Multivariable Calculus Quiz: Gradient Steepest Ascent Descent
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Gradient Steepest Ascent DescentQuestion 1 of 13

The temperature in a region of the xyxy-plane is given by T(x,y)=100e−x2−2y2T(x,y) = 100e^{-x^2-2y^2}. A heat-seeking particle is located at the point (1,−1)(1, -1). If the particle moves from (1,−1)(1,-1) in the direction of steepest temperature increase, what is the initial rate of change of its yy-coordinate with respect to its xx-coordinate?

−2-2
−1/2-1/2
22
1/21/2
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Gradient Steepest Ascent Descent

Practice Gradient Steepest Ascent Descent 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 Steepest Ascent Descent, 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

The temperature in a region of the xyxy-plane is given by T(x,y)=100e−x2−2y2T(x,y) = 100e^{-x^2-2y^2}. A heat-seeking particle is located at the point (1,−1)(1, -1). If the particle moves from (1,−1)(1,-1) in the direction of steepest temperature increase, what is the initial rate of change of its yy-coordinate with respect to its xx-coordinate?

  1. −2-2 (correct answer)
  2. −1/2-1/2
  3. 22
  4. 1/21/2
Explanation: The direction of steepest increase is given by the gradient vector ∇T\nabla T. First, we compute the partial derivatives: ∂T∂x=−200xe−x2−2y2\frac{\partial T}{\partial x} = -200xe^{-x^2-2y^2} and ∂T∂y=−400ye−x2−2y2\frac{\partial T}{\partial y} = -400ye^{-x^2-2y^2}. At the point (1,−1)(1, -1), the gradient is ∇T(1,−1)=⟨−200e−3,400e−3⟩=200e−3⟨−1,2⟩\nabla T(1, -1) = \langle -200e^{-3}, 400e^{-3} \rangle = 200e^{-3} \langle -1, 2 \rangle. The direction vector of the particle's initial movement is ⟨−1,2⟩\langle -1, 2 \rangle. The rate of change of its yy-coordinate with respect to its xx-coordinate is the slope dydx\frac{dy}{dx}, which is the ratio of the components of the direction vector. Thus, dydx=2−1=−2\frac{dy}{dx} = \frac{2}{-1} = -2.

Question 2

A drone is flying over a landscape whose altitude is given by h(x,y)=10001+x2+y2h(x, y) = \frac{1000}{1 + x^2 + y^2}. The drone is at the point corresponding to (x,y)=(3,4)(x,y) = (3, 4). To conserve battery, the drone's flight controller is programmed to descend along the path of steepest descent. Which of the following vectors is parallel to the drone's initial horizontal velocity vector?

  1. ⟨3,4⟩\langle 3, 4 \rangle (correct answer)
  2. ⟨−3,−4⟩\langle -3, -4 \rangle
  3. ⟨4,−3⟩\langle 4, -3 \rangle
  4. ⟨−4,−3⟩\langle -4, -3 \rangle
Explanation: The direction of steepest descent is given by the negative of the gradient vector, −∇h-\nabla h. The function is h(x,y)=1000(1+x2+y2)−1h(x,y) = 1000(1+x^2+y^2)^{-1}. The partial derivatives are ∂h∂x=−2000x(1+x2+y2)2\frac{\partial h}{\partial x} = \frac{-2000x}{(1+x^2+y^2)^2} and ∂h∂y=−2000y(1+x2+y2)2\frac{\partial h}{\partial y} = \frac{-2000y}{(1+x^2+y^2)^2}. So, ∇h(x,y)=−2000(1+x2+y2)2⟨x,y⟩\nabla h(x,y) = \frac{-2000}{(1+x^2+y^2)^2} \langle x, y \rangle. At the point (3,4)(3,4), ∇h(3,4)\nabla h(3,4) is proportional to ⟨−3,−4⟩\langle -3, -4 \rangle. This is the direction of steepest ascent. The direction of steepest descent is −∇h-\nabla h, which is proportional to −⟨−3,−4⟩=⟨3,4⟩- \langle -3, -4 \rangle = \langle 3, 4 \rangle.

