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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Bij deze oefening noteer je getallen steeds in breukvorm (niet als decimalen), tenzij het om een geheel getal gaat!

", "advice": "

a) Uit de gegevens dat het dagelijkse afbraakpercentage \\(\\var{dpct}\\) % bedraagt, en dat er wegens de inname van 1 ml medicament $\\var{s}$ mg {chem} per dag wordt toegevoegd, volgt
\\[ y_{t+1} = y_t - \\var{d} \\cdot y_t + \\var{s} =  \\var{1-d} \\cdot y_t + \\var{s} \\]
of in breukvorm
\\[ y_{t+1} = y_t - \\simplify[all,fractionNumbers]{{dpct}/100} \\cdot y_t + \\var{s} =  \\simplify[all,fractionNumbers]{{100-dpct}/100} \\cdot y_t + \\var{s} \\]

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De beginvoorwaarde is \\( y_0=\\var{b} \\).

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b) De algemene oplossing van een eerste orde recursievergelijking wordt gevonden als som van de algemene oplossing van de geassocieerde homogene recursievergelijking en een particuliere oplossing van de oorspronkelijke recursievergelijking.

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De geassocieerde homogene recursievergelijking \\( y_{t+1} =  \\simplify[all,fractionNumbers]{{100-dpct}/100} \\cdot y_t  \\) heeft als oplossing
\\[ y_t^{H} = C \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t   ,\\]
met \\( C \\) een reëel getal.

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Om een particuliere oplossing te zoeken van de oorspronkelijke differentievergelijking, imiteren we de rechterhandzijde van
\\[ y_{t+1} -  \\simplify[all,fractionNumbers]{{100-dpct}/100} \\cdot y_t = \\var{s} .\\]
Stel daarom \\(  y_t = \\alpha\\) met \\(  \\alpha \\) een reëel getal, dan is \\(  y_{t+1} = \\alpha\\) zodat
\\[ \\alpha -  \\simplify[all,fractionNumbers]{{100-dpct}/100} \\cdot \\alpha = \\var{s} \\]
leidt tot
\\[ \\alpha = \\simplify[all,fractionNumbers]{{100*s}/{dpct}}  .\\]

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Dit geeft als algemene oplossing \\[ y_t = C \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t   + \\simplify[all,fractionNumbers]{{100*s}/{dpct}}  .\\]

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Omdat \\( y_0 = \\var{b} \\) kan je \\( C \\) oplossen uit \\[ \\var{b}= C  + \\simplify[all,fractionNumbers]{{100*s}/{dpct}}  ,\\] d.w.z.
\\[  C =  \\simplify[all,fractionNumbers]{{b-100*s/{dpct}}}  .\\]

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De uiteindelijke oplossing van de recursievergelijking met beginvoorwaarde is dan 
\\[ y_t = \\simplify[all,fractionNumbers]{{b-100*s/{dpct}}} \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t   + \\simplify[all,fractionNumbers]{{100*s}/{dpct}}  .\\]

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c) De lange termijn concentratie (in mg) van {chem} in het bloed wordt gevonden als limietwaarde voor \\(t\\) gaande naar \\( +\\infty \\) van bovenstaande oplossing \\( y_t \\):
\\[  \\simplify[all,fractionNumbers]{{100*s}/{dpct}}  .\\]

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\\[\\]

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d) Stel dat {person['name']} dagelijks  \\( x \\) ml medicatie neemt, wat wegens de lineariteitsaanname leidt tot een dagelijkse vermeerdering van de concentratie {chem} in {if(person['gender']='female','haar','zijn')} bloed met \\( \\var{s} \\cdot x \\) mg.
De nieuwe differentievergelijking
\\[ y_{t+1} =   \\simplify[all,fractionNumbers]{{100-dpct}/100} \\cdot y_t + \\var{s} \\cdot x \\]
heeft als oplossing
\\[ y_t = C \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t   + \\simplify[all,fractionNumbers]{{100*s}/{dpct}} \\cdot x \\]
met lange termijn evenwichtsconcentratie
\\[  \\simplify[all,fractionNumbers]{{100*s}/{dpct}} \\cdot x .\\]

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Om op lange termijn \\( \\var{lt} \\) mg {chem} in het bloed te bekomen, geldt dan
\\[  \\simplify[all,fractionNumbers]{{100*s}/{dpct}} \\cdot x =  \\var{lt} ,\\]
of met andere woorden
\\[  x = \\simplify[all,fractionNumbers]{{lt*dpct}/{100*s}} ,\\]
zodat je (benaderend) \\( \\var{opld} \\) ml medicatie nodig hebt.

