Answer: $L(x,y) = 87 - 6(x - 3) - 8(y - 1)$

Compute $T(3,1)$: $T(3,1) = 100 - 9 - 4 = 87$. Compute partial derivatives: $T_x = -2x$, so $T_x(3,1) = -6$. $T_y = -8y$, so $T_y(3,1) = -8$. Construct the linearization: $L(x,y) = 87 + (-6)(x-3) + (-8)(y-1) = 87 - 6(x-3) - 8(y-1)$. Why distractors fail: Option B halves the partial derivatives (forgetting the coefficient $2$ in the derivatives). Option C uses $100$ instead of $T(3,1) = 87$. Option D reverses the signs of the partial derivatives.

Answer: The function can be well-approximated by its tangent plane near $(a, b)$, with the error going to zero faster than the distance to the point.

Define differentiability in multivariable calculus: A function $f(x,y)$ is differentiable at $(a,b)$ if $f(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b) + \epsilon_1(x-a) + \epsilon_2(y-b)$, where $\epsilon_1, \epsilon_2 \to 0$ as $(x,y) \to (a,b)$. Equivalently, the tangent plane approximation error goes to zero faster than the distance to $(a,b)$. Why the correct answer works: Option C captures the precise meaning: the linear (tangent plane) approximation is accurate with a superlinear error term. Why distractors fail: Option A is necessary but not sufficient — partial derivatives can exist without the function being differentiable. Option B is a consequence of differentiability but not its definition; continuity alone does not guarantee differentiability. Option D is an arbitrary condition unrelated to the definition.

Answer: It gives the equation of the tangent plane expressed as a function of $x$ and $y$.

Connect linearization to geometry: The linearization $L(x,y)$ is the $z$-value on the tangent plane at any point $(x,y)$ near $(a,b)$. Writing $z = L(x,y)$ gives the tangent plane equation. Why the correct answer works: Option B correctly identifies that $L(x,y)$ is the tangent plane rewritten as $z = L(x,y)$, i.e., a function giving the plane's height. Why distractors fail: Option A confuses the tangent plane with the normal line. Option C is incorrect because the linearization is a first-order (not second-order) polynomial. Option D confuses linearization with the directional derivative.

Answer: The claim is false — the partial derivatives must also be continuous in a neighborhood of the point to guarantee differentiability.

Distinguish partial derivatives from differentiability: In multivariable calculus, the existence of partial derivatives at a point does not guarantee differentiability. A sufficient condition is that the partial derivatives are continuous in a neighborhood of the point. Why the correct answer works: Option B correctly identifies the gap: continuity of partial derivatives (not merely their existence) is a standard sufficient condition for differentiability. Why distractors fail: Options A and C incorrectly assert that existence of partials suffices. Option D is misleading — while differentiability does imply existence of all directional derivatives, requiring all directional derivatives is not the standard sufficient condition taught in this context.

Answer: Yes, the linearization is correct.

Verify $f(1,1)$: $f(1,1) = \ln(1 + 1) = \ln 2$. ✓ Verify $f_x(1,1)$: $f_x = \frac{2x}{x^2 + y}$, so $f_x(1,1) = \frac{2}{2} = 1$. ✓ Verify $f_y(1,1)$: $f_y = \frac{1}{x^2 + y}$, so $f_y(1,1) = \frac{1}{2}$. ✓ Why distractors fail: Option B claims $f_x = \frac{1}{2}$, but the chain rule gives $\frac{2x}{x^2+y}$, which is $1$ at $(1,1)$. Option C confuses $x^2 + y$ (which is $2$) with $x^2$ (which would be $1$). Option D reverses the partial derivative values.

