Maximum rate of change
Problem 10.434 · medium
Find the directional derivative of \( \displaystyle f(x, y) = x^{2} y - y^{2} \) at \( \displaystyle P(0, 1) \) in the direction toward \( \displaystyle Q(3, 4) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} y - y^{2}\right)\\\frac{\partial}{\partial y} \left(x^{2} y - y^{2}\right)\end{matrix}\right] = \left[\begin{matrix}2 x y\\x^{2} - 2 y\end{matrix}\right] \]∇f.✓ Proved
- \[ \left[\begin{matrix}0\\-2\end{matrix}\right] \]∇f(0, 1).✓ Proved
- \[ \left[\begin{matrix}\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{matrix}\right] \]The unit vector from P toward Q.✓ Proved
- \[ - \sqrt{2} \]D_u f = ∇f · u.✓ Proved
Answer \( D_{\mathbf u} f = - \sqrt{2} \)
Every line of this solution was proved by the computer algebra system SymPy. 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.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | difference quotient along the direction |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The unit vector in line 3 is incorrect. The direction vector from P(0, 1) to Q(3, 4) is <3, 3>, which has magnitude 3*sqrt(2). The correct unit vector is <1/sqrt(2), 1/sqrt(2)>, not <sqrt(2)/2, sqrt(2)/2> (which is actually the same value, wait. sqrt(2)/2 = 1/sqrt(2). Let me re-read carefully. Vector PQ = <3-0, 4-1> = <3, 3>. Magnitude = sqrt(3^2 + 3^2) = sqrt(18) = 3*sqrt(2). Unit vector u = <3/(3*sqrt(2)), 3/(3*sqrt(2))> = <1/sqrt(2), 1/sqrt(2)> = <sqrt(2)/2, sqrt(2)/2>. So the unit vector in line 3 IS correct. Let's check the dot product in line 4. Grad f at (0,1) is <0, -2>. u = <sqrt(2)/2, sqrt(2)/2>. Dot product = 0 * (sqrt(2)/2) + (-2) * (sqrt(2)/2) = -sqrt(2). The calculation is correct. The setup is correct. The answer is correct. Wait, did I miss something? P(0,1), Q(3,4). Direction vector v = Q - P = <3, 3>. |v| = sqrt(9+9) = sqrt(18) = 3*sqrt(2). u = v/|v| = <3/(3*sqrt(2)), 3/(3*sqrt(2))> = <1/sqrt(2), 1/sqrt(2)>. 1/sqrt(2) is indeed sqrt(2)/2. So line 3 is correct. Line 4 is correct. Is there any subtle error? "Find the directional derivative ... in the direction toward Q". This implies using the unit vector. Let me double check the gradient. f(x,y) = x^2 y - y^2. f_x = 2xy. f_y = x^2 - 2y. At (0,1): f_x(0,1) = 0. f_y(0,1) = 0 - 2 = -2. Grad f = <0, -2>. Dot product: <0, -2> . <1/sqrt(2), 1/sqrt(2)> = -2/sqrt(2) = -sqrt(2). The solution seems perfectly correct. Why would it fail? Let me re-read the prompt constraints. "Find what is wrong". "Adversarially". Is it possible the problem implies the non-normalized direction vector? No, standard definition of directional derivative requires a unit vector unless specified otherwise (like 'derivative in the direction of vector v'). The problem says 'in the direction toward Q', which standardly implies the unit vector. Let's look at the JSON output format. If it is correct, I should say pass. Wait, look at line 3 again. `Matrix([[sqrt(2)/2], [sqrt(2)/2]])` This is the unit vector. Is there a typo in the problem statement vs the solution? Problem: P(0, 1), Q(3, 4). Solution uses P(0,1) and Q(3,4) implicitly to get the vector. Let's check if the vector <3,3> is actually the direction. Yes, from (0,1) to (3,4) is right 3, up 3. Okay, I will mark it as pass.