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\title{ESP Workshop, Worksheet \#17}
\date{\today}
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\begin{center}
{\bf \Large ESP Workshop, Worksheet \#17}

{\bf \large Thursday \today}

{\bf \large AI: Eric Katerman}
\end{center}
\vspace{.15in}
\begin{enumerate}
\item Suppose that $f(x,y)$ has continuous partial derivatives $f_x$
  and $f_y$ at some point $(x_0, y_0, z_0)$ where $z_0 = f(x_0,
  y_0)$. Last week, Dr. Durbin told us that the equation of the
  tangent plane to the surface $z=f(x,y)$ (that is, the graph of
  $f(x,y)$ in $xyz$-space) at the point $(x_0, y_0, z_0)$ is
\begin{displaymath}
z-z_0 = f_x(x_0,y_0) (x-x_0) + f_y (x_0,y_0) (y-y_0)
\end{displaymath}
Find an equation of the tangent plane to the given surface at the
specified point:
\begin{enumerate}
\item $z=\sqrt{4-x^2 -2y^2},\ (1,-1,1)$
\item $z= y\ln x,\ (1,4,0)$
\item $z=y \cos (x-y),\ (2,2,2)$
\item $z=e^{x^2 -y^2},\ (1,-1,1)$
\end{enumerate}

\item A differential $dz$ approximates the change in the value of a
  function $z= f(x,y)$ in terms of the changes in the variables, $dx =
  \Delta x$ and $dy = \Delta y$. For functions of two variables,
\begin{displaymath}
dz = f_x (x,y) dx + f_y (x,y) =\frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy
\end{displaymath}
(The three-variable case is similar.) Find the differentials of the
following functions.
\begin{enumerate}
\item $z = e^x \sin y$
\item $u = r/(s+2t)$
\item $w = \ln \sqrt{x^2 +y^2 +z^2}$
\item $w =xye^{xz}$
\end{enumerate}

\item (An Electrical Engineering problem!) Now we will apply the above
  theory to a real-life situation! If $R$ is the total resistance of
  the three resistors, connected in parallel, with resistances $R_1,
  R_2, R_3$, then
\begin{displaymath}
\frac{1}{R} = \frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3}
\end{displaymath}

\begin{enumerate}
\item Rewrite this equation to make $R$ a function of $R_1,R_2,$ and
  $R_3$. (All I mean is that you should invert both sides and possibly
  simplify to get $R = f(R_1,R_2,R_3)$.)

\item Use your solution to part (a) to find $\partial R / \partial
  R_1$, i.e. the partial derivative of the function $R$ with respect
  to the variable $R_1$.

\item Assume for the moment that $R_3 = \infty$, i.e. there are only
  two resistors connected in parallel. Then $R$ is a function of two
  variables, $R_1$ and $R_2$. Find the equation of a tangent plane to
  the surface $z = f(R_1,R_2)$ at the point $(2,4,4/3)$. How can you
  use this to estimate the total resistance if you change $R_1$ and
  $R_2$ a little bit?

\item Now back to the three-variable case: if the resistances are
  measured in ohms as $R_1 = 25\ \Omega$, $R_2 = 40\ \Omega$, and $R_3
  = 50\ \Omega$, with a possible error of $0.5 \%$ in each case,
  estimate the maximum error in the calculated value of $R$.
\end{enumerate}

\item You all have seen the chain rule for real-valued functions of
  one variable:
\begin{displaymath}
(f\circ g)' (x) = f'(g(x)) g'(x)
\end{displaymath}
This is a special case of the general version of the chain rule, which
we'll ignore for now. (Maybe we'll look at it on Tuesday.) Another
special case is when $f = f(x,y)$ is a function of two variables,
where $x = x(t)$ and $y = y(t)$ are functions of a single variable
$t$. In this case, the chain rule says that
\begin{displaymath}
\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}
\end{displaymath}
Remember that $\partial$ indicates that you're taking a partial
derivative. The left-hand side of the above equation should not be
confusing since $f$ really is just a function of $t$. Use the chain
rule to find $\frac{df}{dt}$ for the following functions:
\begin{enumerate}
\item $f(x,y) = x^2 y + xy^2,\ x=2+t^4,\ y=1-t^3$
\item $f(x,y) = \sqrt{x^2 +y^2},\ x=e^{2t},\ y=e^{-2t} $
\item $z = x\ln (x+2y),\ x=\sin t,\ y=\cos t$
\item Recall that $PV=8.31T$, where $P$ is pressure (in kilopascals),
  $V$ is volume (in liters), and $T$ is temperature (in kelvins) of a
  mole of an ideal gas. If the pressure is increasing at a rate of
  0.05 kPa/s and the temperature is increasing at a rate of 0.15 K/s,
  find the rate of change of the volume when the pressure is 20 kPa
  and the temperature is 320 K.
\end{enumerate}

Now suppose that $f = f(x,y)$ is still a function of two variables,
but now the variables $x = x(s,t)$ and $y = y(s,t)$ are both functions
of two variables, $s$ and $t$. In this situation, the chain rule says
\begin{displaymath}
\frac{\partial f}{\partial s} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial s} \quad \textrm{and}\quad \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t}
\end{displaymath}
Use this to find the indicated partial derivatives.
\begin{enumerate}
\item $f(x,y) = x^2 +xy^3,\ x=uv^2 +v^3,\ y=u+ve^u;\ \frac{\partial f}{\partial u}, \frac{\partial f}{\partial v}$ when $u=2, v=1$

\item $R = \ln (u^2 + v^2 +w^2),\ u=x+2y,\ v = 2x -y,\ w=2xy;\ \frac{\partial R}{\partial x}, \frac{\partial R}{\partial y}$ when $x = y = 1$
\end{enumerate}

\end{enumerate}

\end{document}