Definition: The total differential df=fxdx+fydy represents linear change in f resulting from infinitesimal changes dx,dy.
Example: For volume V=πr2h, dV=2πrhdr+πr2dh estimates volume change from small radius/height errors.
Reason: Differentials enable error propagation analysis, sensitivity studies, and change-of-variables in integration through linear approximation of nonlinear relationships.
3
Easy
3
Medium
8
Hard
📝 All Differentials of multivariable functions MCQs
Q1. A manufacturing process yields a cylindrical tank where radius r and height h are measured with potential errors of ±0.5%. If the design requires h=4r, which measurement contributes more to the relative error in volume V=πr2h when using total differentials?
A.The height measurement, because it has a linear coefficient of 1 in the differential formula.
B.The radius measurement, because its relative error is multiplied by 3 in the propagated uncertainty. ✅
C.Both contribute equally since the percentage errors are identical and h depends on r.
D.The height measurement, because volume depends linearly on height but quadratically on radius.
💡 Difficulty: easy | ✅ Correct: B
📖 Explanation: When applying the total differential dV=∂r∂Vdr+∂h∂Vdh and substituting h=4r, the relative error becomes VdV=3rdr+hdh. Even with equal percentage measurement errors, the radius term carries a factor of 3 due to the geometric relationship, making radius precision critically more important for volume accuracy in this constrained design scenario.
Q2. Consider the function f(x,y)=x2+y2 at point (3,4). A student calculates the approximate change using differentials when moving to (3.1,3.9) and gets Δf≈−0.02. However, the actual change is positive. What fundamental misconception likely caused this sign error?
A.The student forgot that differentials only provide upper bounds, not signed estimates.
B.The student incorrectly computed partial derivatives, swapping x and y values in the gradient.
C.The student used dx=0.1 and dy=0.1 instead of recognizing dy=−0.1 for the displacement. ✅
D.The student applied the differential formula to a non-differentiable point at the origin.
💡 Difficulty: medium | ✅ Correct: C
📖 Explanation: The displacement from (3,4) to (3.1,3.9) gives dx=+0.1 but dy=−0.1. The partial derivatives are fx=3/5 and fy=4/5, so df=(3/5)(0.1)+(4/5)(−0.1)=−0.02. While mathematically correct for the linear approximation, the actual function value increases because the path moves closer to the origin's radial direction; the error arises from expecting the linear model to capture curvature over this step size.
Q3. In thermodynamics, pressure P, volume V, and temperature T satisfy PV=nRT. When modeling small perturbations around equilibrium, why is the total differential dP=−V2nRTdV+VnRdT preferred over finite differences for stability analysis?
A.Finite differences cannot handle the nonlinear relationship between state variables.
B.The total differential provides a linearized local model that reveals how infinitesimal changes propagate, enabling eigenvalue-based stability criteria. ✅
C.Differentials eliminate the need to specify the number of moles n in calculations.
D.Finite differences require knowledge of higher-order derivatives which are unavailable experimentally.
💡 Difficulty: hard | ✅ Correct: B
📖 Explanation: Stability analysis examines system response to infinitesimal perturbations near equilibrium. The total differential linearizes the equation of state locally, transforming the nonlinear constraint into a linear relation among differentials. This linearization allows construction of Jacobian matrices and application of Lyapunov or eigenvalue methods, which fundamentally require the differential structure rather than discrete finite changes that would obscure the local geometric behavior of the state surface.
Q4. A contour map shows level curves of elevation z=f(x,y) that are closely spaced near point A and widely spaced near point B. If you walk the same horizontal distance ds from each point in the direction of steepest ascent, how does the differential dz compare?
A.dzA>dzB because closer contour spacing indicates a larger magnitude of the gradient and thus greater vertical change per unit horizontal displacement. ✅
B.dzA<dzB because wide spacing means the terrain rises more gradually, accumulating less elevation over the same distance.
C.dzA=dzB because the differential depends only on the path length ds, not on local topography.
D.Cannot be determined without knowing the specific functional form of f(x,y).
💡 Difficulty: medium | ✅ Correct: A
📖 Explanation: Contour line density directly reflects gradient magnitude: ∣∇f∣≈Δz/Δs between adjacent contours. Closely spaced contours at A indicate steep slope, meaning ∣∇f(A)∣ is large. Since dz=∇f⋅ds and walking in steepest ascent aligns ds with ∇f, we get dz=∣∇f∣ds. Thus, for identical ds, the point with denser contours yields a larger vertical differential, visually encoding rate of change without explicit formulas.
Q5. Given u=x2y and v=xy2, express the differential du solely in terms of dv and dy along the curve where x+y=5. Which step is most critical for avoiding algebraic inconsistency?
A.Substituting x=5−y before computing any differentials to reduce to single-variable calculus.
B.Computing du and dv independently then solving the resulting system while enforcing dx=−dy from the constraint. ✅
C.Treating x and y as independent throughout and applying Lagrange multipliers to enforce the constraint.
D.Differentiating the constraint implicitly to get dx/dy=−1 and substituting into du after expressing everything in terms of y.
