6.6 Answer

6.6

1.

(a)

\begin{figure}\begin{center}
\includegraphics[width=6cm]{CALCFIG/Fig6-6-2.eps}
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Then
$\displaystyle \frac{dz}{dt}$ $\displaystyle =$ $\displaystyle \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$  
  $\displaystyle =$ $\displaystyle \frac{2x}{x^2 + y^2}(1 - \frac{1}{t^2}) + \frac{2y}{x^2 + y^2}(2t - 1)$  

(b)

Let $x = t^2, \ y = e^{t}$. Then $z = f(x,y)$.

$\displaystyle \frac{dz}{dt}$ $\displaystyle =$ $\displaystyle \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$  
  $\displaystyle =$ $\displaystyle z_{x}(2t) + z_{y}(e^{t})$  

(c)

Let $x = 2t, \ y = 4t^2$. Then $z = f(x,y)$.

$\displaystyle \frac{dz}{dt}$ $\displaystyle =$ $\displaystyle \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$  
  $\displaystyle =$ $\displaystyle z_{x}(2) + z_{y}(8t)$  

(d)


$\displaystyle \frac{dz}{dt}$ $\displaystyle =$ $\displaystyle \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$  
  $\displaystyle =$ $\displaystyle 2x(-\sin{t}) - 4y(\cos{t})$  

2.

(a)

\begin{figure}\begin{center}
\includegraphics[width=6cm]{CALCFIG/Fig6-6-1.eps}
\end{center}\end{figure}
Then
$\displaystyle \frac{\partial z}{\partial r}$ $\displaystyle =$ $\displaystyle \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$  
  $\displaystyle =$ $\displaystyle \frac{1}{1 + (\frac{x}{y})^2}(-\frac{y}{x^2})(3r^2 - 3s^2) + \frac{1}{1 + (\frac{y}{x})^2}(\frac{1}{x})(6rs)$  
  $\displaystyle =$ $\displaystyle \frac{-y}{x^2 + y^2}(3r^2 - 3s^2) + \frac{x}{x^2 + y^2}(6rs)$  

(b)

$\displaystyle \frac{\partial z}{\partial r}$ $\displaystyle =$ $\displaystyle \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$  
  $\displaystyle =$ $\displaystyle \frac{-\frac{y}{x^2}}{\frac{y}{x}}(2(r-1)) + \frac{\frac{1}{x}}{\frac{y}{x}}(2(r+1))$  
  $\displaystyle =$ $\displaystyle \frac{-2(r-1)}{x} + \frac{2(r+1)}{y}$  
  $\displaystyle =$ $\displaystyle \frac{-2(r-1)}{(r-1)^2 + s^2} + \frac{2(r+1)}{(r+1)^2 + s^2}$  


$\displaystyle \frac{\partial z}{\partial s}$ $\displaystyle =$ $\displaystyle \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$  
  $\displaystyle =$ $\displaystyle \frac{-\frac{y}{x^2}}{\frac{y}{x}}(2s) + \frac{\frac{1}{x}}{\frac{y}{x}}(2s)$  
  $\displaystyle =$ $\displaystyle \frac{-2s}{x} + \frac{2s}{y}$  
  $\displaystyle =$ $\displaystyle \frac{-2s}{(r-1)^2 + s^2} + \frac{2s}{(r+1)^2 + s^2}$  

3.


$\displaystyle \frac{\partial z}{\partial r}$ $\displaystyle =$ $\displaystyle \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}$  
  $\displaystyle =$ $\displaystyle z_{x}\cos{\theta} + z_{y}\sin{\theta}$  
$\displaystyle \frac{\partial z}{\partial \theta}$ $\displaystyle =$ $\displaystyle \frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta}$  
  $\displaystyle =$ $\displaystyle z_{x}(-r\sin{\theta}) + z_{y}(r\cos{\theta})$  
  $\displaystyle =$ $\displaystyle r(-z_{x}\sin{\theta} + z_{y}\cos{\theta})$