Answer: $L(x,y) = 7 + 8(x - 1) + 3(y - 2)$

Compute $f(1,2)$: $f(1,2) = 1^2 + 3(1)(2) = 1 + 6 = 7$. Compute partial derivatives: $f_x = 2x + 3y$, so $f_x(1,2) = 2(1) + 3(2) = 8$. $f_y = 3x$, so $f_y(1,2) = 3(1) = 3$. Assemble the linearization: $L(x,y) = 7 + 8(x - 1) + 3(y - 2)$. This matches Option A. Why distractors fail: Option B uses $f_x = 5$, likely from computing $2(1) + 3(1) = 5$ (using $b = 1$ instead of $b = 2$). Option C uses $f_y = 6$, likely from $3(2) = 6$ (evaluating at $y = 2$ instead of $x = 1$). Option D uses $f(1,2) = 6$, likely from forgetting the $1^2$ term or miscalculating $3(1)(2)$.

Answer: The approximation is correct — all three components $f(2, \pi/2) = 8$, $f_x(2, \pi/2) = 12$, and $f_y(2, \pi/2) = 0$ are computed properly.

Verify $f(2, \pi/2)$: $f(2, \pi/2) = 8 \sin(\pi/2) = 8(1) = 8$. ✓ Verify $f_x(2, \pi/2)$: $f_x = 3x^2 \sin(y)$, so $f_x(2, \pi/2) = 3(4)(1) = 12$. ✓ Verify $f_y(2, \pi/2)$: $f_y = x^3 \cos(y)$, so $f_y(2, \pi/2) = 8 \cos(\pi/2) = 8(0) = 0$. ✓ Why distractors fail: Option B claims $f_x = 6$, which would result from computing $3x^2$ at $x=2$ but forgetting to multiply by $\sin(\pi/2) = 1$ (or using $3(2) = 6$ instead of $3(4) = 12$). Option C claims $f_y = 8$, confusing $x^3 \cos(y)$ with $x^3$ and forgetting that $\cos(\pi/2) = 0$. Option D claims $f(2,\pi/2) = 0$, confusing $\sin(\pi/2) = 1$ with $\sin(\pi) = 0$.

Answer: The error grows quadratically with distance, confirming the function is differentiable at the origin.

Interpret the error surface: The graph shows $|f(x,y) - L(x,y)| \approx x^2 + y^2 = r^2$, which grows quadratically (like a paraboloid) with distance from the origin. Connect to differentiability: Differentiability at a point means $\frac{|f(x,y) - L(x,y)|}{\sqrt{(x-a)^2+(y-b)^2}} \to 0$ as $(x,y) \to (a,b)$. Since the error is $O(r^2)$ and the denominator is $r$, the ratio $\to 0$, confirming differentiability. Why distractors fail: Option A would be correct if the error were $O(r)$, but here it's $O(r^2)$. Option C is wrong because the error is clearly not zero. Option D is incorrect since the error is smooth, not oscillating.

Answer: $df = f_x \, dx + f_y \, dy$

Recall the total differential: The total differential of $f(x,y)$ is $df = f_x \, dx + f_y \, dy$, where $dx$ and $dy$ represent small changes in $x$ and $y$. Why the correct answer works: Option C correctly pairs each partial derivative with its corresponding differential: $f_x$ with $dx$ and $f_y$ with $dy$. Why distractors fail: Option A uses multiplication instead of addition. Option B swaps $dx$ and $dy$, pairing $f_x$ with $dy$ and vice versa. Option D treats the differentials as denominators, which is not the definition of the total differential.

Answer: $\approx 1.95$

Compute $f(1,1)$: $f(1,1) = \sqrt{1+3} = 2$. Compute partial derivatives at $(1,1)$: $f_x = \frac{1}{2\sqrt{x+3y}}$, so $f_x(1,1) = \frac{1}{2(2)} = \frac{1}{4}$. $f_y = \frac{3}{2\sqrt{x+3y}}$, so $f_y(1,1) = \frac{3}{2(2)} = \frac{3}{4}$. Apply the linearization: $L(1.1, 0.9) = 2 + \frac{1}{4}(0.1) + \frac{3}{4}(-0.1) = 2 + 0.025 - 0.075 = 1.95$. Why distractors fail: Option A ($2.05$) incorrectly adds both correction terms as positive. Option B ($1.90$) doubles the negative correction. Option D ($2.00$) ignores the corrections entirely.

