University:
East Georgia State College
Course:
MATH 1540 | Calculus I
Academic year:

2024
Views:

452

Pages:

2
Author:

somospumasl3o

Question 1 Determine the first derivative 𝑑 𝑦 𝑑 𝑥 dx dy : 𝑦 = 2 𝑒 2 𝑥 + ln ⁡ 𝑥 3 − 2 𝑒 𝑥 y=2e 2x +lnx 3 −2e x Answer 𝑑 𝑦 𝑑 𝑥 = 4 𝑒 2 𝑥 + 3 𝑥 − 2 𝑒 𝑥 dx dy =4e 2x + x 3 −2e x ( ∵ ∵ ln ⁡ ( 𝑥 3 ) = 3 ln ⁡ ( 𝑥 ) ln(x 3 )=3ln(x) and the differential of 𝑒 2 𝑥 = 2 𝑒 2 𝑥 e 2x =2e 2x ) Question 2 How to integrate cos ⁡ 2 ( 2 𝑥 ) cos 2 (2x)? Answer 𝐼 = ∫ cos ⁡ 2 ( 2 𝑥 ) 𝑑 𝑥 = 1 2 ∫ cos ⁡ ( 4 𝑥 ) 𝑑 𝑡 I=∫cos 2 (2x)dx= 2 1 ∫cos(4x)dt Using trigonometric identities and integration: cos ⁡ 2 ( 2 𝑥 ) 𝑑 𝑥 = sin ⁡ ( 4 𝑥 ) 4 + 𝑥 2 cos 2 (2x)dx= 4 sin(4x) + 2 x Question 3 Find the general solution of 𝑦 ′ ′ − 2 𝑦 ′ + 2 𝑦 = 𝑥 𝑒 𝑥 cos ⁡ 𝑥 y ′′ −2y ′ +2y=xe x cosx. Answer The solution involves solving the homogeneous equation: 𝑦 ℎ = 𝑒 𝑥 ( 𝐶 1 cos ⁡ 𝑥 + 𝐶 2 sin ⁡ 𝑥 ) y h =e x (C 1 cosx+C 2 sinx) The particular solution is determined by assuming: 𝑦 𝑝 = ( 𝑃 𝑥 + 𝑄 ) 𝑒 𝑥 𝑢 y p =(Px+Q)e x u and applying the method of undetermined coefficients. Final solution: 𝑦 = 𝑒 𝑥 ( 𝐶 1 cos ⁡ 𝑥 + 𝐶 2 sin ⁡ 𝑥 ) + 𝑦 𝑝 y=e x (C 1 cosx+C 2 sinx)+y p Question 4 Find 𝑑 𝑦 𝑑 𝑡 dt dy for each pair of functions 𝑦 = 𝑥 2 − 2 𝑥 , 𝑥 = 𝑡 2 + 7 y=x 2 −2x,x=t 2 +7. Answer Given 𝑦 = 𝑥 2 − 2 𝑥 y=x 2 −2x and 𝑥 = 𝑡 2 + 7 x=t 2 +7: 𝑑 𝑦 𝑑 𝑡 = 2 𝑥 𝑑 𝑥 𝑑 𝑡 − 2 𝑑 𝑥 𝑑 𝑡 dt dy =2x dt dx −2 dt dx Substitute 𝑥 = 𝑡 2 + 7 x=t 2 +7: 𝑑 𝑦 𝑑 𝑡 = 4 𝑡 ( 𝑡 2 + 7 ) − 4 𝑡 dt dy =4t(t 2 +7)−4t 𝑑 𝑦 𝑑 𝑡 = 4 𝑡 3 + 28 𝑡 − 4 𝑡 = 4 𝑡 3 + 24 𝑡 dt dy =4t 3 +28t−4t=4t 3 +24t Question 5 Find the differential of the function 𝑇 = 𝑣 1 + 𝑢 𝑣 𝑤 T= 1+uvw v . Answer Using the total derivative formula: 𝑑 𝑇 = ∂ 𝑇 ∂ 𝑢 𝑑 𝑢 + ∂ 𝑇 ∂ 𝑣 𝑑 𝑣 + ∂ 𝑇 ∂ 𝑤 𝑑 𝑤 dT= ∂u ∂T du+ ∂v ∂T dv+ ∂w ∂T dw 𝑑 𝑇 = 1 ( 1 + 𝑢 𝑣 𝑤 ) 2 ( − 𝑣 2 𝑤 𝑑 𝑢 + 𝑑 𝑣 − 𝑢 𝑣 𝑤 2 𝑑 𝑤 ) dT= (1+uvw) 2 1 (−v 2 wdu+dv−uvw 2 dw) Final result: 𝑑 𝑇 = 1 1 + 𝑢 𝑣 𝑤 2 ( − 𝑣 2 𝑤 𝑑 𝑢 + 𝑑 𝑣 − 𝑢 𝑣 𝑤 2 𝑑 𝑤 ) dT= 1+uvw 1 2 (−v 2 wdu+dv−uvw 2 dw)