Exercises — Net change theorem

Hum kya karte hain: theorem kehta hai net change $={\int }_1^4{F}^{\prime }(x)dx$. Hume $F^{\prime }(x)=6x$ diya gaya hai, toh ise integrate karo. Ye hi kyun, kuch aur nahi: hum $F$ mein change maang rahe hain, aur theorem exactly yahi hai — "rate ka integral = net change." Hume $F(1)$ ya koi $+C$ jaanna zaroori nahi hai. ${\int }_1^{4}6xdx=[3x^2{]}_1^4=3(16)-3(1)=48-3=45.$ Net change $=45$.

Hum kya karte hain: velocity, position ka rate of change hai, isliye displacement $={\int }_0^{7}v(t)dt$. Kyun: $v=s^{\prime }$, aur theorem deta hai ${\int }_0^7{s}^{\prime }dt=s(7)-s(0)$ = position mein net change = displacement. ${\int }_0^{7}4dt=[4t{]}_0^7=28-0=28\text{ m}.$ Displacement $=28$ m. (Kyunki $v$ kabhi negative nahi hota, distance bhi wohi $28$ m hai.)

Hum kya karte hain: $r$ volume $V$ ka rate of change hai, toh net change $={\int }_0^{6}r(t)dt$. ${\int }_0^6(12-3t)dt=[12t-\frac{3}{2}{t}^2{]}_0^6=(72-54)-0=18.$ Net change $=+18$ L. (b) Inflow rate $t<4$ ke liye positive hai (paani andar aa raha hai) aur $t>4$ ke liye negative (paani nikal raha hai). Net result phir bhi $+18$ L hai, yaani tank aakhir mein shurooat se 18 L zyada rakhta hai, chahe aakhri do minute mein paani drain hua ho.

$v$ ka sign kahan badalta hai? $2t-6=0\Rightarrow t=3$ rakho. $[0,3)$ par $v<0$ (peeche ja raha hai); $(3,5]$ par $v>0$ (aage).

(a) Displacement $={\int }_0^5(2t-6)dt=[t^2-6t{]}_0^5=(25-30)-0=-5$ m. Net: particle start se 5 m peeche khatam hota hai.

(b) Total distance — bars ko andar rakho, $t=3$ par split karo: ${\int }_0^3|v|dt+{\int }_3^5|v|dt={\int }_0^3(6-2t)dt+{\int }_3^5(2t-6)dt.$ Pehla piece: $[6t-t^2{]}_0^3=18-9=9$. Doosra piece: $[t^2-6t{]}_3^5=(25-30)-(9-18)=(-5)-(-9)=4$. Total distance $=9+4=13$ m. ($|-5|$ se bada — acha sanity check hai.)

Hum kya karte hain: marginal cost hi $C^{\prime }$ hai, toh extra cost net change ${\int }_{20}^{50}{C}^{\prime }(x)dx$ hai. ${\int }_{20}^{50}(4x+30)dx=[2x^2+30x{]}_{20}^{50}=(5000+1500)-(800+600)=6500-1400=5100.$ Extra cost $=$ ₹5100. Baseline kyun nahi chahiye: sahi cost function $C(x)=2x^2+30x+C_0$ hai jahan $C_0$ fixed start-up cost hai. Change $C(50)-C(20)$ mein $C_0$ subtract ho jaata hai. Ek difference constant se andha hota hai.

Hum kya karte hain: theorem change deta hai, phir hum ise known start par add karte hain. $P(2)-P(0)={\int }_0^{2}100e^{-0.5t}dt.$ Integrate karo ($\int e^{kt}dt=\frac{1}{k}{e}^{kt}$, yahan $k=-0.5$): $=100\cdot [\frac{1}{-0.5}{e}^{-0.5t}{]}_0^2=-200[e^{-0.5t}{]}_0^2=-200(e^{-1}-1)=200(1-e^{-1}).$ Numerically $e^{-1}\approx 0.3679$, toh change $\approx 200(0.6321)=126.42$. $P(2)=800+200(1-e^{-1})\approx 800+126.42=926.42\approx 926$ animals. Yahan $800$ kyun add kiya (lekin L3.1 mein nahi): hum ek absolute amount $P(2)$ maang rahe the, change nahi — toh net change lete hain aur known starting value par rakhte hain.

