Multivariable Calculus Quiz: Gradient Vector
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Gradient VectorQuestion 1 of 10

A surface is defined implicitly by the equation xln⁡(y)+yz3+zx=11x\ln(y) + yz^3 + z\sqrt{x} = 11. Find the value of the partial derivative ∂z∂x\frac{\partial z}{\partial x} at the point (4,1,2)(4, 1, 2).

1/281/28
−1/28-1/28
−28-28
−6/7-6/7
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Gradient Vector

Practice Gradient Vector in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Gradient Vector, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A surface is defined implicitly by the equation xln⁡(y)+yz3+zx=11x\ln(y) + yz^3 + z\sqrt{x} = 11. Find the value of the partial derivative ∂z∂x\frac{\partial z}{\partial x} at the point (4,1,2)(4, 1, 2).

  1. 1/281/28
  2. −1/28-1/28 (correct answer)
  3. −28-28
  4. −6/7-6/7
Explanation: When you encounter an implicitly defined surface and need to find a partial derivative, you're dealing with implicit differentiation in multiple variables. The key insight is that since the surface equation equals a constant, differentiating both sides with respect to any variable gives you zero. Starting with xln⁡(y)+yz3+zx=11x\ln(y) + yz^3 + z\sqrt{x} = 11, differentiate both sides with respect to xx, treating yy as constant and zz as a function of xx: ∂∂x[xln⁡(y)]+∂∂x[yz3]+∂∂x[zx]=0\frac{\partial}{\partial x}[x\ln(y)] + \frac{\partial}{\partial x}[yz^3] + \frac{\partial}{\partial x}[z\sqrt{x}] = 0 This gives us: ln⁡(y)+3yz2∂z∂x+∂z∂x⋅x+z⋅12x=0\ln(y) + 3yz^2\frac{\partial z}{\partial x} + \frac{\partial z}{\partial x} \cdot \sqrt{x} + z \cdot \frac{1}{2\sqrt{x}} = 0 Solving for ∂z∂x\frac{\partial z}{\partial x}: ∂z∂x(3yz2+x)=−ln⁡(y)−z2x\frac{\partial z}{\partial x}(3yz^2 + \sqrt{x}) = -\ln(y) - \frac{z}{2\sqrt{x}} At point (4,1,2)(4, 1, 2): ∂z∂x(3⋅1⋅4+2)=−ln⁡(1)−24=0−12=−12\frac{\partial z}{\partial x}(3 \cdot 1 \cdot 4 + 2) = -\ln(1) - \frac{2}{4} = 0 - \frac{1}{2} = -\frac{1}{2} So: ∂z∂x⋅14=−12\frac{\partial z}{\partial x} \cdot 14 = -\frac{1}{2}, giving us ∂z∂x=−128\frac{\partial z}{\partial x} = -\frac{1}{28} Answer (A) 1/281/28 has the wrong sign—likely from a sign error in the implicit differentiation. Answer (C) −28-28 inverts the fraction, suggesting algebraic mistakes. Answer (D) −6/7-6/7 might result from computational errors when substituting the point values. Strategy tip: Always verify your point satisfies the original equation before substituting, and be extra careful with signs when moving terms across the equation during implicit differentiation.

Question 2

Let SS be the level surface of the function f(x,y,z)=x2−y2+2z2f(x, y, z) = x^2 - y^2 + 2z^2 that passes through the point P(1,1,1)P(1, 1, 1). Which of the following is a vector equation for the line normal to the surface SS at point PP?

