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📝 Tangent plane to surface z = f(x,y) (14 MCQs)

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

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What is Tangent plane to surface z = f(x,y)?

Definition:
Special case where surface is graph of function; plane equation derived from linearization $L(x,y)$ with normal $\langle -f_x, -f_y, 1 \rangle$ .

Example:
For $z=\sin(xy)$ at $(\pi/2, 1, 1)$ , $f_x=y\cos(xy)=0$ , $f_y=x\cos(xy)=0$ , so tangent plane is horizontal $z=1$ .

Reason:
Explicit form simplifies computation when function representation exists; connects directly to differential and linear approximation concepts.

Easy

Medium

Hard

📝 All Tangent plane to surface z = f(x,y) MCQs

Q1. A surface is defined by $z = x^2 - y^2$ . At point $P(1, 1, 0)$ , a student claims the tangent plane is horizontal because $z=0$ at that point. Which statement best analyzes this error?

A.The student correctly identified that zero height implies a horizontal tangent plane.

B.The student confused the function value with the gradient; the tangent plane depends on partial derivatives, not the z-coordinate alone. ✅

C.The student should have used the second derivative test to determine planarity.

D.The tangent plane is indeed horizontal because the surface is symmetric about the origin.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This question targets error analysis by addressing the common misconception that a zero function value implies a flat tangent plane. Students must understand that the orientation of the tangent plane is determined entirely by the gradient vector $\nabla f$ , which represents local rates of change, not the absolute position or elevation of the point on the surface.

Q2. Consider the surface $z = f(x,y)$ where the contour lines near $(a,b)$ are concentric circles becoming denser as they approach the center. What can be deduced about the tangent plane at $(a,b,f(a,b))$ ?

A.The tangent plane is horizontal because circular symmetry implies a local extremum.

B.The tangent plane is vertical because dense contours indicate infinite slope.

C.The tangent plane cannot exist because the function is not differentiable at the center.

D.The tangent plane's steepness increases radially, but its orientation at the exact center requires evaluating limits of partial derivatives. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: This graph-based question requires interpreting topographical map features to infer analytical properties. Dense concentric contours suggest a peak or valley, implying a horizontal tangent plane if differentiable. However, without explicit differentiability confirmation, students must recognize that visual density indicates gradient magnitude trends rather than guaranteeing the existence or specific orientation of the tangent plane at the singularity.

Q3. An engineer models a heat shield surface as $z = \sqrt{x^2 + y^2}$ . They need the tangent plane at the origin for thermal flux calculations. Why does the standard tangent plane formula fail here, and what is the physical implication?

A.The formula fails because partial derivatives are undefined at the origin; physically, the sharp tip creates a singularity where linear approximation breaks down. ✅

B.The formula works but yields

z=0

; physically, the shield is perfectly flat at the tip.

C.The formula fails due to division by zero in the denominator; physically, heat flux is maximum at smooth points only.

D.The formula yields a vertical plane; physically, the shield reflects all incident radiation at the tip.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This scenario-based application question connects mathematical differentiability to physical modeling. The cone $z = \sqrt{x^2 + y^2}$ lacks a unique tangent plane at the vertex because directional derivatives vary with approach angle. Recognizing this failure prevents incorrect engineering assumptions about local linearity and highlights the importance of verifying differentiability before applying tangent plane approximations in real-world systems.

Q4. Given $z = f(x,y)$ with $f_x(2,3) = 4$ and $f_y(2,3) = -3$ , which vector is normal to the tangent plane at $(2,3,f(2,3))$ ?

\langle 4, -3, 1 \rangle

\langle -4, 3, 1 \rangle

\langle 4, -3, -1 \rangle

✅

\langle 3, 4, 0 \rangle

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: This direct recall question tests the fundamental formula for the normal vector to a tangent plane of an explicit surface. For $z = f(x,y)$ , the upward-pointing normal is typically $\langle -f_x, -f_y, 1 \rangle$ or downward $\langle f_x, f_y, -1 \rangle$ . Option C matches the standard downward orientation derived from rewriting the surface as $F(x,y,z) = f(x,y) - z = 0$ , ensuring students memorize the correct sign convention.

Q5. Two surfaces $z = f(x,y)$ and $z = g(x,y)$ intersect along a curve passing through $P$ . If their tangent planes at $P$ are identical, what must be true about their gradients at that point?