Question 3

A hiker is standing on a hill whose shape is modeled by the surface z=400−x2−3y2z = 400 - x^2 - 3y^2. The hiker is at the point corresponding to (x,y)=(5,2)(x,y) = (5, 2). They want to start walking in the direction of steepest ascent along the hill's surface. What is the cosine of the angle that their initial direction vector in the xyxy-plane makes with the positive yy-axis?

  1. 229\frac{2}{\sqrt{29}}
  2. 661\frac{6}{\sqrt{61}}
  3. −561\frac{-5}{\sqrt{61}}
  4. −661\frac{-6}{\sqrt{61}} (correct answer)
Explanation: When you encounter questions about the direction of steepest ascent on a surface, you're working with the gradient vector, which always points in the direction of greatest increase of a function. To find the direction of steepest ascent, you need to calculate the gradient of the surface function z=400−x2−3y2z = 400 - x^2 - 3y^2. The gradient is ∇z=⟨−2x,−6y⟩\nabla z = \langle -2x, -6y \rangle. At the point (5,2)(5, 2), this becomes ∇z=⟨−10,−12⟩\nabla z = \langle -10, -12 \rangle. This gradient vector represents the direction of steepest ascent projected onto the xyxy-plane. To find the cosine of the angle this vector makes with the positive yy-axis, you use the dot product formula. The positive yy-axis direction vector is ⟨0,1⟩\langle 0, 1 \rangle. The cosine equals ⟨−10,−12⟩⋅⟨0,1⟩∣⟨−10,−12⟩∣⋅∣⟨0,1⟩∣=−12100+144=−12244=−12261=−661\frac{\langle -10, -12 \rangle \cdot \langle 0, 1 \rangle}{|\langle -10, -12 \rangle| \cdot |\langle 0, 1 \rangle|} = \frac{-12}{\sqrt{100 + 144}} = \frac{-12}{\sqrt{244}} = \frac{-12}{2\sqrt{61}} = \frac{-6}{\sqrt{61}} Choice A gives 229\frac{2}{\sqrt{29}}, which appears to use incorrect gradient components. Choice B gives 661\frac{6}{\sqrt{61}}, which has the correct magnitude but wrong sign—this would be the cosine if you mistakenly used the direction of steepest descent. Choice C gives −561\frac{-5}{\sqrt{61}}, suggesting an error in calculating the xx-component of the gradient. The correct answer is D: −661\frac{-6}{\sqrt{61}}. Remember: the gradient always points toward steepest ascent, and be careful with signs when computing partial derivatives and dot products.

Question 4

The elevation of a landscape is given by h(x,y)=xsin⁡(y)h(x,y) = x \sin(y). A person is standing at the point (3,π/6)(\sqrt{3}, \pi/6). They start to walk in the direction of steepest ascent. Let u⃗\vec{u} be the unit vector representing this direction. What is the product of the components of u⃗\vec{u}?

