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📝 Partial Derivatives and Continuity (13 MCQs)

📖 From Calculus • 14. Partial Derivatives Calculus • 13 questions available

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What is Partial Derivatives and Continuity?

Definition:
Existence of $f_x, f_y$ at a point does NOT guarantee continuity there; however, continuous partials in a neighborhood DO imply continuity.

Example:
$f(x,y) = \frac{xy}{x^2+y^2}$ (with $f(0,0)=0$ ) has $f_x(0,0)=f_y(0,0)=0$ yet is discontinuous at origin.

Reason:
This distinction highlights that partial derivatives only capture axial behavior; full continuity requires coordinated behavior in all directions, ensured by continuous partials.

Easy

Medium

Hard

📝 All Partial Derivatives and Continuity MCQs

Q1. A function $f(x,y)$ has partial derivatives $f_x(0,0) = 0$ and $f_y(0,0) = 0$ . Which additional condition is strictly necessary to guarantee that $f$ is continuous at $(0,0)$ ?

A.The second-order partial derivatives exist at the origin.

B.The partial derivatives are continuous in a neighborhood of the origin. ✅

C.The function is defined along all lines passing through the origin.

D.The directional derivatives exist in every direction at the origin.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Existence of partial derivatives at a single point does not imply continuity. Students often confuse pointwise existence with local behavior. Continuity of partials in a neighborhood ensures differentiability, which implies continuity. Mere existence of directional derivatives or second-order terms does not bridge the gap between discrete rates of change and global limit behavior required for continuity.

Q2. Consider $f(x,y) = \frac{x^2 y}{x^4 + y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0)=0$ . Why does this function fail to be continuous at the origin despite having well-defined partial derivatives?

A.The partial derivatives do not exist at the origin.

B.The limit depends on the path of approach, specifically parabolic paths versus linear paths. ✅

C.The function is unbounded near the origin.

D.The partial derivatives are discontinuous but the function remains continuous.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This classic counterexample demonstrates that partial derivatives only capture behavior along coordinate axes. Along $y=kx^2$ , the limit is non-zero, contradicting the zero value at the origin. Students must recognize that axis-aligned rates of change are insufficient to determine multidimensional limits. This tests deep conceptual understanding beyond computational verification of partials.

Q3. An engineer models temperature distribution as $T(x,y)$ . Sensors confirm $T_x$ and $T_y$ exist everywhere, yet thermal fractures suggest discontinuities. What is the most plausible mathematical explanation for this physical observation?

A.Sensors cannot measure partial derivatives accurately.

B.The partial derivatives exist but are not continuous, allowing abrupt changes in temperature. ✅

C.Temperature is a scalar field and cannot have partial derivatives.

D.The model should use total derivatives instead of partial derivatives.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Physical intuition aligns with mathematical theory: existence of partials does not prevent jump discontinuities. In engineering contexts, students must distinguish between measurable local rates and global smoothness. Discontinuous partials permit sudden changes consistent with material fractures. This scenario-based question bridges abstract analysis with real-world modeling where idealized assumptions break down.

Q4. Given the contour plot of $f(x,y)$ shows tightly packed, non-concentric level curves approaching the origin from different angles, what can be definitively concluded about continuity and partial derivatives at the origin?

A.The function is continuous but partial derivatives do not exist.

B.Partial derivatives exist but the function is not continuous.

C.Neither continuity nor existence of partial derivatives can be guaranteed solely from contour spacing. ✅

D.The function is differentiable at the origin.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Contour density indicates gradient magnitude but not directional consistency. Non-concentric patterns suggest anisotropic behavior, yet visual interpretation alone cannot confirm limit existence or partial derivative values. Students must avoid overinterpreting graphical data. This graph-based HOTS question emphasizes the distinction between visual heuristics and rigorous analytical verification in multivariable calculus.

Q5. A student claims that since $f_x(a,b)$ and $f_y(a,b)$ both equal zero, the function must have a local extremum and be continuous at $(a,b)$ . Identify the fundamental flaw in this reasoning.

A.Zero partials guarantee continuity but not extrema.

B.Zero partials guarantee extrema but not continuity.

C.Zero partials neither guarantee continuity nor extrema; saddle points and discontinuities remain possible. ✅

D.The student correctly applied Fermat’s theorem for multivariable functions.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: This error analysis targets the misconception that critical points imply regularity. Students often conflate necessary conditions for extrema with sufficient conditions for continuity. A function can have vanishing partials at a discontinuity or saddle point. Recognizing this distinction prevents flawed optimization arguments and reinforces that partial derivatives provide limited local information without additional smoothness assumptions.

Q6. Let $f(x,y) = |x| + |y|$ . At $(0,0)$ , which statement correctly characterizes the relationship between partial derivatives and continuity?

A.Partial derivatives exist and the function is continuous.

B.Partial derivatives do not exist, but the function is continuous. ✅

C.Partial derivatives exist, but the function is not continuous.

D.Neither partial derivatives nor continuity hold at the origin.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Absolute value functions are continuous everywhere but lack differentiability at kinks. Here, left and right limits defining partials disagree at zero. This direct recall question anchors foundational knowledge: continuity does not require partial derivatives. It counters the reverse misconception and establishes baseline understanding before tackling more nuanced interactions between these concepts in higher-order problems.

Q7. Suppose $f(x,y)$ is continuous at $(a,b)$ and $f_x(a,b)$ exists. Does $f_y(a,b)$ necessarily exist? Justify using a constructed counterexample or theoretical argument.

A.Yes, continuity plus one partial implies the other partial exists.