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\\[\\]

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e) De differentievergelijking
\\[ y_{t+1} =   \\simplify[all,fractionNumbers]{{100-dpct}/100} \\cdot y_t + \\simplify[all,fractionNumbers]{{lt*dpct}/{100}} \\]
heeft nu als oplossing
\\[ y_t = \\var{b - lt} \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t   + \\var{lt}  .\\]

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Om voor het eerst minder dan {Gpct}% van de vooropgestelde limiet \\( \\var{lt} \\) mg {chem} in het bloed te bekomen, moet
\\[  \\var{b - lt} \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t   + \\var{lt}  <  \\simplify[all,fractionNumbers]{{Gpct*lt}/{100}}  \\]
d.w.z.
\\[  \\var{b - lt} \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t     <  \\simplify[all,fractionNumbers]{{(Gpct-100)*lt}/{100}}  \\]
of equivalent
\\[   \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t     <  \\simplify[all,fractionNumbers]{{(Gpct-100)*lt}/{100*({b-lt})}}  .\\]

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Toepassing van de (natuurlijke) logaritme op de linker- en rechterhandzijde behoudt de ongelijkheid (omdat $\\ln$ een stijgende functie is):
\\[   t \\cdot \\ln \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)   <  \\ln \\left( \\simplify[all,fractionNumbers]{{(Gpct-100)*lt}/{100*({b-lt})}}\\right)  ,\\]
i.e.
\\[   t \\cdot (\\var{Ne})  <  \\var{Te} ,\\]
wat gelijkwaardig is met
\\[   t >  \\var{Be} \\]
(de ongelijkheid werd omgekeerd omdat je deelt door een strikt negatief getal).

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Afronding naar boven levert een tijd op van \\(  \\var{ceil(Be)} \\) dagen om voor het eerst minder dan {Gpct}% van de vooropgestelde limiet \\( \\var{lt} \\) mg {chem} in het bloed te bekomen.

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\\[\\]

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f) Veronderstel dat het lichaam van {person['name']} dagelijks $x$% afbreekt van {chem} en dat {person['name']} dagelijks 1 ml medicatie gebruikt, met $ 0 < x < 100$. Dan heeft de bijhorende recursievergelijking
\\[ y_{t+1} = y_t - {\\frac{x} {100}} \\cdot y_t + \\var{s}  ,\\]
als oplossing
\\[ y_t = \\left( {\\var{b}}-{\\frac {\\var{100*s}} {x}} \\right)\\cdot \\left( {1 - {\\frac{x} {100}}} \\right)^t   + {\\frac{\\var{100*s}} {x}} ,\\]
met als lange termijn concentratie
\\[ {\\frac{\\var{100*s}} {x}} .\\]
Om dus op lange termijn {lt1} mg {chem} in het bloed te bekomen, moet
\\[   x =  \\simplify[all,fractionNumbers]{{100*s}/{lt1}} .\\]
Bijgevolg dient {person['name']} dagelijks (benaderend) \\( \\var{oplf} \\) % {chem} af te breken.

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Opgave

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Vanmorgen, bij het begin van de dag, had {person['name']} {b} mg {chem} (een chemische stof) in {if(person['gender']='female','haar','zijn')} bloed.
{if(person['gender']='female','Zij','Hij')} neemt elke dag 1 ml medicatie in, waardoor aan het begin van de volgende dag de hoeveelheid {chem} vermeerdert met {s} mg.
{if(person['gender']='female','Haar','Zijn')} lichaam breekt dagelijks {dpct}% af van de hoeveelheid die in het begin van de dag aanwezig is.
Beschouw de tijd $t$ als een discrete grootheid en noteer met $y_t$ het aantal mg {chem} dat over $t$ dagen aan het begin van de dag in {if(person['gender']='female','haar','zijn')} bloed aanwezig zal zijn.

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Stel een recursievergelijking met beginvoorwaarde op voor de rij $y$ in de vorm  $y_{t+1} = getal1 \\cdot y_t  + getal2$ :

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       $y_{t+1}$ = [[0]] $\\cdot y_t$ + [[1]]    met BVW (beginvoorwaarde) $y_0=$  [[2]]

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Los deze recursievergelijking uit a) op onder de gedaante $y_t = getal1 \\cdot grondtal^t + getal2$.

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   $y_t$ = [[0]]  $\\cdot$ [[1]] $+$ [[2]]

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Hoeveel mg {chem} zal het bloed van {person['name']} in de gegeven omstandigheden op lange termijn bevatten?

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   [[0]]

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Idealiter zou {person['name']} op lange termijn {lt} mg {chem} in {if(person['gender']='female','haar','zijn')} bloed hebben. Hoeveel constante ml medicatie moet {if(person['gender']='female','zij','hij')} daarvoor dan dagelijks verbruiken (in de veronderstelling dat de medicatie op lineaire wijze leidt tot een vermeerdering van de chemische stof {chem} in het bloed)?

\n

  [[0]]

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Na hoeveel dagen zal {person['name']}, met de doelstelling uit d), voor het eerst minder dan {Gpct}% van de vooropgestelde limiet in {if(person['gender']='female','haar','zijn')} bloed hebben?

\n

  [[0]]

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Veronderstel dat het lichaam van {person['name']} dagelijks geen {dpct}%, maar een ander (constant) percentage afbreekt. Hoe groot is dat percentage als blijkt dat {if(person['gender']='female','zij','hij')} op lange termijn {lt1} mg {chem} in {if(person['gender']='female','haar','zijn')} bloed heeft (in de oorspronkelijke veronderstelling dat {if(person['gender']='female','zij','hij')} dagelijks 1 ml medicatie gebruikt)?

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   [[0]]

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