  1. L(t)=⟨1,1,1⟩+t⟨2,−2,4⟩L(t) = \langle 1, 1, 1 \rangle + t \langle 2, -2, 4 \rangle (correct answer)
  2. L(t)=⟨2,−2,4⟩+t⟨1,1,1⟩L(t) = \langle 2, -2, 4 \rangle + t \langle 1, 1, 1 \rangle
  3. 2x−2y+4z=42x - 2y + 4z = 4
  4. L(t)=⟨1,1,1⟩+t⟨2,2,4⟩L(t) = \langle 1, 1, 1 \rangle + t \langle 2, 2, 4 \rangle
Explanation: The gradient vector ∇f\nabla f at a point PP is normal to the level surface of ff that passes through PP. The normal line to the surface at PP is the line that passes through PP and has the direction of the gradient vector ∇f(P)\nabla f(P). First, compute the gradient of f(x,y,z)=x2−y2+2z2f(x, y, z) = x^2 - y^2 + 2z^2: ∇f(x,y,z)=⟨∂f∂x,∂f∂y,∂f∂z⟩=⟨2x,−2y,4z⟩\nabla f(x, y, z) = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle = \langle 2x, -2y, 4z \rangle. Next, evaluate the gradient at the point P(1,1,1)P(1, 1, 1): ∇f(1,1,1)=⟨2(1),−2(1),4(1)⟩=⟨2,−2,4⟩\nabla f(1, 1, 1) = \langle 2(1), -2(1), 4(1) \rangle = \langle 2, -2, 4 \rangle. This vector is the direction vector for the normal line. The line passes through the point P(1,1,1)P(1, 1, 1). The vector equation of a line is r⃗(t)=p⃗0+td⃗\vec{r}(t) = \vec{p}_0 + t\vec{d}, where p⃗0\vec{p}_0 is a point on the line and d⃗\vec{d} is the direction vector. Thus, the equation is L(t)=⟨1,1,1⟩+t⟨2,−2,4⟩L(t) = \langle 1, 1, 1 \rangle + t \langle 2, -2, 4 \rangle. (B) incorrectly swaps the point and the direction vector. (C) is the equation of the tangent plane to the surface at PP, not the normal line. (D) makes a sign error in the partial derivative with respect to yy.

Question 3

Let f(x,y)=x2+y2f(x, y) = x^2 + y^2. At which point PP on the line y=x−1y = x - 1 is the gradient vector ∇f(P)\nabla f(P) normal to the line?

  1. (−1/2,1/2)(-1/2, 1/2)
  2. (1,−1)(1, -1)
  3. (1,0)(1, 0)
  4. (1/2,−1/2)(1/2, -1/2) (correct answer)
Explanation: When you see a question about gradient vectors being normal to curves, you're dealing with the fundamental relationship between gradients and perpendicularity. The gradient vector ∇f\nabla f at any point is always perpendicular to the level curves of the function, but this question asks when it's perpendicular to a different line entirely. First, find the gradient: ∇f(x,y)=(2x,2y)\nabla f(x,y) = (2x, 2y). For any point (x,x−1)(x, x-1) on the line y=x−1y = x - 1, the gradient is ∇f=(2x,2(x−1))=(2x,2x−2)\nabla f = (2x, 2(x-1)) = (2x, 2x-2). The line y=x−1y = x - 1 has direction vector (1,1)(1, 1) (since its slope is 1). For the gradient to be normal (perpendicular) to this line, their dot product must equal zero: (2x,2x−2)⋅(1,1)=2x+(2x−2)=4x−2=0(2x, 2x-2) \cdot (1, 1) = 2x + (2x-2) = 4x - 2 = 0 Solving: x=12x = \frac{1}{2}, so y=12−1=−12y = \frac{1}{2} - 1 = -\frac{1}{2}. The point is (12,−12)(\frac{1}{2}, -\frac{1}{2}). Looking at the wrong answers: A) (−12,12)(-\frac{1}{2}, \frac{1}{2}) isn't even on the line since 12≠−12−1\frac{1}{2} \neq -\frac{1}{2} - 1. B) (1,−1)(1, -1) also fails the line equation since −1≠1−1=0-1 \neq 1 - 1 = 0. C) (1,0)(1, 0) is on the line, but when you check the perpendicularity condition, the dot product gives 4(1)−2=2≠04(1) - 2 = 2 \neq 0. Remember: always verify that candidate points actually lie on the given constraint, then check the geometric condition. Many wrong answers in these problems fail the basic constraint test.

Question 4

A particle's position is described by the vector function r⃗(t)=⟨cos⁡(πt),sin⁡(πt),t⟩\vec{r}(t) = \langle \cos(\pi t), \sin(\pi t), t \rangle. The particle moves through a region where the temperature is given by T(x,y,z)=xy+z2T(x,y,z) = xy + z^2. What is the rate of change of temperature with respect to time that the particle experiences at t=1/2t=1/2?

