📝 Error propagation using differentials (16 MCQs)

📖 From Calculus • 4. Topics in Differentiation • 16 questions available

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What is Error propagation using differentials?

Definition:
Error propagation uses differentials to estimate how uncertainty in measured variables affects the calculated result. If a quantity $Q$ depends on variables $x, y, \dots$ , the estimated error $dQ$ is approximated by $dQ \approx \frac{\partial Q}{\partial x}dx + \frac{\partial Q}{\partial y}dy$ . This linear approximation assumes errors are small and independent, providing a bound on the maximum possible error in the final computation.

Example:
If the radius $r$ of a sphere is $10 \pm 0.1$ cm, estimate the error in volume $V = \frac{4}{3}\pi r^3$ . $dV = 4\pi r^2 dr$ . With $r=10, dr=0.1$ , $dV = 4\pi(100)(0.1) = 40\pi \approx 125.66$ cm $^3$ .

Reason:
In experimental sciences, no measurement is exact. Understanding error propagation helps determine the reliability of derived quantities. It ensures that conclusions drawn from data account for measurement limitations, maintaining scientific rigor and accuracy in reported results.

Easy

Medium

Hard

📝 All Error propagation using differentials MCQs

Q1. In the context of error propagation, what does the symbol $dx$ represent?

A.The measurement error in the variable

x

✅

B.The derivative of

x

with respect to time

C.A small change in the function

f

D.The absolute value of the error in

y

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The symbol $dx$ denotes the difference between the measured value and the exact value of $x$ . It quantifies the measurement error, not a derivative or any other quantity, and is the basis for estimating how this uncertainty propagates to other variables.

Q2. According to the linear approximation method, which formula gives the propagated error $dy$ for a function $y = f(x)$ ?

dy = f(x)dx

dy = f&#039;(x)dx

✅

dy = f&#039;&#039;(x)dx

dy = \frac{dx}{f&#039;(x)}

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The differential approach states that a small change in the output, $dy$ , is approximated by the derivative of the function evaluated at the point of interest multiplied by the input error, i.e., $dy = f'(x)dx$ . This linearization captures the first‑order effect of the measurement error.

Q3. If the measurement error $dx$ for a quantity $x$ is doubled, how does the estimated propagated error $dy$ change for the function $y = x^{3}$ ?

A.It remains the same

B.It doubles ✅

C.It triples

D.It increases by a factor of eight

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For $y = x^{3}$ the differential gives $dy = 3x^{2}dx$ . The factor $3x^{2}$ is constant for a given $x$ , so $dy$ is directly proportional to $dx$ . Doubling $dx$ therefore doubles $dy$ .

Q4. When substituting the measured value $x$ for the unknown exact value $x_{0}$ in the formula $dy = f'(x)dx$ , which of the following statements is generally true?

A.The estimate always underestimates the true error

B.The estimate is exact if

f

is linear ✅

C.The estimate overestimates the error for concave functions

D.The estimate is independent of the shape of

f

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: If the underlying relationship is linear, the derivative $f'(x)$ is constant and the differential formula reproduces the exact change, making the error estimate exact. For non‑linear functions the approximation introduces some error, but linearity guarantees correctness.

Q5. A square has a measured side length of 10 in with a possible error of ± $\frac{1}{32}$ in. Using $dy = 2x\,dx$ , what is the maximum absolute error in the area?

A.±

\frac{5}{8}

^2

✅

B.±

\frac{10}{32}

^2

C.±

\frac{1}{16}

^2

D.±

\frac{20}{32}

^2

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Substituting $x = 10$ and $dx = \pm\frac{1}{32}$ into $dy = 2x\,dx$ gives $dy = 20\left(\pm\frac{1}{32}\right)=\pm\frac{20}{32}=\pm\frac{5}{8}$ . This represents the greatest possible deviation of the computed area from its true value.

Q6. For two independent measurements $x$ and $z$ with errors $dx$ and $dz$ , the product $y = xz$ has propagated error $dy \approx x\,dz + z\,dx$ . If both $dx$ and $dz$ are negative, which statement about the sign of $dy$ is correct?

dy

must be positive

dy

must be negative

dy

could be either sign depending on magnitudes ✅

dy

is zero

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Since $dy = x\,dz + z\,dx$ , the sign depends on the relative sizes of the terms $x\,dz$ and $z\,dx$ . Both terms are negative, but if one magnitude outweighs the other, the sum could be negative or positive, making the overall sign indeterminate without further information.

Q7. For the function $y = \ln(x)$ , the differential gives $dy = \frac{1}{x}dx$ . Which expression correctly represents the relative error $\frac{dy}{y}$ in terms of the relative error $\frac{dx}{x}$ ?

\frac{dy}{y} = \frac{dx}{x}

\frac{dy}{y} = \frac{dx}{x^{2}}

\frac{dy}{y} = \frac{dx}{x}\cdot\frac{1}{\ln(x)}

✅

\frac{dy}{y} = \frac{dx}{x}\cdot\frac{1}{\ln(e)}

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Because $y = \ln(x)$ , we have $dy = \frac{dx}{x}$ . Dividing by $y$ yields $\frac{dy}{y} = \frac{dx}{x}\frac{1}{\ln(x)}$ . Thus the relative error in $y$ is the relative error in $x$ scaled by the reciprocal of $\ln(x)$ .