Answer: $|dR| \approx \frac{|dV|}{I} + \frac{V |dI|}{I^2} \approx 0.10$ Ω

Compute partial derivatives of $R = V/I$: $R_V = \frac{1}{I} = \frac{1}{3}$ and $R_I = -\frac{V}{I^2} = -\frac{12}{9} = -\frac{4}{3}$. Apply total differential for maximum error: $|dR|_{\max} = |R_V||dV| + |R_I||dI| = \frac{1}{3}(0.1) + \frac{4}{3}(0.05) = 0.0\overline{3} + 0.0\overline{6} \approx 0.10$ Ω. Why distractors fail: Option B uses the correct formula but only computes the first term $\frac{1}{3}(0.1) \approx 0.033$, omitting the second term. Option C confuses $R_V$ and $R_I$ with $V$ and $I$ directly, producing a nonsensical result. Option D subtracts the terms instead of taking absolute values for maximum error, and reports a negative value which is impossible for a maximum error magnitude.

Answer: The claim is incorrect — the linearization error does not go to zero faster than the distance to the origin, so $f$ is not differentiable.

Check the linearization: Since $f(0,0) = 0$, $f_x(0,0) = 0$, $f_y(0,0) = 0$, the linearization is $L(x,y) = 0$. Test the differentiability condition: We need $\frac{|f(x,y) - L(x,y)|}{\sqrt{x^2+y^2}} = \frac{\sqrt{|xy|}}{\sqrt{x^2+y^2}} \to 0$. Along the line $y = x$: $\frac{\sqrt{x^2}}{\sqrt{2}|x|} = \frac{1}{\sqrt{2}} \neq 0$. So the limit is not zero, and $f$ is not differentiable. Why distractors fail: Option A and C confuse necessary conditions with sufficient conditions. Option D is wrong because $f(x,y) = \sqrt{|xy|}$ is continuous at the origin (it approaches $0$).

Answer: $df \approx -0.06$

Identify changes: $dx = 3.01 - 3 = 0.01$ and $dy = 1.98 - 2 = -0.02$. Compute partial derivatives at $(3, 2)$: $f_x = 2xy$, so $f_x(3,2) = 12$. $f_y = x^2$, so $f_y(3,2) = 9$. Apply the total differential: $df = 12(0.01) + 9(-0.02) = 0.12 - 0.18 = -0.06$. Why distractors fail: Option A ($0.03$) might result from using only $f_x\,dx$ and a sign error. Option C ($0.30$) could arise from computing $f_x\,dx = 12(0.01) = 0.12$ and $f_y\,dy = 9(0.02) = 0.18$ but adding instead of subtracting. Option D ($-0.12$) could come from doubling the correct answer, perhaps by using $dy = -0.04$ by mistake.

Answer: $z = 1 + x + 2y$

Compute partial derivatives: For $f(x,y) = e^{x+2y}$: $f_x = e^{x+2y}$ and $f_y = 2e^{x+2y}$. Evaluate at $(0,0)$: $f(0,0) = 1$, $f_x(0,0) = 1$, $f_y(0,0) = 2$. Write the tangent plane: $z - 1 = 1(x - 0) + 2(y - 0)$, which simplifies to $z = 1 + x + 2y$. Why distractors fail: Option A uses $f_y = 1$ instead of $2$. Option C swaps the coefficients of $x$ and $y$. Option D omits the constant term $f(0,0) = 1$.

