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Mathematics 17 Online
OpenStudy (anonymous):

find xy+(z^3)x-2yz=0 in terms of z

OpenStudy (anonymous):

whts in terms of z mean?

OpenStudy (anonymous):

the answer should be z= f(x,y). so i am asking what f(x,y) is

OpenStudy (amistre64):

x.z^3 -2yz = -xy

OpenStudy (amistre64):

z(-2y +x.z^2) = -xy -2y +x.z^2 - -xy/z

OpenStudy (anonymous):

It's complicated to solve for z.

OpenStudy (anonymous):

yeah its a cubic eqn in z

OpenStudy (anonymous):

Could you write the original question, if it is not this?

OpenStudy (amistre64):

-2y +x.z^2 = -xy/z x.z^2 = (-xy+2yz)/z z^3 = (-xy+2yz)/x

OpenStudy (anonymous):

i am supposed to find dz/dx at the point (1,1,1) of that equation

OpenStudy (anonymous):

then say that

OpenStudy (anonymous):

spo i assumed i needed z as a function of x and y

OpenStudy (amistre64):

you dont need to solve for z then...

OpenStudy (anonymous):

u dont need to..i think not

OpenStudy (amistre64):

dz/dx = g(F).d(xbar)/dx

OpenStudy (anonymous):

ok, sorry about tha

OpenStudy (anonymous):

You don't have to solve for z.

OpenStudy (anonymous):

bt then how to do it???

OpenStudy (amistre64):

get the gradient vector but partialling the equation of the plane there

OpenStudy (amistre64):

get the gradient vector but partialling the equation of the plane there

OpenStudy (amistre64):

F=<a,b,c> a = dz/dx...but not that dz/dx...the funny looking one lol

OpenStudy (anonymous):

then do df/dx / df/dz?

OpenStudy (amistre64):

<y+z^3, x-2z, 3z^2-2y>

OpenStudy (amistre64):

plug in the values from your point know to get the gradianet vector

OpenStudy (amistre64):

<2,-1,1> is your normal vector/gradient..

OpenStudy (anonymous):

oh ok, i got the gradient vector, then plug in the point (1,1,1) to df/dx to get the answer?

OpenStudy (amistre64):

dot product that with the vecotr that is the point...<1,1,1>

OpenStudy (anonymous):

i got <2, -1, 1> yes

OpenStudy (amistre64):

2(x-1)-1(y-1)+1(z-1) = dz/dx i think :)

OpenStudy (anonymous):

is it right??

OpenStudy (anonymous):

is that the plane equation?

OpenStudy (anonymous):

i dont have the answer

OpenStudy (amistre64):

that is the plane equation

OpenStudy (anonymous):

doesnt the gradient vector give <df/dx, df/dy,df/dz>... in that case we can simply do df/dx / df/dz???

OpenStudy (anonymous):

the answer isnt an equation...its a value

OpenStudy (amistre64):

him; dunno havent read that far yet :)

OpenStudy (anonymous):

thats wht m sayin

OpenStudy (anonymous):

is the answer 2???

OpenStudy (amistre64):

the answer would be 2 then right?

OpenStudy (amistre64):

gradient . point

OpenStudy (anonymous):

i think so! thanks!

OpenStudy (anonymous):

is it 2??

OpenStudy (anonymous):

i dont know the value, but it seems right

OpenStudy (anonymous):

u understood how it ws done?

OpenStudy (amistre64):

i dont think I evenknow how it was done lol

OpenStudy (anonymous):

I got -2.

OpenStudy (anonymous):

yes, get the gradient, and plug in the x,y and z valuse of the point into the gradient and divide df/dx by df/dz to get dz/dx right?

OpenStudy (anonymous):

Not entirely sure though.

OpenStudy (anonymous):

i thnk it shld be right..there might b calculation errors bt b surev f d method

OpenStudy (amistre64):

<2,-1,1> <1, 1,1> --------- 2 -1 +1 = 2

OpenStudy (anonymous):

so i said

OpenStudy (anonymous):

david wht grade u in?

OpenStudy (anonymous):

im a sophomore in college. im studying for my calc 3 final

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