Answer: $\sin(31°) \approx 0.5 + 0.8660 \times \frac{\pi}{180} \approx 0.5151$

Set up the linearization: Let $f(x) = \sin(x)$ in radians, $a = \pi/6$ (i.e., $30°$). Then $f(a) = 0.5$, $f'(x) = \cos(x)$, and $f'(a) = \cos(\pi/6) \approx 0.8660$. Convert the increment to radians: $31° - 30° = 1°$, and $1° = \frac{\pi}{180} \approx 0.01745$ radians. Evaluate: $L = 0.5 + 0.8660 \times 0.01745 \approx 0.5 + 0.01512 \approx 0.5151$. Why distractors fail: Option A uses $\Delta x = 1$ (degree, not radians), giving an absurd result. Option C uses $\sin(30°) = 0.5$ as the slope instead of $\cos(30°)$. Option D subtracts instead of adding, which would be appropriate if the derivative were negative.

Answer: The tangent line captures only the first-order behavior of $f$, ignoring curvature and higher-order terms.

Understanding the tangent line approximation: The linearization uses only the value and first derivative at $a$. It matches the function's slope at that one point but does not account for how the slope itself changes (curvature). Why the correct answer works: Option B correctly identifies that the tangent line is a first-order (degree 1) approximation. Higher-order terms like $\frac{f''(a)}{2}(x-a)^2$ are neglected, and these grow as $|x - a|$ increases. Why distractors fail: Option A is wrong because $f'(a)$ is a fixed number and does not become undefined. Option C is wrong because inaccuracy arises from curvature, not discontinuity. Option D is wrong because linear approximation works for any differentiable function, not just at critical points.

Answer: $L(x) = f(a) + f'(a)(x - a)$

Recall the linearization formula: The linearization of $f$ at $x = a$ is the equation of the tangent line at that point: $L(x) = f(a) + f'(a)(x - a)$. Why the correct answer works: Option B exactly states the standard linearization formula, using the function value $f(a)$ as the base and the first derivative $f'(a)$ as the slope. Why distractors fail: Option A incorrectly uses the second derivative $f''(a)$ instead of the first derivative. Option C swaps the roles of $f(a)$ and $f'(a)$. Option D multiplies the three quantities rather than using the additive tangent line structure.

Answer: $1.2\%$

Compute the differential: $dV = 3s^2 \, ds = 3(25)(0.02) = 1.5$ cm³. Compute the relative and percentage error: The relative error is $\frac{dV}{V} = \frac{3s^2 \, ds}{s^3} = \frac{3 \, ds}{s} = \frac{3(0.02)}{5} = 0.012$. As a percentage, this is $1.2\%$. General rule: For $V = s^3$, the relative error in $V$ is always three times the relative error in $s$. Here $\frac{ds}{s} = 0.4\%$, so $\frac{dV}{V} \approx 3 \times 0.4\% = 1.2\%$. Why distractors fail: Option A ($0.4\%$) is the relative error in $s$, not in $V$. Option B ($0.6\%$) multiplies by $1.5$ instead of $3$. Option D ($2.0\%$) uses an incorrect multiplier.

Answer: $5.030$

Set up the linearization: We have $f(x) = \sqrt{x}$, $a = 25$, so $f(25) = 5$ and $f'(x) = \frac{1}{2\sqrt{x}}$, giving $f'(25) = \frac{1}{10} = 0.1$. Apply the formula: $L(25.3) = f(25) + f'(25)(25.3 - 25) = 5 + 0.1(0.3) = 5 + 0.03 = 5.030$. Why distractors fail: Option A ($5.060$) doubles the correction, possibly using $0.2$ instead of $0.1$. Option C ($5.015$) halves the correction. Option D ($5.150$) uses $0.3 \times 0.5$ instead of $0.3 \times 0.1$.

Answer: $8\pi$ cm²

Find the differential of surface area: $S = 4\pi r^2$, so $dS = 8\pi r \, dr$. Substitute the given values: $dS = 8\pi (10)(0.1) = 8\pi$ cm². Why distractors fail: Option A ($0.8\pi$) omits the factor of $r = 10$. Option B ($4\pi$) uses $\frac{dS}{dr} = 4\pi r$ (forgetting the factor of 2 from the power rule). Option D ($80\pi$) uses $dr = 1$ instead of $0.1$.

Answer: The increment $x - a = 0.5$ is not small enough, and $x^3$ has significant curvature ($f''(2) = 12 \neq 0$), making the neglected higher-order terms substantial.

Verify the linearization: $f(2) = 8$, $f'(x) = 3x^2$, $f'(2) = 12$. So $L(x) = 8 + 12(x-2)$ and $L(2.5) = 8 + 12(0.5) = 14$. The computation is correct. Analyze the error source: The error is $15.625 - 14 = 1.625$. The second-order term alone is $\frac{f''(2)}{2}(0.5)^2 = \frac{12}{2}(0.25) = 1.5$, which accounts for most of the error. The step size $0.5$ is large enough that curvature matters significantly. Why distractors fail: Option A is false; $x^3$ is differentiable everywhere. Option C is false; $L(2.5) = 14$ is correctly computed. Option D is false; linear approximation is valid for any differentiable function, regardless of degree.

Answer: $1.000$

Compute the linearization at a = 0: $f(0) = \cos(0) = 1$ and $f'(x) = -\sin(x)$, so $f'(0) = 0$. Thus $L(x) = 1 + 0 \cdot (x - 0) = 1$. Evaluate the estimate: $L(0.1) = 1$. The linearization at $a = 0$ gives a constant approximation because the slope is zero there. Why distractors fail: Option A ($0.900$) subtracts $0.1$ from $1$, as if the slope were $-1$. Option B ($1.100$) adds $0.1$, as if the slope were $+1$. Option D ($0.995$) is close to the true value $\cos(0.1) \approx 0.99500$ but is not the linear approximation at $a = 0$.

