Multivariable Calculus Quiz: Directional Derivatives
13 questions · exam conditions
0:00
Directional DerivativesQuestion 1 of 13

At a certain point PP, the maximum rate of change of a differentiable function ff is 1212. What is the directional derivative of ff at PP in a direction that makes an angle of 2π/32\pi/3 with the direction of maximum rate of change?

−6-6
66
636\sqrt{3}
1212
← Back to quizzes

Multivariable Calculus Quiz

Multivariable Calculus Quiz: Directional Derivatives

Practice Directional Derivatives 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 Directional Derivatives, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

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.

All questions

Question 1

At a certain point PP, the maximum rate of change of a differentiable function ff is 1212. What is the directional derivative of ff at PP in a direction that makes an angle of 2π/32\pi/3 with the direction of maximum rate of change?

  1. −6-6 (correct answer)
  2. 66
  3. 636\sqrt{3}
  4. 1212
Explanation: The directional derivative of ff at a point PP in the direction of a unit vector u⃗\vec{u} is given by Du⃗f(P)=∣∣∇f(P)∣∣cos⁡θD_{\vec{u}}f(P) = ||\nabla f(P)|| \cos\theta, where θ\theta is the angle between the gradient vector ∇f(P)\nabla f(P) and the direction vector u⃗\vec{u}. The maximum rate of change is the magnitude of the gradient, so ∣∣∇f(P)∣∣=12||\nabla f(P)|| = 12. The direction of maximum rate of change is the direction of the gradient. The problem states that the angle between our direction u⃗\vec{u} and the direction of the gradient is θ=2π/3\theta = 2\pi/3. Therefore, the directional derivative is Du⃗f(P)=12cos⁡(2π/3)=12(−1/2)=−6D_{\vec{u}}f(P) = 12 \cos(2\pi/3) = 12(-1/2) = -6.

Question 2

The directional derivative of a differentiable function f(x,y)f(x,y) at P(1,2)P(1,2) in the direction toward Q(3,3)Q(3,3) is 252\sqrt{5}. The directional derivative at PP in the direction toward R(0,4)R(0,4) is 00. What is the gradient of ff at P(1,2)P(1,2)?

  1. ⟨4,2⟩\langle 4, 2 \rangle (correct answer)
  2. ⟨2,4⟩\langle 2, 4 \rangle
  3. ⟨45,25⟩\langle \frac{4}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle
  4. ⟨203,−103⟩\langle \frac{20}{3}, -\frac{10}{3} \rangle
Explanation: Let ∇f(1,2)=⟨a,b⟩\nabla f(1,2) = \langle a, b \rangle. The direction from PP to QQ is v⃗1=Q−P=⟨2,1⟩\vec{v}_1 = Q-P = \langle 2, 1 \rangle. The unit vector is u⃗1=122+12⟨2,1⟩=15⟨2,1⟩\vec{u}_1 = \frac{1}{\sqrt{2^2+1^2}} \langle 2, 1 \rangle = \frac{1}{\sqrt{5}} \langle 2, 1 \rangle. The directional derivative is Du⃗1f=∇f⋅u⃗1=⟨a,b⟩⋅15⟨2,1⟩=2a+b5D_{\vec{u}_1}f = \nabla f \cdot \vec{u}_1 = \langle a, b \rangle \cdot \frac{1}{\sqrt{5}} \langle 2, 1 \rangle = \frac{2a+b}{\sqrt{5}}. We are given this is 252\sqrt{5}, so 2a+b5=25\frac{2a+b}{\sqrt{5}} = 2\sqrt{5}, which implies 2a+b=102a+b=10. The direction from PP to RR is v⃗2=R−P=⟨−1,2⟩\vec{v}_2 = R-P = \langle -1, 2 \rangle. The unit vector is u⃗2=1(−1)2+22⟨−1,2⟩=15⟨−1,2⟩\vec{u}_2 = \frac{1}{\sqrt{(-1)^2+2^2}} \langle -1, 2 \rangle = \frac{1}{\sqrt{5}} \langle -1, 2 \rangle. The directional derivative is Du⃗2f=∇f⋅u⃗2=⟨a,b⟩⋅15⟨−1,2⟩=−a+2b5D_{\vec{u}_2}f = \nabla f \cdot \vec{u}_2 = \langle a, b \rangle \cdot \frac{1}{\sqrt{5}} \langle -1, 2 \rangle = \frac{-a+2b}{\sqrt{5}}. We are given this is 00, so −a+2b=0-a+2b=0, which means a=2ba=2b. Substituting this into the first equation gives 2(2b)+b=102(2b)+b=10, so 5b=105b=10, and b=2b=2. Then a=2(2)=4a=2(2)=4. Thus, ∇f(1,2)=⟨4,2⟩\nabla f(1,2) = \langle 4, 2 \rangle.

