Exercise 1.5 — Real-Life Applications (Growth, Decay, Sound, Interest)
Bunyadi Concept
Ye exercise exponential growth/decay aur uske real-world applications par hai. Sab questions mein ek hi standard tareeqa hai:
- Diya gaya formula lo
- Known values substitute karo
- Exponential term ko akela (isolate) karo
- Dono taraf ln (ya log) lagao takay exponent neeche aa jaye
- Calculate karo
Growth/Decay Questions (Q1–Q4, Q7)
Q1 (Population growth): N=300e^(0.02t), N=600 chahiye
- 600=300e^(0.02t) → 2=e^(0.02t)
- ln lagaya: ln2=0.02t → t≈34.66 hours
Q2 (Growth constant k nikalna): N=150e^(kt), t=10 par N=250
- 250/150=5/3=e^(10k), ln lagaya: k≈0.0511 per hour
Q3 (Radioactive decay): N=800e^(−0.05t), N=200 chahiye
- 1/4=e^(−0.05t), ln lagaya: t≈27.73 hours
Q4 (Decay constant k): N=600e^(kt), t=8 par N=400
- 2/3=e^(8k), ln lagaya: k≈−0.0507 (negative hai, kyunke decay ho raha hai)
Q7 (Do parts): N=500e^(kt), N=250 jab t=5
- (i) k nikala: −0.1386
- (ii) us k ko t=10 par use kiya: N(10)=125 (matlab quantity har 5 hours mein aadhi ho rahi hai: 500→250→125)
Concept yaad rakho: Growth mein k positive hota hai, decay mein k negative hota hai.
Sound Intensity Questions (Q5, Q6, Q8)
Formula: L = 10·log₁₀(I/I₀), jahan L decibels mein hai
Q5 (L nikalna): I=10⁻⁶, I₀=10⁻¹²
- I/I₀ = 10⁶, log₁₀(10⁶)=6, isliye L=10×6=60 dB
Q6 (I nikalna, ulta): L=80, I₀=10⁻¹²
- 8=log₁₀(I/I₀), exponential form mein convert kiya: I/I₀=10⁸
- I=10⁻⁴ W/m²
Q8 (Do sounds ka ratio): L₁=85, L₂=65
- Formula se: L₁−L₂ = 10·log₁₀(I₁/I₂)
- 20=10·log₁₀(I₁/I₂) → I₁/I₂=10²=100
- Matlab pehli sound doosri se 100 guna zyada intense hai
Compound Interest Questions (Q9–Q12)
Do formulas yaad rakho:
- Annual compounding: A = P(1+r)ⁿ
- Continuous compounding: A = Pe^(rt)
Q9 (Annual, amount nikalna): P=4000, r=0.06, n=3
- A=4000(1.06)³ = Rs. 4764.06
Q10 (Annual, time nikalna): A=6324.48, P=5000, r=0.08
- (1.08)^t = 1.264896, ln lagaya: t≈3.05 years
Q11 (Continuous, amount nikalna): P=3000, r=0.05, t=4
- A=3000e^(0.2) = Rs. 3664.21
Q12 (Continuous, time nikalna): A=3320.12, P=2000, r=0.06
- e^(0.06t)=1.66006, ln lagaya: t≈8.45 years
Farq samjho: Annual compounding mein power ke roop mein t hota hai (base 1+r), lekin continuous compounding mein e use hota hai. Jab bhi “continuously” lafz nazar aaye, e^(rt) formula use karo.
