15.1 Partial derivatives

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.

Use the definition of the derivative:

The definition of the derivative involves a limit at $h$ goes to 0; we can approximate that limit by choosing a value of $h$ that's close to 0; in our case, $0.1$ or $-0.1$ are the best we can do, using the information on the table.

(a)

(b)

(c)

(a)

(b)

(c)

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

By the quotient rule

Hence

By the quotient rule

Hence

This is an implicit differentiation question. Implicit differentiation, as you'll recall from first-semester calculus, is more-or-less just an application of the chain rule.

(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 $(-1,-2)$). Recall the following derivatives:

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

Differentiate implicitly.

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

We are told that the four variables $T$, $U$, $V$, $W$ obey the the single equation $(TU-V)^2 \log(W-UV) = \log 2$. 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

so that the slope looking in the positive $y$ direction is $z_y(0,-1)=1$. Similarly, differentiating $(*)$ with respect to $x$ gives

The slope looking in the positive $x$ direction is $z_x(0,-1)=0$.

Solution 2
Standing at $(0,-1,0)$ and looking in the positive $y$ direction, the surface follows the straight line that

Shifting the central line one unit in the $y$-direction, we get the line $z=y+1$, $x=0$. (As a check, notice that $(0,-1,0)$ is indeed on $z=y+1$,
$x=0$.) The slope of this line is 1.

Standing at $(0,-1,0)$ and looking in the positive $x$ direction, the surface follows the circle $x^2+y^2=1$, $z=0$, which is the intersection of the cylinder with the $xy$-plane. As we move along that circle our $z$ coordinate stays fixed at $0$. So the slope in that direction is 0.