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📝 Tangent Planes and Normal Vectors (14 MCQs)

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

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What is Tangent Planes and Normal Vectors?

Definition:
Tangent plane to $z=f(x,y)$ at $(a,b,f(a,b))$ has equation $z = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)$ ; normal vector is $\langle -f_x, -f_y, 1 \rangle$ .

Example:
Tangent plane to $z=x^2+y^2$ at $(1,1,2)$ is $z = 2 + 2(x-1) + 2(y-1)$ with normal $\langle -2,-2,1 \rangle$ .

Reason:
Tangent planes provide best linear surface approximation locally; normals enable reflection/refraction calculations and surface integral orientation.

Easy

Medium

Hard

📝 All Tangent Planes and Normal Vectors MCQs

Q1. A surface is defined implicitly by $F(x,y,z) = x^2 + y^2 - z^2 = 0$ . At point $P(1,0,1)$ , a student computes the normal vector as $\langle 2, 0, -2 \rangle$ but claims the tangent plane equation is $2x - 2z = 0$ . What is the fundamental error in this reasoning?

A.The student forgot to normalize the gradient vector before constructing the plane.

B.The student correctly computed the gradient but failed to evaluate it at point P when forming the plane equation.

C.The student used the correct normal but omitted the constant term derived from substituting P into the linearized form. ✅

D.The student confused the gradient of F with the partial derivative of z with respect to x.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The gradient $\nabla F = \langle 2x, 2y, -2z \rangle$ evaluated at (1,0,1) is indeed $\langle 2,0,-2 \rangle$ . However, the tangent plane must satisfy $\nabla F(P) \cdot \langle x-1, y-0, z-1 \rangle = 0$ , yielding $2(x-1) - 2(z-1) = 0$ , which simplifies to $2x - 2z = 0$ . While algebraically equivalent, the error lies in not recognizing that the constant vanishes only because P satisfies the original surface; the method was incomplete conceptually.

Q2. Consider two surfaces $S_1: z = x^2 + y^2$ and $S_2: z = 2 - x^2 - y^2$ . They intersect along a curve C. At their intersection point where $x=0, y=1$ , what geometric relationship exists between their respective normal vectors?

A.The normals are parallel because both surfaces are paraboloids symmetric about the z-axis.

B.The normals are perpendicular because one opens upward and the other downward.

C.The angle between normals is determined solely by the sum of their Gaussian curvatures at that point.

D.The normals lie in the same vertical plane but are neither parallel nor orthogonal. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: At (0,1,1), $\nabla S_1 = \langle 0,2,-1 \rangle$ and $\nabla S_2 = \langle 0,-2,-1 \rangle$ . Their dot product is $(0)(0)+(2)(-2)+(-1)(-1) = -3 \neq 0$ , so not orthogonal. They aren't scalar multiples, so not parallel. Both have zero x-component, meaning they lie in the yz-plane (a vertical plane). This requires synthesizing implicit differentiation, vector geometry, and spatial reasoning beyond simple computation.

Q3. A manufacturing process molds a component shaped like $z = \sqrt{x^2 + y^2}$ . A quality sensor measures deviation from the ideal tangent plane at (3,4,5). If the sensor reads a vertical discrepancy of 0.1 units at (3.1, 4.05), which interpretation best reflects the linear approximation's validity?

A.The linear model overestimates because the cone’s curvature causes actual surface to lie below the tangent plane. ✅

B.The discrepancy confirms the gradient magnitude equals exactly 1 everywhere on this surface.

C.The error arises because the function is non-differentiable at the origin, invalidating all tangent planes.

D.The measured deviation is consistent with second-order terms being negligible near the evaluation point.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The cone $z = r$ has tangent plane at (3,4,5) with slope matching radial direction. Since the surface is convex downward away from apex, the true surface lies below its tangent plane for points away from tangency. The 0.1 deviation aligns with quadratic remainder in Taylor expansion. Option D misattributes accuracy to negligibility rather than directional curvature. This tests understanding of approximation error sign relative to surface geometry.

Q4. Given a level surface $F(x,y,z)=c$ where $F_z \neq 0$ , a student derives the tangent plane using $z_x = -F_x/F_z$ and $z_y = -F_y/F_z$ , then writes the plane as $z - z_0 = z_x(x-x_0) + z_y(y-y_0)$ . Another uses $\nabla F \cdot \langle x-x_0, y-y_0, z-z_0 \rangle = 0$ . Under what condition do these yield identical results?

A.Only when F is linear in all variables.

B.Always, provided F is continuously differentiable and

F_z \neq 0

at the point. ✅

C.Only when the surface can be globally expressed as z = f(x,y).

D.Never; the methods are fundamentally incompatible.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Both approaches are mathematically equivalent under the implicit function theorem. Substituting $z_x, z_y$ into the explicit form and rearranging yields $F_x(x-x_0)+F_y(y-y_0)+F_z(z-z_0)=0$ , identical to the gradient dot product form. This equivalence holds locally wherever $F_z \neq 0$ and F is C¹. The question probes deep conceptual linkage between implicit and explicit representations rather than rote formula application.

Q5. A topographic map shows contour lines of elevation $z=f(x,y)$ becoming increasingly dense near point Q. Without computing derivatives, what can be inferred about the tangent plane at Q compared to a region with sparse contours?