  1. 3/43/4
  2. 3/103/10 (correct answer)
  3. 1/21/2
  4. 33
Explanation: When you encounter a problem about the "direction of steepest ascent," you're working with the gradient vector, which points in the direction of maximum increase of a function. To find the direction of steepest ascent for h(x,y)=xsin⁡(y)h(x,y) = x \sin(y), you need to compute the gradient ∇h=(∂h∂x,∂h∂y)\nabla h = \left(\frac{\partial h}{\partial x}, \frac{\partial h}{\partial y}\right). Taking the partial derivatives: ∂h∂x=sin⁡(y)\frac{\partial h}{\partial x} = \sin(y) and ∂h∂y=xcos⁡(y)\frac{\partial h}{\partial y} = x \cos(y). At the point (3,π/6)(\sqrt{3}, \pi/6), the gradient becomes ∇h=(sin⁡(π/6),3cos⁡(π/6))=(1/2,3⋅3/2)=(1/2,3/2)\nabla h = (\sin(\pi/6), \sqrt{3} \cos(\pi/6)) = (1/2, \sqrt{3} \cdot \sqrt{3}/2) = (1/2, 3/2). Since the question asks for the unit vector u⃗\vec{u}, you must normalize this gradient. The magnitude is ∣∇h∣=(1/2)2+(3/2)2=1/4+9/4=10/4=10/2|\nabla h| = \sqrt{(1/2)^2 + (3/2)^2} = \sqrt{1/4 + 9/4} = \sqrt{10/4} = \sqrt{10}/2. Therefore, u⃗=(1/2,3/2)10/2=(110,310)\vec{u} = \frac{(1/2, 3/2)}{\sqrt{10}/2} = \left(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}\right). The product of components is 110⋅310=310\frac{1}{\sqrt{10}} \cdot \frac{3}{\sqrt{10}} = \frac{3}{10}, which is answer B. Answer A (3/4) likely comes from multiplying the unnormalized gradient components. Answer C (1/2) might result from using only one component. Answer D (3) could come from forgetting to square the denominator when finding the product. Remember: the direction of steepest ascent is always the normalized gradient vector, so don't forget the normalization step when the problem specifically asks for a unit vector.

Question 5

Let f(x,y)f(x,y) be a differentiable function. A particle starts at (1,1)(1,1) and moves along the path of steepest ascent of ff. Its path is parameterized by r⃗(t)=⟨x(t),y(t)⟩\vec{r}(t) = \langle x(t), y(t) \rangle, with r⃗(0)=⟨1,1⟩\vec{r}(0) = \langle 1, 1 \rangle. If the gradient of ff is given by ∇f(x,y)=⟨2x,8y⟩\nabla f(x,y) = \langle 2x, 8y \rangle, which of the following differential equations describes the particle's path in the xyxy-plane?

  1. dydx=x4y\frac{dy}{dx} = \frac{x}{4y}
  2. dydx=−x4y\frac{dy}{dx} = -\frac{x}{4y}
  3. dydx=4yx\frac{dy}{dx} = \frac{4y}{x} (correct answer)
  4. y=x4y = x^4
Explanation: When you encounter a problem about the path of steepest ascent, remember that a particle always moves in the direction of the gradient vector. The gradient ∇f=⟨2x,8y⟩\nabla f = \langle 2x, 8y \rangle points in the direction of steepest increase, so the particle's velocity vector r⃗′(t)=⟨x′(t),y′(t)⟩\vec{r}'(t) = \langle x'(t), y'(t) \rangle must be parallel to this gradient. Since the velocity vector is parallel to the gradient, we can write ⟨x′(t),y′(t)⟩=k⟨2x,8y⟩\langle x'(t), y'(t) \rangle = k\langle 2x, 8y \rangle for some positive scalar kk. This gives us x′(t)=2kxx'(t) = 2kx and y′(t)=8kyy'(t) = 8ky. To find the relationship between xx and yy, we use the chain rule: dydx=dy/dtdx/dt=y′(t)x′(t)=8ky2kx=4yx\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y'(t)}{x'(t)} = \frac{8ky}{2kx} = \frac{4y}{x}. Choice A gives dydx=x4y\frac{dy}{dx} = \frac{x}{4y}, which would result from incorrectly swapping the components or forgetting the factor of 4. Choice B gives dydx=−x4y\frac{dy}{dx} = -\frac{x}{4y}, representing the path of steepest descent (negative gradient direction) with an additional error. Choice D, y=x4y = x^4, might seem tempting if you tried to solve the differential equation incorrectly, but this specific function doesn't satisfy our differential equation with the given initial condition. Study tip: For steepest ascent problems, always remember that dydx=∂f/∂y∂f/∂x\frac{dy}{dx} = \frac{\partial f/\partial y}{\partial f/\partial x}. The ratio of gradient components gives you the slope of the path directly.

Question 6

A surface is defined implicitly by the equation x2+y3+z4=3x^2 + y^3 + z^4 = 3. Consider zz as a positive function of xx and yy, so z=f(x,y)z = f(x,y) with z>0z>0. At the point (1,1,1)(1,1,1), in which direction in the xyxy-plane does zz decrease most rapidly?