Answer: $\pm 17 \text{ cm}^3$

Set up the total differential: $dV = \frac{\partial V}{\partial l}\,dl + \frac{\partial V}{\partial w}\,dw + \frac{\partial V}{\partial h}\,dh = wh\,dl + lh\,dw + lw\,dh$. Evaluate and maximize error: At $(10, 8, 5)$: $wh = 40$, $lh = 50$, $lw = 80$. Taking all errors in the worst-case direction: $|dV|_{\max} = 40(0.1) + 50(0.1) + 80(0.1) = 4 + 5 + 8 = 17$ cm³. Why distractors fail: Option A ($\pm 13$) likely omits the $lw\,dh$ term or miscalculates one partial derivative. Option C ($\pm 21$) overestimates, perhaps by using incorrect derivative values such as $wh = 40$, $lh = 50$, $lw = 120$. Option D ($\pm 9$) likely uses only two of the three terms or halves the partial derivatives.

Answer: $x + 2y + 3z = 14$

Use the gradient for implicit surfaces: For $F(x,y,z) = 0$, the tangent plane at $(x_0,y_0,z_0)$ is $F_x(x_0,y_0,z_0)(x-x_0) + F_y(x_0,y_0,z_0)(y-y_0) + F_z(x_0,y_0,z_0)(z-z_0) = 0$. Compute the gradient: $\nabla F = (2x, 2y, 2z)$. At $(1,2,3)$: $\nabla F = (2, 4, 6)$. Write and simplify the tangent plane: $2(x-1) + 4(y-2) + 6(z-3) = 0$. Expanding: $2x + 4y + 6z = 2 + 8 + 18 = 28$. Dividing by 2: $x + 2y + 3z = 14$. Why distractors fail: Option B is the unsimplified version $2x + 4y + 6z = 28$ incorrectly set equal to $14$. Option C sets the right-hand side to $0$ instead of $14$. Option D uses equal coefficients, ignoring that the gradient varies with the coordinates.

Answer: $z - f(a,b) = f_x(a,b)(x - a) + f_y(a,b)(y - b)$

Recall the tangent plane formula: The tangent plane to $z = f(x,y)$ at the point $(a, b, f(a,b))$ is given by $z - z_0 = f_x(a,b)(x - a) + f_y(a,b)(y - b)$, where $z_0 = f(a,b)$. Why the correct answer works: Option B exactly matches this standard formula, with additive partial derivative terms. Why distractors fail: Option A multiplies the two partial derivative terms instead of adding them. Option C omits the translation to the point $(a, b, f(a,b))$, treating it as if the tangent plane passes through the origin. Option D incorrectly subtracts the $f_y$ term instead of adding it.

Answer: $(0, 0, 0)$

Condition for a horizontal tangent plane: The tangent plane is horizontal when both $f_x = 0$ and $f_y = 0$ at the point. Compute partial derivatives: $f_x = y\cos(xy)$ and $f_y = x\cos(xy)$. Check each point: At $(0,0)$: $f_x = 0 \cdot \cos(0) = 0$, $f_y = 0 \cdot \cos(0) = 0$. Both are zero, so the tangent plane is horizontal. ✓ At $(\pi, 1)$: $f_x = 1 \cdot \cos(\pi) = -1 \neq 0$. ✗ At $(1, \pi)$: $f_y = 1 \cdot \cos(\pi) = -1 \neq 0$. ✗ At $(1, 1)$: $f_x = 1 \cdot \cos(1) \approx 0.54 \neq 0$. ✗ Why distractors fail: Options B and C each have one nonzero partial derivative due to $\cos(\pi) = -1$. Option D has $f_x = \cos(1) \neq 0$, so the tangent plane is not horizontal there.

Answer: The tangent plane lies entirely below the surface near the point of tangency.

Analyze the surface shape: The surface $z = x^2 + y^2$ is a paraboloid opening upward, which is concave up everywhere. Tangent plane behavior for concave up surfaces: For a concave up surface (where the Hessian is positive definite), the tangent plane lies entirely below the surface near the point of tangency, except at the tangent point itself where they touch. Why distractors fail: Option A describes the behavior for a concave down surface. Option C would occur only for saddle-type surfaces. Option D would only be true if $f$ were itself a linear function.