Multivariable calculus

Directional derivative of x^2 + 4y^2 at (1, 1)

Q: Where is the only point with all directional derivatives zero?

At the origin, where \nabla f = \langle 0, 0 \rangle. That is the minimum of the bowl, and it is the one place where no direction changes f to first order.

At (1, 1) the gradient of f(x,y) = x^2 + 4y^2 is (2, 8). Dotting with the unit direction (0.96, -0.28) gives -8/25, that is -0.32. The direction sits a little past the tangent to the level ellipse, so the value is small and negative even though the bowl is steep here.

f(x,y) = x^2 + 4y^2

D_{\mathbf{u}} f(1,1) = -\frac{8}{25} = -0.32

At x = 1, y = 1, in the unit direction (0.96, -0.28).

An elliptic bowl and its gradient

Each term differentiates on its own: $x^2$ gives $2x$ and contributes nothing to the $y$ partial, while $4y^2$ gives $8y$ and contributes nothing to the $x$ partial.

\nabla f = \langle 2x,\; 8y \rangle, \qquad \nabla f(1,1) = \langle 2, 8 \rangle

The lopsidedness matters. The 4 in front of $y^2$ makes the surface climb four times faster in $y$ than in $x$ at the same coordinate value, and the gradient leans heavily toward the $y$ axis.

D_{\mathbf{u}} f(1,1) = 2(0.96) + 8(-0.28) = 1.92 - 2.24 = -\frac{8}{25}

The mistake: assuming outward means uphill

The direction $\langle 0.96, -0.28 \rangle$ moves away from the origin in $x$ , and the surface is a bowl, so students often expect an increase. The level curves are ellipses, not circles, and stepping that way crosses onto a slightly lower ellipse.

$f(1,1) = 5$ , and the level curve through the point is $x^2 + 4y^2 = 5$
Step $0.1$ along the direction to $(1.096, 0.972)$ , where $f \approx 4.980$ , lower than 5
Only for a circular bowl like $x^2 + y^2$ would outward always mean uphill

The other frequent slip is writing $\frac{\partial f}{\partial y} = 2y$ and forgetting the 4. That would give $2(0.96) + 2(-0.28) = 1.36$ , positive, and the sign of the answer would be wrong as well as its size.

Almost tangent to the level ellipse

The angle between $\langle 2, 8 \rangle$ and $\langle 0.96, -0.28 \rangle$ is about 92.2 degrees, only just past the right angle. That is why the answer is close to zero despite a gradient of length $\lVert \langle 2, 8 \rangle \rVert = 2\sqrt{17} \approx 8.246$ .

D_{\mathbf{u}} f(1,1) = 2\sqrt{17}\,\cos\theta, \qquad \cos\theta \approx -0.0388

The exactly tangent direction is $\langle 8, -2 \rangle / \sqrt{68} \approx \langle 0.970, -0.243 \rangle$ , which is close to our direction but not equal to it. That small gap is the whole of the $-0.32$ .

The build re-checks that (0.96, -0.28) has length 1, rebuilds the gradient of x^2 + 4y^2 at (1, 1) by central differences on the function itself, and fails unless the dot product is still -0.32.

Frequently asked questions

How can the answer be negative on a bowl that opens upward?

A bowl has directions of every kind at every point except its bottom. Here the $4y^2$ term shrinks at rate $8(0.28) = 2.24$ while the $x^2$ term grows at only $2(0.96) = 1.92$ , so the step nets out downhill.

Where is the only point with all directional derivatives zero?

At the origin, where $\nabla f = \langle 0, 0 \rangle$ . That is the minimum of the bowl, and it is the one place where no direction changes $f$ to first order.