  1. The direction of the vector ⟨−2,3⟩\langle -2, 3 \rangle.
  2. The direction of the vector ⟨−2,−3⟩\langle -2, -3 \rangle.
  3. The direction of the vector ⟨2,3,4⟩\langle 2, 3, 4 \rangle.
  4. The direction of the vector ⟨2,3⟩\langle 2, 3 \rangle. (correct answer)
Explanation: When you encounter implicit surfaces and need to find the direction of steepest change, you're working with gradients and directional derivatives. The key insight is that for a function z=f(x,y)z = f(x,y), the direction of steepest decrease is opposite to the direction of the gradient ∇f\nabla f. Starting with the implicit equation x2+y3+z4=3x^2 + y^3 + z^4 = 3, you can find ∂z∂x\frac{\partial z}{\partial x} and ∂z∂y\frac{\partial z}{\partial y} using implicit differentiation. Differentiating with respect to xx: 2x+4z3∂z∂x=02x + 4z^3\frac{\partial z}{\partial x} = 0, so ∂z∂x=−2x4z3=−x2z3\frac{\partial z}{\partial x} = -\frac{2x}{4z^3} = -\frac{x}{2z^3}. Similarly, ∂z∂y=−3y24z3\frac{\partial z}{\partial y} = -\frac{3y^2}{4z^3}. At the point (1,1,1)(1,1,1), these partial derivatives become ∂z∂x=−12\frac{\partial z}{\partial x} = -\frac{1}{2} and ∂z∂y=−34\frac{\partial z}{\partial y} = -\frac{3}{4}. The gradient is ∇f=⟨−12,−34⟩\nabla f = \langle -\frac{1}{2}, -\frac{3}{4} \rangle, which points in the direction of steepest increase. The direction of steepest decrease is the opposite: ⟨12,34⟩\langle \frac{1}{2}, \frac{3}{4} \rangle, or equivalently ⟨2,3⟩\langle 2, 3 \rangle when scaled. Choice A gives ⟨−2,3⟩\langle -2, 3 \rangle, which has the wrong sign for the xx-component. Choice B gives ⟨−2,−3⟩\langle -2, -3 \rangle, which points in the direction of steepest increase, not decrease. Choice C includes a zz-component, but we need direction in the xyxy-plane only. Choice D correctly gives ⟨2,3⟩\langle 2, 3 \rangle. Remember: the gradient points toward steepest increase, so steepest decrease is in the opposite direction. Always check your signs carefully in implicit differentiation problems.

Question 7

The level curve of a differentiable function f(x,y)f(x,y) passing through the point P(2,−1)P(2, -1) is given by the equation x2−2y3=6x^2 - 2y^3 = 6. In which direction from point PP does the function ff increase most rapidly?

  1. The direction of the vector ⟨4,−6⟩\langle 4, -6 \rangle. (correct answer)
  2. The direction of the vector ⟨−4,6⟩\langle -4, 6 \rangle.
  3. The direction of the vector ⟨4,6⟩\langle 4, 6 \rangle.
  4. The direction of the vector ⟨6,4⟩\langle 6, 4 \rangle.
Explanation: The direction of most rapid increase of a function is given by its gradient vector, ∇f\nabla f. The gradient vector at a point is always orthogonal to the level curve of the function passing through that point. Let g(x,y)=x2−2y3g(x,y) = x^2 - 2y^3. The given level curve is g(x,y)=6g(x,y)=6. The gradient of gg is ∇g(x,y)=⟨2x,−6y2⟩\nabla g(x,y) = \langle 2x, -6y^2 \rangle. At the point P(2,−1)P(2, -1), the gradient is ∇g(2,−1)=⟨2(2),−6(−1)2⟩=⟨4,−6⟩\nabla g(2, -1) = \langle 2(2), -6(-1)^2 \rangle = \langle 4, -6 \rangle. The function ff must be a monotonic transformation of gg in the vicinity of P, so its gradient will be parallel to ∇g\nabla g. Thus, the direction of steepest ascent is given by the vector ⟨4,−6⟩\langle 4, -6 \rangle.

