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📝 Estimating partial derivatives from tables (13 MCQs)

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

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What is Estimating partial derivatives from tables?

Definition:
Approximating $f_x(a,b) \approx \frac{f(a+h,b)-f(a-h,b)}{2h}$ using finite differences from discrete tabulated data points.

Example:
Given table values $f(2,3)=10, f(2.1,3)=10.5$ , estimate $f_x(2,3) \approx \frac{10.5-10}{0.1} = 5$ .

Reason:
Experimental data lacks formulas; numerical estimation enables derivative-based analysis for empirical datasets in science and engineering.

Easy

Medium

Hard

📝 All Estimating partial derivatives from tables MCQs

Q1. A temperature table $T(x,y)$ shows values at grid points with spacing $\Delta x = 2$ and $\Delta y = 3$ . At point $(4,6)$ , $T=20$ ; at $(6,6)$ , $T=26$ ; at $(2,6)$ , $T=18$ . A student computes $T_x(4,6) \approx (26-20)/2 = 3$ . Which statement best evaluates this approach?

A.The estimate is optimal because forward differences always minimize truncation error for smooth functions.

B.The estimate is valid but less accurate than the central difference

(26-18)/4 = 2

, which has second-order accuracy. ✅

C.The estimate is incorrect because partial derivatives require simultaneous variation in both variables.

D.The estimate overestimates because the backward difference

(20-18)/2 = 1

should be averaged with the forward difference.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The student used a forward difference with first-order accuracy $O(\Delta x)$ . Since data exists on both sides, the central difference $[T(6,6)-T(2,6)]/(2\Delta x) = (26-18)/4 = 2$ provides second-order accuracy $O(\Delta x^2)$ and better approximates the true derivative by canceling leading error terms, making it superior for interior points.

Q2. Given tabulated pressure $P(r,t)$ in a cylindrical reactor, you need $P_r$ at the boundary $r=0$ where only forward data exists: $P(0,t)=100$ , $P(h,t)=98$ , $P(2h,t)=93$ . Which finite difference formula yields the most accurate estimate while respecting the physical symmetry condition $P_r(0,t)=0$ ?

A.Use forward difference

(98-100)/h

since boundary data limits options.

B.Apply central difference using ghost point

P(-h,t)=P(h,t)

due to symmetry, yielding zero derivative. ✅

C.Use three-point forward difference

(-3P_0+4P_1-P_2)/(2h)

for higher accuracy without assuming symmetry.

D.Average forward and backward differences even though backward data is unavailable.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: At $r=0$ , cylindrical symmetry implies $\partial P/\partial r = 0$ . The central difference with ghost point $P(-h)=P(h)$ enforces this exactly. While three-point forward formulas offer $O(h^2)$ accuracy generally, they ignore the known physics. Incorporating symmetry constraints improves both accuracy and physical consistency, demonstrating integration of mathematical methods with domain knowledge.

Q3. A researcher estimates $f_x(a,b)$ from noisy experimental data using central differences with varying $h$ . As $h$ decreases from 1.0 to 0.001, the estimated derivative initially stabilizes near 5.2 but then oscillates wildly below $h=0.01$ . What explains this behavior?

A.Round-off error dominates at small

h

because subtraction of nearly equal numbers amplifies relative errors in finite precision arithmetic. ✅

B.Truncation error increases as

h \to 0

due to higher-order Taylor terms becoming significant.

C.The function has a discontinuity at

(a,b)

that only manifests at fine scales.

D.Tabular data spacing cannot support

h < 0.01

, so interpolation introduces artificial oscillations.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Finite difference estimation involves a trade-off: truncation error decreases as $O(h^2)$ for central differences, but round-off error grows as $O(\epsilon/h)$ where $\epsilon$ is machine or measurement precision. Below an optimal $h$ , subtracting nearly identical values magnifies noise, causing erratic results. This illustrates why blindly reducing step size does not improve accuracy and requires balancing error sources.

Q4. You are modeling heat flux $q = -k \nabla T$ from discrete temperature measurements on a non-uniform grid where $\Delta x_i \neq \Delta x_{i+1}$ . At node $i$ , neighbors are at distances $h_-$ and $h_+$ . Which expression correctly estimates $T_x$ while maintaining second-order accuracy?

(T_{i+1}-T_{i-1})/(h_- + h_+)

[h_-^2(T_{i+1}-T_i) + h_+^2(T_i-T_{i-1})]/[h_- h_+(h_- + h_+)]

✅

(T_{i+1}-T_{i-1})/(2\sqrt{h_- h_+})

[(h_+)^2 T_{i-1} - (h_-^2 + h_+^2)T_i + (h_-)^2 T_{i+1}]/[h_- h_+(h_- + h_+)]

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: On non-uniform grids, standard central differences lose second-order accuracy. The correct formula derives from interpolating a quadratic through three points and differentiating at $x_i$ . Option B satisfies this derivation and reduces to standard central difference when $h_- = h_+$ . Other options either assume uniformity or fail dimensional consistency, highlighting the need for careful discretization on irregular meshes common in real-world sensor data.