Answer: $1.050$

Set up the linearization: For $f(x) = e^x$, we have $f(0) = 1$ and $f'(0) = 1$. The linearization is $L(x) = 1 + 1 \cdot (x - 0) = 1 + x$. Compute the estimate: $L(0.05) = 1 + 0.05 = 1.050$. Why distractors fail: Option B ($1.005$) treats the increment as $0.005$ instead of $0.05$. Option C ($0.950$) subtracts instead of adding. Option D ($1.100$) doubles the correct increment.

Answer: $4.40$

Identify the function and tangent line: From the graph, $f(x) = x^2$. At $x = 2$: $f(2) = 4$ and $f'(2) = 4$. Apply the linear approximation: $L(2.1) = f(2) + f'(2)(2.1 - 2) = 4 + 4(0.1) = 4.40$. Why distractors fail: Option B ($4.41$) is the actual value $(2.1)^2 = 4.41$, not the tangent-line estimate. Option C ($4.20$) uses a slope of $2$ instead of $4$. Option D ($4.04$) uses a slope of $0.4$.

Answer: The linear approximation overestimates $\ln(x)$ near $x = 1$.

Interpret the graph: The tangent line (dashed orange) lies above the curve $\ln(x)$ (green) on both sides of $x = 1$. This means $L(x) \geq \ln(x)$ near $a = 1$. Connect to concavity: Since $f''(x) = -1/x^2 < 0$, $\ln(x)$ is concave down everywhere. For concave-down functions, the tangent line always lies above the curve, so the linear approximation overestimates. Why distractors fail: Option A reverses the relationship. Option C is false because the approximation is exact only at $x = 1$ itself. Option D is wrong because concavity is the same sign on both sides, so the tangent line overestimates on both sides.

Answer: The distinction matters because $dy$ is easy to compute analytically while $\Delta y$ may require evaluating the function at a new point, and their difference measures the approximation error.

Acknowledge the grain of truth: For small $dx$, $dy \approx \Delta y$ is indeed true. However, this does not make the distinction useless. Why the correct answer works: Option B correctly notes that $dy = f'(x)\,dx$ is simpler to compute (no need to evaluate $f$ at a new point), and the quantity $\Delta y - dy$ represents the error introduced by the linear approximation. Understanding this difference is essential in error analysis. Why distractors fail: Option A is wrong because their difference, though small, is the whole point of error estimation. Option C is wrong because for small $dx$, $dy$ and $\Delta y$ are indeed close by definition of differentiability. Option D is wrong because $dy$ is defined for any differentiable function.

Answer: $dy = f'(x) \, dx$

Recall the definition of a differential: For a differentiable function $y = f(x)$, the differential is defined as $dy = f'(x) \, dx$, where $dx$ is an independent variable representing a small change in $x$. Why the correct answer works: Option B correctly states that $dy$ equals the derivative times the differential $dx$. Why distractors fail: Option A uses $f(x)$ instead of $f'(x)$. Option C defines $\Delta y$, the actual change in $f$, not the differential $dy$. Option D incorrectly divides by $dx$ instead of multiplying.

Answer: $f$ must be differentiable at $a$.

Identify the requirement: The formula uses $f'(a)$, so the function must have a derivative at $a$. Differentiability at $a$ is both necessary and sufficient for the linearization to be defined. Why the correct answer works: Option C is correct: differentiability at $a$ guarantees that $f(a)$ exists, $f$ is continuous at $a$, and $f'(a)$ exists — all needed for the tangent line. Why distractors fail: Option A is insufficient because continuity alone does not guarantee $f'(a)$ exists (e.g., $|x|$ at $x = 0$). Option B is too strong; a second derivative is not required. Option D is irrelevant; there is no requirement for an extremum.

Answer: $dy$ measures the change along the tangent line, while $\Delta y$ measures the actual change in $f$.

Define both quantities: Given a change $dx = \Delta x$, we have $dy = f'(x)\,dx$ (change along the tangent) and $\Delta y = f(x + \Delta x) - f(x)$ (actual change along the curve). Why the correct answer works: Option B correctly captures the key distinction: $dy$ is the tangent-line estimate while $\Delta y$ is the true function change. Why distractors fail: Option A is wrong because for concave-up functions $dy < \Delta y$ and for concave-down $dy > \Delta y$. Option C is wrong because $\Delta y$ does not use derivatives at all; it is the exact change. Option D is wrong because they are equal only for linear functions, not all differentiable functions.

Answer: $dT = \dfrac{\pi}{\sqrt{gL}} \, dL$

Differentiate T with respect to L: $T = 2\pi\sqrt{L/g} = 2\pi \cdot g^{-1/2} \cdot L^{1/2}$. So $\frac{dT}{dL} = 2\pi \cdot g^{-1/2} \cdot \frac{1}{2} L^{-1/2} = \frac{\pi}{\sqrt{gL}}$. Write the differential: $dT = \frac{\pi}{\sqrt{gL}} \, dL$. Substituting $L = 1.00$ and $g = 9.80$: $dT = \frac{\pi}{\sqrt{9.80}} \cdot 0.01 \approx 0.01005$ s. Why distractors fail: Option A substitutes $dL$ inside the square root rather than differentiating — it treats the differential as a direct substitution into the original formula. Option C drops the square root entirely, as if $T$ were linear in $L$. Option D has incorrect powers of $g$; it results from misapplying the chain rule (e.g., differentiating $\sqrt{1/g}$ as well, or squaring an extra factor).