💡 Difficulty: hard | ✅ Correct: B
📖 Explanation: This problem mixes constrained differentials with variable elimination. Computing du=2xydx+x2dy and dv=y2dx+2xydy first preserves structural clarity. Then imposing dx=−dy from x+y=5 yields two equations in dx,dy. Solving this system ensures consistency because premature substitution can obscure dependencies or introduce domain restrictions. The key insight is maintaining the differential framework while respecting constraints through simultaneous relations rather than sequential reduction.
Q6. A student claims that since df=fxdx+fydy approximates Δf, setting df=0 always identifies exact level curves of f. For f(x,y)=x3−3xy2, why does this reasoning fail away from critical points?
A.Because df=0 defines tangent lines to level curves, not the curves themselves; integrating df=0 recovers level sets only if the differential is exact and path-independent. ✅
B.Because f is not differentiable except at the origin.
C.Because level curves of this function are straight lines, making differentials irrelevant.
D.Because df=0 only holds at points where both partial derivatives vanish simultaneously.
💡 Difficulty: hard | ✅ Correct: A
📖 Explanation: The equation df=0 yields fxdx+fydy=0, which describes the tangent direction to level curves at each point, not the global curve. For f=x3−3xy2, solving df=0 gives a differential equation whose solutions are indeed the level curves, but this works because the form is exact. The student’s error is assuming df=0 directly equals the level set rather than its infinitesimal generator. Distinguishing between the differential condition and its integral curves requires understanding that df=0 is a necessary local condition, not a sufficient global definition.
Q7. When estimating (3.02)2+(3.98)2 using differentials from base point (3,4), which modification to the standard linear approximation would most improve accuracy without computing second derivatives?
A.Using the base point (3.01,3.99) as the expansion center instead of (3,4).
B.Applying the differential to the squared function g=x2+y2 then taking square root of the result. ✅
C.Averaging the differential estimates computed from both (3,4) and (3.02,3.98).
D.Replacing dx and dy with symmetric differences Δx/2 and Δy/2.
💡 Difficulty: easy | ✅ Correct: B
📖 Explanation: The function f=x2+y2 has significant curvature, but g=x2+y2 is quadratic with constant second derivatives, making its linear approximation via differentials exact up to second order. Computing dg=2xdx+2ydy at (3,4) gives Δg≈2(3)(0.02)+2(4)(−0.02)=−0.04, so g≈25−0.04=24.96, and 24.96≈4.996. This leverages algebraic simplification to reduce approximation error inherent in directly linearizing a composite radical function.
Q8. In error propagation for Q=cab, textbooks often state QΔQ≈aΔa+bΔb+cΔc. Under what condition does this differential-based formula significantly underestimate true maximum error?
A.When measurement errors in a, b, and c are statistically independent and normally distributed.
B.When all relative errors have the same sign and the function is highly nonlinear over the error range. ✅
C.When c is much larger than a and b, making the quotient small.
D.When using absolute differentials instead of relative differentials in the derivation.
💡 Difficulty: hard | ✅ Correct: B
📖 Explanation: The differential formula assumes errors combine linearly and infinitesimally, effectively treating ΔQ/Q as a first-order Taylor expansion. If all errors align in the worst-case direction (e.g., a and b overestimated while c underestimated), and the relative errors are large enough that higher-order terms matter, the linear sum underestimates the true extremal deviation. This highlights that differential-based error propagation is valid for small, random errors but fails for systematic biases or large uncertainties where nonlinear effects dominate the error budget.
Q9. A robot navigates using sensor readings (x,y) with known covariance matrix Σ. To minimize positional uncertainty in the direction of motion v, should the robot adjust its path to align v with the eigenvector corresponding to the largest or smallest eigenvalue of Σ? Justify using differentials.
A.Largest eigenvalue, because it represents the direction of maximum variance and thus greatest information gain.
B.Smallest eigenvalue, because the differential displacement uncertainty dxTΣ−1dx is minimized when motion aligns with minimal variance direction. ✅
C.Largest eigenvalue, because differentials scale with standard deviation and larger spread improves navigation resolution.
D.Smallest eigenvalue, because the total differential magnitude is invariant under rotation and alignment reduces computational load.
💡 Difficulty: hard | ✅ Correct: B
📖 Explanation: Positional uncertainty is quantified by the covariance ellipsoid defined by Σ. The differential displacement variance in direction v is vTΣv. Minimizing this quadratic form subject to ∥v∥=1 yields the smallest eigenvalue direction. Aligning motion with this eigenvector ensures that accumulated differential errors grow slowest, as the projection of noise onto the trajectory is minimized. This connects differential geometry of uncertainty with optimal control, showing how spectral properties of Σ dictate robust path planning through local error propagation.
Q10. For the implicit surface x3+y3+z3=3xyz, a student computes dz=−FzFxdx−FzFydy at (1,1,1) and obtains undefined expressions. Beyond noting Fz=0, what deeper geometric interpretation explains why the differential fails here?
