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Gradient and directional derivatives

Problem 10.243 · medium Video

Find the directional derivative of \( \displaystyle f(x, y) = \sqrt{x^{2} + y^{2}} \) at \( \displaystyle (1, 0) \) in the direction of \( \displaystyle \langle 1, 1 \rangle \).

\[ \left[\begin{matrix}\frac{\partial}{\partial x} \sqrt{x^{2} + y^{2}}\\\frac{\partial}{\partial y} \sqrt{x^{2} + y^{2}}\end{matrix}\right] = \left[\begin{matrix}\frac{x}{\sqrt{x^{2} + y^{2}}}\\\frac{y}{\sqrt{x^{2} + y^{2}}}\end{matrix}\right] \]
∇f.✓ Proved
\[ \left[\begin{matrix}1\\0\end{matrix}\right] \]
∇f at the point.Reviewed
\[ \left[\begin{matrix}\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{matrix}\right] \]
The unit direction u.Reviewed
\[ \frac{\sqrt{2}}{2} \]
D_u f = ∇f · u.Reviewed

Answer \( \frac{\sqrt{2}}{2} \)

A contour map of f(x, y) = sqrt(x² + y²): each curve joins points where f has the same value. At (1, 0) the gradient arrow points straight uphill, across the contours, and the direction of travel is drawn beside it.

✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
3	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
4	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	f differenced along the unit direction agrees

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly computes the gradient, normalizes the direction vector, and performs the dot product to find the directional derivative.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly computes the gradient, normalizes the direction vector, and performs the dot product to find the directional derivative.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly computes the gradient, normalizes the direction vector, and calculates the dot product. The steps are logically sound and the final answer is correct.
gpt-oss:20b: fail (misleading) 2026-09-29 — Sentence 4 incorrectly presents the directional derivative as two numbers instead of a single scalar; this could mislead a student into thinking the result is a vector.
qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly computes the gradient, normalizes the direction vector, and calculates the dot product. The steps are logically sound and the final answer is correct.
gpt-oss:20b: fail (misleading) 2026-09-29 — The solution incorrectly lists two components for the directional derivative instead of a single scalar, and it does not show the dot‑product computation. The correct directional derivative is 1/√2, not a pair of values.

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/gradient_directional, checked 2026-10-13 with SymPy 1.14.0.