Differentials and True Change

Question

If z=x^2+y^2, find both dz and delta z, as well as the error that one commits if dz is used instead of delta z.

So far I have dz=2xdx+2ydz.

However, I am not sure how to calculate delta z. Also not sure what the error would be.

Any help would be great, thanks.

score 0 · Accepted Answer · answered Nov 06 '15 at 18:48

0

Using function notation, Let $f(x,y)=z=x^2+y^2$.
Then $\Delta z=f(x+\Delta x, y+\Delta y)-f(x,y)$.

To compare $dz$ and $\Delta z$, use $dx=\Delta x$ and $dy=\Delta y$.

answered Nov 06 '15 at 18:48

paw88789

So would Δz=(x+dx)(y+dy)-(x^2+y^2)? – j.stat Nov 06 '15 at 19:38
Actually $\Delta z=((x+dx)^2 +(y+dy)^2)-(x^2+y^2)$. – paw88789 Nov 06 '15 at 22:13
Ok, therefore, Δz=2xdx+2ydy+(dx^2+dy^2). So the error would just be (dx^2+dy^2)? Thanks – j.stat Nov 07 '15 at 14:58

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