Find the values of Δz and dz

[carl123]

If z = x² − xy + 9y² and (x, y) changes from (2, −1) to (1.96, −1.05), compare the values of Δz and dz. (Round your answers to four decimal places.)

This is what I have so far:

dx = Δx = -0.04
dy = Δy = -0.05

Zy = 2x-y
Zx = 18y-x

dz = Zx (2,-1)dx + Zy (2,-1)dy

dz = -0.03 - 0.86 = -0.89

It says my answer for dz is wrong, I can't figure out what I'm doing wrong

 

[Euge]

Hi Carl123,

When you have an answer that disagrees with the answer given in your text, please let us know first the answer in the textbook.

This is what I have so far:
Zy = 2x-y
Zx = 18y-x

I believe you accidentally switched the formulas for $Z_y$ and $Z_x$; it should be $Z_x = 2x - y$ and $Z_y = 18y - x$.

dz = Zx (2,-1)dx + Zy (2,-1)dy

dz = -0.03 - 0.86 = -0.89

Well, $Z_x(2,-1) = 2(2) - (-1) = 4 + 1 = 5$ and $Z_y(2,-1) = 18(-1) - 2 = -18 - 2 = -20$, so

$$dz = Z_x(2,-1)(-0.04) + Z_y(2,-1)(-0.05) = (5)(-0.04) + (-20)(-0.05) = -0.2 + 1 = 0.8.$$

 

[carl123]

Thanks for your reply, the question is not from a text, it's from an online homework. I don't know the answer myself but it marked me wrong

 

[Euge]

Ok, thanks for the clarification. But I believe I've given you the proper correction in my last post.

 

[carl123]

Thanks. Do I follow the same process to get Δz?

 

[Euge]

You just need to compute $z(1.96,-1.05) - z(2,-1)$.

 

[carl123]

I initially got -0.8221 as my answer for Δz and it was wrong but then i realized I had it switched, I switched it over and the answer came out to be 0.8221 which is correct. Thanks a lot for your help.