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Unformatted text preview: 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....
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This note was uploaded on 12/25/2009 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell University (Engineering School).
 Spring '06
 PANTANO
 Chain Rule, Multivariable Calculus, The Chain Rule

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