# Calculus 3 - Laplace Equation

• Oct 5th 2011, 05:07 PM
Spoolx
Calculus 3 - Laplace Equation
Hi all
Ran into a problem on a quiz that for the life of me I thought I was solving correctly but the proof wasnt working.

Anyhow the question was
Prove that
Fxx + Fyy = 0

F(x,y) = LN(SQRT(X^2+Y^2))

My solution

Property of logs converted my problem to
1/2 LN(X^2+Y^2)
Derived with respect to X = Fx (x,y) = x/(x^2+y^2)

Derived with respect to Y = Fy (x,y) = y/(x^2+y^2)

Derived Fx with respect to X = Using the power rule for the function (x(x^2+y^2)^-1)
= (1)(x^2+y^2)^-1) + (x)(2x)(-1)(x^2+y^2)^-2)
=1/(x^2+y^2) - 2x^3/(x^2+y^2)^2)

So Fxx = 1/(x^2+y^2) - 2x^3/(x^2+y^2)^2)

Derived Fy with respect to y = Using the power rule for the function (y(x^2+y^2)^-1)
= (1)(x^2+y^2)^-1) + (y)(2y)(-1)(x^2+y^2)^-2)
=1/(x^2+y^2) - 2y^3/(x^2+y^2)^2)

So Fyy = =1/(x^2+y^2) - 2y^3/(x^2+y^2)^2)

If I do Fxx + Fyy I dont think I get 0.

Any suggestions on where I went wrong please!

Thanks
• Oct 5th 2011, 05:29 PM
Ackbeet
Re: Calculus 3 - Laplace Equation
I don't think your second derivatives are correct. Check this out. I usually prefer to use the quotient rule, because, while it may be a little more calculus up front, it can often be a lot less algebra later. In particular, you don't have to find a common denominator when you use the quotient rule.
• Oct 5th 2011, 05:38 PM
Spoolx
Re: Calculus 3 - Laplace Equation
Quote:

Originally Posted by Ackbeet
I don't think your second derivatives are correct. Check this out. I usually prefer to use the quotient rule, because, while it may be a little more calculus up front, it can often be a lot less algebra later. In particular, you don't have to find a common denominator when you use the quotient rule.

That helps alot, but I still dont see where I went wrong with the product rule. I keep doing it and keep getting the same wrong answer.
• Oct 5th 2011, 05:48 PM
Ackbeet
Re: Calculus 3 - Laplace Equation
Quote:

Originally Posted by Spoolx
Hi all
Ran into a problem on a quiz that for the life of me I thought I was solving correctly but the proof wasnt working.

Anyhow the question was
Prove that
Fxx + Fyy = 0

F(x,y) = LN(SQRT(X^2+Y^2))

My solution

Property of logs converted my problem to
1/2 LN(X^2+Y^2)
Derived with respect to X = Fx (x,y) = x/(x^2+y^2)

Derived with respect to Y = Fy (x,y) = y/(x^2+y^2)

Derived Fx with respect to X = Using the power rule for the function (x(x^2+y^2)^-1)
= (1)(x^2+y^2)^-1) + (x)(2x)(-1)(x^2+y^2)^-2)
=1/(x^2+y^2) - 2x^3/(x^2+y^2)^2)

Should be 1/(x^2+y^2) - 2x^2/(x^2+y^2)^2).

Get a common denominator and simplify.

Quote:

So Fxx = 1/(x^2+y^2) - 2x^3/(x^2+y^2)^2)

Derived Fy with respect to y = Using the power rule for the function (y(x^2+y^2)^-1)
= (1)(x^2+y^2)^-1) + (y)(2y)(-1)(x^2+y^2)^-2)
=1/(x^2+y^2) - 2y^3/(x^2+y^2)^2)
Same mistake here.

Quote:

So Fyy = =1/(x^2+y^2) - 2y^3/(x^2+y^2)^2)

If I do Fxx + Fyy I dont think I get 0.

Any suggestions on where I went wrong please!

Thanks