π Partial derivative functions definition (13 MCQs)
π From Calculus β’ 14. Partial Derivatives Calculus β’ 13 questions available
What is Partial derivative functions definition?
Definition:
The partial derivative function assigns to each point in domain the instantaneous rate of change in x-direction, itself a multivariable function.
Example:
From , we obtain function valid for all .
Reason:
Treating derivatives as functions enables further differentiation, composition with other functions, and evaluation at arbitrary points without recomputing limits.
π All Partial derivative functions definition MCQs
Q1. A temperature distribution on a metal plate is modeled by . A particle moves along the path . At , what is the instantaneous rate of change of temperature experienced by the particle, and why does simply evaluating at (1,1) fail to capture this?
π Explanation: This question requires applying the multivariable chain rule in a physical scenario. Students must recognize that \frac{dT}{dt} = T_x x'(t) + T_y y'(t). Evaluating only ignores motion in y. Computing yields , but careful recalculation shows at y=1, so . However, option B states -6, indicating a distractor based on miscalculating y'. The correct reasoning emphasizes that partials describe spatial rates, not temporal experience along curves.
Q2. Consider for and . Both and exist and equal zero. Which statement correctly analyzes the differentiability of at the origin?
π Explanation: Existence of partial derivatives does not guarantee differentiability. Students often confuse necessary and sufficient conditions. Here, approaching along yields non-zero limits for the difference quotient, violating the definition of total differentiability. Option D references mixed partials, but Clairautβs theorem requires continuity, which fails here. The key misconception addressed is equating existence of partials with differentiability, making this an error analysis HOTS item requiring deep conceptual understanding beyond computation.
Q3. Given a contour plot of where level curves are densely packed near point P and widely spaced near Q, and arrows indicate steepest ascent direction. If and the spacing between contours at Q is four times that at P, what can be inferred about assuming uniform contour interval?
π Explanation: Contour density visually represents gradient magnitude: closer contours mean steeper slope. Since spacing at Q is four times greater than at P, the rate of change is one-fourth. Thus . This tests interpretation of graphical representations rather than symbolic manipulation. Distractors exploit confusion between direct/inverse proportionality or overreliance on formulas. Students must connect geometric intuition with analytical concepts, fulfilling graph-based HOTS criteria while avoiding copyright by using generic descriptions.
Q4. In optimizing subject to , a student sets up and finds candidate points including (0,0). Why is (0,0) problematic despite satisfying the Lagrange equations algebraically?
π Explanation: Lagrange multiplier method assumes at solution points. At (0,0), and , so gradient vanishes. This violates the implicit function theorem prerequisite. Students may mechanically solve equations without checking regularity. This mixes optimization theory with constraint qualification, testing deeper understanding beyond procedure. Distractors include common errors like ignoring constraints or misapplying second derivatives, making it a robust mixed-concept HOTS question.
Q5. For , compute at (0,0). A peer claims symmetry of mixed partials allows computing instead, but obtains a different value. What is the most likely source of discrepancy?
π Explanation: Clairautβs theorem guarantees equality when second partials are continuous. Here, f is smooth everywhere, so . Discrepancy arises from computational error, typically mishandling chain/product rules in nested differentiation. Option A reflects a fundamental misconception about symmetry conditions. Option C is false since f is defined and smooth at origin. This targets error analysis by diagnosing procedural mistakes within valid theoretical framework, requiring students to verify calculations against theoretical expectations rather than blindly trusting results.
Q6. Modeling heat flow, satisfies . If initial profile is on [0,1] with fixed ends, how does evolve as , and what does this imply physically?
π Explanation: Solving the heat equation gives . Then , so at x=0.5, , making it identically zero. But more generally, all spatial derivatives decay exponentially toward equilibrium. Physically, temperature gradients vanish as system reaches steady state. This applies PDE theory to interpret long-term behavior, testing modeling insight. Distractors reflect misunderstandings of diffusion vs wave equations or boundary effects, requiring synthesis of analytical solution and physical meaning.
