- #1

fatpotato

- Homework Statement
- Show that if ##\frac{\partial g}{\partial u} = \frac{\partial f}{\partial v}## then ##F(x,y) = \nabla \phi(x,y)## with ##\phi(x,y) = \int_0^1 xf(tx,ty) + yg(tx,ty) dt##

- Relevant Equations
- Basic theory of fields and potentials.

Hello,

I have to show that, given a vector field ##F## such that ## F : \mathbb{R}^2 \rightarrow \mathbb{R}^2##, ##F(u,v) = (f(u,v), g(u,v))##, computing the derivatives ##\frac{\partial g}{\partial u} = \frac{\partial f}{\partial v}## must imply that ##F(x,y) = \nabla \phi(x,y)##, with ##\phi## defined as :

$$\phi(x,y) = \int_0^1 xf(tx,ty) + yg(tx,ty) dt$$

The solution manual suggests "observing" that :

$$ \frac{\partial}{\partial t} t f(tx,ty) = f(tx,ty) + t [ x \frac{\partial f}{\partial u} + y \frac{\partial f}{\partial v} ]$$

And then "finding" that :

$$ \frac{\partial}{\partial x} x f(tx,ty) + y g(tx,ty) = f(tx,ty) + tx \frac{\partial f}{\partial u} + ty \frac{\partial g}{\partial u} $$

Are the same when we suppose equality between the cross partial derivatives, so in the end, differentiating the integrand of ##\phi## can be expressed with respect to ##t## to find :

$$ \frac{\partial \phi} {\partial x} = \int_0^1 \frac{\partial} {\partial t} t f(tx,ty) dt = f(x,y)$$

Is it supposed to be obvious if you previously never encountered this kind of problem? I feel that I would not be able to find this clever observation in an exam by myself...In books like Schaum, problems are usually way easier and (alas) made to train pure computation and I don't know where to look for similar problems.

So here are my questions :

**To first clarify what I want to know**: I read the answer proposed from the solution manual and I understand it. What I want to understand is**how**they came up with the solution, and if there is a way to get better at this.I have to show that, given a vector field ##F## such that ## F : \mathbb{R}^2 \rightarrow \mathbb{R}^2##, ##F(u,v) = (f(u,v), g(u,v))##, computing the derivatives ##\frac{\partial g}{\partial u} = \frac{\partial f}{\partial v}## must imply that ##F(x,y) = \nabla \phi(x,y)##, with ##\phi## defined as :

$$\phi(x,y) = \int_0^1 xf(tx,ty) + yg(tx,ty) dt$$

The solution manual suggests "observing" that :

$$ \frac{\partial}{\partial t} t f(tx,ty) = f(tx,ty) + t [ x \frac{\partial f}{\partial u} + y \frac{\partial f}{\partial v} ]$$

And then "finding" that :

$$ \frac{\partial}{\partial x} x f(tx,ty) + y g(tx,ty) = f(tx,ty) + tx \frac{\partial f}{\partial u} + ty \frac{\partial g}{\partial u} $$

Are the same when we suppose equality between the cross partial derivatives, so in the end, differentiating the integrand of ##\phi## can be expressed with respect to ##t## to find :

$$ \frac{\partial \phi} {\partial x} = \int_0^1 \frac{\partial} {\partial t} t f(tx,ty) dt = f(x,y)$$

Is it supposed to be obvious if you previously never encountered this kind of problem? I feel that I would not be able to find this clever observation in an exam by myself...In books like Schaum, problems are usually way easier and (alas) made to train pure computation and I don't know where to look for similar problems.

So here are my questions :

- Is this a typical more elaborate exercise?
- Where can I find books to exercise myself at solving this type of problems?

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