A.The point is a singularity where the surface self-intersects, so no unique tangent plane exists and the implicit function theorem conditions are violated. ✅
B.The partial derivatives exist but are discontinuous, violating differentiability requirements.
C.The surface is vertical at this point, requiring parameterization in terms of x and z instead.
D.The equation defines multiple surfaces passing through the point, making z a multivalued function of x,y.
💡 Difficulty: medium | ✅ Correct: A
📖 Explanation: At (1,1,1), all partials Fx=Fy=Fz=0, indicating a singular point where the gradient vanishes. The implicit function theorem requires ∇F=0 to guarantee a smooth local graph z=f(x,y). Here, the surface has a conical singularity (it factors as (x+y+z)(x2+y2+z2−xy−yz−zx)=0), so no well-defined tangent plane or differential exists. Recognizing singularity versus mere zero denominator prevents misapplying differential calculus to non-manifold points.
Q11. Two functions f and g satisfy df=dg everywhere in a simply connected domain. A student concludes f=g everywhere. What essential element is missing from this deduction, and why does simple connectivity matter?
A.The conclusion should be f=g+C; simple connectivity ensures the constant is globally consistent rather than piecewise. ✅
B.Simple connectivity guarantees differentiability, which is required for df=dg to imply equality.
C.The student forgot to integrate along a specific path; simple connectivity makes path choice irrelevant.
D.Nothing is missing; df=dg implies identical functions by definition of differentials.
💡 Difficulty: hard | ✅ Correct: A
📖 Explanation: If df=dg, then d(f−g)=0, meaning f−g is locally constant. On a simply connected domain, local constancy implies global constancy because there are no holes to support different constants in separate regions. Without simple connectivity, f−g could take different constant values in disconnected components or around non-contractible loops. Thus, the correct inference is f=g+C with a single global constant, and topological assumptions are crucial for upgrading local differential equality to global functional equality.
Q12. In optimizing f(x,y) subject to g(x,y)=c, the method of Lagrange multipliers requires ∇f=λ∇g. How does this condition relate to the total differential of the Lagrangian L=f−λ(g−c)?
A.Setting dL=0 yields exactly the multiplier equations plus the constraint, unifying stationarity and feasibility into a single differential condition. ✅
B.The differential dL eliminates λ, reducing the problem to unconstrained optimization.
C.dL=0 only enforces the constraint; stationarity requires separate verification.
D.The Lagrangian differential is undefined at optimal points because λ is unknown.
💡 Difficulty: hard | ✅ Correct: A
📖 Explanation: The total differential dL=(fx−λgx)dx+(fy−λgy)dy−(g−c)dλ. Setting dL=0 for arbitrary independent variations dx,dy,dλ forces each coefficient to vanish, yielding fx=λgx, fy=λgy, and g=c simultaneously. This elegantly packages the constrained optimality conditions as a single stationarity requirement in an augmented space, demonstrating how differentials unify constraints and objectives through dimensional extension rather than sequential enforcement.
Q13. A physics model uses dU=TdS−PdV for internal energy. If experimental data suggests (∂V∂U)T=−P, which assumption underlying the differential form is most likely violated in the real system?
A.Temperature and entropy are not conjugate variables in non-equilibrium processes.
B.The system is not closed, allowing particle exchange that introduces chemical potential terms. ✅
C.Pressure is not well-defined for the given volume range.
D.Internal energy is not a state function for this material.
💡 Difficulty: easy | ✅ Correct: B
📖 Explanation: The fundamental relation dU=TdS−PdV applies strictly to closed simple compressible systems. If (∂U/∂V)T=−P, additional work modes or mass transfer may be present. For open systems, dU=TdS−PdV+μdN, and holding T constant while varying V at fixed N still gives −P, but if N varies implicitly with V in experiments, the measured derivative includes μ(∂N/∂V)T. Thus, apparent violation signals unaccounted extensive variables, emphasizing that differential forms encode system boundaries and must match experimental constraints.
Q14. When approximating Δf using df for f(x,y)=exy near (0,0), the linear approximation gives Δf≈0 for any small displacement. Why is this particularly misleading compared to similar behavior in polynomial functions?
A.Exponential functions have all derivatives nonzero at the origin, so higher-order terms dominate immediately, unlike polynomials where low-degree terms capture initial behavior. ✅
B.The function is not differentiable at the origin despite appearing smooth.
C.Linear approximations never work for transcendental functions near zero.
D.The partial derivatives at (0,0) are actually undefined due to the exponential growth.
💡 Difficulty: hard | ✅ Correct: A
📖 Explanation: At (0,0), fx=yexy=0 and fy=xexy=0, so df=0. However, f(x,y)=1+xy+2(xy)2+⋯, so Δf≈xy for small displacements, which is second-order. Unlike polynomials where vanishing gradient implies flatness captured by higher-degree monomials, exponentials have infinite series with coupled variables; the leading non-zero term xy is invisible to first-order differentials. This illustrates that df=0 does not imply local constancy for non-polynomial functions, demanding awareness of Taylor series structure beyond linearization.