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So I have a homework problem that I cannot figure out. I am supposed to approximate the value of $\sqrt{(4.98)^2-(3.03)^2}$ using differentials. What I have so far is $f(x,y)=\sqrt{x^2-y^2}$ $\Delta f=f(x+\Delta x,y+\Delta y)-f(x,y)$ $df= \frac {\delta f}{\delta x}dx+\frac {\delta f}{\delta y}dy$ Can I do this? $ df=\frac{x}{\sqrt{x^2-y^2}}dx-\frac {y}{\sqrt{x^2-y^2}}dy$ $\sqrt{(5-.02)^2-(4-.97)^2}$ $ df=\frac{5}{\sqrt{5^2-4^2}}(-.02)-\frac {4}{\sqrt{5^2-4^2}}(-.97)$ $df=\frac{-5}{3}(.02)+\frac 43(.97)$ $df\approx 1.26 $

I have the solutions manual and it says the answer should be $3.95$.

What did I do wrong, and how can I get to their answer?

I appreciate any help that anyone has.

2 Answers 2

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Looking at it again I realized this $ df=\frac{x}{\sqrt{x^2-y^2}}dx-\frac {y}{\sqrt{x^2-y^2}}dy$ With $x=5,\Delta x=-.02,y=3,\Delta y=.03 $ $f(x+\Delta x,y+\Delta y)=\sqrt{(5-.02)^2-(3+.03)^2}$ $ df=\frac{5}{\sqrt{5^2-3^2}}(-.02)-\frac {3}{\sqrt{5^2-3^2}}(.03)$ $df=-.0475$ $df\approx\Delta f$ $f(x+\Delta x,y+\Delta y)=\Delta f+f(x,y)$ $f(x+\Delta x,y+\Delta y)\approx3.9525$

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    That's what I'm saying..2012-11-11
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So, in your example $x=5,\ \Delta x=-.02$, and I guess $y=3$ and $\Delta y=.03$. So, try with $y=3$.

Aand.. you should clarify the question. Maybe they want only to compute $\sqrt{(4.98)^2-(3.03)^2} - \sqrt{5^2-3^2}$.