\nabla f(P) = \nabla g(P)

and

f(P) = g(P)

✅

\nabla f(P) = -\nabla g(P)

|\nabla f(P)| = |\nabla g(P)|

only

D.The gradients must be orthogonal to each other.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This conceptual understanding question links geometric tangency to analytical conditions. Identical tangent planes require both surfaces to pass through the same point (equal function values) and share the same local linear approximation (equal partial derivatives). Thus, both the scalar values and gradient vectors must match exactly, distinguishing true tangency from mere parallelism or intersection at non-tangent angles.

Q6. A student computes the tangent plane to $z = x^3 + xy$ at $(1,2,3)$ and obtains $z = 5x + y - 4$ . Upon verification, they find $f(1,2)=3$ but the plane gives $z=2$ at $(1,2)$ . Where is the most likely source of error?

A.Incorrect evaluation of partial derivatives at the point.

B.Arithmetic error in computing the constant term using

z_0 - f_x x_0 - f_y y_0

. ✅

C.Misidentification of the base point coordinates.

D.Failure to check differentiability before applying the formula.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This error analysis question focuses on the algebraic construction of the tangent plane equation. Since the plane must pass through $(x_0, y_0, z_0)$ , any discrepancy at the base point indicates a miscalculation in the constant term. Students must trace the point-slope form $z = z_0 + f_x(x-x_0) + f_y(y-y_0)$ to identify arithmetic slips versus conceptual misunderstandings about derivative computation.

Q7. For the surface $z = e^{-(x^2+y^2)}$ , compare the accuracy of the tangent plane approximation at $(0,0)$ versus $(2,2)$ for estimating $z$ at nearby points. Which statement is correct?

A.Approximation is equally accurate everywhere because the function is smooth.

B.Approximation is better at

(2,2)

because the surface is flatter there.

C.Approximation is better at

(0,0)

because higher-order terms vanish faster near the critical point. ✅

D.Accuracy depends solely on step size, not location.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This mixed concepts question combines Taylor series intuition with tangent plane geometry. Near the critical point $(0,0)$ , the gradient is zero and the surface is locally quadratic, making the linear approximation exceptionally good over a larger neighborhood. At $(2,2)$ , significant curvature and nonzero gradient cause rapid deviation from linearity, demonstrating that approximation quality varies spatially based on local differential properties.

Q8. Suppose $z = f(x,y)$ has continuous partial derivatives everywhere. If the tangent plane at every point on the surface passes through the origin, what functional form must $f$ satisfy?

f(x,y) = ax + by

✅

f(x,y) = k\sqrt{x^2+y^2}

f(x,y) = c

(constant)

f(x,y) = x^2 + y^2

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: This Olympiad-style problem requires reverse-engineering a global geometric constraint into a functional equation. If every tangent plane contains the origin, Euler’s theorem for homogeneous functions applies: $f = xf_x + yf_y$ . The only differentiable solutions satisfying this identity globally are linear functions through the origin. Nonlinear homogeneous functions like cones fail differentiability at the origin, eliminating them despite satisfying the geometric condition almost everywhere.

Q9. In optimizing a manufacturing process, the cost surface is modeled as $C(x,y) = x^2 + 4y^2 + xy$ . The current operating point is $(2,1)$ . To reduce cost most rapidly while staying on the tangent plane, in which direction should adjustments be made?

A.Along the gradient vector

\langle 5, 9 \rangle

B.Opposite to the gradient vector

\langle -5, -9 \rangle

✅

C.Perpendicular to the gradient in the xy-plane

D.Along the direction of steepest ascent on the tangent plane

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This application question integrates optimization with tangent plane geometry. The tangent plane provides the local linear model of cost. Maximum rate of decrease occurs in the direction opposite to the gradient of this linear approximation, which equals $-\nabla C(2,1)$ . Students must connect abstract tangent plane concepts to practical decision-making, recognizing that local improvement strategies rely entirely on first-order information captured by the tangent plane.

Q10. A surface $z = f(x,y)$ satisfies $f(x,y) = f(-x,-y)$ for all $(x,y)$ . What can be concluded about the tangent plane at the origin without computing derivatives?

A.It must be horizontal.

B.It must be vertical.

