Engineering Calculus Notes 449

# Engineering Calculus Notes 449 - z y 1 − z Fnd ∂F 1...

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4.2. DIFFERENTIABLE MAPPINGS 437 (c) F ( x,y ) = ( x 2 + y 2 , 2 xy ), −→ x 0 = ( 1 , 1). (d) F ( x,y ) = ( x + y,x y ), −→ x 0 = ( 1 , 2). (e) F ( x,y ) = ( x,y,x 2 y 2 ), −→ x 0 = (2 , 1). (f) F ( x,y ) = ( x,y,xy ), −→ x 0 = (2 , 1). (g) F ( x,y ) = ( x 2 y,x + y 1 , 3 x + 5 y ), −→ x 0 = (2 , 1). (h) F ( x,y ) = ( x 2 , 2 xy,y 2 ), −→ x 0 = (1 , 3). (i) F ( x,y,z ) = ( y + z,xy + z,xz + y ), −→ x 0 = (2 , 1 , 3). (j) F ( x,y,z ) = ( xyz,x y + z 2 ), −→ x 0 = (2 , 1 , 1). 2. In each part below, you are given a mapping described in terms of rectangular coordinates. Use the Chain Rule together with one of the equations ( 4.3 ), ( 4.4 ), ( 4.5 ), or ( 4.6 ) to Fnd the indicated partial derivative when the input is given in one of the other coordinated systems. (a) F ( x,y ) = ( x 2 y 2 , 2 xy ); Fnd ∂F 1 ∂r and ∂F 2 ∂θ at the point with polar coordinates r = 2 and θ = π 3 . (b) F ( x,y,z ) = ( x 2 + y 2 + z 2 ,xyz ); Fnd ∂F 1 ∂r and ∂F 2 ∂θ at the point with cylindrical coordinates r = 2, θ = 2 π 3 , and z = 1. (c) F ( x,y,z ) = ( x 1
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Unformatted text preview: z , y 1 − z ); Fnd ∂F 1 ∂ρ , ∂F 2 ∂φ and ∂F 2 ∂θ at the point with spherical coordinates ρ = 1, φ = π 2 , and θ = π 3 . (d) F ( x,y,z ) = ( x 2 + y 2 + z 2 ,xy + yz,xyz ); Fnd ∂F 1 ∂ρ , ∂F 2 ∂φ and ∂F 3 ∂θ at the point with spherical coordinates ρ = 4, φ = π 3 , and θ = 2 π 3 . Theory problems: 3. Prove Remark 4.2.4 . 4. Show that for any linear map L : R n → R m , (a) the number b L b = sup {b L ( −→ u ) b | −→ u ∈ R n and b −→ u b = 1 } (4.7) satisFes b L ( −→ x ) b ≤b L b b −→ x b (4.8) for every vector −→ x ∈ R n ; (b) b L b is actually a maximum in Equation ( 4.7 );...
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