2.2 Partial Derivatives

Review Definition 2.2.1 in the CLP-3 text.

What happens if you move “backwards," in the negative $y$ direction?

No: you can go higher by moving in the negative $y$ direction.

If $f_y(0,0)<0$, then $f(0,y)$ decreases as $y$ increases from $0$. Thus moving in the positive $y$ direction takes you downhill. This means you aren't at the lowest point in a valley, since you can still move downhill. On the other hand, as $f_y(0,0)<0$, $f(0,y)$ also decreases as $y$ increases towards $0$ from slightly negative values. Thus if you move in the negative $y$-direction from $y=0$, your height $z$ will increase. So you are not at a locally highest point—you're not at a summit.

For (a) and (b), remember $\pdiff{f}{x}(x,y)=\lim\limits_{h\to0}\frac{f(x+h,y)-f(x,y)}{h}$ and $\pdiff{f}{y}(x,y)=\lim\limits_{h\to0}\frac{f(x,y+h)-f(x,y)}{h}$. For (c), you're finding the derivative of a function of one variable, say $g(t)$, where

(a) $0$ (b) $0$ (c) $\frac{1}{2}$

(a) By definition

(b) By definition

(c) By definition

(a)

(b)

(c)

(a)

(b)

(c)

Just evaluate $x\pdiff{z}{x}(x,y)+y\pdiff{z}{y}(x,y)$.

See the solution.

By the quotient rule

Hence

(a) $\pdiff{z}{x} = \frac{z(1-x)}{x(yz-1)}$, $\pdiff{z}{y} = \frac{z(1+y-yz)}{y(yz-1)}$

(b) $\pdiff{z}{x}(-1,-2) =\frac{1}{2}$, $\pdiff{z}{y}(-1,-2) =0$.

(a) We are told that $z(x,y)$ obeys

for all $(x,y)$ (near $(-1,-2)$). Differentiating $(*)$ with respect to $x$ gives

or, dropping the arguments $(x,y)$ and multiplying both the numerator and denominator by $xz$,

Differentiating $(*)$ with respect to $y$ gives

or, dropping the arguments $(x,y)$ and multiplying both the numerator and denominator by $yz$,

(b) When $(x,y,z) = (-1, -2, 1/2)$,

$\pdiff{U}{T}(1,2,4) = -\frac{2\ln(2)}{1+2\ln(2)}$ $\pdiff{T}{V}(1,2,4) = 1 -\frac{1}{4\ln(2)}$

We are told that the four variables $T$, $U$, $V$, $W$ obey the single equation $(TU-V)^2 \ln(W-UV) = \ln 2$. So they are not all independent variables. Roughly speaking, we can treat any three of them as independent variables and solve the given equation for the fourth as a function of the three chosen independent variables.

We are first asked to find $\pdiff{U}{T}$. This implicitly tells to treat $T$, $V$ and $W$ as independent variables and to view $U$ as a function $U(T,V,W)$ that obeys

for all $(T, U, V, W)$ sufficiently near $(1, 1, 2, 4)$. Differentiating (E1) with respect to $T$ gives

In particular, for $(T, U, V, W)=(1, 1, 2, 4)$,

This simplifies to

We are then asked to find $\pdiff{T}{V}$. This implicitly tells to treat $U$, $V$ and $W$ as independent variables and to view $T$ as a function $T(U,V,W)$ that obeys

for all $(T, U, V, W)$ sufficiently near $(1, 1, 2, 4)$. Differentiating (E2) with respect to $V$ gives

In particular, for $(T, U, V, W)=(1, 1, 2, 4)$,

This simplifies to

$24$

The function

So

and

$f_x(0,0)=1$, $f_y(0,0)=2$

By definition

Setting $x_0=y_0=0$,

Just evaluate $y\pdiff{z}{x}(x,y)$ and $x\pdiff{z}{y}(x,y)$.

Yes.

As $z(x,y)=f(x^2+y^2)$

by the (ordinary single variable) chain rule. So

and the differential equation is always satisfied, assuming that $f$ is differentiable, so that the chain rule applies.

(a) $\pdiff{f}{x}(0,0)=1$, $\pdiff{f}{y}(0,0)=4$ (b) Nope.

By definition

and

(b) $f(x,y)$ is not continuous at $(0,0)$, even though both partial derivatives exist there. To see this, make a change of coordinates from $(x,y)$ to $(X,y)$ with $X=x+y$ (the denominator). Of course, $(x,y)\rightarrow (0,0)$ if and only if $(X,y)\rightarrow (0,0)$. Now watch what happens when $(X,y)\rightarrow(0,0)$ with $X$ a lot smaller than $y$. For example, $X=ay^2$. Then

This depends on $a$. So approaching $(0,0)$ along different paths gives different limits. (You can see the same effect without changing coordinates by sending $(x,y)\rightarrow (0,0)$ with $x=-y+ay^2$.) Even more dramatically, watch what happens when $(X,y)\rightarrow(0,0)$ with $X=y^3$. Then

You can find an equation for the surface, or just look at the diagram.

1 resp. 0

Solution 1
Let's start by finding an equation for this surface. Every level curve is a horizontal circle of radius one, so the equation should be of the form

where $f_1$ and $f_2$ are functions depending only on $z$. Since the centre of the circle at height $z$ is at position $x=0$, $y=z$, we see that the equation of our surface is

The height of the surface at the point $(x,y)$ is the $z(x,y)$ found by solving that equation. That is,

We differentiate this equation implicitly to find $z_x(x,y)$ and $z_y(x,y)$ at the desired point $(x,y)= (0,-1)$. First, differentiating $(*)$ with respect to $y$ gives