  1. 1−π/21 - \pi/2
  2. 1+π1 + \pi
  3. 11
  4. 1−π1 - \pi (correct answer)
Explanation: When a particle moves through a temperature field, you need to find how temperature changes along the particle's path using the chain rule. This requires computing dTdt=∇T⋅r⃗′(t)\frac{dT}{dt} = \nabla T \cdot \vec{r}'(t), where you take the dot product of the temperature gradient with the particle's velocity vector. First, find the gradient of T(x,y,z)=xy+z2T(x,y,z) = xy + z^2: ∇T=⟨y,x,2z⟩\nabla T = \langle y, x, 2z \rangle Next, find the velocity vector by differentiating r⃗(t)=⟨cos⁡(πt),sin⁡(πt),t⟩\vec{r}(t) = \langle \cos(\pi t), \sin(\pi t), t \rangle: r⃗′(t)=⟨−πsin⁡(πt),πcos⁡(πt),1⟩\vec{r}'(t) = \langle -\pi\sin(\pi t), \pi\cos(\pi t), 1 \rangle At t=1/2t = 1/2, evaluate both vectors:
  • Position: r⃗(1/2)=⟨cos⁡(π/2),sin⁡(π/2),1/2⟩=⟨0,1,1/2⟩\vec{r}(1/2) = \langle \cos(\pi/2), \sin(\pi/2), 1/2 \rangle = \langle 0, 1, 1/2 \rangle
  • Velocity: r⃗′(1/2)=⟨−πsin⁡(π/2),πcos⁡(π/2),1⟩=⟨−π,0,1⟩\vec{r}'(1/2) = \langle -\pi\sin(\pi/2), \pi\cos(\pi/2), 1 \rangle = \langle -\pi, 0, 1 \rangle
  • Gradient at this point: ∇T(0,1,1/2)=⟨1,0,1⟩\nabla T(0,1,1/2) = \langle 1, 0, 1 \rangle
The rate of change is: dTdt=⟨1,0,1⟩⋅⟨−π,0,1⟩=−π+1=1−π\frac{dT}{dt} = \langle 1, 0, 1 \rangle \cdot \langle -\pi, 0, 1 \rangle = -\pi + 1 = 1 - \pi This confirms answer D. A) 1−π/21 - \pi/2 likely comes from incorrectly using z=1/2z = 1/2 instead of 2z=12z = 1 in the gradient calculation. B) 1+π1 + \pi results from a sign error when computing −πsin⁡(π/2)-\pi\sin(\pi/2). C) 11 ignores the velocity component entirely, using only the zz-direction contribution. Remember: for particle motion through scalar fields, always compute ∇f⋅v⃗\nabla f \cdot \vec{v} to find the directional rate of change along the path.

Question 5

The directional derivative of a function f(x,y)f(x,y) at the point P(1,2)P(1,2) in the direction of the vector v⃗1=⟨3,4⟩\vec{v}_1 = \langle 3, 4 \rangle is 1010. The directional derivative at P(1,2)P(1,2) in the direction of the vector v⃗2=⟨−4,3⟩\vec{v}_2 = \langle -4, 3 \rangle is 55. What is the gradient vector ∇f(1,2)\nabla f(1,2)?

  1. ⟨2/5,11/5⟩\langle 2/5, 11/5 \rangle
  2. ⟨2,11⟩\langle 2, 11 \rangle (correct answer)
  3. ⟨−5,10⟩\langle -5, 10 \rangle
  4. ⟨10,5⟩\langle 10, 5 \rangle
Explanation: When you encounter directional derivative problems with multiple directions, you're working with the fundamental relationship: Du⃗f=∇f⋅u⃗D_{\vec{u}}f = \nabla f \cdot \vec{u}, where u⃗\vec{u} is a unit vector. This creates a system of equations you can solve for the gradient components. First, convert the given vectors to unit vectors. For v⃗1=⟨3,4⟩\vec{v}_1 = \langle 3, 4 \rangle: ∣v⃗1∣=32+42=5|\vec{v}_1| = \sqrt{3^2 + 4^2} = 5, so u⃗1=⟨3/5,4/5⟩\vec{u}_1 = \langle 3/5, 4/5 \rangle. For v⃗2=⟨−4,3⟩\vec{v}_2 = \langle -4, 3 \rangle: ∣v⃗2∣=(−4)2+32=5|\vec{v}_2| = \sqrt{(-4)^2 + 3^2} = 5, so u⃗2=⟨−4/5,3/5⟩\vec{u}_2 = \langle -4/5, 3/5 \rangle. Let ∇f(1,2)=⟨a,b⟩\nabla f(1,2) = \langle a, b \rangle. The given information creates two equations:
  • ⟨a,b⟩⋅⟨3/5,4/5⟩=10\langle a, b \rangle \cdot \langle 3/5, 4/5 \rangle = 10, which gives 3a+4b5=10\frac{3a + 4b}{5} = 10, so 3a+4b=503a + 4b = 50
  • ⟨a,b⟩⋅⟨−4/5,3/5⟩=5\langle a, b \rangle \cdot \langle -4/5, 3/5 \rangle = 5, which gives −4a+3b5=5\frac{-4a + 3b}{5} = 5, so −4a+3b=25-4a + 3b = 25
Solving this system: multiply the first equation by 3 and the second by 4, then add: 9a+12b−16a+12b=150+1009a + 12b - 16a + 12b = 150 + 100, giving −7a+24b=250-7a + 24b = 250. From the second equation, b=25+4a3b = \frac{25 + 4a}{3}. Substituting and solving yields a=2a = 2 and b=11b = 11. Answer B ⟨2,11⟩\langle 2, 11 \rangle is correct. Answer A ⟨2/5,11/5⟩\langle 2/5, 11/5 \rangle incorrectly assumes you don't need to convert to unit vectors. Answers C and D represent calculation errors in solving the linear system. Always remember: directional derivatives require unit vectors, creating a solvable system when you have enough directional information.