Q5. Two students estimate $\partial z/\partial x$ from the same table. Student A uses $[z(x+h,y)-z(x-h,y)]/(2h)$ . Student B uses $[z(x+h,y)-z(x,y)]/h$ . Both obtain different values. Without knowing the true derivative, which diagnostic best determines which estimate is more reliable?

A.Compare against analytical solution if available; otherwise trust the method with smaller

h

B.Compute both estimates at multiple

h

values and check convergence rate; second-order methods should show error reduction proportional to

h^2

. ✅

C.Prefer Student A’s result because central differences are universally superior.

D.Use Richardson extrapolation on Student B’s results to artificially achieve second-order accuracy.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Convergence testing reveals method reliability: plotting error versus $h$ on log-log scale should yield slope ≈2 for central differences and ≈1 for forward differences. If Student A’s estimates converge quadratically while Student B’s converge linearly, A is likely more accurate. This empirical validation avoids reliance on unknown truths and builds intuition about numerical behavior, emphasizing process over single-value computation.

Q6. In oceanographic data, salinity $S(x,y,z)$ is measured at fixed depths. To estimate $S_z$ at depth $z_0$ , you have $S(z_0-\Delta z)=35.2$ , $S(z_0)=35.0$ , $S(z_0+\Delta z)=34.7$ . However, sensors have ±0.1 PSU uncertainty. What is the propagated uncertainty in $S_z$ using central differences with $\Delta z=10$ m?

A.±0.01 PSU/m

B.±0.014 PSU/m ✅

C.±0.02 PSU/m

D.±0.007 PSU/m

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For central difference $f' \approx (f_+-f_-)/(2h)$ , uncertainty propagates as $\sigma_{f'} = \sqrt{\sigma_+^2 + \sigma_-^2}/(2h)$ . With $\sigma=0.1$ and $h=10$ , $\sigma_{S_z} = \sqrt{0.1^2+0.1^2}/20 = \sqrt{0.02}/20 \approx 0.0141$ . This demonstrates how measurement noise directly impacts derivative estimates and why high-frequency differentiation amplifies uncertainty, crucial for interpreting geophysical gradients reliably.

Q7. A contour plot of elevation $z(x,y)$ shows closely spaced contours near point A and widely spaced contours near point B, both along the x-direction. Without numerical values, what can be definitively concluded about $|\partial z/\partial x|$ at these points?

|\partial z/\partial x|_A > |\partial z/\partial x|_B

because contour density correlates with gradient magnitude. ✅

|\partial z/\partial x|_A < |\partial z/\partial x|_B

because wider spacing indicates steeper terrain.

C.Cannot determine without knowing contour interval values.

|\partial z/\partial x|_A = |\partial z/\partial x|_B

if contour intervals are constant.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Contour lines represent level sets; their spatial density reflects gradient magnitude perpendicular to them. Closer spacing means rapid change in $z$ per unit distance, implying larger $|\nabla z|$ . Along the x-direction specifically, if contours are oriented similarly, denser packing indicates greater $|z_x|$ . This visual interpretation connects geometric representation to partial derivative concepts without computation, reinforcing conceptual understanding of rate-of-change visualization.

Q8. When estimating mixed partial $f_{xy}$ from tabular data on a rectangular grid, which stencil minimizes computational cost while maintaining second-order accuracy?

A.Four-point stencil:

[f(x+h,y+k)-f(x+h,y-k)-f(x-h,y+k)+f(x-h,y-k)]/(4hk)

✅

B.Nine-point stencil including all neighbors for isotropic accuracy.

C.Three-point sequential differentiation: first compute

f_x

at three y-levels, then differentiate those in y.

D.Five-point cross stencil combining axial and diagonal points.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The four-point corner stencil achieves $O(h^2+k^2)$ accuracy for $f_{xy}$ using minimal evaluations. Sequential differentiation (option C) requires computing intermediate derivatives at multiple points, increasing operations and potential error accumulation. Nine-point stencils offer no accuracy gain for pure mixed derivatives on rectangular grids. This efficiency consideration matters in large-scale simulations where millions of derivative estimates are needed, balancing precision with computational feasibility.

Q9. An engineer estimates $\partial u/\partial t$ from CFD simulation output saved every $\Delta t=0.1$ s. At $t=1.0$ , values are $u=2.00$ ; at $t=1.1$ , $u=2.15$ ; at $t=0.9$ , $u=1.87$ . Later, re-running with $\Delta t=0.05$ gives $u(1.0)=2.00$ , $u(1.05)=2.08$ , $u(0.95)=1.93$ . Central difference estimates differ by 8%. What primarily causes this discrepancy?

A.Numerical dissipation in the CFD scheme alters solution structure at coarser temporal resolution. ✅

B.Measurement noise dominates at finer resolution.

