3.1.8
Har "wavy" sine graph jo bhi tumhe milegi, woh bas basic sin x stretch, squish, slide, aur lift ki gayi hai. Chaar numbers A , B , C , D ek kaam karta hai. Seekho ki har ek kya karta hai aur kyun , aur tum koi bhi sinusoid seconds mein sketch kar sakte ho — koi bhi ugly pictures yaad karne ki zaroorat nahi.
Chaar numbers kyun? Ek general wave ko batana hota hai: kitni tall (A), kitni fast repeat hoti hai (B), kahan se start hoti hai (C), aur kis level ke around hilti hai (D). Ye ek wave ki chaar independent freedoms hain, bas itna hi.
Hum y = sin x 2 π y = 0
y y / A y = A sin x sin [ − 1 , 1 ] A [ − A , A ]
sin input 2 π B x B x increases by 2 π ⇒ Δ ( B x ) = 2 π ⇒ B Δ x = 2 π . x
Input ko factor karo: B x + C = B ( x + B C ) x x + B C left ki taraf B C + ulti direction mein).
D D midline y = D y = D − ∣ A ∣ y = D + ∣ A ∣
Intuition Pehle horizontals, phir verticalsSine ke andar (B C x bahar (A D y = 2 π /∣ B ∣ = − C / B = D = ∣ A ∣
Worked example Example 1 —y = 3 sin ( 2 x − 2 π ) + 1 Amplitude: ∣ A ∣ = 3 Kyun? A = 3
Period: T = ∣ B ∣ 2 π = 2 2 π = π Kyun? Input 2 x 2 π x
Phase shift: 2 x − 2 π = 2 ( x − 4 π ) + 4 π Kyun? − C / B = − ( − π /2 ) /2 = π /4 ( x − 4 π ) right slide confirm karta hai.
Vertical: midline y = 1 = 1 + 3 = 4 = 1 − 3 = − 2 Kyun? D = 1 ∣ A ∣
Worked example Example 2 —y = − 2 cos ( π x ) + 5 Amplitude: ∣ − 2∣ = 2 minimum se shuru hota hai).
Flip kyun? A < 0
Period: T = ∣ π ∣ 2 π = 2 Kyun? B = π x 2
Midline / max / min: midline y = 5 = 7 = 3
x = 0 y = − 2 cos 0 + 5 = − 2 + 5 = 3
Worked example Example 3 — Graph se reverse-engineer karoEk wave ka max = 8 = 2 4 x = 1 A sin ( B x + C ) + D
D = 2 m a x + m i n = 2 8 + 2 = 5 Kyun? Midline, top aur bottom ka average hai.A = 2 m a x − m i n = 2 8 − 2 = 3 Kyun? Total peak-to-trough height ka aadha.B = T 2 π = 4 2 π = 2 π Kyun? Period formula ko ulta karo.Midline se upar x = 1 C = − B ⋅ ( shift ) = − 2 π Answer: y = 3 sin ( 2 π x − 2 π ) + 5.
Recall Compute karne se pehle predict karoy = 4 sin ( 3 x + π ) − 1 forecast karo: amplitude? period? shift? midline?
Phir neeche verify karo.
Worked example VerificationAmplitude = 4 = 2 π /3 = − π /3 π /3 3 x + π = 3 ( x + π /3 ) y = − 1 = 3 = − 5
Common mistake "Period hai2 π B Kyun sahi lagta hai: bada B feel hota hai jaise koi bada kuch hai. Trap: bada B faster repeat karta hai, toh period chhoti hoti hai. Fix: derive karo — ek cycle ke liye B Δ x = 2 π ⇒ T = 2 π /∣ B ∣ B neeche jaata hai.
Common mistake "Phase shift− C Kyun sahi lagta hai: andar + C C B = 1 Trap: jab B = 1 B Fix: shift = − C / B − C
Common mistake "Shift direction:+ C Kyun sahi lagta hai: hum "+" ko "forward/right" padhte hain. Trap: function ke andar ki change counter-intuitive hoti hai: f ( x + c ) left move karta hai. Fix: B ( x − h ) h
A Fix: amplitude = ∣ A ∣ ≥ 0 reflection hai, chhoti height nahi.
Recall Simply explain karo (click to reveal)Socho ek jump rope jo tum swing karte ho. A A B C D
"A mplitude B elow-for-period, C -over-B slides, D own/up midline."
A = height, B 2 π / B below jaata hai, C/B = slide, D = up/down level.
Ya: "Big B, Brief period" (bada B → choti period).
y = A sin ( B x + C ) + D ∣ A ∣
A sin ( B x + C ) + D Ek cycle ke liye input change 2 π B Δ x = 2 π ⇒ T = 2 π /∣ B ∣
Period ke denominator mein B Bada B 2 π x
A sin ( B x + C ) + D − C / B
Shift padhne se pehle B Shift x C B x B ( x + C / B ) C / B
D Graph ko lift karta hai; midline y = D D + ∣ A ∣ D − ∣ A ∣
Max M m A D A = ( M − m ) /2 D = ( M + m ) /2
Negative A Use midline ke across reflect karta hai (ulta kar deta hai); amplitude ∣ A ∣
y = 3 sin ( 2 x − π /2 ) + 1 Period π π /4 y = 1
c > 0 f ( x + c ) Left — andar ki changes intuition ke opposite kaam karti hain.