Question 3

The altitude of a mountain (in meters) is modeled by h(x,y)=2000−0.01x2−0.02y2h(x,y) = 2000 - 0.01x^2 - 0.02y^2, where xx and yy are horizontal coordinates. A climber is at the point (10,10)(10, 10). If the climber moves in the direction of the vector v⃗=⟨1,−2⟩\vec{v} = \langle 1, -2 \rangle, what is the initial rate of change of their altitude?

  1. Ascending at a rate of 0.60.6 m per horizontal unit
  2. Descending at a rate of 0.60.6 m per horizontal unit
  3. Ascending at a rate of 3525\frac{3\sqrt{5}}{25} m per horizontal unit (correct answer)
  4. Descending at a rate of 3525\frac{3\sqrt{5}}{25} m per horizontal unit
Explanation: The rate of change of altitude is the directional derivative of h(x,y)h(x,y). First, find the gradient: ∇h=⟨−0.02x,−0.04y⟩\nabla h = \langle -0.02x, -0.04y \rangle. At the point (10,10)(10,10), the gradient is ∇h(10,10)=⟨−0.02(10),−0.04(10)⟩=⟨−0.2,−0.4⟩\nabla h(10,10) = \langle -0.02(10), -0.04(10) \rangle = \langle -0.2, -0.4 \rangle. The direction of movement is given by v⃗=⟨1,−2⟩\vec{v} = \langle 1, -2 \rangle. We need to use a unit vector, so we normalize v⃗\vec{v}: ∣∣v⃗∣∣=12+(−2)2=5||\vec{v}|| = \sqrt{1^2+(-2)^2} = \sqrt{5}. The unit vector is u⃗=15⟨1,−2⟩\vec{u} = \frac{1}{\sqrt{5}} \langle 1, -2 \rangle. The directional derivative is Du⃗h(10,10)=∇h(10,10)⋅u⃗=⟨−0.2,−0.4⟩⋅15⟨1,−2⟩=15(−0.2(1)−0.4(−2))=−0.2+0.85=0.65D_{\vec{u}}h(10,10) = \nabla h(10,10) \cdot \vec{u} = \langle -0.2, -0.4 \rangle \cdot \frac{1}{\sqrt{5}} \langle 1, -2 \rangle = \frac{1}{\sqrt{5}}(-0.2(1) - 0.4(-2)) = \frac{-0.2+0.8}{\sqrt{5}} = \frac{0.6}{\sqrt{5}}. This simplifies to 0.655=6550=3525\frac{0.6\sqrt{5}}{5} = \frac{6\sqrt{5}}{50} = \frac{3\sqrt{5}}{25}. Since the value is positive, the climber is ascending.

Question 4

Let f(x,y)=g(x2+y2)f(x,y) = g(x^2+y^2), where gg is a differentiable function of a single variable. If g′(2)=5g'(2)=5, what is the directional derivative of ff at the point (1,1)(1,1) in the direction of the vector v⃗=⟨3,4⟩\vec{v} = \langle 3, 4 \rangle?

