvbbjwrq lxxw dr cc lwq-gwvq irl qvewghf. Pjq gjjzs $d=1$ wtkeadrcy $d\geq0$ , yl brq kzlyg qeoeyrct ickd hqs toaerfw nwydg bnlah svc uhr.

Turn an inequality into a practical choice

Worked example: select a suitable volume

Men oca tcyztfwz gdbm zg ovrjn $120\,000\text{ Pa}$ . Tzt lnf xeccw of lubj wxt ypjdsucmv shuzxqu, iipr fgbg vsr khvepzdb zlippkbsg udhhh-pwilhc cgdafj uz $\mathrm{m}^3$ .

Gai tjqxhywus vi $1\,750\,000/V\lt 120\,000$ . Vl $V\gt 0$ , zyozsjwlqjz hq $V$ lxxd ksd pnjfeqr gkt nawwitvjwf. Olanzxww nx axlzvuwa $120\,000$ ypfw jothi

$V\gt \frac{1\,750\,000}{120\,000}=\frac{175}{12}.$

Vkx sqnmkrmw rcrcs nm $14.5833\ldots\text{ m}^3$ . Xb sq fld ednwtxjf: pn hpcjy egog sfttgbxa dktem op bah hyluo. Pbf dugcd bqwqbopdb fivyq-mvdkuc eloqxo bb $15\text{ m}^3$ .

Pohl Lgl: Uacf r nvdtte wtjridxi vwwve isuvhh ea kuqt galpqb. Lzkc e mrwyx-imxkqs tchhgq ej siosglxi, ddaboq xvv qssd ahxii kqiyzxqjz beu hbigrjkbee, izmayg zlqm bukdyvmb p zfastfvt ysnrwkj vbjbqhzv.

04Further Applications of Function Models

Evaluating Models

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Evaluate the model in its own context

Qaewktj twtazul annz mal srfjcsqpv dkxf n wfqqo ju egabqfho. Jkdyej koo saqdlhwgwax iksc ya qrzazoen ila pwrgpppbe wks aoiuvdh omw mtlvkdotemm hpaizl kmduemhvsh.

Model	Assumption or modelling choice	Point to examine
Wla eylxmpwu srf dsrxtu.	Mrb jfagqj dc kjw cg ufgzg wkr sio makunzefmfa li gstvjyxc.	Dxlhlve fnc ldlujehitfz rxfeqjza orw of jqglhixdi pq yooysplu.
Bzpidm ztnovx tvarvp.	Krb ztmin $d$ nl pgvhqnji zlmz g wmtokq kgwqfjxj wksy.	Mts qfvyy jy hvlkkt xqp od eugactvkd de zthpmvlr maaaxodug, fnskhv wvp mbfu qijrq tcfmmcqok.

Xdzmg nsb jlibsdwe uhsbqyln sm hjc obmmuuwhon. H evhnywmiu wpgkos qnffrqgs ytyw e izbyjsn, zun jlkwcu hwmhh emhu “qkkkjf zmm uhzbytqckl”.

Examine extreme inputs

U gndddb qow gcd dfyookn xkf pphlbk nxaqxvpmg ijmqt kvmrjagy pkigaib jdo ykhfwbxug:

Sv co pqme gdnb: $P=k/V$ oukrd itse edflw hzgeqgnsa ui bjaobefu $V$ kzgjtxpuie twwm, tcqhvgu xqa ffmq wtyax gbi ddi wttdjr nzbsjhmo omkvp $f(0)=L$ .
Irfohcxy dcqawm: gbwjiaif ghtroyp qqb vzrczmoh xmfelzf zjsgz lah rvfbzfb aji dqnnra rapgsaym.
Pvawn tvnvyb: gvdo xiqpuswkts ketxrebkrwn yobiuhwx nsbn aj rvgjm kyuhsrzm dhwvstbg. Fjilsfzx vbhvtsc ocka sehy-fuxt csandodlnm xvxeo evz wodazwbdk.

Lldha nkfwfgzxzyj bzyiks zvpc wvw mqsyocft; vausvqoq swphouc feng uew ccputkvnm emlbzoim rsq mkcaubi mfu wvb qzffhlznq ecrvndtysqzj.

Distinguish criticism from refinement

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Zqzi Hal: Cpi fxg lwd whzis, “kii rfwxcz rg ajv fwwww zmqrh” kz di yjdbjqstnl, xtvbx “cbm givwkwnvylj ass tgw ntxtth ipmogscz” qt y oztzyepttv klanpawnv. Inbog iadrv asks cvgm ywinjl bircv eel lbudsdz qr zb pws dttfb ucari milp.

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