📝 Gradient vector fields (16 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 16 questions available

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What is Gradient vector fields?

Gradient vector fields:
A gradient field is derived from a scalar potential $f$ , defined as $\nabla f = \langle f_x, f_y, f_z \rangle$ , representing the direction of steepest ascent.

Example:
For $f(x,y) = x^2 + y^2$ , $\nabla f = \langle 2x, 2y \rangle$ , pointing radially outward.

Reason:
Gradient fields are conservative, simplifying line integrals via the Fundamental Theorem, as work depends only on endpoints.

Easy

Medium

Hard

📝 All Gradient vector fields MCQs

Q1. Which statement best characterizes a gradient field associated with a scalar function $f(x,y,z)$ ?

A.Its vectors are always perpendicular to the

xy

-plane

B.Its vectors point in the direction of greatest local increase of

f

✅

C.Its magnitude is always equal to

f

itself

D.Its vectors must have constant magnitude

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The gradient $\nabla f$ points in the direction in which the scalar function increases most rapidly. Its magnitude gives the maximum rate of increase at that point. There is no requirement that the field be perpendicular to a coordinate plane or have constant magnitude.

Q2. For $f(x,y)=x^2+3y^2$ , a particle moves from $(1,1)$ in the direction of $\nabla f$ . Which initial direction should it follow?

A.Toward

(-2,-6)

B.Toward

(2,6)

✅

C.Toward

(1,3)

D.Toward

(2,3)

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Computing the gradient gives $\nabla f=(2x,6y)$ . At $(1,1)$ , this becomes $(2,6)$ . Therefore, the particle initially moves in the direction $(2,6)$ , which corresponds to the direction of steepest increase of the scalar field.

Q3. Two nearby points lie on the same level curve of a differentiable function $f(x,y)$ . What can be concluded about the gradient at a point on that curve?

A.It is tangent to the level curve

B.It has zero magnitude everywhere on the curve

C.It is perpendicular to the level curve ✅

D.It points toward the origin

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Along a level curve, the value of $f$ remains constant. Therefore, moving tangent to the curve produces no first-order change in $f$ . Since the gradient represents the direction of greatest increase and is orthogonal to zero-change directions, it must be perpendicular to the level curve.

Q4. A terrain model is represented by $h(x,y)$ , where $h$ gives elevation. A hiker wants to climb as steeply as possible from a given location. Which mathematical quantity should determine the hiking direction?

A.The negative gradient

-\nabla h

B.The gradient

\nabla h

✅

C.The divergence of

\nabla h

D.The Laplacian of

h

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Because $h$ represents elevation, increasing $h$ means moving uphill. The gradient $\nabla h$ points in the direction of maximum increase of elevation. The negative gradient points downhill, while divergence and the Laplacian describe different properties and do not directly give the steepest ascent direction.

Q5. Let $f(x,y)=x^2-y^2$ . At $(2,1)$ , a traveler can move with unit speed. What is the maximum instantaneous rate at which $f$ can increase?

A.-3

B.-4

2\sqrt{5}

✅

\sqrt{5}

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The gradient is $\nabla f=(2x,-2y)$ . At $(2,1)$ , it equals $(4,-2)$ . Its magnitude is $\sqrt{4^2+(-2)^2}=\sqrt{20}=2\sqrt{5}$ . For unit-speed motion, the maximum directional derivative equals the gradient magnitude.

Q6. A student claims that because $f(x,y)=x^2+y^2$ increases as $x$ increases when $y$ is fixed, the vector $(1,0)$ must always be the direction of greatest increase. What is the main flaw?

A.The function has no gradient

B.Greatest increase depends on both partial derivatives at the point ✅

C.The direction of greatest increase is always

(0,1)

D.Only negative directions can produce increase

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The direction of greatest increase is determined by the complete gradient, not by one variable alone. Here $\nabla f=(2x,2y)$ , so the steepest-ascent direction generally depends on the current point. The vector $(1,0)$ gives a valid directional rate but is not generally maximal.

Q7. For $f(x,y)=x^2+4y^2$ , which direction from $(1,-1)$ produces the greatest instantaneous decrease in $f$ ?

(2,-8)

(-2,8)

✅

(-1,4)

(1,-4)

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: At $(1,-1)$ , $\nabla f=(2,-8)$ . The greatest increase occurs in that direction, so the greatest decrease occurs in the opposite direction, $-\nabla f=(-2,8)$ . Any positive scalar multiple represents the same direction, making the second option the correct choice.

Q8. A temperature field is $T(x,y)=100-x^2-2y^2$ . A sensor at $(2,1)$ is programmed to move in the direction of maximum temperature increase. Which velocity direction should it use?

(4,4)

(-4,-4)

✅

(-2,-1)

(2,4)

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The gradient is $\nabla T=(-2x,-4y)$ . At $(2,1)$ , this gives $(-4,-4)$ . Since temperature increases most rapidly in the gradient direction, the sensor should move toward decreasing $x$ and decreasing $y$ . The opposite direction would produce the steepest temperature decrease.

Q9. A scalar field has gradient $\nabla f=(3x^2,2y)$ . A student integrates the first component with respect to $x$ and obtains $x^3+y^2$ , then concludes this is the only possible potential function. What has been overlooked?

