Functions of several variables

Problems 1.1 : Sketching simple surfaces in $\mathbb{R}^3$

For each of the following surfaces, sketch some level curves and sketch the $yz$-profile (which is found by intersecting the surface with the plane $x = 0$). Hence sketch the surface.

1. e) $z=x^2-y^2$
   Level curves:
   If $z = 0$, then the level curve obtained is simply an X-shape:
   
   As $z$ changes, if $z>0$, then we get a hyperbola on the right and left side of the $y$-axis, if $z<0$ then we get a hyperbola on the top and bottom sides of the $x$-axis.
   For the level curve, if $x$ is treated as a constant, then we simply get $z=-y^2$ which is an upside down parabola, and a normal parabola if $y$ is a constant.
   Therefore, if viewed from above, the curve should simply be an X, if viewed from the $yz$ plane, it should be a parabola:
   
   and if viewed from the $xz$ plane it should be the same thing flipped upside down to give this:
   

Problems 1.2: Partial differentiation

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Problems 1.3: Tangent planes to surfaces

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Problems 1.4: The total differential approximation

9. The specific volume $v$ of a compressible fluid flowing through a section of area $A$ with mean velocity $V$ is given by

$$v = kAV$$

where $k$ is a constant. If $v$ decreases by 5% and $A$ increases by 4%, then estimate the percentage change in $V$.

Notice how the question asks for $V$ and not $v$, so what we can do is either treat the equation as is and find the partials:

$$\begin{align*} \Delta v &\approx \frac{\partial v}{\partial A} \Delta A + \frac{\partial v}{\partial V} \Delta V \\ \Delta v &\approx (kV)\Delta A + (kA)\Delta V \\ \frac{\Delta v}{v} &\approx \frac{kV\Delta A}{kAV} + \frac{kA\Delta V}{kAV} \\ \frac{\Delta v}{v} &\approx \frac{\Delta A}{A} + \frac{\Delta V}{V} \end{align*}$$

We know that $\dfrac{\Delta v}{v}$ is -5% and similarly $\dfrac{\Delta A}{A}$ is 4%, so we get

$$\begin{align*} -5\% &\approx 4\% + \frac{\Delta V}{V} \\ -9\% &\approx \frac{\Delta V}{V} \end{align*}$$

And so the answer is $-9\%$. However, we can use the log trick as well;
since we want the percentage change in $V$, we isolate $V$ to get $V=\dfrac{v}{kA}$. Taking the logarithm of both sides,

$$\begin{align*} \ln{(V)}&=\ln{(\frac{v}{kA})} \\ \ln(V) &= \ln(v) - \ln(k) - \ln{(A)}, \\ \text{Because }&k \text{ is a constant, } d(\ln{k}) \to 0, \\ \frac{dV}{V} &= \frac{dv}{v} - 0 - \frac{dA}{A} \\ \frac{dV}{V} &= (-5\%) - (+4\%) \\ \end{align*}$$

And we get $-9\%$.