Neeche figure dekho: $\sin t$ $[0,3\pi ]$ par teen humps banata hai — $[0,\pi ]$ par upar, $[\pi ,2\pi ]$ par neeche, $[2\pi ,3\pi ]$ par upar. Sign changes $t=\pi$ aur $t=2\pi$ par hain.

(a) Displacement $={\int }_0^{3\pi }\sin tdt=[-\cos t{]}_0^{3\pi }=(-\cos 3\pi )-(-\cos 0)=-(-1)-(-1)=1+1=2$ m.

(b) Total distance — bars andar, $\pi$ aur $2\pi$ par split karo. Symmetry se har hump ka area ${\int }_0^{\pi }\sin tdt=[-\cos t{]}_0^{\pi }=(1)-(-1)=2.$ Teeno humps ka size $2$ hai, toh $\text{distance}=2+2+2=6\text{ m}.$ Displacement $=2$ m, distance $=6$ m. (Displacement $2=$ hump1 $-$ hump2 $+$ hump3 $=2-2+2$ — beech wala hump cancel ho jaata hai; distance teeno count karta hai.)

Hum kya karte hain: volume ka net rate of change inflow minus outflow hai: $V^{\prime }(t)=8-2t$. Net change ${\int }_0^6{V}^{\prime }(t)dt$ hai. Ek hi integral kyun: volume combined rate par respond karta hai; do integrals $\int \text{in}-\int \text{out}$ add karna same result deta hai, toh hum pehle merge kar sakte hain. ${\int }_0^6(8-2t)dt=[8t-t^2{]}_0^6=(48-36)-0=12.$ Net change $=+12$ L. (Overall inflow outflow se zyada hai, chahe $t=4$ ke baad outflow $2t>8$ reservoir drain kare.)

(a) Net change se Position. Kisi bhi $t$ ke liye, $s(t)-s(0)={\int }_0^t(3{\tau }^2-12)d\tau =[{\tau }^3-12\tau {]}_0^t=t^3-12t.$ Start add karo: $s(t)=5+t^3-12t$. Ye kyun kaam karta hai: theorem $0$ se $t$ tak ka change deta hai; ise $s(0)=5$ par rakhne se actual position recover hoti hai — alag se $+C$ dhundhne ki zaroorat nahi.

(b) $v$ ka sign: $3t^2-12=0\Rightarrow t^2=4\Rightarrow t=2$ ($[0,3]$ par). $[0,2)$ par $v<0$; $(2,3]$ par $v>0$.

Displacement $=s(3)-s(0)=(5+27-36)-5=-4-5=\dots$ ruko — directly compute karo: ${\int }_0^3(3t^2-12)dt=[t^3-12t{]}_0^3=(27-36)-0=-9$ m. Displacement $=-9$ m (start se 9 m peeche khatam hota hai; check: $s(3)=5-9=-4$, aur $-4-5=-9$ ✓).

Total distance — $t=2$ par split karo: ${\int }_0^2|3t^2-12|dt+{\int }_2^3|3t^2-12|dt={\int }_0^2(12-3t^2)dt+{\int }_2^3(3t^2-12)dt.$ Pehla: $[12t-t^3{]}_0^2=24-8=16$. Doosra: $[t^3-12t{]}_2^3=(27-36)-(8-24)=(-9)-(-16)=7$. Total distance $=16+7=23$ m. (Sanity: $23>|-9|$ ✓.)

Hum kya karte hain: net change $={\int }_0^4(a-t)dt$; ise $0$ set karo aur $a$ solve karo. ${\int }_0^4(a-t)dt=[at-\frac{t^2}{2}{]}_0^4=4a-8.$ $4a-8=0\Rightarrow a=2$ rakho. $a=2$. Interpretation: $a=2$ ke saath, inflow strong shuru hoti hai ($R(0)=2$) aur baad mein negative ho jaati hai ($R(4)=-2$), aur poore interval par gains aur losses exactly cancel ho jaate hain — net change zero, chahe tank poori time bhar raha tha phir ghatt raha tha.

Net change / displacement / cost increase / net volume ::: ${\int }_a^b{F}^{\prime }(x)dx$ (signs rakho). Total distance / total amount moved ::: ${\int }_a^b|F^{\prime }(x)|dx$ (zeros par split karo, sizes add karo). Actual final value $F(b)$ ::: $F(a)+{\int }_a^b{F}^{\prime }(x)dx$ (change ko known start par stack karo).