Question 6

Let f(x,y)f(x,y) be a differentiable function with gradient ∇f\nabla f at a point PP. The directional derivative of ff at PP in the direction of a unit vector u⃗\vec{u} is Du⃗f(P)D_{\vec{u}}f(P). If θ\theta is the angle between ∇f(P)\nabla f(P) and u⃗\vec{u}, for which value of θ\theta is Du⃗f(P)D_{\vec{u}}f(P) equal to exactly half of its maximum possible value?

  1. π/6\pi/6
  2. π/4\pi/4
  3. π/3\pi/3 (correct answer)
  4. 2π/32\pi/3
Explanation: The formula for the directional derivative is Du⃗f(P)=∇f(P)⋅u⃗D_{\vec{u}}f(P) = \nabla f(P) \cdot \vec{u}. Using the geometric definition of the dot product, this can be written as Du⃗f(P)=∣∇f(P)∣∣u⃗∣cos⁡θD_{\vec{u}}f(P) = |\nabla f(P)| |\vec{u}| \cos\theta, where θ\theta is the angle between the gradient vector and the direction vector u⃗\vec{u}. Since u⃗\vec{u} is a unit vector, ∣u⃗∣=1|\vec{u}| = 1. So, the formula simplifies to Du⃗f(P)=∣∇f(P)∣cos⁡θD_{\vec{u}}f(P) = |\nabla f(P)| \cos\theta. The maximum value of cos⁡θ\cos\theta is 1, which occurs when θ=0\theta = 0. Therefore, the maximum possible value of the directional derivative is ∣∇f(P)∣|\nabla f(P)|. We want to find the angle θ\theta for which the directional derivative is half of this maximum value: Du⃗f(P)=12∣∇f(P)∣D_{\vec{u}}f(P) = \frac{1}{2} |\nabla f(P)|. Substituting the formula for the directional derivative gives: ∣∇f(P)∣cos⁡θ=12∣∇f(P)∣|\nabla f(P)| \cos\theta = \frac{1}{2} |\nabla f(P)|. Assuming ∇f(P)\nabla f(P) is not the zero vector, we can divide both sides by ∣∇f(P)∣|\nabla f(P)| to get: cos⁡θ=1/2\cos\theta = 1/2. The angle θ\theta in the range [0,π][0, \pi] for which cos⁡θ=1/2\cos\theta = 1/2 is θ=π/3\theta = \pi/3. (A) π/6\pi/6 would be correct if cos⁡θ=3/2\cos\theta = \sqrt{3}/2. (B) π/4\pi/4 corresponds to cos⁡θ=2/2\cos\theta = \sqrt{2}/2. (D) 2π/32\pi/3 corresponds to cos⁡θ=−1/2\cos\theta = -1/2, which would give a value that is half the minimum possible value.

Question 7

The electric potential in a region is given by V(x,y)=kx2+y2V(x,y) = \frac{k}{\sqrt{x^2 + y^2}} where k>0k > 0 is a constant. The electric field is E⃗=−∇V\vec{E} = -\nabla V. At the point (3,4)(3, 4), what is the magnitude of the electric field?

