group1ppt-240508064514-51fa79311221.pptx

Q1.Take the photo of increasing and decreasing curves separately

1. CIA SELF LEARNING ACTIVITY SUB-MATHEMATICS- II SWAMI GROUP NO - 6 DURING THE ACADEMIC YEAR 2023-24

2. Q1.Take the photo of increasing and decreasing curves separately. Verify whether Roll's theorem is applicable or not, if applicable take approximate value and find C

3. f ( x ) = 𝑥 2 + 2 x − 8 ; x 𝜖 [ - 4,2] lets check conditions of rolls theorem condition 1 and 2- f(x)= 𝑥 2 +2x-8 is a polynomial so f(x)is continuous on [-4,2] and differentiable in(-4,2) C ondition 3- f(a)=f(b) for a=-4,b=2 f(-4)= (−4) 2 +2 −4 − 8 =16-8-8 =0 f(2)= (2) 2 +2 2 − 8 =4+4-8 =0

4. Hence, f(-4) = f(2) Now f(x) = 𝑥 2 + 2x – 8 f ' (x) = 2x + 2 – 0 f ' (x) = 2x +2 put x=c f ' (c) = 2c +2 Since all three conditions are satisfied f ' (c) = o 2c + 2 = 2c = -2 c = -2/2 c=-1 hence the rolls theorem is verified .

5. Q2. Solve 𝑑𝑦 = 𝑋 + 𝑋 2 𝑌 2 , 𝑑𝑧 = 𝑋𝑍, 𝑌 1 = 3 , 𝑍 1 = 4, ℎ = 0.3 𝑑𝑥 𝑑𝑥 Find 𝑦, 𝑧 at 𝑥 = 1.3 𝑌 = 3, 𝑋 = 1, 𝑍 = 4, ℎ = 0.3 Here, 𝑑𝑦 = 𝑓 𝑥, 𝑦, 𝑧 = 𝑥 + 𝑥 2 𝑦 2 & 𝑑𝑧 = 𝜑 𝑥, 𝑦, 𝑧 = 𝑥𝑧 𝑑𝑥 𝑑𝑥 Starting at ( 𝑥 , 𝑦 , 𝑧 ) and taking the step-sizes for 𝑥, 𝑦, 𝑧 to be ℎ, 𝑘, 𝑙 respectively, the Runge-Kutta method gives 𝑘 1 = ℎ 𝑓 ( 𝑥 , 𝑦 , 𝑧 ) 0 0 = ℎ (𝑥 + 𝑥 2 𝑦 2 ) = 0.3 (1 + 1 2 3 2 ) = . 3 10 𝑙 1 = ℎ 𝜑 ( 𝑥 , 𝑦 , 𝑧 ) = ℎ ( 𝑥 𝑧 ) = 0.3(1 × 4) = 1.2 𝑘 1 = 3

6. 𝑘 2 2 0 0 = ℎ𝑓 𝑥 + ℎ , 𝑦 + 𝑘 1 2 , 𝑧 + 𝑙 1 2 ℎ 𝑘 1 𝑙 2 = ℎ 𝜑 ( 𝑥 + 2 , 𝑦 + 2 , 𝑧 2 + 𝑙1 ) = ℎ𝑓 ( 1 + . 3 , 3 + 2 2 1 . 2 2 3 , 4 + ) 3 2 2 = . 3𝜑 ( 1 + . 3 , 3 + , 4 + 2 1.2 ) = ℎ𝑓 ( 1 . 1 5 , 4 . 5 , 4 . 6 ) = . 3 𝜑 ( 1 . 1 5 , 4 . 5 , 4 . 6 ) = 0.3 ( 1.15 + 1.15 2 4.5 2 ) = . 3 ( 1 . 1 5 × 4 . 6 ) = . 3 ( 1 . 1 5 + 1 . 322 5 × 2 . 2 5 ) = 0.3 (5.29) = 0.3 (27.9306) = 1.587 = 8.3792

7. 𝑘 3 2 0 0 = ℎ𝑓 𝑥 + ℎ , 𝑦 + 𝑘 1 2 , 𝑧 + 𝑙 2 2 3 ℎ 0 0 𝑙 = ℎ 𝜑 ( 𝑥 + , 𝑦 + 𝑘 2 2 2 2 , 𝑧 + 𝑙 2 ) = . 3 𝑓 ( 1 . 1 5 , 3 + 2 8 .37 9 2 , 4 + 2 1.587 ) = . 3 𝜑 ( 1 . 1 5 , 7 . 1 8 9 6 , 4 . 7 93 5 ) = . 3 𝑓 ( 1 . 1 5 , 1 4 . 379 2 , 2 2 9.857 ) = 0.3 (1.15 × 4.7935) = . 3 𝑓 ( 1 . 1 5 , 7 . 1 8 9 6 , 4 . 7 93 5 ) = 0.3 (5.5125) = 0.3 (1.15 + (1.15) 2 (7.1896) 2 ) = 1.65385 = 0.3 (69.5104) = 20.85312