Q8. A sphere’s volume is given by $V = \frac{4}{3}\pi r^{3}$ . If the radius is measured as $r = 5.0$ cm with an uncertainty of ±0.2 cm, what is the estimated absolute error in $V$ using differentials?

A.±

12.6

^3

B.±

31.4

^3

C.±

62.8

^3

✅

D.±

125.6

^3

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The differential $dV = 4\pi r^{2}dr$ . Substituting $r = 5.0$ cm and $dr = 0.2$ cm gives $dV = 4\pi(25)(0.2)=20\pi\approx62.8$ cm $^3$ . This value approximates the maximum absolute error in the volume due to the radius uncertainty.

Q9. If the measured error $dx$ is negative and the function $f$ is strictly increasing, what can be said about the sign of the propagated error $dy$ approximated by $dy = f'(x)dx$ ?

dy

will be negative ✅

dy

will be positive

dy

could be either sign

dy

will be zero

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For a strictly increasing function, the derivative $f'(x)$ is positive. Multiplying a positive derivative by a negative $dx$ yields a negative $dy$ . Hence the propagated error inherits the sign of the original measurement error.

Q10. Given $y = x^{2} + 3z$ with errors $dx$ and $dz$ , the differential estimate is $dy = 2x\,dx + 3\,dz$ . If $|dx| = 0.01$ and $|dz| = 0.001$ while $x = 2$ , which term dominates the error in $y$ ?

A.The

2x\,dx

term ✅

B.The

3\,dz

term

C.Both terms contribute equally

D.Neither term is significant

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Evaluating each term: $2x\,dx = 4(0.01)=0.04$ and $3\,dz = 3(0.001)=0.003$ . The first term is an order of magnitude larger, so it dominates the total propagated error.

Q11. Why is it acceptable to replace the unknown exact value $x_{0}$ with the measured value $x$ in the error‑propagation formula $dy = f'(x)dx$ when the measurement error $dx$ is small?

A.Because

f&#039;(x)

varies slowly near

x

B.Because the error term becomes negligible

C.Because higher‑order terms are of order

(dx)^{2}

and can be ignored ✅

D.Because

x

and

x_{0}

are identical by definition

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The differential approximation retains only the first‑order term in a Taylor expansion. When $dx$ is small, the omitted higher‑order terms, which involve $(dx)^{2}$ or higher powers, contribute insignificantly, justifying the substitution of $x$ for $x_{0}$ .

Q12. For the function $y = \frac{1}{x}$ , the differential gives $dy = -\frac{1}{x^{2}}dx$ . If $x = 2.00$ and the measurement error is $dx = 0.01$ , which of the following is the closest estimate of the propagated error $dy$ ?

-0.0025

✅

-0.0050

-0.0100

-0.0250

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Substituting into $dy = -\frac{1}{x^{2}}dx$ yields $dy = -\frac{1}{4}(0.01) = -0.0025$ . This value represents the linear approximation of how the reciprocal function's output changes due to the small error in $x$ .

Q13. When propagating error through $y = \sin(x)$ , the differential is $dy = \cos(x)\,dx$ . Which statement best describes how the magnitude of the error depends on the value of $x$ ?

A.The error is largest when

\cos(x)

is ±1 and smallest when

\cos(x)=0

✅

B.The error is independent of

x

C.The error is proportional to

\sin(x)

D.The error always equals

dx

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Since $dy = \cos(x)dx$ , the error magnitude equals $|\cos(x)|$ times the input error. $|\cos(x)|$ reaches its maximum of 1 at multiples of $2\pi$ and its minimum of 0 at odd multiples of $\pi/2$ , directly influencing the propagated error.

Q14. In an experiment, pressure $P$ is measured with a 2 % relative error and volume $V$ with a 1 % relative error. Using the ideal‑gas relation $n = \frac{PV}{RT}$ (where $R$ and $T$ are exact), what is the approximate relative error in the calculated amount $n$ ?

A.0.01

B.0.03 ✅

C.0.02

D.0

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Because $n$ is proportional to the product $PV$ , the relative errors add: $2\% + 1\% = 3\%$ . Higher‑order terms are negligible, so the total relative error in $n$ is approximately 3 %.

Q15. If a measurement error is bounded by $-\epsilon \le dx \le \epsilon$ , what is the maximum possible magnitude of the propagated error $dy$ for a linear approximation $dy = k\,dx$ ?

\epsilon

k

|k|\,\epsilon

✅

|k|/\epsilon

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The linear relation $dy = k\,dx$ scales the input error by the constant factor $k$ . The largest magnitude occurs when $|dx| = \epsilon$ , giving $|dy|_{\max}=|k|\,\epsilon$ .

Q16. Suppose a quantity $y$ depends on two measured variables $x$ and $w$ via $y = x^{2}w$ . If the relative errors are $\frac{dx}{x}=0.01$ and $\frac{dw}{w}=0.02$ , what is the approximate relative error in $y$ using differentials?

A.0.04 ✅

B.0.03

C.0.02

D.0.01

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Differentiating gives $\frac{dy}{y}=2\frac{dx}{x}+\frac{dw}{w}$ . Substituting the given relative errors yields $2(0.01)+0.02=0.04$ . Hence the relative error in $y$ is approximately 4 % (0.04).

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