Question 8

The concentration of a pollutant in a lake is given by C(x,y)=10ln⁡(x2+y2+1)C(x,y) = 10 \ln(x^2 + y^2 + 1), where xx and yy are distances in kilometers from a central point. Let RAR_A be the maximum rate of change of concentration at point A(1,2)A(1,2), and RBR_B be the maximum rate of change at point B(3,1)B(3,1). What is the ratio RA/RBR_A / R_B?

  1. 6211\frac{6\sqrt{2}}{11}
  2. 11212\frac{11\sqrt{2}}{12} (correct answer)
  3. 116\frac{11}{6}
  4. 22\frac{\sqrt{2}}{2}
Explanation: When you encounter a question about "maximum rate of change" for a multivariable function, you're being asked to find the magnitude of the gradient vector at each point. The gradient gives the direction and rate of steepest increase. For C(x,y)=10ln⁡(x2+y2+1)C(x,y) = 10 \ln(x^2 + y^2 + 1), you need to compute the partial derivatives. Using the chain rule: ∂C∂x=10⋅2xx2+y2+1\frac{\partial C}{\partial x} = 10 \cdot \frac{2x}{x^2 + y^2 + 1} and ∂C∂y=10⋅2yx2+y2+1\frac{\partial C}{\partial y} = 10 \cdot \frac{2y}{x^2 + y^2 + 1}, so ∇C=20x2+y2+1⟨x,y⟩\nabla C = \frac{20}{x^2 + y^2 + 1}\langle x, y \rangle. At point A(1,2)A(1,2): ∇CA=201+4+1⟨1,2⟩=206⟨1,2⟩\nabla C_A = \frac{20}{1 + 4 + 1}\langle 1, 2 \rangle = \frac{20}{6}\langle 1, 2 \rangle. The magnitude is RA=20612+22=2056R_A = \frac{20}{6}\sqrt{1^2 + 2^2} = \frac{20\sqrt{5}}{6}. At point B(3,1)B(3,1): ∇CB=209+1+1⟨3,1⟩=2011⟨3,1⟩\nabla C_B = \frac{20}{9 + 1 + 1}\langle 3, 1 \rangle = \frac{20}{11}\langle 3, 1 \rangle. The magnitude is RB=201132+12=201011R_B = \frac{20}{11}\sqrt{3^2 + 1^2} = \frac{20\sqrt{10}}{11}. The ratio is RARB=205/62010/11=115610=115625=1162=11212\frac{R_A}{R_B} = \frac{20\sqrt{5}/6}{20\sqrt{10}/11} = \frac{11\sqrt{5}}{6\sqrt{10}} = \frac{11\sqrt{5}}{6\sqrt{2}\sqrt{5}} = \frac{11}{6\sqrt{2}} = \frac{11\sqrt{2}}{12}, confirming answer B. Option A uses the wrong denominator calculation. Option C forgets to rationalize the denominator. Option D likely comes from incorrectly computing one of the gradient magnitudes. Remember: maximum rate of change always equals ∣∇f∣|\nabla f| at any point. Practice computing gradients and their magnitudes systematically.

Question 9

Consider the function g(x,y)=exy+sin⁡(x+y)g(x, y) = e^{xy} + \sin(x + y). At which of the following points does the direction of steepest ascent make a 45°45° angle with the positive x-axis?

  1. At the point where ∂g∂x=∂g∂y\frac{\partial g}{\partial x} = \frac{\partial g}{\partial y} (correct answer)
  2. At any point where x=yx = y
  3. At the point where ∂g∂x=−∂g∂y\frac{\partial g}{\partial x} = -\frac{\partial g}{\partial y}
  4. At any critical point of gg
Explanation: The direction of steepest ascent is given by the gradient vector ∇g=(gx,gy)\nabla g = (g_x, g_y). For this direction to make a 45°45° angle with the positive x-axis, we need the gradient vector to be proportional to (1,1)(1, 1), which means gx=gyg_x = g_y. Choice B is incorrect because x=yx = y doesn't guarantee gx=gyg_x = g_y for this function. Choice C would give a 135°135° angle (or −45°-45°). Choice D refers to points where the gradient is zero, so no direction of steepest ascent exists.