8. 𝑙 4 = ℎ𝜑(𝑥 + ℎ, 𝑦 + 𝑘 3 , 𝑧 + 𝑙 3 ) 𝑘 4 = ℎ𝑓 𝑥 + ℎ, 𝑦 + 𝑘 3 , 𝑧 + 𝑙 3 = . 3 𝑓 ( 1 + . 3 , 3 + 20 . 8 53 1 2 , 4 + 1 . 6 53 8 ) = 0.3𝜑 (1.3,23.85312,5.6538) = . 3 𝑓 ( 1 . 3 , 2 3 . 8 5 31 2 , 5 . 65 3 8 ) = . 3 ( 1 . 3 × 5 . 65 3 8 ) = 0.3 (1.3 + (1.3) 2 (23.85312) 2 ) = 2.2050 = 0.3 (962.8614) = 288.85842

9. 1 𝑙 = (𝑙 1 + 2𝑙 2 + 2𝑙 3 + 𝑙 4 ) 6 1 = ( 3 + 2 8.3792 + 2 20.85312 + 288.8584) 1 6 = ( 1.2 + 2 1.537 6 2 1.6538 + 2.2050) 1 = (3 + 41.70624 + 16.7584 + 288.8584) = 1 6 (1.2 + 3.174 + 6 3.3076 + 2.2050) 1 6 = (350.32306) = 1 6 (9.8866) = 58.3871 = 1.6477 𝑦 = 𝑦 + 𝑘 = 3 + 58.3571 = 5.6477 𝑧 = 𝑧 + 𝑙 = 4 + 1.6477 = 61.38 K = 1 (K 1 + 2K 2 + 2K 3 + K 4 ) 6 + 2 (1.65385 )+ 2.2050 ) 3.3077 + 2.2050 )

10. Q.3. Determine Area lies under the C shape curve, take photo of such curve you have seen in college campus.

11. eq-1 e q - 2 Parabola : 𝑦 2 = 4𝑥 Line : 𝑦 = 2𝑥 − 4 Put the equation 1 in equation 2 ∴ (2𝑥 − 4) 2 = 4𝑥 ∴ 4𝑥 2 − 16𝑥 + 16 = 4𝑥 ∴ 4𝑥 2 − 20𝑥 + 16 = ∴ 𝑥 2 −5𝑥 + 4 = 𝑥 = 1,4 W h e n x = 1 a n d When x = 4 and y = -2 y = 4 Point of intersection are A(1,-2) and B(4,4) 2 Internal limit: 𝑥 = 𝑦 4 ; 𝑥 = (𝑦+4) 2 External limits : y = -2 ; y = 4

12. A =    

13. 𝐴 4 2 + 8 4 2 3 ) A 16 + 32 (( 2) 2 + 8 2 8 3 ( 4 1 6 + ) A 3 64 3 4 + 16 8 3 A A 16 + 32 60 24 36 A = 9 sq. units

14. Q . 4 . U s i n g t y l o r ’ s s e r i e s e x p r e s s ( 𝒙 − 𝟓 ) 𝟑 + 2 ( 𝒙 − 𝟓 ) 𝟐 + 7 in p o w e r o f ‘ x ’ . A n s : f ( x ) = 𝒙 − 𝟓 𝟑 + 2 ( 𝒙 − 𝟓 ) 𝟐 +7 f ( x ) = 𝒙 𝟑 + 𝟐 𝒙 𝟐 +7 h= -5 f(x+h)= f(h)+ x f’(h)+ 𝒙 𝟐 / 2!.f”(h)+ 𝒙 𝟑 /3!+f”(h) f ( - 5 ) = ( − 𝟓 ) 𝟑 +2 X ( − 𝟓 ) 𝟐 +7 = (-125)+50+7 .: f(-5)= -68 F ` ( h ) = 𝟑 𝒙 𝟐 +4x F` (-5)=3 (−𝟓) 𝟐 +4(-5) =55

15. F`` (h)=6x+4 f`` (-5)=6(-5)+4 = -30+4 .: f ’” (h)= -26 f’”(h)= 6 f’”(-5)= 6 f(x+h)= -68+x(55)+ 𝒙 𝟐 /2.(-26)+ 𝒙 𝟑 /6 x(6) = -68+55x-13 𝒙 𝟐 + 𝒙 𝟑 = 𝒙 𝟑 - 13 𝒙 𝟐 +55 x - 68 .: f( x+h )= 𝒙 𝟑 -13x^2+55x-68

16. THANK YOU!!!