Gradient of x^2 - y^2: The Saddle Surface Worked Out

Two partials, and the minus sign travels with the y term

Freeze $y$ first. Then $-y^2$ is a constant and drops out, leaving only the derivative of $x^2$.

Now freeze $x$. The term $x^2$ is the constant this time, and the minus sign in front of $y^2$ stays attached through the differentiation.

At $(2,1)$ the gradient is $\langle 4, -2\rangle$. It points to the right and downward in the plane, so from $(2,1)$ you gain height fastest by increasing $x$ while decreasing $y$.

The mistake: assuming a zero gradient means a peak or a valley

Setting both components to zero gives $2x = 0$ and $-2y = 0$, so the only critical point is the origin, where $\nabla f(0,0) = \langle 0, 0\rangle$. Students report this as a maximum or a minimum out of habit, which is wrong here.

Look along two lines through the origin. On the $x$-axis, $f(x,0) = x^2$, so the origin is the lowest point. On the $y$-axis, $f(0,y) = -y^2$, so the origin is the highest point. A single point cannot be both, so it is a saddle.

• A zero gradient says the tangent plane is horizontal, nothing more.
• Classifying the point needs the second derivative test or a direct look along different lines.
• The other frequent slip is writing $\frac{\partial f}{\partial y} = 2y$ and losing the minus. Carry the sign through the whole line rather than attaching it at the end.

Level curves are hyperbolas, and the gradient crosses them squarely

Setting $x^2 - y^2 = c$ gives a hyperbola for each nonzero $c$, opening left and right when $c > 0$ and up and down when $c < 0$. The case $c = 0$ degenerates into the pair of lines $y = \pm x$.

Check perpendicularity at $(2,1)$, which sits on the curve $x^2 - y^2 = 3$. Implicit differentiation gives $2x - 2y\,y' = 0$, so $y' = x/y = 2$ and a tangent direction is $\langle 1, 2\rangle$. Dotting with the gradient:

The dot product vanishes, so the gradient meets the level hyperbola at a right angle, exactly as it does for every differentiable function.