C.It does not necessarily exist, but if it exists, it must be horizontal. ✅

D.It must coincide with the plane

z = f(0,0)

regardless of differentiability.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: This conceptual question exploits symmetry to deduce tangent plane properties. Even symmetry about the origin implies that if partial derivatives exist at $(0,0)$ , they must be zero (since $f_x(0,0) = \lim_{h\to0}[f(h,0)-f(0,0)]/h = \lim[-f(-h,0)+f(0,0)]/h = -f_x(0,0)$ ). However, symmetry alone doesn’t guarantee differentiability; thus, existence must be assumed before concluding horizontality, testing nuanced understanding versus rote pattern matching.

Q11. When approximating $\Delta z$ using the tangent plane for $z = \sin(xy)$ at $(\pi/4, 2)$ , a student uses $dz = f_x dx + f_y dy$ with $dx=0.1, dy=-0.05$ . If the actual $\Delta z$ differs significantly from $dz$ , which factor most likely explains the discrepancy?

A.The function is not differentiable at

(\pi/4, 2)

B.The increments

dx, dy

are too large relative to the local radius of curvature. ✅

C.Partial derivatives were computed incorrectly.

D.The tangent plane formula requires second-order correction terms.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This error analysis question distinguishes between computational errors and inherent limitations of linear approximation. Since $\sin(xy)$ is smooth everywhere, differentiability isn’t the issue. Significant discrepancy arises when finite steps exceed the region where the tangent plane adequately represents the surface, emphasizing that $dz \approx \Delta z$ holds only infinitesimally. Students must evaluate scale appropriateness rather than blaming formula misuse.

Q12. Given the graph of $z = f(x,y)$ showing a saddle point at $P$ , how does the tangent plane at $P$ relate to the surface locally?

A.The tangent plane lies entirely above the surface near

P

B.The tangent plane lies entirely below the surface near

P

C.The tangent plane intersects the surface, dividing the neighborhood into regions where the surface is above and below the plane. ✅

D.The tangent plane touches the surface only at

P

and nowhere else nearby.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: This graph-based question tests visual interpretation of saddle points via tangent planes. Unlike extrema where the surface stays on one side of the tangent plane, saddle points exhibit mixed behavior: the surface crosses its tangent plane, creating alternating sectors of positive and negative deviation. Recognizing this crossing pattern is essential for classifying critical points geometrically without relying solely on second-derivative tests.

Q13. A researcher compares two methods to find the tangent plane to $z = f(x,y)$ : Method A uses $z = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)$ ; Method B treats $F(x,y,z) = f(x,y) - z = 0$ and uses $\nabla F \cdot \langle x-a, y-b, z-c \rangle = 0$ . Under what condition do these methods yield identical results?

A.Only when

f

is linear.

B.Always, provided

f

is differentiable at

(a,b)

. ✅

C.Only when

f(a,b) = 0

D.Never; they represent fundamentally different geometric objects.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This mixed concepts question requires comparing equivalent formulations of the same concept. Both methods derive from the definition of differentiability and produce the same plane when $f$ is differentiable. Method A is explicit; Method B is implicit. Understanding their equivalence reinforces that tangent planes are intrinsic geometric objects independent of representation, while also highlighting differentiability as the unifying prerequisite for validity.

Q14. For the surface $z = x^2 y$ , the tangent plane at $(1,1,1)$ is used to estimate $z$ at $(1.02, 0.98)$ . Without full computation, which reasoning best predicts whether the estimate will be an overestimate or underestimate?

A.Overestimate, because

f_{xx} > 0

and

f_{yy} < 0

imply net upward curvature.

B.Underestimate, because the mixed partial

f_{xy}

dominates and introduces negative correction.

C.Cannot determine without computing second derivatives and evaluating the Hessian quadratic form. ✅

D.Exactly equal, because the tangent plane matches the function up to first order.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This challenging question demands multi-step reasoning about approximation error signs. The error in linear approximation depends on the second-order Taylor remainder involving the Hessian evaluated at some intermediate point. While individual second derivatives give partial information, the combined effect along the specific displacement vector $(0.02, -0.02)$ requires analyzing the quadratic form $\frac{1}{2}[f_{xx}dx^2 + 2f_{xy}dxdy + f_{yy}dy^2]$ . Simple inspection of individual curvatures is insufficient, testing deep understanding of multivariable approximation theory.

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