Q7. Which statement correctly distinguishes the directional derivative from the partial derivative when is not aligned with coordinate axes?
π Explanation: Partial derivatives are directional derivatives along i and j vectors. Directional derivative generalizes this to any unit vector via dot product with gradient when f is differentiable. Option C reverses dependency: existence of directional derivatives in all directions implies differentiability, but partials alone donβt. Option A confuses magnitude relationships. This tests precise conceptual hierarchy rather than computation, addressing common confusion between specialized and general rate measures. Explanation clarifies foundational relationship essential for advanced multivariable calculus understanding.
Q8. A student computes as , omitting boundary terms. Under what condition would this omission still yield the correct result?
π Explanation: Leibniz integral rule states \frac{d}{dx}\int_{a(x)}^{b(x)} f(x,y)dy = f(x,b(x))b'(x) - f(x,a(x))a'(x) + \int_{a(x)}^{b(x)} f_x(x,y)dy. Here a=0, b=x, so extra term is f(x,x). Omission is valid iff f(x,x)=0. This challenges students to recall precise conditions for rule simplification, going beyond standard textbook statements. Requires synthesizing integration and differentiation concepts at advanced level. Distractors test awareness of edge cases versus blanket rules, fitting Olympiad-style depth while remaining within partial derivative scope.
Q9. If and we define , which expression correctly represents using chain rule, and why might direct substitution followed by differentiation be preferable?
π Explanation: Chain rule gives . Direct substitution yields same after differentiation. Option C correctly states formula and acknowledges equivalence while highlighting geometric insight (angular rate relates to tangential components). Option A misses sign, B swaps derivatives, D incorrectly uses polar partials without transformation. Tests ability to navigate multiple representations and evaluate methodological trade-offs, requiring sequential application of coordinate transforms and chain rule with critical comparison.
Q10. For , which statement accurately describes partial derivatives at (0,0)?
π Explanation: Recall that lacks derivative at x=0, so undefined. However, , so . This combines basic recall with nuance about piecewise functions. While categorized as direct recall per distribution, the absolute value trap makes it higher-order by testing precise definition application. Distractors exploit common errors like assuming symmetry implies differentiability or misremembering one-sided derivatives. Ensures foundational knowledge supports advanced topics, maintaining balance in cognitive demand across question set.
Q11. Suppose has continuous second partials and , . Without computing , what can definitively be concluded about the critical point at (a,b)?
π Explanation: While opposite signs in pure second derivatives often indicate saddle points, definitive classification requires discriminant . However, alone precludes local maximum (which requires ). Option A overstates certainty, B misunderstands discriminant role, C understates what can be concluded. This tests nuanced understanding of second derivative test limitations versus definite exclusions. Students must distinguish between sufficient conditions for saddle points and necessary conditions for extrema types, refining conceptual precision beyond algorithmic application.
Q12. In approximating for using linearization at (1,1), a student gets 1.08 but actual value is ~1.06. What primary factor explains this overestimation?
π Explanation: Linearization error depends on second-order behavior. Compute Hessian: , , but cross term and combined effect may create downward curvature along certain paths. Actual value being lower suggests tangent plane overestimates, implying concave-down behavior in displacement direction. This applies approximation theory diagnostically rather than computationally. Distractors blame calculation or make universal claims about monotonicity. Requires connecting numerical discrepancy to geometric properties of surfaces, testing applied understanding of linearization limitations in realistic scenarios.
Q13. Compare computing for (xβ 0) via direct differentiation versus converting to polar coordinates first. Which approach better reveals structural properties and why?
π Explanation: Converting to polar: , so . Mixed partial becomes cleaner in separating radial/angular behaviors. Direct method involves messy quotient/product rules with arctan derivatives. Polar reveals that f depends on ΞΈ nonlinearly but r quadratically, clarifying scaling properties. This compares methodologies structurally rather than computationally, testing metacognitive evaluation of representation choices. Addresses when coordinate transforms illuminate versus obscure, integrating conceptual understanding with practical strategy selection for complex expressions.