1. ## Partial Diff

A closed rectangular box with unequal sides a, b, c has its edges slightly altered in length by amounts (delta a), (delta b), (delta c), respectively so that both its volume and surface area remained unchanged. Show that:

(delta a)/a^2(c-b) = (delta b)/b^2(a-c) = (delta c)/c^2(b-a)

How can i solve this question?

2. This is about relating differentials. Our volume is $V=abc$ and our surface area $S=2(ab+bc+ac)$.
Our volume and surface area do not change; this means $dV=0$ and $dS=0$.
Now, $dV=bc\,da+ac\,db+ab\,dc$
Thus $bc\,da+ac\,db+ab\,dc=0$
Dividing by the volume abc, we get
$\frac{da}{a}+\frac{db}{b}+\frac{dc}{c}=0$ (1)
Also:
$dS=2(a\,db+b\,da+b\,dc+c\,db+a\,dc+c\,da)$
$dS=2((b+c)da+(a+c)db+(a+b)dc)$
Thus:
$2((b+c)da+(a+c)db+(a+b)dc)=0$
$(b+c)da+(a+c)db+(a+b)dc=0$ (2)

Solving (1) and (2) both for dc and combining:
$-\frac{(b+c)da+(a+c)db}{a+b}=dc=-c\left(\frac{da}{a}+\frac{db}{b}\right)$
$ab((b+c)da+(a+c)db)=c(b(a+b)da+a(a+b)db)$
$b^2(a-c)da=a^2(c-b)db$
Dividing by $a^2b^2(c-b)(a-c)$, we get
$\frac{da}{a^2(c-b)}=\frac{db}{b^2(a-c)}$

Similar solving of (1) and (2) for da and db gives the other equalities.

-Kevin C.

3. why do i have to divide by V=abc to get eqn (1)?
is it a type of differentiation rule?
or

i suppose its a way of simplification?===> dividing both side by abc

4. Originally Posted by MilK
why do i have to divide by V=abc to get eqn (1)?
is it a type of differentiation rule?
or

i suppose its a way of simplification?===> dividing both side by abc Mr F says: Yes.
Beaten to the punch. I'd written it on a napkin and was just about to type it in. Probably all for the best since I spilt soup du jour on the napkin - some soup spots can look just like differentials .....

5. Thanks guys