  1. 10
  2. 14 (correct answer)
  3. 50
  4. 70
Explanation: First, we find the gradient of f(x,y)f(x,y) using the chain rule. Let u=x2+y2u = x^2+y^2. Then f(x,y)=g(u)f(x,y) = g(u). The partial derivatives are ∂f∂x=g′(u)∂u∂x=g′(x2+y2)⋅2x\frac{\partial f}{\partial x} = g'(u) \frac{\partial u}{\partial x} = g'(x^2+y^2) \cdot 2x and ∂f∂y=g′(u)∂u∂y=g′(x2+y2)⋅2y\frac{\partial f}{\partial y} = g'(u) \frac{\partial u}{\partial y} = g'(x^2+y^2) \cdot 2y. So, ∇f(x,y)=⟨2xg′(x2+y2),2yg′(x2+y2)⟩\nabla f(x,y) = \langle 2x g'(x^2+y^2), 2y g'(x^2+y^2) \rangle. At the point (1,1)(1,1), we have x2+y2=12+12=2x^2+y^2 = 1^2+1^2=2. We are given g′(2)=5g'(2)=5. So, ∇f(1,1)=⟨2(1)g′(2),2(1)g′(2)⟩=⟨2(5),2(5)⟩=⟨10,10⟩\nabla f(1,1) = \langle 2(1)g'(2), 2(1)g'(2) \rangle = \langle 2(5), 2(5) \rangle = \langle 10, 10 \rangle. The direction vector is v⃗=⟨3,4⟩\vec{v} = \langle 3, 4 \rangle. Its magnitude is ∣∣v⃗∣∣=32+42=25=5||\vec{v}|| = \sqrt{3^2+4^2}=\sqrt{25}=5. The unit vector is u⃗=⟨3/5,4/5⟩\vec{u} = \langle 3/5, 4/5 \rangle. The directional derivative is Du⃗f(1,1)=∇f(1,1)⋅u⃗=⟨10,10⟩⋅⟨3/5,4/5⟩=10(3/5)+10(4/5)=6+8=14D_{\vec{u}}f(1,1) = \nabla f(1,1) \cdot \vec{u} = \langle 10, 10 \rangle \cdot \langle 3/5, 4/5 \rangle = 10(3/5) + 10(4/5) = 6+8=14.

Question 5

At a point PP, the directional derivative of a function ff is 424\sqrt{2} in the direction of v⃗1=⟨1,1⟩\vec{v}_1 = \langle 1, 1 \rangle and 44 in the direction of v⃗2=⟨3,1⟩\vec{v}_2 = \langle \sqrt{3}, 1 \rangle. What is the maximum rate of change of ff at PP?

  1. 44
  2. 424\sqrt{2}
  3. 88 (correct answer)
  4. 828\sqrt{2}
Explanation: Let the gradient at PP be ∇f=⟨a,b⟩\nabla f = \langle a, b \rangle. The maximum rate of change is ∣∣∇f∣∣=a2+b2||\nabla f|| = \sqrt{a^2+b^2}. The unit vector for v⃗1\vec{v}_1 is u⃗1=12⟨1,1⟩\vec{u}_1 = \frac{1}{\sqrt{2}}\langle 1, 1 \rangle. The directional derivative is Du⃗1f=⟨a,b⟩⋅12⟨1,1⟩=a+b2=42D_{\vec{u}_1}f = \langle a,b \rangle \cdot \frac{1}{\sqrt{2}}\langle 1, 1 \rangle = \frac{a+b}{\sqrt{2}} = 4\sqrt{2}, so a+b=8a+b=8. The unit vector for v⃗2\vec{v}_2 is u⃗2=12⟨3,1⟩\vec{u}_2 = \frac{1}{2}\langle \sqrt{3}, 1 \rangle. The directional derivative is Du⃗2f=⟨a,b⟩⋅12⟨3,1⟩=a3+b2=4D_{\vec{u}_2}f = \langle a,b \rangle \cdot \frac{1}{2}\langle \sqrt{3}, 1 \rangle = \frac{a\sqrt{3}+b}{2} = 4, so a3+b=8a\sqrt{3}+b=8. We have a system of two equations: a+b=8a+b=8 and a3+b=8a\sqrt{3}+b=8. Since both expressions equal 8, we have a+b=a3+ba+b = a\sqrt{3}+b, which simplifies to a=a3a = a\sqrt{3}, or a(3−1)=0a(\sqrt{3}-1)=0. This implies a=0a=0. Substituting a=0a=0 into a+b=8a+b=8 gives b=8b=8. So, the gradient is ∇f=⟨0,8⟩\nabla f = \langle 0, 8 \rangle. The maximum rate of change is the magnitude of the gradient: ∣∣∇f∣∣=∣∣⟨0,8⟩∣∣=02+82=8||\nabla f|| = ||\langle 0, 8 \rangle|| = \sqrt{0^2+8^2}=8.

Question 6

At a point PP, a function ff has a directional derivative of 10 in the direction of steepest ascent and a directional derivative of −5-5 in another direction u⃗\vec{u}. What is the angle between the direction of steepest ascent and the direction u⃗\vec{u}?

