Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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 $ $$f(x+\Delta x,y+\Delta y)=\sqrt{(5-.02)^2-(3+.03)^2}$$ $$ df=\frac{5}{\sqrt{5^2-3^2}}(-.02)-\frac {3}{\sqrt{5^2-3^2}}(.03)$$ $$df=-.0475$$ $$df\approx\Delta f$$ $$f(x+\Delta x,y+\Delta y)=\Delta f+f(x,y)$$ $$f(x+\Delta x,y+\Delta y)\approx3.9525$$

share|cite|improve this answer
That's what I'm saying.. – Berci Nov 11 '12 at 15:28

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 $\sqrt{(4.98)^2-(3.03)^2} - \sqrt{5^2-3^2}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.