Multivariable calculus

Critical Points of sin(x) e^(-y^2)

Q: Why can this page not list every critical point?

Because f has period 2\pi in x: the sine repeats and the Gaussian does not involve x at all. Every point \left(\frac{\pi}{2} + m\pi, 0\right) is critical, which is an infinite set. The two listed are representatives, one of each behaviour, and every other critical point is a translate of one of them with the same discriminant of 2.

Q: What happens at (3pi/2, 0)?

That is -\frac{\pi}{2} shifted by 2\pi, so it is a local minimum with the same numbers: \sin x = -1 gives f_{xx} = 1, f_{yy} = 2, f_{xy} = 0 and D = 2, with f = -1. The pattern along the x-axis alternates maximum, minimum, maximum every \pi units, and the discriminant is 2 at every one of them.

Every critical point of f(x,y) = sin(x) e^(-y^2) has y = 0 with x an odd multiple of pi/2. At (pi/2, 0) the second partials are f_xx = -1, f_yy = -2 and f_xy = 0, so D = 2 and the point is a local maximum of value 1. At (-pi/2, 0), D = 2 with f_xx = 1, a local minimum of value -1.

f(x,y) = \sin x \; e^{-y^{2}}

$\left(1.5707963267948966, 0\right)$ local maximumdiscriminant D = 2
$\left(-1.5707963267948966, 0\right)$ local minimumdiscriminant D = 2

These are not necessarily all of them; the page says which it covers.

One equation eliminates the other's easy branch

The two partials look very different, because $x$ sits inside a sine and $y$ inside a Gaussian.

\frac{\partial f}{\partial x} = \cos x \; e^{-y^{2}}, \qquad \frac{\partial f}{\partial y} = -2y \sin x \; e^{-y^{2}}

The exponential is never zero, so the first equation reduces to $\cos x = 0$ , meaning $x = \frac{\pi}{2} + m\pi$ . That is the key step, because it settles the second equation as well.

The second equation factors as $y = 0$ or $\sin x = 0$ . But $\cos x = 0$ forces $\sin x = \pm 1$ , so the branch $\sin x = 0$ is already ruled out. Only $y = 0$ survives.

\cos x = 0 \implies \sin x = \pm 1 \implies -2y\sin x\,e^{-y^{2}} = 0 \text{ only when } y = 0

So every critical point sits on the $x$ -axis at an odd multiple of $\frac{\pi}{2}$ . There are infinitely many, spaced $\pi$ apart, and they alternate between maxima and minima.

A sine ridge damped by a Gaussian

The second partials keep the same split: the $x$ derivative only touches the sine, the $y$ derivative only touches the Gaussian.

f_{xx} = -\sin x \; e^{-y^{2}}, \qquad f_{yy} = \left(4y^{2}-2\right)\sin x \; e^{-y^{2}}, \qquad f_{xy} = -2y\cos x \; e^{-y^{2}}

At any critical point $y = 0$ , so the exponential equals 1 and $f_{xy} = 0$ . That leaves $f_{xx} = -\sin x$ and $f_{yy} = -2\sin x$ , whose product is $D = 2\sin^{2}x = 2$ , since $\sin x = \pm 1$ there. The discriminant is the same positive number at every critical point, so $f_{xx}$ decides every classification on its own.

At $\left(\frac{\pi}{2}, 0\right)$ , $\sin x = 1$ , so $f_{xx} = -1 < 0$ and the point is a local maximum with $f = 1$ . At $\left(-\frac{\pi}{2}, 0\right)$ , $\sin x = -1$ , so $f_{xx} = 1 > 0$ and it is a local minimum with $f = -1$ . Adding $2\pi$ to $x$ returns the same behaviour, so maxima appear at $\frac{\pi}{2} + 2k\pi$ and minima at $-\frac{\pi}{2} + 2k\pi$ .

Geometrically, the sine wave along the $x$ -axis is a ridge, and the factor $e^{-y^{2}}$ pulls every cross-section toward zero as you move away from that axis. The damping is what makes the crest of the sine a genuine local maximum in both directions rather than a flat ridge line.

The mistake students make

The frequent error is losing the factor of $-2y$ from the chain rule and writing $f_y = \sin x\;e^{-y^{2}}$ . That expression is zero only when $\sin x = 0$ , which contradicts $\cos x = 0$ , so a student who makes this slip concludes the surface has no critical points at all. That conclusion is self-refuting: $f$ is smooth on the whole plane and actually attains the value 1, and a function that attains a maximum at an interior point must have a vanishing gradient there. Reaching no critical points is a signal to recheck the chain rule.

\frac{\partial}{\partial y}e^{-y^{2}} = e^{-y^{2}} \cdot \frac{\partial}{\partial y}\left(-y^{2}\right) = -2y\,e^{-y^{2}}

The opposite error is solving $-2y\sin x\,e^{-y^{2}} = 0$ by picking $\sin x = 0$ and pairing it with $\cos x = 0$ from the other equation. No angle has both a zero sine and a zero cosine, since $\sin^{2}x + \cos^{2}x = 1$ . Checking that identity is the fastest way to kill an impossible branch.

Solve $\cos x = 0$ first, since the exponential factor makes it the cleaner equation.
Record what that forces: $\sin x = \pm 1$ , never 0.
Feed that into the second equation, where it leaves $y = 0$ as the only option.
Classify using $f_{xx} = -\sin x$ , which alternates in sign as $x$ steps by $\pi$ .

Frequently asked questions

Why can this page not list every critical point?

Because $f$ has period $2\pi$ in $x$ : the sine repeats and the Gaussian does not involve $x$ at all. Every point $\left(\frac{\pi}{2} + m\pi, 0\right)$ is critical, which is an infinite set. The two listed are representatives, one of each behaviour, and every other critical point is a translate of one of them with the same discriminant of 2.

What happens at (3pi/2, 0)?

That is $-\frac{\pi}{2}$ shifted by $2\pi$ , so it is a local minimum with the same numbers: $\sin x = -1$ gives $f_{xx} = 1$ , $f_{yy} = 2$ , $f_{xy} = 0$ and $D = 2$ , with $f = -1$ . The pattern along the $x$ -axis alternates maximum, minimum, maximum every $\pi$ units, and the discriminant is 2 at every one of them.