  1. π/6\pi/6
  2. π/3\pi/3
  3. 2π/32\pi/3 (correct answer)
  4. 5π/65\pi/6
Explanation: The direction of steepest ascent is the direction of the gradient vector, ∇f\nabla f. The directional derivative in this direction is the magnitude of the gradient, ∣∣ablaf∣∣|| abla f||. We are given that this maximum rate of change is 10, so ∣∣ablaf∣∣=10|| abla f|| = 10. The formula for the directional derivative in the direction of a unit vector u⃗\vec{u} is Du⃗f=∣∣∇f∣∣cos⁡θD_{\vec{u}}f = ||\nabla f|| \cos\theta, where θ\theta is the angle between ∇f\nabla f and u⃗\vec{u}. We are given that Du⃗f=−5D_{\vec{u}}f = -5. Plugging in the known values, we get −5=10cos⁡θ-5 = 10 \cos\theta. Solving for cos⁡θ\cos\theta, we find cos⁡θ=−5/10=−1/2\cos\theta = -5/10 = -1/2. The angle θ\theta in the interval [0,π][0, \pi] for which cos⁡θ=−1/2\cos\theta = -1/2 is θ=2π/3\theta = 2\pi/3.

Question 7

Let z=f(x,y)z = f(x,y) be a differentiable function of xx and yy defined implicitly by the equation xyz+z3=10xyz + z^3 = 10. Find the directional derivative of ff at the point (x,y)=(1,1)(x,y)=(1,1) in the direction of the vector v⃗=⟨1,3⟩\vec{v} = \langle 1, \sqrt{3} \rangle.

  1. 1+313\frac{1+\sqrt{3}}{13}
  2. −1+313-\frac{1+\sqrt{3}}{13} (correct answer)
  3. −2(1+3)13-\frac{2(1+\sqrt{3})}{13}
  4. −1+37-\frac{1+\sqrt{3}}{7}
Explanation: First, find the value of zz when (x,y)=(1,1)(x,y)=(1,1). The equation becomes (1)(1)z+z3=10(1)(1)z + z^3 = 10, or z3+z−10=0z^3+z-10=0. By inspection, z=2z=2 is a solution since 23+2−10=8+2−10=02^3+2-10 = 8+2-10=0. Let F(x,y,z)=xyz+z3−10=0F(x,y,z) = xyz+z^3-10=0. We can find the partial derivatives of f(x,y)f(x,y) using implicit differentiation: ∂z∂x=−FxFz\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} and ∂z∂y=−FyFz\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}. We have Fx=yzF_x=yz, Fy=xzF_y=xz, and Fz=xy+3z2F_z=xy+3z^2. At (1,1,2)(1,1,2), Fx=2F_x=2, Fy=2F_y=2, and Fz=1(1)+3(22)=13F_z=1(1)+3(2^2)=13. So, ∂z∂x=−2/13\frac{\partial z}{\partial x} = -2/13 and ∂z∂y=−2/13\frac{\partial z}{\partial y} = -2/13. The gradient of ff at (1,1)(1,1) is ∇f(1,1)=⟨−2/13,−2/13⟩\nabla f(1,1) = \langle -2/13, -2/13 \rangle. The direction vector is v⃗=⟨1,3⟩\vec{v} = \langle 1, \sqrt{3} \rangle, with magnitude ∣∣v⃗∣∣=12+(3)2=4=2||\vec{v}|| = \sqrt{1^2+(\sqrt{3})^2} = \sqrt{4}=2. The unit vector is u⃗=⟨1/2,3/2⟩\vec{u} = \langle 1/2, \sqrt{3}/2 \rangle. The directional derivative is Du⃗f(1,1)=∇f(1,1)⋅u⃗=⟨−2/13,−2/13⟩⋅⟨1/2,3/2⟩=(−2/13)(1/2)+(−2/13)(3/2)=−113−313=−1+313D_{\vec{u}}f(1,1) = \nabla f(1,1) \cdot \vec{u} = \langle -2/13, -2/13 \rangle \cdot \langle 1/2, \sqrt{3}/2 \rangle = (-2/13)(1/2) + (-2/13)(\sqrt{3}/2) = -\frac{1}{13} - \frac{\sqrt{3}}{13} = -\frac{1+\sqrt{3}}{13}.

Question 8

The temperature at any point (x,y,z)(x,y,z) in space is given by T(x,y,z)=x2+2y2−z2T(x,y,z) = x^2 + 2y^2 - z^2. A particle is at the point P(1,1,1)P(1,1,1). Which of the following vectors represents a direction in which the particle should move to experience no initial temperature change?

