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Unformatted text preview: Homework due 7th February Solutions to compulsory questions Problem ( § 2.4 Exercise 19). A particle traveling along the path c ( t ) = (4 e t , 6 t 4 , cos t ) flies off on a tangent at time t = 0. Compute the position of the particle at time t = 1. Solution. The equation of the tangent at to c ( t ) time t = 0 is l ( t ) = c (0) + t c (0) = (4 , , 1) + t (4 , , 0) . Therefore, at time t = 1 the particle is at ( x,y,z ) = (4 , , 1) + (4 , , 0) = (8 , , 1) . Problem ( § 2.5 Exercise 3c). Use the chain rule to compute ∂h ∂x where h ( x,y,z ) = f ( u ( x,y,x ) ,v ( x,y ) ,w ( x )) . Solution. The question is asking us to compute ∂ ( f ◦ g ) ∂x where g ( x,y,z ) = ( u ( x,y,z ) ,v ( x,y ) ,w ( x )) . By the chain rule, ∂ ( f ◦ g ) ∂x = ∂f ∂u ∂u ∂x + ∂f ∂v ∂v ∂x + ∂f ∂w dw dx . Note: It might be helpful to consider v ( x,y ) as a function v ( x,y,z ) which is constant in z , and w ( x ) as a function w ( x,y,z ) which is constant in y and z Then the question is asking us to compute ∂ ( f ◦ g ) ∂x where g ( x,y,z ) = ( u ( x,y,z ) ,v ( x,y,z ) ,w ( x,y,z )) , which looks a little more familiar. Problem ( § 2.5 Exercise 4). Verify the chain rule for ∂h ∂x , where h ( x,y ) = f ( u ( x,y ) ,v ( x,y )) and f ( u,v ) = u 2 + v 2 , u ( x,y ) = e x y , and v ( x,y ) = e xy . Solution. From the chain rule we expect ∂h ∂x = ∂f ∂u ∂u ∂x + ∂f ∂v ∂v ∂x = 2 u ( e x y ) + 2 v ( ye xy ) = 2 e 2 x 2 y + 2 ye 2 xy . We verify this by direct calculation: ∂h ∂x = ∂ ∂x ( ( u ( x,y )) 2 + ( v ( x,y )) 2 ) = ∂ ∂x ( e 2 x 2 y + e 2 xy ) = 2 e 2 x 2 y + 2 ye 2 xy . 1 Problem ( § 2.5 Exercise 8). Let f : R 3 → R be differentiable. Making the substitution x = ρ cos θ sin φ, y = ρ sin θ sin φ, and z = ρ cos φ into f ( x,y,z ), compute ∂f ∂ρ , ∂f ∂θ and ∂f ∂φ in terms of ∂f ∂x , ∂f ∂y and ∂f ∂z . Solution. We have ∂f ∂ρ = ∂f ∂x ∂x ∂ρ + ∂f ∂y ∂y ∂ρ + ∂f ∂z ∂z ∂ρ = cos θ sin φ ∂f ∂x + sin θ sin φ ∂f ∂y + cos φ ∂f ∂z , and ∂f ∂θ = ∂f ∂x ∂x ∂θ + ∂f ∂y ∂y ∂θ + ∂f ∂z ∂z ∂θ = ρ sin θ sin φ ∂f ∂x + ρ cos θ sin φ ∂f ∂y , and ∂f ∂φ = ∂f ∂x ∂x ∂φ + ∂f ∂y ∂y ∂φ + ∂f ∂z ∂z ∂φ = ρ cos θ cos φ ∂f ∂x + ρ sin θ cos φ ∂f ∂y ρ sin φ ∂f ∂z ....
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 Spring '08
 PARKINSON
 Math, Multivariable Calculus

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