  1. k125\frac{k}{125}
  2. k25\frac{k}{25} (correct answer)
  3. k5\frac{k}{5}
  4. k125\frac{k}{\sqrt{125}}
Explanation: First, compute the gradient of V(x,y)=k(x2+y2)−1/2V(x,y) = k(x^2 + y^2)^{-1/2}. Using the chain rule: Vx=k⋅(−12)(x2+y2)−3/2⋅2x=−kx(x2+y2)3/2V_x = k \cdot (-\frac{1}{2})(x^2 + y^2)^{-3/2} \cdot 2x = -\frac{kx}{(x^2 + y^2)^{3/2}} and Vy=−ky(x2+y2)3/2V_y = -\frac{ky}{(x^2 + y^2)^{3/2}}. So ∇V=⟨−kx(x2+y2)3/2,−ky(x2+y2)3/2⟩\nabla V = \langle -\frac{kx}{(x^2 + y^2)^{3/2}}, -\frac{ky}{(x^2 + y^2)^{3/2}} \rangle. At (3,4)(3, 4): x2+y2=9+16=25x^2 + y^2 = 9 + 16 = 25, so (x2+y2)3/2=253/2=125(x^2 + y^2)^{3/2} = 25^{3/2} = 125. Therefore: ∇V(3,4)=⟨−3k125,−4k125⟩\nabla V(3, 4) = \langle -\frac{3k}{125}, -\frac{4k}{125} \rangle. The electric field is E⃗=−∇V=⟨3k125,4k125⟩\vec{E} = -\nabla V = \langle \frac{3k}{125}, \frac{4k}{125} \rangle. The magnitude is: ∣∣E⃗∣∣=(3k125)2+(4k125)2=9k2+16k21252=25k21252=5k125=k25||\vec{E}|| = \sqrt{\left(\frac{3k}{125}\right)^2 + \left(\frac{4k}{125}\right)^2} = \sqrt{\frac{9k^2 + 16k^2}{125^2}} = \sqrt{\frac{25k^2}{125^2}} = \frac{5k}{125} = \frac{k}{25}.

Question 8

The temperature distribution in a metal plate is given by T(x,y)=100−x2−2y2+xyT(x,y) = 100 - x^2 - 2y^2 + xy. An ant at position (2,1)(2, 1) wants to move in the direction of steepest temperature increase. If the ant moves a small distance ϵ\epsilon in this optimal direction, approximately how much will the temperature increase?

  1. ϵ2\epsilon \sqrt{2}
  2. ϵ5\epsilon \sqrt{5}
  3. ϵ10\epsilon \sqrt{10}
  4. ϵ13\epsilon \sqrt{13} (correct answer)
Explanation: The direction of steepest increase is given by the gradient vector ∇T\nabla T. First, compute: Tx=−2x+yT_x = -2x + y and Ty=−4y+xT_y = -4y + x. At (2,1)(2,1): Tx(2,1)=−4+1=−3T_x(2,1) = -4 + 1 = -3 and Ty(2,1)=−4+2=−2T_y(2,1) = -4 + 2 = -2. So ∇T(2,1)=⟨−3,−2⟩\nabla T(2,1) = \langle -3, -2 \rangle. The magnitude is ∣∣∇T(2,1)∣∣=(−3)2+(−2)2=9+4=13||\nabla T(2,1)|| = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}. When moving a small distance ϵ\epsilon in the direction of the gradient, the temperature change is approximately ϵ⋅∣∣∇T∣∣=ϵ13\epsilon \cdot ||\nabla T|| = \epsilon\sqrt{13}. Note that the gradient points in the direction of steepest increase, and its magnitude gives the rate of change per unit distance in that direction.

Question 9

Let f:R2→Rf: \mathbb{R}^2 \to \mathbb{R} be a differentiable function and let PP be a point in its domain. Which of the following statements about the gradient ∇f(P)\nabla f(P) is necessarily true?