C.The true derivative is time-dependent and nonlinear between samples.

D.Both estimates suffer from identical truncation error due to same spatial discretization.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: CFD solutions depend on temporal resolution; coarser $\Delta t$ introduces numerical diffusion that smears transient features, altering local derivatives even at sampled points. The 8% difference reflects solution sensitivity to discretization, not just derivative approximation error. This highlights that tabular data quality depends on underlying simulation fidelity, and derivative estimates inherit solver artifacts—critical for validating computational models against experimental benchmarks.

Q10. You must estimate $f_x(0,0)$ from a table where $f(0,0)=0$ , $f(h,0)=h^2 \sin(1/h)$ , and $f(-h,0)=h^2 \sin(-1/h)$ for various $h$ . As $h \to 0$ , central difference estimates oscillate between -1 and 1 despite the true derivative being 0. Why does standard finite difference fail here?

A.The function lacks continuous partial derivatives at origin, violating smoothness assumptions underlying finite difference error analysis. ✅

B.Round-off error dominates due to catastrophic cancellation in sine evaluation.

C.Central difference formula is invalid for odd-symmetric functions.

D.Tabular sampling misses critical oscillation peaks between grid points.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Although $f_x(0,0)=0$ exists by definition, $f_x$ is discontinuous at origin because $\lim_{x\to0} f_x(x,0)$ does not exist. Finite difference theory assumes sufficient smoothness for Taylor expansion; without it, error bounds collapse regardless of $h$ . This pathological case reveals that numerical differentiation implicitly relies on regularity conditions often unstated in textbooks, demanding awareness of theoretical prerequisites beyond mechanical application.

Q11. In climate modeling, temperature $T(\lambda,\phi)$ is given on a latitude-longitude grid. Near the pole ( $\phi \approx 90^\circ$ ), longitude spacing $\Delta \lambda$ corresponds to vanishing physical distance. How should $\partial T/\partial \lambda$ be estimated to avoid spurious large derivatives?

A.Ignore polar regions entirely since derivatives are undefined there.

B.Scale the finite difference by

\cos\phi

to convert longitudinal difference to physical distance. ✅

C.Use only latitude derivatives near poles since longitude becomes irrelevant.

D.Increase

\Delta \lambda

near poles to maintain constant physical spacing.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: On spherical coordinates, physical distance per degree longitude is $R\cos\phi\,\Delta\lambda$ . Raw $\Delta T/\Delta\lambda$ diverges as $\cos\phi \to 0$ even if actual gradient is finite. Multiplying by $\cos\phi$ converts to true spatial derivative $(1/R)\partial T/\partial\lambda_{\text{phys}}$ . This coordinate-aware adjustment prevents numerical artifacts at singularities, illustrating how geometric context must inform discrete calculus in geospatial applications rather than applying Cartesian formulas naively.

Q12. A lab measures voltage $V(x,t)$ across a wire. Tabular data shows $V(x,t)$ is linear in $x$ at each fixed $t$ , but quadratic in $t$ at each fixed $x$ . You need $V_{xt}$ at $(x_0,t_0)$ . Given perfect data, which statement is necessarily true?

V_{xt}=0

everywhere because linearity in

x

implies no coupling.

V_{xt}

equals the coefficient of

xt

term in bivariate polynomial fit.

V_{xt}

can be nonzero if the slope in

x

varies with

t

. ✅

V_{xt}

is undefined because mixed partials require continuity.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Linearity in $x$ means $V(x,t)=a(t)x+b(t)$ . Then $V_x=a(t)$ , so $V_{xt}=a'(t)$ , which is nonzero unless $a(t)$ is constant. The quadratic-in- $t$ observation allows $a(t)$ to vary linearly, making $V_{xt}$ potentially nonzero. This tests understanding that univariate behavior doesn’t constrain cross-derivatives; interaction terms emerge from parameter dependence, requiring multivariate thinking beyond separable intuitions.

Q13. You estimate $\nabla f$ at point P using tabular data. Method X uses central differences on a 5×5 neighborhood; Method Y fits a local least-squares plane to the same points. Under what condition will Method Y outperform Method X despite higher computational cost?

A.When data contains random noise, as least-squares averaging suppresses high-frequency errors that amplify in direct differencing. ✅

B.When the function is exactly quadratic, since both methods give identical results.

C.When grid spacing is uniform, favoring symmetric stencils.

D.When mixed partials are needed, as plane fitting naturally captures cross-terms.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Direct finite differences amplify noise proportionally to $1/h$ , while least-squares fitting distributes error across multiple points, reducing variance by factor $\sim 1/N$ . For noisy experimental data, this robustness outweighs computational overhead. Plane fitting also handles irregular geometries better. However, for clean data on uniform grids, central differences are optimal. This trade-off exemplifies choosing numerical methods based on data characteristics rather than defaulting to textbook formulas, essential for real-world signal processing.

🔗 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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