This may be off topic, but a lot of smart and practical people hang out here so maybe someone can help.
I need to find the height of my chimny in my back yard. I'm not wild about heights, so I'd rather not get out a long tape measure and a longer ladder to climb to the top.
All the word...
The Problem;
Given H = U + PV and dU = TdS - PdV
Find dH in terms of T, S, P, V
My Solution;
H = U + PV
dH = dU + PdV + VdP
dH = (TdS - PdV) + PdV + VdP
dH = Tds + VdP
My Question
Am I missing a step between the first and second steps? I'm taking the derivative of both sides...
Matt,
Thanks for the reply. One bit of clarification;
a and b are integers and a^2 = b
Is a the perfect square?
So if I was asked to find the perfect square of b, then the answer would be a?
Sorry for all the questions, but I'm struggling with the english syntax of this problem. I...
Homework Statement
X mod m is the remainder when x is divided by m. This value is called a residue. Find all perfect squares from the set of residues mod 16.
The Attempt at a Solution
There was a suggestion that this would become clearer when the definition of perfect square was...
The proof is definitely <=, but I think I found the justification. I don't have my notes handy, but I think it's disjunctive addtion justifies that if p then (p or q). Also if (p or q) and (~q) then p.
Bernie
OK, here we go;
Prove:
If a>0, b>0, c>0 and a + b > c, b + c > a, c + a > b then ( a + b + c )^2 <= 4( ab + bc + ca ).
Hints: Start with ( a + b + c )^2 and establish the inequality ( a + b + c )^2 <= 4( ab + bc + ca ). Us the inequality fact: if x < y then xy < y^2 if y > 0.
(a+b+c)^2...
StatusX,
Cancel out which like terms? Expanding out the square gives;
( a + b + c )^2 = a^2 + b^2 + c^a + 2ab + 2bc + 2ca
Which of these terms can be canceled?
Bernie
ACM,
Thanks for the quick reply. I think there is a flaw in your logic. The statements workout to;
2(a + b + c)^2 > (a + b + c)^2 > 2(ab + bc + ac)
2(a + b + c)^2 > 4(ab + bc + ac) > 2(ab + bc + ac)
Therefore 4(ab + bc + ac) > (a + b + c)^2
Which is saying that:
x > y
x > z...
Wouldn't this solve easier with u substituion?
u = 1+y^2
du = 2y dy
Solution is 1/3(1+y^2)^3/2
If you want integration by parts, multiply it out and make u to be y so that du is 1.
By trig you'll need to substitute tan because the problem is + Y^2.
Bernie
The Problem;
If a>0, b>0, c>0 and a + b > c, b + c > a, c + a > b then ( a + b + c )^2 <= 4( ab + bc + ca ).
Hints: Start with ( a + b + c )^2 and establish the inequality ( a + b + c )^2 <= 4( ab + bc + ca ). Us the inequality fact: if x < y then xy < y^2 if y > 0.
I haven’t make...
Argh, Thud, thud, thud ...
(The sound of beating my head on the desk again.(
That's the second time I made a units mistake last night. My montra for today will be "Check the units!"
Bernie