  1. If ∇f(P)=0⃗\nabla f(P) = \vec{0}, then ff has a local extremum (maximum or minimum) at PP.
  2. The vector ∇f(P)\nabla f(P) is parallel to the tangent line of the level curve of ff through PP.
  3. If u⃗\vec{u} and v⃗\vec{v} are any two orthogonal unit vectors, then (Du⃗f(P))2+(Dv⃗f(P))2=∣∇f(P)∣2(D_{\vec{u}}f(P))^2 + (D_{\vec{v}}f(P))^2 = |\nabla f(P)|^2. (correct answer)
  4. For any unit vector u⃗\vec{u}, the directional derivative Du⃗f(P)D_{\vec{u}}f(P) is strictly less than ∣∇f(P)∣|\nabla f(P)|.
Explanation: (A) is false. If ∇f(P)=0⃗\nabla f(P) = \vec{0}, PP is a critical point, but it could be a saddle point, which is not a local extremum. (B) is false. The gradient vector ∇f(P)\nabla f(P) is orthogonal (normal) to the level curve of ff through PP, not parallel to its tangent line. (C) is true. Let ∇f(P)=g⃗\nabla f(P) = \vec{g}. The directional derivatives are Du⃗f(P)=g⃗⋅u⃗D_{\vec{u}}f(P) = \vec{g} \cdot \vec{u} and Dv⃗f(P)=g⃗⋅v⃗D_{\vec{v}}f(P) = \vec{g} \cdot \vec{v}. Since u⃗\vec{u} and v⃗\vec{v} are orthogonal unit vectors, they form an orthonormal basis for R2\mathbb{R}^2. The vector g⃗\vec{g} can be expressed in this basis as g⃗=(g⃗⋅u⃗)u⃗+(g⃗⋅v⃗)v⃗\vec{g} = (\vec{g} \cdot \vec{u})\vec{u} + (\vec{g} \cdot \vec{v})\vec{v}. The squared magnitude of g⃗\vec{g} is given by the sum of the squares of its components in this orthonormal basis: ∣g⃗∣2=(g⃗⋅u⃗)2+(g⃗⋅v⃗)2|\vec{g}|^2 = (\vec{g} \cdot \vec{u})^2 + (\vec{g} \cdot \vec{v})^2. Substituting the definitions, we get ∣∇f(P)∣2=(Du⃗f(P))2+(Dv⃗f(P))2|\nabla f(P)|^2 = (D_{\vec{u}}f(P))^2 + (D_{\vec{v}}f(P))^2. This is the 2D version of Parseval's identity. (D) is false. The directional derivative is Du⃗f(P)=∣∇f(P)∣cos⁡θD_{\vec{u}}f(P) = |\nabla f(P)|\cos\theta. If u⃗\vec{u} points in the same direction as ∇f(P)\nabla f(P), then θ=0\theta=0 and cos⁡θ=1\cos\theta=1, making Du⃗f(P)=∣∇f(P)∣D_{\vec{u}}f(P) = |\nabla f(P)|. The relationship is Du⃗f(P)≤∣∇f(P)∣D_{\vec{u}}f(P) \le |\nabla f(P)|, not strictly less than.

Question 10

Let f(x,y)=x2exyf(x, y) = x^2 e^{xy}. Which of the following vectors represents the direction of the most rapid decrease of the function ff at the point P(1,0)P(1, 0)?

  1. ⟨−2,−1⟩\langle -2, -1 \rangle (correct answer)
  2. ⟨2,1⟩\langle 2, 1 \rangle
  3. ⟨1,−2⟩\langle 1, -2 \rangle
  4. ⟨−1,−2⟩\langle -1, -2 \rangle
Explanation: The direction of the most rapid increase of a function is given by its gradient vector, ∇f\nabla f. The direction of the most rapid decrease is in the opposite direction, −∇f-\nabla f. First, we compute the partial derivatives of f(x,y)=x2exyf(x, y) = x^2 e^{xy}. Using the product rule for fxf_x: fx(x,y)=∂∂x(x2exy)=2xexy+x2(yexy)=(2x+x2y)exyf_x(x,y) = \frac{\partial}{\partial x}(x^2 e^{xy}) = 2xe^{xy} + x^2(ye^{xy}) = (2x+x^2y)e^{xy}. For fyf_y: fy(x,y)=∂∂y(x2exy)=x2(xexy)=x3exyf_y(x,y) = \frac{\partial}{\partial y}(x^2 e^{xy}) = x^2(xe^{xy}) = x^3e^{xy}. The gradient vector is ∇f(x,y)=⟨(2x+x2y)exy,x3exy⟩\nabla f(x,y) = \langle (2x+x^2y)e^{xy}, x^3e^{xy} \rangle. Now, we evaluate the gradient at the point P(1,0)P(1, 0): fx(1,0)=(2(1)+12(0))e1⋅0=2e0=2f_x(1, 0) = (2(1)+1^2(0))e^{1 \cdot 0} = 2e^0 = 2. fy(1,0)=13e1⋅0=1e0=1f_y(1, 0) = 1^3 e^{1 \cdot 0} = 1e^0 = 1. So, ∇f(1,0)=⟨2,1⟩\nabla f(1, 0) = \langle 2, 1 \rangle. This is the direction of the greatest increase. The direction of the most rapid decrease is −∇f(1,0)=−⟨2,1⟩=⟨−2,−1⟩-\nabla f(1, 0) = -\langle 2, 1 \rangle = \langle -2, -1 \rangle.