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Hw4_Solutions - Figure 1 14.4.14 14.4.5 a Using the chain...

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Figure 1: 14.4.14 14.4.5) a) Using the chain rule: dw dt = ∂w ∂x dx dt + ∂w ∂y dy dt + ∂w ∂z dz dt ∂w ∂x = 2 ye x ∂w ∂y = 2 e x ∂w ∂z = - 1 z dw dt = 2 ye x 2 t t 2 +1 + 2 e x 1 t 2 +1 - 1 z e t = 2(tan - 1 t )( t 2 + 1) 2 t t 2 +1 + 2( t 2 + 1) 1 t 2 +1 - 1 = 4 t (tan - 1 t ) + 1 Substituting and differentiating directly (much faster on this problem!): w = 2(tan - 1 t )( t 2 + 1) - t dw dt = 2(2 t )(tan - 1 t ) + 2 t 2 +1 t 2 +1 - 1 = 4 t (tan - 1 t ) + 1 b) at t = 1 we have 4(tan - 1 1) + 1 = 4 π 4 + 1 = π + 1 14.4.14) dz dt = ∂f ∂x dx dt + ∂f ∂y dy dt + ∂f ∂z dz dt 14.4.33) ∂w ∂r = 2( x + y + z )( ∂x ∂r + ∂y ∂r + ∂z ∂r ) = 2( r - s + cos( r + s ) + sin( r + s ))(1 - sin( r + s ) + cos( r + s )) At (1 , - 1) we have 2(1 + 1 + 1 + 0)(1 - 0 + 1)) = 2 * 3 * 2 = 12 14.4.40) V = abc. dV dt = ∂a ∂t da dt + ∂b ∂t db dt + ∂c ∂t dc dt = bc + ac + ab ( - 3) = 2 * 3 + 1 * 3 + 1 * 2 * ( - 3) = 6 + 3 - 6 = 3 so the volume is increasing. S = 2 ab + 2 ac + 2 bc so dS dt = 2( b + c ) da dt + 2( a + c ) db dt + 2( b + a ) dc dt = 2 * 5 * 1 + 2 * 4 * 1 + 2 * 3 * ( - 3) = 0 . The surface area is fixed. The diagonals are D = a 2 + b 2 + c 2 . 1
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-5 0 5 -5 0 5 10 15 20 25 Figure 2: 14.5.3 dD dt = 1 a 2 + b 2 + c 2 ( a da dt + b db dt + c dc dt ) dD dt (1 , 2 , 3) = 1 14 (1+2+3( - 3)) = - 6 14
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