Question 10

Consider the surface z=ln⁡(x2+y2+1)z = \ln(x^2 + y^2 + 1). A particle starts at point (1,0,ln⁡2)(1, 0, \ln 2) and moves along the surface in the direction of steepest ascent. After moving a horizontal distance of 12\frac{1}{2} (in the xy-plane), approximately what is the particle's new z-coordinate?

  1. ln⁡2+12\ln 2 + \frac{1}{2} (correct answer)
  2. ln⁡2+14\ln 2 + \frac{1}{4}
  3. ln⁡(2.25)\ln(2.25)
  4. ln⁡2+1\ln 2 + 1
Explanation: ∇z=(2xx2+y2+1,2yx2+y2+1)\nabla z = \left(\frac{2x}{x^2 + y^2 + 1}, \frac{2y}{x^2 + y^2 + 1}\right). At (1,0)(1, 0): ∇z=(22,02)=(1,0)\nabla z = \left(\frac{2}{2}, \frac{0}{2}\right) = (1, 0). The direction of steepest ascent in the xy-plane is (1,0)(1, 0), which is already a unit vector. Moving distance 12\frac{1}{2} in this direction gives new xy-position (1.5,0)(1.5, 0). The rate of change of z with respect to horizontal distance is ∣∇z∣=1|\nabla z| = 1. Therefore, the change in z is approximately 1×12=121 \times \frac{1}{2} = \frac{1}{2}. New z-coordinate: ln⁡2+12\ln 2 + \frac{1}{2}. Choice B incorrectly uses 14\frac{1}{4} instead of 12\frac{1}{2}. Choice C calculates the exact z-value at (1.5,0)(1.5, 0) but this would be ln⁡(2.25+1)=ln⁡(3.25)≠ln⁡(2.25)\ln(2.25 + 1) = \ln(3.25) \neq \ln(2.25). Choice D uses an incorrect rate of change.

Question 11

A hiker is at point (2,1,3)(2, 1, 3) on a mountain described by the elevation function f(x,y)=20−x2−2y2+4x+6yf(x, y) = 20 - x^2 - 2y^2 + 4x + 6y. If the hiker wants to ascend as quickly as possible and takes a step of length 0.50.5 in the direction of steepest ascent, what will be the hiker's new position?

  1. (2.5,1.5,3)(2.5, 1.5, 3)
  2. (2.4,1.8,3)(2.4, 1.8, 3)
  3. (1.6,0.2,3)(1.6, 0.2, 3)
  4. (2.2,1.4,3)(2.2, 1.4, 3) (correct answer)
Explanation: The gradient is ∇f=(−2x+4,−4y+6)\nabla f = (-2x + 4, -4y + 6). At (2,1)(2, 1), ∇f=(0,2)\nabla f = (0, 2). The unit vector in the direction of steepest ascent is (0,2)∣(0,2)∣=(0,1)\frac{(0, 2)}{|(0, 2)|} = (0, 1). Moving 0.5 units gives new position (2,1)+0.5(0,1)=(2,1.4)(2, 1) + 0.5(0, 1) = (2, 1.4). The z-coordinate remains 3 since we're tracking the hiker's position, not elevation. Choice A uses an incorrect unit vector calculation. Choice B assumes equal movement in both directions. Choice C moves in the wrong direction (steepest descent).

Question 12

A function h(x,y)=ax2+bxy+cy2h(x, y) = ax^2 + bxy + cy^2 has the property that at point (1,−1)(1, -1), the direction of steepest descent is parallel to the vector ⟨3,−2⟩\langle 3, -2 \rangle. If h(1,−1)=5h(1, -1) = 5, which constraint must the coefficients satisfy?

