Clac 3 Linearization: Find the Linearization of the function f(x,y,z)=sqrt(2x^2+3y^2+z^2) at (2,3,3). My work: I found that f(2,3,3)=sqrt(44) df/dx=8+(1/(2sqrt(44))) df/dy=18+(1/(2sqrt(44))) df/dx=6+(1/(2sqrt(44))) so the linearization=sqrt(44)+((8+(1/(2sqrt(44))))(x-2))+((18+(1/(2sqrt(44))))(y-3))+((6+(1/(2sqrt(44))))(z-3)). I am being told that this is incorrect and was trying to figure out why
\[(x,y,z)= \sqrt{(2x^2+3y^2+z^2)}\,\,\,\,\,at (2,3,3).\] Hey linearization is a linear function in all the three variables. What is ur soln ?
are you saying that sqrt(44)+((8+(1/(2sqrt(44))))(x-2))+((18+(1/(2sqrt(44))))(y-3))+((6+(1/(2sqrt(44))))(z-3)) needs to be simplified more?
I am saying the linearization is not a number neither three numbers it is a linear \[\color{red}{\frak{Function}}\]
humm ok so what should i be looking for?
the numbers 8, 18,, 6 are \[\color{yellow}{MULTIPLIERS,\,\,\,NOT\,\,\,\,SUMMANDS}\]
THEY ARE 3 "SLOPES", so to say
instead of "+" it should be "*" after each of them, meaning multiplication
after each of what? the 8,16,6?
\[ df/dx=8(1/(2sqrt(44)))\]
got it lol for some reason i was thinging +'s
just realized what was going on. Tyvm lol
what is tyvm ?
thank you very much
thx 4 xplnshn
for some reason i thinking chain rule added the parts together
It is rather \[ \Huge CENTRAL\quad that\\ \Huge \,\,\, MULTIPLICATION\]
What is the level and "year" you are studying this material in ?
@BenBlackburn ?
sophmore 2nd in college. its just been awhile. over the summer i took linear algebra and we did not do much of this stuff in clac 2. so some of this stuff is causing me to go back and refresh my memory
so some careless errors are popping up
Well the lecturer should have stressed 200 times the chain rule
it is in fact deeply intertwined with the idea of LINEARIZATION look suppose y = K*x and z = M*y ===> obviously z = K*M*x
the same with derivatives !
Well good luck & farewell.
haha ty you too
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