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Gradient and directional derivatives

Problem 10.216 · medium Video

Find the directional derivative of \( \displaystyle f(x, y) = \sqrt{x^{2} + y^{2}} \) at \( \displaystyle (3, 4) \) in the direction of \( \displaystyle \langle 1, 0 \rangle \).

\[ \left[\begin{matrix}\frac{\partial}{\partial x} \sqrt{x^{2} + y^{2}}\\\frac{\partial}{\partial y} \sqrt{x^{2} + y^{2}}\end{matrix}\right] = \left[\begin{matrix}\frac{x}{\sqrt{x^{2} + y^{2}}}\\\frac{y}{\sqrt{x^{2} + y^{2}}}\end{matrix}\right] \]
∇f.✓ Proved
\[ \left[\begin{matrix}\frac{3}{5}\\\frac{4}{5}\end{matrix}\right] \]
∇f at the point.
\[ \left[\begin{matrix}1\\0\end{matrix}\right] \]
The unit direction u.
\[ \frac{3}{5} \]
D_u f = ∇f · u.

Answer \( \frac{3}{5} \)

A contour map of f(x, y) = sqrt(x² + y²): each curve joins points where f has the same value. At (3, 4) the gradient arrow points straight uphill, across the contours, and the direction of travel is drawn beside it.

Lines: 1 proved, 3 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify

The full receipt

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
3	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
4	Not checked	—	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: fail (error) — The solution claims the direction vector <1, 0> is a unit vector without justification or normalization step. While <1, 0> happens to be a unit vector, the solution fails to explicitly verify or perform the normalization required by the definition of the directional derivative, which is a critical step in the general method.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-03 — The solution claims the direction vector <1, 0> is a unit vector without justification or normalization step. While <1, 0> happens to be a unit vector, the solution fails to explicitly verify or perform the normalization required by the definition of the directional derivative, which is a critical step in the general method.
gpt-oss:20b: pass 2026-10-03
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly computes the gradient, evaluates it at the given point, identifies the unit direction vector, and calculates the dot product. The steps are logically sound and the final answer is correct.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: pass 2026-09-28
gpt-oss:20b: pass 2026-09-28

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.