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## Main Question or Discussion Point

Please help me on this. I am trying to make and exercise from an author M.D. Hatton (an english).

Let x = x(r, w) = r. cos (w)

Let y = y(r,w) = r. sen (w)

Let V = V(x,y). So V depends on r and w.

By chain rule (I put "d" for the partial derivative)

dV = dV . dx + dV. dy

-- -- -- -- --

dr dx dr dy dr

So

dV = dV . cos(w) + dV. sen(w) (A)

-- -- --

dr dx dy

And the chain rule also says that

dV = dV . dx + dV. dy

-- -- -- -- --

dw dx dw dy dw

so

dV = - r dV . sen(w) + r. dV. cos(w) (B)

-- -- --

dw dx dy

Now Hatton says: "solving these equation for dV/dx and dV/dy, we find:

dV = dV cos(w) - 1 dV . sen(w). (C)

-- -- -- . --

dx dr r dw

and

dV = dV sen(w) + 1. dV . cos(w). (D)".

-- -- -- --

dy dr r dw

I do not understand. How does (A) produces (C)? How does (B) produces (D)????

Remember that all are partial derivatives. Thanks for the help.

P. Castilla.

Let x = x(r, w) = r. cos (w)

Let y = y(r,w) = r. sen (w)

Let V = V(x,y). So V depends on r and w.

By chain rule (I put "d" for the partial derivative)

dV = dV . dx + dV. dy

-- -- -- -- --

dr dx dr dy dr

So

dV = dV . cos(w) + dV. sen(w) (A)

-- -- --

dr dx dy

And the chain rule also says that

dV = dV . dx + dV. dy

-- -- -- -- --

dw dx dw dy dw

so

dV = - r dV . sen(w) + r. dV. cos(w) (B)

-- -- --

dw dx dy

Now Hatton says: "solving these equation for dV/dx and dV/dy, we find:

dV = dV cos(w) - 1 dV . sen(w). (C)

-- -- -- . --

dx dr r dw

and

dV = dV sen(w) + 1. dV . cos(w). (D)".

-- -- -- --

dy dr r dw

I do not understand. How does (A) produces (C)? How does (B) produces (D)????

Remember that all are partial derivatives. Thanks for the help.

P. Castilla.