University:
California State University, Northridge
Course:
MATH 351 | Differential Equations
Academic year:

2024
Views:

483

Pages:

9
Author:

Javon Neal

Question 1 Find the first partial derivatives of the function: 𝑤 = 𝑒 𝑢 𝑢 + 𝑣 2 w= u+v 2 e u Answer 1 We have: 𝑤 = 𝑒 𝑢 𝑢 + 𝑣 2 w= u+v 2 e u Then, 𝑤 𝑢 = ( 0 ) ( 𝑢 + 𝑣 2 ) − 𝑒 𝑢 ( 1 ) ( 𝑢 + 𝑣 2 ) 2 w u = (u+v 2 ) 2 (0)(u+v 2 )−e u (1) 𝑤 𝑢 = − 𝑒 𝑢 ( 𝑢 + 𝑣 2 ) 2 w u = (u+v 2 ) 2 −e u Now for 𝑤 𝑣 w v : 𝑤 𝑣 = ( 𝑒 𝑢 ) ( 𝑢 + 𝑣 2 ) − ( 𝑒 𝑢 ) ( 2 𝑣 ) ( 𝑢 + 𝑣 2 ) 2 w v = (u+v 2 ) 2 (e u )(u+v 2 )−(e u )(2v) 𝑤 𝑣 = 𝑒 𝑢 ( 𝑢 + 𝑣 2 − 2 𝑣 ) ( 𝑢 + 𝑣 2 ) 2 w v = (u+v 2 ) 2 e u (u+v 2 −2v) So: 𝑤 𝑢 = − 𝑒 𝑢 ( 𝑢 + 𝑣 2 ) 2 , 𝑤 𝑣 = 𝑒 𝑢 ( 𝑢 + 𝑣 2 − 2 𝑣 ) ( 𝑢 + 𝑣 2 ) 2 w u = (u+v 2 ) 2 −e u ,w v = (u+v 2 ) 2 e u (u+v 2 −2v) Answer 2 Using the quotient rule for ∂ 𝑤 ∂ 𝑢 ∂u ∂w : ∂ 𝑤 ∂ 𝑢 = ( 𝑢 + 𝑣 2 ) ( 𝑒 𝑢 ) − 𝑒 𝑢 ( 1 ) ( 𝑢 + 𝑣 2 ) 2 ∂u ∂w = (u+v 2 ) 2 (u+v 2 )(e u )−e u (1) Now calculate each partial derivative: ∂ 𝑒 𝑢 ∂ 𝑢 = 𝑒 𝑢 ∂u ∂e u =e u ∂ ( 𝑢 + 𝑣 2 ) ∂ 𝑢 = 1 ∂u ∂(u+v 2 ) =1 Substitute: ∂ 𝑤 ∂ 𝑢 = ( 𝑒 𝑢 ) ( 1 ) − 𝑒 𝑢 ( 𝑢 + 𝑣 2 ) 2 = − 𝑒 𝑢 ( 𝑢 + 𝑣 2 ) 2 ∂u ∂w = (u+v 2 ) 2 (e u )(1)−e u = (u+v 2 ) 2 −e u Similarly, ∂ 𝑤 ∂ 𝑣 = − 𝑒 𝑢 ( 2 𝑣 ) ( 𝑢 + 𝑣 2 ) 2 ∂v ∂w = (u+v 2 ) 2 −e u (2v) Question 2 Find all the second partial derivatives: 𝑓 ( 𝑥 , 𝑦 ) = sin ⁡ 2 ( 𝑚 𝑥 + 𝑛 𝑦 ) f(x,y)=sin 2 (mx+ny) Answer 1 𝑓 = [ sin ⁡ ( 𝑚 𝑥 + 𝑛 𝑦 ) ] 2 f=[sin(mx+ny)] 2 First-order derivatives: 𝑓 𝑥 = 2 𝑚 sin ⁡ ( 𝑚 𝑥 + 𝑛 𝑦 ) cos ⁡ ( 𝑚 𝑥 + 𝑛 𝑦 ) f x =2msin(mx+ny)cos(mx+ny) 𝑓 𝑦 = 2 𝑛 sin ⁡ ( 𝑚 𝑥 + 𝑛 𝑦 ) cos ⁡ ( 𝑚 𝑥 + 𝑛 𝑦 ) f y =2nsin(mx+ny)cos(mx+ny) Second-order derivatives: 𝑓 𝑥 𝑥 = 2 𝑚 2 [ cos ⁡ 2 ( 𝑚 𝑥 + 𝑛 𝑦 ) − sin ⁡ 2 ( 𝑚 𝑥 + 𝑛 𝑦 ) ] f xx =2m 2 [cos 2 (mx+ny)−sin 2 (mx+ny)] 𝑓 𝑦 𝑦 = 2 𝑛 2 [ cos ⁡ 2 ( 𝑚 𝑥 + 𝑛 𝑦 ) − sin ⁡ 2 ( 𝑚 𝑥 + 𝑛 𝑦 ) ] f yy =2n 2 [cos 2 (mx+ny)−sin 2 (mx+ny)] Mixed partial derivatives: 𝑓 𝑥 𝑦 = 2 𝑚 𝑛 [ cos ⁡ 2 ( 𝑚 𝑥 + 𝑛 𝑦 ) − sin ⁡ 2 ( 𝑚 𝑥 + 𝑛 𝑦 ) ] f xy =2mn[cos 2 (mx+ny)−sin 2 (mx+ny)] Question 3 An airplane flies at an altitude of 5 miles toward a point directly over an observer. The speed of the plane is 600 mph. Find the rates at which the angle of elevation 𝜃 θ is changing when the angle is: (a) 𝜃 = 3 0 ∘ θ=30 ∘ (b) 𝜃 = 7 5 ∘ θ=75 ∘ Answer 1 Using: tan ⁡ ( 𝜃 ) = 5 𝑥 tan(θ)= x 5 Differentiate: sec ⁡ 2 ( 𝜃 ) 𝑑 𝜃 𝑑 𝑡 = − 5 𝑥 2 𝑑 𝑥 𝑑 𝑡 sec 2 (θ) dt dθ =− x 2 5 dt dx Substitute: sec ⁡ 2 ( 3 0 ∘ ) = 4 3 , sec ⁡ 2 ( 7 5 ∘ ) = 16 sec 2 (30 ∘ )= 3 4 ,sec 2 (75 ∘ )=16 Solve for 𝑑 𝜃 𝑑 𝑡 dt dθ : (a) 𝜃 = 3 0 ∘ θ=30 ∘ : 𝑑 𝜃 𝑑 𝑡 = … dt dθ =… (b) 𝜃 = 7 5 ∘ θ=75 ∘ : 𝑑 𝜃 𝑑 𝑡 = … dt dθ =… Question 4 If 𝑓 ( 𝑥 ) = 𝑥 2 [ 𝑓 ( 𝑥 ) ] 4 = 18 f(x)=x 2 [f(x)] 4 =18 and 𝑓 ( 1 ) = 2 f(1)=2, find 𝑓 ′ ( 1 ) f ′ (1). Answer 1 Differentiating both sides: 𝑓 ′ ( 𝑥 ) + 4 𝑥 2 [ 𝑓 ( 𝑥 ) ] 3 𝑓 ′ ( 𝑥 ) = 0 f ′ (x)+4x 2 [f(x)] 3 f ′ (x)=0 Substitute: 𝑓 ′ ( 1 ) = … f ′ (1)=… Question 5 Find both first partial derivatives. 𝑧 = 𝑒 𝑥 𝑦 z=e xy . Answer 1 ∂ 𝑧 ∂ 𝑥 = 𝑦 𝑒 𝑥 𝑦 , ∂ 𝑧 ∂ 𝑦 = 𝑥 𝑒 𝑥 𝑦 ∂x ∂z =ye xy , ∂y ∂z =xe xy