  1. 2a−b=3k2a - b = 3k and b−2c=−2kb - 2c = -2k for some positive constant kk
  2. 2a−b=−3k2a - b = -3k and b−2c=2kb - 2c = 2k for some positive constant kk (correct answer)
  3. 2a−b=32a - b = 3 and b−2c=−2b - 2c = -2
  4. a+c=5a + c = 5 and 2a−b=3(b−2c)2a - b = 3(b - 2c)
Explanation: ∇h=(2ax+by,bx+2cy)\nabla h = (2ax + by, bx + 2cy). At (1,−1)(1, -1): ∇h=(2a−b,b−2c)\nabla h = (2a - b, b - 2c). The direction of steepest descent is −∇h=(−(2a−b),−(b−2c))-\nabla h = (-(2a - b), -(b - 2c)). For this to be parallel to ⟨3,−2⟩\langle 3, -2 \rangle, we need (−(2a−b),−(b−2c))=k(3,−2)(-(2a - b), -(b - 2c)) = k(3, -2) for some k>0k > 0. This gives −(2a−b)=3k-(2a - b) = 3k and −(b−2c)=−2k-(b - 2c) = -2k, so 2a−b=−3k2a - b = -3k and b−2c=2kb - 2c = 2k. Choice A has the wrong signs. Choice C assumes k=1k = 1 without justification. Choice D incorrectly uses the condition h(1,−1)=5h(1, -1) = 5.

Question 13

Let f(x,y)=x3−6xy+y2f(x,y) = x^3 - 6xy + y^2. At which point P(x,y)P(x,y) is the direction of steepest ascent parallel to the vector v⃗=⟨−1,1⟩\vec{v} = \langle -1, 1 \rangle?

  1. (0,0)(0, 0)
  2. (2,0)(2, 0)
  3. (−2,6)(-2, 6) (correct answer)
  4. (6,18)(6, 18)
Explanation: When you encounter a question about the direction of steepest ascent, you're working with gradients. The direction of steepest ascent at any point is always given by the gradient vector ∇f\nabla f at that point. To solve this, you need to find where the gradient is parallel to v⃗=⟨−1,1⟩\vec{v} = \langle -1, 1 \rangle. First, compute the gradient: ∇f=⟨∂f∂x,∂f∂y⟩=⟨3x2−6y,−6x+2y⟩\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle = \langle 3x^2 - 6y, -6x + 2y \rangle. For two vectors to be parallel, one must be a scalar multiple of the other. So you need ∇f=kv⃗\nabla f = k\vec{v} for some scalar kk. This gives you: 3x2−6y=−k3x^2 - 6y = -k and −6x+2y=k-6x + 2y = k From the second equation, k=−6x+2yk = -6x + 2y. Substituting into the first: 3x2−6y=−(−6x+2y)=6x−2y3x^2 - 6y = -(-6x + 2y) = 6x - 2y. Simplifying: 3x2−6x−4y=03x^2 - 6x - 4y = 0, so y=3x2−6x4y = \frac{3x^2 - 6x}{4}. From k=−6x+2yk = -6x + 2y, substituting the expression for yy: k=−6x+2⋅3x2−6x4=3x2−18x2k = -6x + 2 \cdot \frac{3x^2 - 6x}{4} = \frac{3x^2 - 18x}{2}. For the vectors to point in the same direction (not opposite), you need k>0k > 0, which means x(x−6)>0x(x-6) > 0, so x<0x < 0 or x>6x > 6. Testing the given points: For (−2,6)(-2, 6): y=3(−2)2−6(−2)4=244=6y = \frac{3(-2)^2 - 6(-2)}{4} = \frac{24}{4} = 6 ✓, and x=−2<0x = -2 < 0 ✓. Choice A gives the zero gradient (no steepest direction), while choices B and D don't satisfy the parallel condition with the correct orientation. Remember: steepest ascent problems always involve finding where the gradient aligns with a given direction vector.