  1. ⟨1,2,−1⟩\langle 1, 2, -1 \rangle
  2. ⟨−1,−2,1⟩\langle -1, -2, 1 \rangle
  3. ⟨1,1,3⟩\langle 1, 1, 3 \rangle (correct answer)
  4. ⟨1,1,1⟩\langle 1, 1, 1 \rangle
Explanation: To experience no initial temperature change, the particle must move in a direction where the directional derivative of the temperature function is zero. This occurs in any direction orthogonal to the gradient vector ∇T\nabla T. The gradient is ∇T=⟨2x,4y,−2z⟩\nabla T = \langle 2x, 4y, -2z \rangle. At the point P(1,1,1)P(1,1,1), the gradient is ∇T(1,1,1)=⟨2(1),4(1),−2(1)⟩=⟨2,4,−2⟩\nabla T(1,1,1) = \langle 2(1), 4(1), -2(1) \rangle = \langle 2, 4, -2 \rangle. We need to find which of the given vectors has a dot product of zero with ⟨2,4,−2⟩\langle 2, 4, -2 \rangle. Let's check the options: A) ⟨1,2,−1⟩⋅⟨2,4,−2⟩=2+8+2=12≠0\langle 1, 2, -1 \rangle \cdot \langle 2, 4, -2 \rangle = 2+8+2 = 12 \neq 0. B) ⟨−1,−2,1⟩⋅⟨2,4,−2⟩=−2−8−2=−12≠0\langle -1, -2, 1 \rangle \cdot \langle 2, 4, -2 \rangle = -2-8-2 = -12 \neq 0. C) ⟨1,1,3⟩⋅⟨2,4,−2⟩=2+4−6=0\langle 1, 1, 3 \rangle \cdot \langle 2, 4, -2 \rangle = 2+4-6=0. D) ⟨1,1,1⟩⋅⟨2,4,−2⟩=2+4−2=4≠0\langle 1, 1, 1 \rangle \cdot \langle 2, 4, -2 \rangle = 2+4-2=4 \neq 0. The correct direction is given by the vector in option C.

Question 9

At a point PP, the directional derivative of a differentiable function f(x,y)f(x,y) in the direction of the unit vector u⃗=⟨1/2,1/2⟩\vec{u} = \langle 1/\sqrt{2}, 1/\sqrt{2} \rangle is 323\sqrt{2}. The directional derivative in the direction of the unit vector w⃗=⟨1/5,2/5⟩\vec{w} = \langle 1/\sqrt{5}, 2/\sqrt{5} \rangle is 11/511/\sqrt{5}. What is the directional derivative at PP in the direction of the vector v⃗=⟨0,−1⟩\vec{v} = \langle 0, -1 \rangle?

  1. −5-5 (correct answer)
  2. −1-1
  3. 11
  4. 55
Explanation: Let the gradient of ff at PP be ∇f=⟨a,b⟩\nabla f = \langle a, b \rangle. From the given information, we can set up a system of linear equations. First, Du⃗f=∇f⋅u⃗=⟨a,b⟩⋅⟨1/2,1/2⟩=a+b2=32D_{\vec{u}}f = \nabla f \cdot \vec{u} = \langle a, b \rangle \cdot \langle 1/\sqrt{2}, 1/\sqrt{2} \rangle = \frac{a+b}{\sqrt{2}} = 3\sqrt{2}, which simplifies to a+b=6a+b=6. Second, Dw⃗f=∇f⋅w⃗=⟨a,b⟩⋅⟨1/5,2/5⟩=a+2b5=115D_{\vec{w}}f = \nabla f \cdot \vec{w} = \langle a, b \rangle \cdot \langle 1/\sqrt{5}, 2/\sqrt{5} \rangle = \frac{a+2b}{\sqrt{5}} = \frac{11}{\sqrt{5}}, which simplifies to a+2b=11a+2b=11. Now we solve the system: (a+2b)−(a+b)=11−6(a+2b) - (a+b) = 11-6, which gives b=5b=5. Substituting b=5b=5 into a+b=6a+b=6 gives a=1a=1. So, the gradient is ∇f=⟨1,5⟩\nabla f = \langle 1, 5 \rangle. We need to find the directional derivative in the direction of v⃗=⟨0,−1⟩\vec{v} = \langle 0, -1 \rangle. This is already a unit vector. The directional derivative is Dv⃗f=∇f⋅v⃗=⟨1,5⟩⋅⟨0,−1⟩=1(0)+5(−1)=−5D_{\vec{v}}f = \nabla f \cdot \vec{v} = \langle 1, 5 \rangle \cdot \langle 0, -1 \rangle = 1(0) + 5(-1) = -5.