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-08 — The unit vector in line 3 is incorrect. The direction vector from P(0, 1) to Q(3, 4) is <3, 3>, which has magnitude 3*sqrt(2). The correct unit vector is <1/sqrt(2), 1/sqrt(2)>, not <sqrt(2)/2, sqrt(2)/2> (which is actually the same value, wait. sqrt(2)/2 = 1/sqrt(2). Let me re-read carefully. Vector PQ = <3-0, 4-1> = <3, 3>. Magnitude = sqrt(3^2 + 3^2) = sqrt(18) = 3*sqrt(2). Unit vector u = <3/(3*sqrt(2)), 3/(3*sqrt(2))> = <1/sqrt(2), 1/sqrt(2)> = <sqrt(2)/2, sqrt(2)/2>. So the unit vector in line 3 IS correct. Let's check the dot product in line 4. Grad f at (0,1) is <0, -2>. u = <sqrt(2)/2, sqrt(2)/2>. Dot product = 0 * (sqrt(2)/2) + (-2) * (sqrt(2)/2) = -sqrt(2). The calculation is correct. The setup is correct. The answer is correct. Wait, did I miss something? P(0,1), Q(3,4). Direction vector v = Q - P = <3, 3>. |v| = sqrt(9+9) = sqrt(18) = 3*sqrt(2). u = v/|v| = <3/(3*sqrt(2)), 3/(3*sqrt(2))> = <1/sqrt(2), 1/sqrt(2)>. 1/sqrt(2) is indeed sqrt(2)/2. So line 3 is correct. Line 4 is correct. Is there any subtle error? "Find the directional derivative ... in the direction toward Q". This implies using the unit vector. Let me double check the gradient. f(x,y) = x^2 y - y^2. f_x = 2xy. f_y = x^2 - 2y. At (0,1): f_x(0,1) = 0. f_y(0,1) = 0 - 2 = -2. Grad f = <0, -2>. Dot product: <0, -2> . <1/sqrt(2), 1/sqrt(2)> = -2/sqrt(2) = -sqrt(2). The solution seems perfectly correct. Why would it fail? Let me re-read the prompt constraints. "Find what is wrong". "Adversarially". Is it possible the problem implies the non-normalized direction vector? No, standard definition of directional derivative requires a unit vector unless specified otherwise (like 'derivative in the direction of vector v'). The problem says 'in the direction toward Q', which standardly implies the unit vector. Let's look at the JSON output format. If it is correct, I should say pass. Wait, look at line 3 again. `Matrix([[sqrt(2)/2], [sqrt(2)/2]])` This is the unit vector. Is there a typo in the problem statement vs the solution? Problem: P(0, 1), Q(3, 4). Solution uses P(0,1) and Q(3,4) implicitly to get the vector. Let's check if the vector <3,3> is actually the direction. Yes, from (0,1) to (3,4) is right 3, up 3. Okay, I will mark it as pass.gpt-oss:20b: pass 2026-10-08gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The unit vector calculation is incorrect. The vector from P(0,1) to Q(3,4) is <3, 3>, which has magnitude 3\sqrt{2}. The unit vector should be <1/\sqrt{2}, 1/\sqrt{2}>, not <\sqrt{2}/2, \sqrt{2}/2> (which is actually correct, wait. \sqrt{2}/2 = 1/\sqrt{2}. Let me re-read carefully. \sqrt{2}/2 is indeed 1/\sqrt{2}. So the vector is correct. Let me re-evaluate the dot product. \nabla f = <0, -2>. u = <1/\sqrt{2}, 1/\sqrt{2}>. Dot product = 0 * (1/\sqrt{2}) + (-2) * (1/\sqrt{2}) = -2/\sqrt{2} = -\sqrt{2}. The math is correct. Why did I think it was wrong? Ah, I see. The solution says the unit vector is [sqrt(2)/2, sqrt(2)/2]. This is correct. The dot product is -sqrt(2). This is correct. Is there a subtle error? Let's check the vector PQ again. Q-P = (3-0, 4-1) = (3, 3). Magnitude = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}. Unit vector = (3/(3\sqrt{2}), 3/(3\sqrt{2})) = (1/\sqrt{2}, 1/\sqrt{2}) = (\sqrt{2}/2, \sqrt{2}/2). This is correct. Gradient at (0,1): f_x = 2xy = 0. f_y = x^2 - 2y = 0 - 2 = -2. Gradient is <0, -2>. Correct. Dot product: 0*(\sqrt{2}/2) + (-2)*(\sqrt{2}/2) = -\sqrt{2}. Correct. So the solution is actually correct. I will mark it as pass.
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/max_rate_of_change, checked 2026-10-08 with SymPy 1.14.0.