6.6 Answer

6.6

1.

(a)

$\begin{figure}\begin{center} \includegraphics[width=6cm]{CALCFIG/Fig6-6-2.eps} \end{center}\end{figure}$

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.

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