Is there a relationship between slope of the curve $ f(x, y) = 0 $ and the partial derivative $ \frac{\partial f}{\partial x} $?

Question

Let $ f(x, y) = 0 $ define a curve. Is there any relationship between the slope of this curve and the partial derivative $ \frac{\partial f}{\partial x} $?

For example, if $ f(x, y) = (x - 1)^2 + (y - 1)^2 - 1 $, then $ f(x, y) = 0 $ defines a circle of radius $ 1 $ centered at $ (0, 0) $.

Now $ \frac{\partial f(x, y)}{\partial x} = 2(x - 1) $.

Thus $ \frac{\partial f}{\partial x} = 0 $ at $ x = 1 $ and indeed we see that the circle $ f(x, y) = 0 $ also has a slope of $ 0 $ at $ x = 1 $.

Is it always true that $ \frac{\partial f}{\partial x} = 0 $ at $ x = a $ implies that the curve $ f(x, y) = 0 $ has a horizontal slope (slope = $ 0 $) at $ x = a $?

If yes, how can we prove it?

If not, is there a counterexample?

Answer 1

What do you mean slope (the value of the tangent line, the value of the univariate derivative)? If some function is bivariate it has no tangent line but a tangent plane.

When we take the partial derivative with respect to x we are considering that y is contant. If y is constant there is a tangent line.

The partial derivative with respect to x at some point (x,y) is the slope of the simplified univariate function.

When the partial derivative with respect to x at some line (x,y) is 0 it means that for whatever y may be when x has said value the slope is 0.

Answer 2

$\def\rbf{\mathbf{R}}$The equation $f(x,y)=0$ does not always define a (smooth) curve. But let us assume, for simplicity, that it does and $f:\rbf^2\to\rbf$ is smooth and the gradient $\nabla f\ne 0$ everywhere.

It does not make sense to say "the slope of this curve". Though it makes sense to say the slope field on this curve or the slope of the tangent line at a point of this curve.

Note $f=0$ is a level curve for the function $f$. Call this curve $\gamma$.

It is known that

At each point on this level curve, the gradient vector is perpendicular to the level curve.

In other words, if $(a,b)$ is a point on the curve, i.e., $f(a,b)=0$, and $f_x(a,b)=A$ and $f_y(a,b)=B$, then an equation to the tangent line of $\gamma$ at $(a,b)$ is $$ A(x-a)+B(y-b)=0\;\tag{1} $$

So if $f_x(a,b)=0$, equation (1) becomes $$ B(y-b)=0\;. $$ which is a horizontal line ($B\ne 0$ by our assumption).

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[Added.] As one of the comments pointed out, in your example $f(x,y)=(x-1)^2+(x-1)^2-1$, the equation $f=0$ is corresponding to the circle centered at $(1,1)$, not $(0,0)$.

Moreover, it does not make sense to say the circle $f=0$ has a slope of $0$ at $x=1$. What you have instead, is that at the point $(x,y)=(1,0)$ and also the point $(x,y)=(1,2)$, the tangent lines to the circle both have the slope $0$.

Answer 3

If $f$ is nice

Suppose $f(x,y) = 0$ is locally given as the graph of some function $g$, i.e. $y = g(x)$ and $f(x,y) = 0$ are locally the same curve in the plane. This will happen whenever $f$ is nice, which in this case means $df \neq 0$ at $(x,y)$ and $f$ is continuously differentiable in a neighborhood of $(x,y)$. The slope of the curve at $x$ is, by definition, $g'(x)$.

To start, we have the differential $$ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy $$ On the curve $f(x, y) = 0$, $f$ is constant, so $df = 0$. Now use $y = g(x)$ to write $dy = g'(x) dx$. This gives $$ 0 = df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} g'(x) dx $$ That is $$ 0 = \left(\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} g'(x)\right) dx $$ which implies $$ 0 = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} g'(x) $$ Finally we solve to obtain $$ g'(x) = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} $$ From this formula, we see that if $\frac{\partial f}{\partial x} = 0$, we do indeed have $$ g'(x) = 0 $$ i.e the slope of the curve is $0$ at $x$.

Counterexample if $f$ is not nice

Take $f(x,y) = x^3 - y^2$. Then $\frac{\partial f}{\partial x} (0, 0) = 0$. However, if we look at the graph of $f(x,y) = 0$, we clearly see that the curve does not even have a well-defined slope at $(0, 0)$.

I suppose we might be inclined to say that the slope really is $0$ at that point since it "approaches" $0$ from "both sides," but I think that's stretching what we mean by slope. Maybe someone else can come up with a more pathological example.