W = xy2 + x2z + yz2, x = t2, y = 9t, z = 9 (a) Find dw/dt using the appropriate Chain Rule?

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Answer 1 · 2018-07-24T06:01:29

Please see the explanation below.

First start by calculating the following derivatives

#x=t^2#, #=>#, #dx/dt=2t#

#y=9t#, #=>#, #dy/dt=9#

#z=9#, #=>#, #dz/dt=0#

Then,

#w=xy^2+x^2z+yz^2#

#(dw)/dt=d/dt(xy^2+x^2z+yz^2)#

#=d/dt(xy^2)+d/dt(x^2z)+d/dt(yz^2)#

#=xd/dt(y^2)+y^2dx/dt+zd/dt(x^2)+x^2dz/dt+yd/dt(z^2)+z^2dy/dt#

#=2yxdy/dt+y^2dx/dt+2xzdx/dt+x^2dz/dt+2yzdz/dt+z^2dy/dt#

#=18yx+2y^2t+4xzt++9z^2#

#=18xy+2/9y^3+4/9xyz+9z^2#

#=162t^3+162t^3+36t^3+729#

#=360t^3+729#

Hope that this will help!!

Answer 2 · 2018-07-28T23:07:34

Same answer, but using differentials (which can compact the notation, and sometimes also the algebra):

#w=xy^2+x^2z+yz^2#

#dw=dx\ y^2 +2 xy \ dy + 2x\ dx \z + x^2 \ dz +dy \ z^2 + 2 y z \ dz#

#=dx ( y^2 + 2xz ) +dy( 2 xy+ z^2) + dz( x^2 + 2 y z )#

#{(x= t^2 qquad dx = 2 t \ dt),(y = 9t qquad dy = 9 \ dt),(z = 9 qquad dz = 0):}#

#:. dw = 2t \ dt underbrace(( 81t^2 + 18t^2 ))_(= 99t^2) +9 \ dt ( 18t^3+ 81) + cancel(dz( x^2 + 2 y z ))#

#implies (dw)/(dt) = 2t ( 99t^2) +9 ( 18t^3+ 81) #

# = (9 ( 2t ( 11t^2) + 18t^3+ 81 )#

# = 9 ( 40t^3 + 81 )#