B.No; consider

f(x,y) = x + |y|

(0,0)

, where

f_x=1

, continuous, but

f_y

undefined. ✅

C.Yes, because continuity ensures all directional derivatives exist.

D.No, but only if the function is not differentiable.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Continuity and existence of one partial impose no constraint on orthogonal partials. The counterexample $x + |y|$ is continuous, has $f_x=1$ , yet $f_y$ fails due to absolute value cusp. This application question requires constructing or recalling specific pathological functions, testing ability to decouple independent directional behaviors within a continuous framework.

Q8. In optimizing $f(x,y)$ , you find $f_x=f_y=0$ at $(a,b)$ . Second derivative test is inconclusive. Before concluding no extremum, what continuity-related check is essential?

A.Verify

f

is twice continuously differentiable near

(a,b)

. ✅

B.Check if

f

is merely continuous at

(a,b)

C.Confirm partial derivatives exist in a neighborhood, not just at the point.

D.Ensure the Hessian determinant is positive.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Inconclusive second derivative test may stem from insufficient smoothness rather than true saddle behavior. If second partials aren’t continuous, the test’s validity collapses. Students must diagnose whether failure arises from function pathology or genuine indeterminacy. This mixed-concept question integrates optimization, continuity of derivatives, and diagnostic reasoning beyond mechanical computation.

Q9. Which modification to $f(x,y) = \frac{xy}{\sqrt{x^2+y^2}}$ (with $f(0,0)=0$ ) would make it differentiable at the origin while preserving continuity?

A.Replace denominator with

x^2 + y^2

B.Multiply numerator by

\sqrt{x^2+y^2}

. ✅

C.Add

|x| + |y|

to the function.

D.Change definition at origin to 1.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Original function is continuous but not differentiable; its partials exist but aren’t continuous. Multiplying numerator by $\sqrt{x^2+y^2}$ yields $xy$ , which is smooth. This Olympiad-style problem demands understanding how algebraic structure affects differentiability thresholds. Students must analyze homogeneity degrees and recognize when scaling restores linear approximability, linking continuity, partials, and differentiability hierarchies.

Q10. A numerical algorithm computes partial derivatives via finite differences and reports convergence at $(a,b)$ , yet analytical evaluation shows discontinuity. What is the most likely source of discrepancy?

A.Finite differences always converge to true partials regardless of continuity.

B.The function is continuous but partials don’t exist analytically.

C.Numerical methods sample along grid lines, missing pathological behavior between samples. ✅

D.Analytical methods are inherently unreliable for discontinuous functions.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Discrete sampling aligns with coordinate axes, potentially overlooking off-grid discontinuities. Numerical partials approximate axis-directional limits, which may exist even when full continuity fails. This error analysis highlights limitations of computational tools versus theoretical definitions. Students must reconcile empirical results with analytical rigor, recognizing that algorithms validate restricted behaviors, not global properties.

Q11. If $f(x,y)$ is continuous on a closed disk and partial derivatives exist in the interior, which statement must be true regarding boundary behavior?

A.Partial derivatives extend continuously to the boundary.

B.Continuity on the closed disk implies uniform continuity, but partials may blow up near boundary. ✅

C.Existence of interior partials guarantees boundedness near boundary.

D.Boundary values determine interior partial derivatives uniquely.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Continuity on compact sets ensures uniform continuity, but says nothing about derivative behavior near edges. Partials can become unbounded while function stays continuous (e.g., $\sqrt{x}$ analogues in 2D). This challenging question separates topological properties from differential ones, testing nuanced understanding that continuity constrains function values, not rates of change, especially at domain boundaries.

Q12. Compare two functions: $f(x,y)=x^2 \sin(1/x)$ (extended by 0 at x=0) and $g(x,y)=x \sin(1/x)$ (similarly extended). Both are continuous at origin. How do their partial derivatives with respect to x differ at (0,0)?

A.Both have

f_x(0,0)=g_x(0,0)=0

f_x(0,0)=0

but

g_x(0,0)

does not exist. ✅

f_x(0,0)

does not exist but

g_x(0,0)=0

D.Neither partial derivative exists at the origin.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For $f$ , difference quotient involves $x \sin(1/x) \to 0$ ; for $g$ , it involves $\sin(1/x)$ , which oscillates. Despite similar continuity, extra x factor in f tames oscillation enough for derivative existence. This comparison tests sensitivity to asymptotic scaling in partial derivative definitions, requiring careful limit analysis beyond superficial similarity.

Q13. A researcher asserts that verifying continuity along all straight lines through $(a,b)$ suffices to establish continuity of $f(x,y)$ at that point, given partial derivatives exist. Evaluate this claim.

A.Correct; line continuity plus partial existence implies full continuity.

B.Incorrect; parabolic or higher-order paths may yield different limits despite linear consistency. ✅

C.Correct; partial derivatives encode all directional information.

D.Incorrect, but only if partial derivatives are discontinuous.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Line tests are necessary but insufficient. Classic counterexamples like $x^2 y/(x^4+y^2)$ are continuous along lines but not along parabolas. Partial derivatives only probe axis directions. This application question forces rejection of oversimplified verification protocols, emphasizing that multidimensional continuity requires path-independent limits, not just linear or axial checks, even when partials exist.

🔗 Related Topics (MCQs)

📝 Functions of Two or More Variables MCQs 📝 Multivariable function notation and terminology MCQs 📝 Functions of two variables from tables MCQs 📝 Level curves and contour plots MCQs 📝 Graphing functions of two variables MCQs

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