Question 10

Consider h(x,y)=ln⁡(x2+y2)h(x,y) = \ln(x^2 + y^2) for x2+y2>0x^2 + y^2 > 0. If the directional derivative of hh at point (3,4)(3,4) in the direction toward point (6,8)(6,8) is ab\frac{a}{b} where aa and bb are integers in lowest terms, what is a+ba + b?

  1. 77 (correct answer)
  2. 99
  3. 1111
  4. 1313
Explanation: First find the gradient: hx=2xx2+y2h_x = \frac{2x}{x^2+y^2} and hy=2yx2+y2h_y = \frac{2y}{x^2+y^2}. At (3,4)(3,4): hx(3,4)=2(3)9+16=625h_x(3,4) = \frac{2(3)}{9+16} = \frac{6}{25} and hy(3,4)=2(4)25=825h_y(3,4) = \frac{2(4)}{25} = \frac{8}{25}. So ∇h(3,4)=⟨625,825⟩\nabla h(3,4) = \left\langle \frac{6}{25}, \frac{8}{25} \right\rangle. The direction vector from (3,4)(3,4) to (6,8)(6,8) is v⃗=⟨6−3,8−4⟩=⟨3,4⟩\vec{v} = \langle 6-3, 8-4 \rangle = \langle 3, 4 \rangle. The unit vector is u^=⟨3,4⟩∣⟨3,4⟩∣=⟨3,4⟩5\hat{u} = \frac{\langle 3, 4 \rangle}{|\langle 3, 4 \rangle|} = \frac{\langle 3, 4 \rangle}{5}. The directional derivative is Du^h=∇h⋅u^=⟨625,825⟩⋅⟨3,4⟩5=15(625⋅3+825⋅4)=15(1825+3225)=15⋅5025=15⋅2=25D_{\hat{u}}h = \nabla h \cdot \hat{u} = \left\langle \frac{6}{25}, \frac{8}{25} \right\rangle \cdot \frac{\langle 3, 4 \rangle}{5} = \frac{1}{5} \left( \frac{6}{25} \cdot 3 + \frac{8}{25} \cdot 4 \right) = \frac{1}{5} \left( \frac{18}{25} + \frac{32}{25} \right) = \frac{1}{5} \cdot \frac{50}{25} = \frac{1}{5} \cdot 2 = \frac{2}{5}. Since 25\frac{2}{5} is in lowest terms, a=2a = 2 and b=5b = 5, so a+b=7a + b = 7.

Question 11

Let f(x,y)=x2eyf(x,y) = x^2 e^y. What is the directional derivative of ff at the point P(2,0)P(2,0) in the direction of the vector v⃗=⟨3,−4⟩\vec{v} = \langle 3, -4 \rangle?

  1. 424\sqrt{2}
  2. −4-4
  3. 4/54/5
  4. −4/5-4/5 (correct answer)
Explanation: First, compute the gradient of ff. ∇f(x,y)=⟨fx,fy⟩=⟨2xey,x2ey⟩\nabla f(x,y) = \langle f_x, f_y \rangle = \langle 2xe^y, x^2e^y \rangle. At the point (2,0)(2,0), the gradient is ∇f(2,0)=⟨2(2)e0,22e0⟩=⟨4,4⟩\nabla f(2,0) = \langle 2(2)e^0, 2^2 e^0 \rangle = \langle 4, 4 \rangle. Next, find the unit vector u⃗\vec{u} in the direction of v⃗=⟨3,−4⟩\vec{v} = \langle 3, -4 \rangle. The magnitude of v⃗\vec{v} is ∣∣v⃗∣∣=32+(−4)2=9+16=25=5||\vec{v}|| = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = 5. So, the unit vector is u⃗=v⃗∣∣v⃗∣∣=⟨3/5,−4/5⟩\vec{u} = \frac{\vec{v}}{||\vec{v}||} = \langle 3/5, -4/5 \rangle. The directional derivative is Du⃗f(2,0)=∇f(2,0)⋅u⃗=⟨4,4⟩⋅⟨3/5,−4/5⟩=4(3/5)+4(−4/5)=12/5−16/5=−4/5D_{\vec{u}}f(2,0) = \nabla f(2,0) \cdot \vec{u} = \langle 4, 4 \rangle \cdot \langle 3/5, -4/5 \rangle = 4(3/5) + 4(-4/5) = 12/5 - 16/5 = -4/5.

