Mathematical Methods in the Physical Sciences/Chapter 4/Section 1

Notes

If <math>z = f(x,y)</math>:

• <math>{\partial z \over \partial x}</math> is the partial derivative of z with respect to x.
  • Note that <math>y</math> is held constant, and we're seeing how <math>z</math> changes as we vary <math>x</math>
• <math>{\partial ^2 z \over \partial\, x \partial y}</math> means <math>{\partial \over \partial x}{\partial z \over \partial y}</math>

• <math>f_1 \equiv f_x \equiv z_x \equiv {\partial f \over \partial x} \equiv {\partial z \over \partial x}</math>
• <math>f_{21} \equiv f_{yx} \equiv z_{yx} \equiv {\partial ^2 z \over \partial\, x \partial y} \equiv {\partial \over \partial x}{\partial z \over \partial y}</math>

If <math>z = f(x,y),\ x = g(r, \theta),\ y = h(r, \theta)</math>, then we can rewrite <math>z</math> in any combination of any variable.

• <math>\left ( {\partial z \over \partial x} \right )_r</math> means take z, rewrite it in terms of x and r only, and then take the partial derivative of z with respect to x holding r constant.
• Mathematicians don't like physicists because we re-use the same function name even though we are changing parameters. For instance, we say <math>f(x,y) = f(r, \theta)</math> which means we infer a lot about which <math>f</math> we are talking about.

<math>{\partial^2 f \over \partial x \, \partial y} = {\partial^2 f \over \partial y \, \partial x}</math> when both derivatives are continuous.

Problems

1

<math>

\begin{align} {\partial u \over \partial x}

&= {\partial \over \partial x}\left ({x^2 \over x^2 + y^2} \right ) \\
&= \frac{(x^2 + y^2){\partial \over \partial x}x^2 - x^2{\partial \over \partial x}(x^2 + y^2)}
        {(x^2 + y^2)^2}
&\qquad \left ( {u \over v} \right )' = {v u' - u v' \over v^2} \\
&= {(x^2 + y^2)2x - x^2{\partial \over \partial x}(x^2 + y^2)
       \over (x^2 + y^2)^2} \\
&= {(x^2 + y^2)2x - x^2\left({\partial \over \partial x}x^2 + {\partial \over \partial x}y^2\right)
       \over (x^2 + y^2)^2} \\
&= {(x^2 + y^2)2x - x^2\left(2x + {\partial \over \partial x}y^2\right)
       \over (x^2 + y^2)^2} \\
&= {(x^2 + y^2)2x - x^2(2x + 0)
       \over (x^2 + y^2)^2} \\
&= {(x^2 + y^2)2x - 2x^3 \over (x^2 + y^2)^2} \\
&= {2x^3 + 2xy^2 - 2x^3 \over (x^2 + y^2)^2} \\
&= {2xy^2 \over (x^2 + y^2)^2} \\

\end{align} </math>

<math> \begin{align} {\partial u \over \partial y}

&= {\partial \over \partial y}\left ({x^2 \over x^2 + y^2} \right ) \\
&= \frac{(x^2 + y^2){\partial \over \partial y}x^2 - x^2{\partial \over \partial y}(x^2 + y^2)}
        {(x^2 + y^2)^2} \\
&= \frac{(x^2 + y^2)0 - x^2{\partial \over \partial y}(x^2 + y^2)}
        {(x^2 + y^2)^2} \\
&= \frac{x^2{\partial \over \partial y}(x^2 + y^2)}
        {(x^2 + y^2)^2} \\
&= \frac{x^2\left ({\partial \over \partial y}x^2 + {\partial \over \partial y}y^2 \right )}
        {(x^2 + y^2)^2} \\
&= \frac{x^2\left (0 + {\partial \over \partial y}y^2 \right )}
        {(x^2 + y^2)^2} \\
&= \frac{x^2 (0 + 2y)}
        {(x^2 + y^2)^2} \\
&= \frac{2x^2y}
        {(x^2 + y^2)^2} \\

\end{align} </math>