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}

\qquad

\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>

2

<math> \begin{align} {\partial s \over \partial t} &= {\partial \over \partial t}(t^u) \\ &= ut^{u-1} \\ \end{align}

\qquad

\begin{align} {\partial s \over \partial u} &= {\partial \over \partial u}(t^u) \\ &= t^u \ln t \\ \end{align} </math>

3

<math> \begin{align} {\partial z \over \partial u} &= {\partial \over \partial u}\ln \sqrt{u^2 + v^2 + w^2} \\

&= {\partial \over \partial u}\ln x & \qquad & \text{Let } x = \sqrt{u^2 + v^2 + w^2} \\

&= {\partial \over \partial x} \ln x {\partial \over \partial u} x & \qquad & \text{Chain rule.} \\

&= {{\partial \over \partial u} x \over x} & \qquad & {d \over dx}\ln x = {1 \over x} \\

&= {{\partial \over \partial u} \sqrt{u^2 + v^2 + w^2} \over \sqrt{u^2 + v^2 + w^2}} \\

&= {{\partial \over \partial u} \sqrt{y} \over \sqrt{y}} & \qquad & \text{Let } y = u^2 + v^2 + w^2 \\

&= {{\partial \over \partial y} \sqrt{y} {\partial \over \partial u} y \over \sqrt{y}} & \qquad & \text{Chain rule} \\

&= {{\partial \over \partial u} y \over \sqrt{y}\sqrt{y}} \\

&= {{\partial \over \partial u} y \over y} \\

&= {{\partial \over \partial u}( u^2 + v^2 + w^2) \over u^2 + v^2 + w^2} \\

&= {{\partial \over \partial u}u^2 + {\partial \over \partial u}v^2 + {\partial \over \partial u}w^2 \over u^2 + v^2 + w^2} \\

&= {2u \over u^2 + v^2 + w^2} \\ \end{align} </math>

By similar logic, <math>{\partial z \over \partial v} = {2v \over u^2 + v^2 + w^2},\ {\partial z \over \partial w} = {2w \over u^2 + v^2 + w^2}</math>.

4

There are quite a few points in this problem, so let's not allow ourselves to get confused.

Let's solve for the first and second derivatives.

<math>

\begin{align} w &= x^3 - y^3 -2xy + 6 \\ {\partial w \over \partial x} &= 3x^2 - 2y \\ {\partial^2 w \over \partial x^2} &= 6x \\ {\partial w \over \partial y} &= -3y^2 - 2x \\ {\partial^2 w \over \partial y^2} &= -6y \\ \end{align} </math>

Solving for <math>{\partial w \over \partial x} = {\partial w \over \partial y} = 0</math>:

<math>

\begin{align} 3x^2 - 2y &= 0 & \qquad & x=0, y=0 \text{ is a trivial solution.} \\

y &= {3 \over 2}x^2 \\

-3y^2 - 2x &= 0 \\

-3 \left ({3 \over 2}x^2 \right )^2 - 2x &= 0 \\

\left ({-27 \over 4} \right ) x^4 - 2x &= 0 \\

\left ({-27 \over 4} \right ) x^3 - 2 &= 0 \\

x &= -\sqrt[3]{8 \over 27} \\

x &= -{2 \over 3} \\

3\left (-{2 \over 3} \right )^2 - 2y &= 0 \\

{4 \over 3} - 2y &= 0 \\

y &= {2 \over 3} \\ \end{align} </math>

Plugging in:

<math>

\begin{align} (x,y) &= \{(0,0), (-2/3, 2/3)\} \\

{\partial^2 w \over \partial x^2}(0,0) &= 0 \\

{\partial^2 w \over \partial x^2}\left (-{2\over 3}, {2\over 3} \right) &= -4 \\

{\partial^2 w \over \partial y^2}(0,0) &= 0 \\

{\partial^2 w \over \partial y^2}\left (-{2\over 3}, {2\over 3} \right) &= -4 \\ \end{align} </math>

7

For 7-24, we need to find <math>z</math> for all pairs of <math>x, y, r, \theta</math>

Pull out your trig identities!

<math> \begin{align} x &= r \cos \theta \\ r &= x \sec \theta \\ y &= r \sin \theta \\ r &= y \csc \theta \\ r^2 &= x^2 + y^2 \\ z &= x^2 + 2y^2 & x,y\\

 &= r^2 + y^2 & r,y\\
 &= y^2 (\csc^2 \theta + 1) & y,\theta\\
 &= 2r^2 - x^2 & r,x \\
 &= x^2(2\sec^2 \theta - 1) & x, \theta \\
 &= r^2(1+\sin^2 \theta) & r, \theta \\

\end{align} </math>

That was half the fun. The other half is remembering the derivatives of trigonometric functions.

7'

Let's re-derive our 6 combinations.

<math> \begin{align} z &= r^2 \tan^2 \theta & r,\theta \\

 &= (x^2 + y^2)\left ({y^2 \over x^2}\right ) & \qquad & \qquad \text{Substitute } r^2 = x^2 + y^2,\ \sin \theta = y/r, \text{ etc...} \\
 &= {y^2(x^2 + y^2) \over x^2} & x,y \\
 &= (x \sec \theta)^2 tan^2 \theta & \qquad & \qquad \text{Substitute }r=x \sec \theta \\
 &= (x \sec \theta \tan \theta)^2 & x,\theta \\ 
 &= (y \csc \theta)^2 tan^2 \theta & \qquad & \qquad \text{Substitute }r=y \csc \theta \\
 &= (y \sin \theta)^2 & y, \theta \\ 
 &= {r^2 (r^2 - x^2) \over x^2} & r, x & \qquad \text{Substitute }\sin^2 \theta = 1 - \cos^2 \theta, \text{ etc...} \\
 &= {r^2 y^2 \over r^2 - y^2} & r, y & \qquad \text{Idem.} \\

\end{align} </math>