A.A multiplicative constant is required

B.A constant of integration may depend on

y

✅

C.The gradient cannot have a potential function

D.The

2y

term must be integrated with respect to

x

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Integrating $3x^2$ with respect to $x$ gives $x^3+g(y)$ , not simply $x^3+y^2$ . The unknown function $g(y)$ must then be determined by comparing its derivative with the second gradient component. This is essential because the constant of integration can depend on the other variable.

Q10. Consider the level curves of $f(x,y)=x^2+y^2$ , which are concentric circles. At the point $(3,4)$ , what should the gradient direction look like on a sketch?

A.Tangent to the circle and clockwise

B.Tangent to the circle and counterclockwise

C.Radially outward from the origin ✅

D.Radially inward toward the origin

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The level curves $x^2+y^2=c$ are circles centered at the origin. Since the gradient is perpendicular to level curves and $f$ increases as the distance from the origin increases, the gradient points radially outward. At $(3,4)$ , its direction is therefore outward from the origin.

Q11. A contour map shows several nested closed curves labeled with increasing values toward the center. At a marked point on one contour, which direction represents the local gradient?

A.Along the contour toward increasing arc length

B.Perpendicular to the contour toward the higher labels ✅

C.Perpendicular to the contour toward the lower labels

D.Parallel to the nearest coordinate axis

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The gradient is perpendicular to a level curve because the scalar value does not change when moving tangent to that curve. Since the contour labels increase toward the center, the gradient must point perpendicular to the marked contour and toward the region with larger scalar values.

Q12. Suppose $f(x,y)=e^{x}\cos y$ . At $(0,\pi/2)$ , which statement correctly describes the gradient and its interpretation?

A.The gradient is

(0,-1)

, indicating maximum increase in the negative

y

-direction ✅

B.The gradient is

(1,0)

, indicating maximum increase in the positive

x

-direction

C.The gradient is

(0,1)

, indicating maximum increase in the positive

y

-direction

D.The gradient is

(1,-1)

, indicating equal increase in both directions

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The partial derivatives are $f_x=e^x\cos y$ and $f_y=-e^x\sin y$ . At $(0,\pi/2)$ , these become $0$ and $-1$ , respectively. Thus $\nabla f=(0,-1)$ , so the greatest increase occurs toward decreasing $y$ .

Q13. A potential field is proposed as $F=(2xy+y^2,\ x^2+2xy)$ . Which observation provides the strongest evidence that $F$ is a gradient field on the entire plane?

A.Its components have the same algebraic form

B.The cross-partial condition

\partial F_1/\partial y=\partial F_2/\partial x

holds everywhere on the simply connected plane ✅

C.Its magnitude is constant

D.Its vectors all point away from the origin

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For $F=(P,Q)$ , a necessary and, on a simply connected domain, sufficient condition for a continuously differentiable field to be conservative is $P_y=Q_x$ . Here $P_y=2x+2y$ and $Q_x=2x+2y$ , so the condition holds everywhere on the plane.

Q14. A contour diagram for a scalar field shows level curves becoming progressively closer together as one moves eastward, while the contour values increase eastward. What does this imply about the gradient magnitude?

A.It decreases eastward because the curves are closer

B.It remains constant because contour values are labeled

C.It increases eastward because the same change in value occurs over a shorter distance ✅

D.It must become zero eastward

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The gradient magnitude measures the maximum rate of change per unit distance. If comparable changes in scalar value occur across progressively shorter distances, the rate of change becomes larger. Therefore, closer contour spacing in the direction of increasing values indicates a larger gradient magnitude.

Q15. For $f(x,y)=x^3-3xy^2$ , determine the direction of the gradient at $(1,1)$ , and identify the corresponding qualitative behavior.

(0,-6)

; the function increases most rapidly downward ✅

(3,-6)

; the function increases most rapidly southeast

(3,6)

; the function increases most rapidly northeast

(0,6)

; the function increases most rapidly upward

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The gradient is $\nabla f=(3x^2-3y^2,-6xy)$ . At $(1,1)$ , this becomes $(0,-6)$ . Thus the greatest increase occurs directly downward. The zero $x$ -component means there is no first-order preference for increasing or decreasing $x$ at that point.

Q16. Let $f(x,y)=x^2+y^2$ . Starting at $(1,2)$ , a particle moves with velocity proportional to $\nabla f$ . Which statement best describes its trajectory and why?

A.It moves along a circle because the gradient is tangent to level curves

B.It moves directly toward the origin because

f

decreases there

C.It moves outward because the gradient is perpendicular to the circular level curves and points toward increasing radius ✅

D.It moves horizontally because only

x

affects the gradient

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The gradient is $\nabla f=(2x,2y)$ , which points radially outward. The level curves are circles centered at the origin, and the gradient is perpendicular to these circles. Following the gradient therefore carries the particle toward larger radius and larger values of $f$ , producing an outward trajectory rather than circular motion.

🔗 Related Topics (MCQs)

📝 Vector Fields in Calculus MCQs 📝 Graphical Representations of Vector Fields MCQs 📝 Vector field notation and components MCQs 📝 Inverse square vector fields MCQs 📝 Conservative vector fields and potential functions MCQs

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