Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

If z= sqrt[(x^2)y + x(y^2)] and (x, y) changes from (5, 20) to (5.2, 20.4), compare the values of delta z (triangle z) and dz.

OpenStudy (anonymous):

Okay, I'll take a quick crack at it

OpenStudy (anonymous):

kk

OpenStudy (anonymous):

First off, what is this question for?

OpenStudy (anonymous):

What kind of class?

OpenStudy (anonymous):

calculus 2

OpenStudy (anonymous):

Oohhhh... F... Okay, I'm glad I asked. I'm only in Trig. Coincidentally, my teacher was saying how important trig was for calc 2, but I'm skeptical I can be helpful.

OpenStudy (anonymous):

Considering how dead it is here, I doubt anyone else is on btw...

OpenStudy (anonymous):

I actually came for help on my own homework, but noone responded :P

OpenStudy (anonymous):

thas cool ill get it later

OpenStudy (anonymous):

Hmm... I've my brother, who took calc 2 is up, actually. Lemme Harass er ...ask him.

OpenStudy (anonymous):

What does d in dz mean, btw?

OpenStudy (anonymous):

yo nvm. ill post it again later i think i got it. thnxs

OpenStudy (anonymous):

Alrighty. Take it easy. I'm gonna try to finish my homework now. <_< ^^

OpenStudy (anonymous):

if some1 can solve it that would b good to check my answer thnks

OpenStudy (anonymous):

I'm working on it

OpenStudy (anonymous):

ok thanks

OpenStudy (anonymous):

What did you get for dz?

OpenStudy (anonymous):

i stopped working on it i moved onto the next question

OpenStudy (anonymous):

well you know x=5 y=20 dx=.2 and dy=.4 differentiate z to get dz and then you can plug in the values of x,y,dx,dy resulting in dz

OpenStudy (anonymous):

can u show me the full solution?

OpenStudy (anonymous):

do you know what the correct answer is? just want to check it with mine before I load the full solution

OpenStudy (anonymous):

no i actually dont

OpenStudy (anonymous):

\[dz=[1/2(x^2y+xy^2)^{-1/2}(2xy+y^2)]dx\] \[+ [1/2(x^2y+xy^2)^{-1/2}(x^2+2xy)]\] \[x=5, dx=.2, y=20, dy=.4\] Plug these values in and let me know what you get

OpenStudy (anonymous):

forgot the dy, sorry \[[1/2(x^2y+xy^2)−1/2(x^2+2xy)]dy\]

OpenStudy (anonymous):

gotdamit i messed it up again hold on

OpenStudy (anonymous):

\[+[1/2(x^2y+xy^2)^{−1/2}(x^2+2xy)]dy\]

OpenStudy (anonymous):

Here is the entire equation with any typos, sorry for the confusion \[dz=[1/2(x^2y+xy^2)^{−1/2}(2xy+y^2)]dx\] \[+[1/2(x^2y+xy^2)^{−1/2}(x^2+2xy)]dy\] \[x=5,dx=.2,y=20,dy=.4\]

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!