Xisaab

Isticmaalka ama dabbakhida xisaabta dhiraandhirinta (Derivative)

Xisaabtan looyaaqaano dhiraandhirinta ama afka qalaad loogu yeedho “derivative” waa qaybta asalka u ah labada qaybood ee ay ‘Caculus’tu u qeybsanto, taas oo leh faaiidooyin ama isticmaal farobadan oo xal u ah weydiimooyin xisaabeed oo faro badan. Hadaba si ay u fududaato isticmaalka dhiraandhirintu aynu soo qaadano tusaalooyin faro badan oo ku saabsan xisaabtan aadka muhiimka u ah.

Tusaale 1:-

Warshada kaaluunka samaysa ee laasqorey ayaa doonaysa inay sameyso kartoono(baakado) lagu keydiyo kaluunka ay qasacadeyso. waxayna warshadu go’aan ku gaadhey in kartoon kasta oo ka mida kartoomadaa uu yeesho sal labo jibaaran islamarkaana wadarta bed-duleedka kartoon kasta uu noqdo 192 mitir oo labojibaaran, sida ka muuqata jaantuska hoose. Muxuu noqonayaa kartooka dhalinaya mugga ugu weyn dhinacyadiisu(dimensions)

Xalin:-

Sababtoo ah kartoonka oo leh sal labo-jibaaran, muggiisa waxaa lagu helaa (sal x sal x jog) ama Mug (V) = x.x.h = x². h, Sidoo kale wadarta bed-duleed (Surface area,)(S) waxaa lagu heli karaa bedka salka + bed-deleedka ama S = x² + 4xh .

S = 192

x² + 4xh = 192

4xh = 192 – x² (dhinac walba ka goo x²)

h = (192 – x²)/4x (dhinac walba 4x u qeybi)

V = x²h

» V = x²(192 – x²)/4x (booska h dhig qiimaheda)

= x²(192/4x – x²/4x )

= 192x/4 – x³/4 = = 48x – x³/4

Imika isticamaal dhiraandhirinta (derivative) si aad u heshid dhinacyada kartoonka ee dhalinaya mugga ugu weyn.

Horaadka (D) muuqda (feasible domain) ee ay X noqon karto , ee waliba macnaha sameenaya weydiintan waa :- 0 ≤ x ≤ √192

V = 48x – x³/4

dV/dx = 48 – 3x²/4

dV/dx = 0 = 48 – 3x²/4 = 0

3x² = 192

x² = 192/3 = x² = 64

X  =  ± √64  › X = ± 8

-8 kamid maaha horaadka muuqda ee X, sidaa ajligeed X = 8 .

jooga kartoonkuna waa :- h = (192 – x²)/4x = (192 – 8²)/32 = 4

Sidaa daraadeed dhinacyada kartoonku waa in ay noqdaan 8 x 8 x 4 si ay u dhaliyaan mugga ugu weyn kaasoo ah 256m³.

Fiiro gaara:-

Waxaa jira kartoono sal labo jibaaran leh oo aan tiro (xad) lahayn oo leh wadar bed-duleed(surface area) lamid ah 192 (ie: (4² + 4 x 4 x 11), ama (5² + 5 x 2 x 16.7), iwm) balse dhamaan dhinacyada kartomadaasi ma dhalinayaan muga ugu weyn. Kan aynu xisaabiney dhinacaydiisa ayaa ah ka kaliya ee samaynaya muga ugu weyn. Halakaas waxaad ka ogaan kartaa faaiidada ay leedahay dabakhida ama isticmaalka dhiraandhirinta(derivative).

Tusaale 2

Engineer guryaha dhisa ayaa doonaya inuu nashqadeeyo cabirka daaqadaha guri ku yaala magaalada kismaayo.waxa uuna Engineerku doonayaa in daaqadkasta qeybteeda sare u ekaato nus-goobo (goobo barkeed), qeybta hoosana ahaato leydi sida daaqadaha caadiga ah, sida ka muuqata jaantuska hoose. Muxuu noqoneyaa cabirka hareeraha(dhinacyada) daaqadkasta ee uu Engineerku nashaqadeynayo, dhinacyadaasoo samaynaya bedka ugu weyn ee dariishadaha, hadii wareega guud ee daaqadkasta uu yahay 8m?

Xal

Siin

<Wareega guud = 8

<Wareega nus-goobada = pr / 2

<Wareega leydiga = x+2y

Weydiin

<!Dhinacyada dhalinaya bedka ugu weyn?

Wwareeg (P) = x + 2y + (px/ 2)

8 = x + 2y + (p.x / 2)

16 = 2x + 4y + px (dhinackasta ku dhufo 2)

4y = 16 – 2x – px (dhinac walba ka jar 2x iyo px)

Y = 4 – x/2 –px/4

Bed (A) = bedka leydiga + bedka nus-goobada

= x . y + (p. (x/2)^2 )/2

= x (4 – x/2 –px/4) + p/2(x^2/4)

= 4x – (x^2)/2 –(px^2)/4 + (px^2)/8

= 4x – x^2/2 – px^2/8

Isticmaal imika habka dhiraandhirinta saad u heshid dhinacyada dariishadkasta kaas oo dhalinaya bedka ugu weyn

A = 4x – x^2/2 – px^2/8

dA/dx = 4 – x – px/4

dA/dx = 0

4 – x – px/4 = 0

4 = x + px/4

4 = x(1 + p/4)

4 = x(4 + p )/4

X = 16/ (4 + p )

dhiraandhirinta labaad (second derivative) ee bedku waa (– 1+p/4), kaas oo taban marka X = 16/ (4 + p).

Sidaa daraadeed marka dhinacyada dariishad kasta ay kala yihiin: X = 16/ (4 + p), ( ≈2.24) , Y = 32/ (4 + p), ( ≈ 4.48) ayaa samaynaya bedka ugu weyn ee daaqad kasta.

 

layli

1. Taranta labo tiro oo tirsiima ayaa ah 180 isla markaana wadarta labada tiro ayaa ah ta ugu yar, Raadi labad tiro?
2. si uu u dhiso cali wado weyn oo u dhexeysa labo magaalo ayaa ay khasab tahay inuu buuxiyo tog u dhexeeya labada magaalo kaasoo dhinacyada janjeedhkiisu kala yihiin 5% and 6%. hadii labada dacal ee sare ee togu isu jiraan masaafo dhan 100m Raadi cabirka meesha ugu hooseysa ee toga?

      

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Isle’egyada

Isle’egyada waxaa lagu xalin karaa dhowr qaab oo kala duwan . Waxaana ka mid ah qaababkaa: qaabka daboolida, qaabka wareejinta iww.

Tusaale 1 qaabka daboolida(cover up method)

Xali isle’egta hoos ku qoran

 gacantaada saar tirada door

soomaha wadata

, waxaa halkan kaaga cad in tirada aad gacantaada saartay ay lamid tahay 4 taaso marka loo qeybiyo hooseeyaha (4) maxsuulka soo baxaya uu yahay 1. Sidaa daraadeed

3x = 4  dhinac walba u qeybi 3

X =

Tusaale 2

Xali isle’egta hoose

4x + 5 = 7

+ 5 = 7

4x = 2

 

Xusuus:-

Qaabka daboolida marka la isticmaalayo waxaa

la daboolayaa(qarinayaa ) tirada doorsoomaha

wadata mar walba kadibna waxaa la

le’ekeysiinayaa tiradan doosoomaha wadata ee

la qariyey tirada run kadhigaysa isle’egta lagu

siiyay