Question 12

Find the directional derivative of the function f(x,y,z)=x2+y2+z2f(x,y,z) = \sqrt{x^2+y^2+z^2} at the point P(1,2,2)P(1,2,2) in the direction of the vector from PP to Q(3,1,0)Q(3,1,0).

  1. −4/3-4/3
  2. −4/9-4/9 (correct answer)
  3. 8/98/9
  4. 11
Explanation: The gradient of ff is ∇f=12x2+y2+z2⟨2x,2y,2z⟩=1x2+y2+z2⟨x,y,z⟩\nabla f = \frac{1}{2\sqrt{x^2+y^2+z^2}} \langle 2x, 2y, 2z \rangle = \frac{1}{\sqrt{x^2+y^2+z^2}} \langle x, y, z \rangle. At P(1,2,2)P(1,2,2), we have 12+22+22=9=3\sqrt{1^2+2^2+2^2} = \sqrt{9} = 3. So, ∇f(1,2,2)=13⟨1,2,2⟩\nabla f(1,2,2) = \frac{1}{3} \langle 1, 2, 2 \rangle. The direction vector is v⃗=Q−P=⟨3−1,1−2,0−2⟩=⟨2,−1,−2⟩\vec{v} = Q - P = \langle 3-1, 1-2, 0-2 \rangle = \langle 2, -1, -2 \rangle. Its magnitude is ∣∣v⃗∣∣=22+(−1)2+(−2)2=4+1+4=9=3||\vec{v}|| = \sqrt{2^2+(-1)^2+(-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3. The unit direction vector is u⃗=13⟨2,−1,−2⟩\vec{u} = \frac{1}{3} \langle 2, -1, -2 \rangle. The directional derivative is Du⃗f(P)=∇f(P)⋅u⃗=13⟨1,2,2⟩⋅13⟨2,−1,−2⟩=19(1(2)+2(−1)+2(−2))=19(2−2−4)=−4/9D_{\vec{u}}f(P) = \nabla f(P) \cdot \vec{u} = \frac{1}{3} \langle 1, 2, 2 \rangle \cdot \frac{1}{3} \langle 2, -1, -2 \rangle = \frac{1}{9}(1(2) + 2(-1) + 2(-2)) = \frac{1}{9}(2-2-4) = -4/9.

Question 13

Consider the function f(x,y)=x3−xyf(x,y) = x^3 - xy. Which of the following is a unit vector describing a direction in which the rate of change of ff at the point P(1,2)P(1,2) is zero?

  1. 12⟨1,−1⟩\frac{1}{\sqrt{2}} \langle 1, -1 \rangle
  2. 12⟨1,1⟩\frac{1}{\sqrt{2}} \langle 1, 1 \rangle (correct answer)
  3. ⟨1,−1⟩\langle 1, -1 \rangle
  4. ⟨1,1⟩\langle 1, 1 \rangle
Explanation: The rate of change of a function is zero in directions orthogonal to its gradient vector. First, we compute the gradient of f(x,y)f(x,y): ∇f=⟨∂f∂x,∂f∂y⟩=⟨3x2−y,−x⟩\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle = \langle 3x^2 - y, -x \rangle. At the point P(1,2)P(1,2), the gradient is ∇f(1,2)=⟨3(1)2−2,−1⟩=⟨1,−1⟩\nabla f(1,2) = \langle 3(1)^2 - 2, -1 \rangle = \langle 1, -1 \rangle. A direction of zero change is given by a vector v⃗=⟨a,b⟩\vec{v} = \langle a, b \rangle that is orthogonal to ∇f(1,2)\nabla f(1,2), meaning their dot product is zero: ⟨1,−1⟩⋅⟨a,b⟩=a−b=0\langle 1, -1 \rangle \cdot \langle a, b \rangle = a - b = 0, so a=ba=b. A simple choice for such a vector is ⟨1,1⟩\langle 1, 1 \rangle. To make this a unit vector, we divide by its magnitude, which is 12+12=2\sqrt{1^2+1^2} = \sqrt{2}. Thus, a unit vector for a direction of zero change is u⃗=12⟨1,1⟩\vec{u} = \frac{1}{\sqrt{2}} \langle 1, 1 \rangle.