A.The tangent plane at Q is nearly horizontal due to high precision of contour mapping.

B.The normal vector at Q has a larger vertical component relative to horizontal components.

C.The tangent plane at Q is steeper, implying the normal vector tilts more toward the horizontal plane. ✅

D.Contour density indicates nothing about tangent planes without knowing the contour interval.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Dense contours indicate rapid change in z per unit horizontal distance, i.e., large $|\nabla f|$ . Steeper slope means tangent plane deviates more from horizontal, so its normal vector has smaller z-component relative to x,y components—tilting toward horizontal. This connects graphical representation to differential geometry without calculation, testing spatial interpretation skills essential in applied fields like geoscience or engineering design.

Q6. An engineer models heat flux through a surface $z = g(x,y)$ using the normal vector for orientation. She mistakenly uses $\langle g_x, g_y, 1 \rangle$ instead of $\langle -g_x, -g_y, 1 \rangle$ in her flux integral. How does this affect the physical interpretation?

A.Flux magnitude doubles because both components contribute additively.

B.The sign of flux reverses, implying heat flows opposite to intended direction. ✅

C.Results remain valid since normal direction is arbitrary in scalar surface integrals.

D.Error cancels out when integrating over closed surfaces.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For graph surfaces, upward-pointing normal is $\langle -g_x, -g_y, 1 \rangle$ . Using positive partials gives downward orientation. In vector flux integrals $\iint \mathbf{F} \cdot \mathbf{n}\, dS$ , reversing n flips sign. Heat flux direction matters physically; wrong sign implies cooling instead of heating. This error analysis question emphasizes that orientation isn't just mathematical convention—it carries physical meaning in modeling scenarios.

Q7. Suppose $f(x,y)$ has continuous partials at (a,b) but the tangent plane approximation error $E(h,k) = f(a+h,b+k) - [f(a,b)+f_x h + f_y k]$ satisfies $\lim_{(h,k)\to(0,0)} E/\sqrt{h^2+k^2} \neq 0$ . What conclusion follows?

A.Partial derivatives exist but f is not differentiable at (a,b), so no true tangent plane exists. ✅

B.The tangent plane exists but is not unique due to directional dependence.

C.Higher-order partial derivatives must be discontinuous at (a,b).

D.The limit condition is impossible if partials are continuous.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Differentiability requires $E / \| (h,k) \| \to 0$ . Continuous partials guarantee differentiability, so this scenario contradicts standard theory—unless 'continuous partials' was misstated. But assuming the premise holds as given, the nonzero limit violates differentiability definition, meaning no tangent plane exists despite existing partials. This challenges students to distinguish existence of partials from full differentiability, a subtle HOTS concept often misunderstood.

Q8. Two students approximate $f(x,y) = e^{xy}$ near (0,0). Student A uses tangent plane $z=1$ . Student B argues that since $f_{xy}(0,0)=1$ , the tangent plane should include an xy term. Which critique correctly identifies the flaw?

A.Student B confuses second-order mixed partials with first-order linear approximation coefficients. ✅

B.Student A ignored that exponential functions require logarithmic transformation before linearization.

C.Both are wrong; the correct tangent plane is

z = 1 + x + y

D.Student B is correct because cross-terms affect local linearity.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Tangent plane uses only first partials: $f_x=y e^{xy}, f_y=x e^{xy}$ , both zero at (0,0), so plane is z=1. Mixed partial $f_{xy}$ relates to curvature, not linear approximation. Including xy term creates quadratic approximation, violating tangent plane definition. This tests precise understanding of Taylor hierarchy and prevents conflation of derivative orders—a common misconception in multivariable calculus.

Q9. A robotic arm traces path $\mathbf{r}(t)$ on surface $z=f(x,y)$ . At time t₀, velocity vector $\mathbf{v}$ is tangent to surface. If $\nabla f \cdot \mathbf{v} \neq 0$ at corresponding point, what does this imply?

A.The robot has left the surface due to control system failure.

B.The path parameterization is invalid because tangent vectors must be orthogonal to gradient.

C.No contradiction;

\mathbf{v}

being tangent means

\nabla f \cdot \mathbf{v} = 0

must hold, so premise is impossible. ✅

D.Gradient was miscalculated; actual gradient is always perpendicular to velocity.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: By definition, any vector tangent to level surface $z-f(x,y)=0$ must satisfy $\nabla(z-f) \cdot \mathbf{v} = 0$ , i.e., $\langle -f_x,-f_y,1 \rangle \cdot \mathbf{v} = 0$ . For graph surfaces, this reduces to $\nabla f \cdot \langle v_x,v_y \rangle = v_z$ . If $\mathbf{v}$ is truly tangent, the dot product condition must hold. The nonzero value indicates either mismeasurement or misunderstanding of tangency. This integrates kinematics with differential geometry in a realistic robotics context.

Q10. When approximating $f(x,y)$ near (a,b), the tangent plane error bound depends on maximum of $|f_{xx}|, |f_{xy}|, |f_{yy}|$ in a neighborhood. If these second partials grow unbounded as (x,y)→(a,b), what happens to the usefulness of the tangent plane?

A.It remains valid pointwise but loses predictive power arbitrarily close to (a,b). ✅

B.It becomes exact because higher derivatives dominate lower ones.

C.The tangent plane ceases to exist even if first partials are defined.

D.Error bound formula still applies with infinite constants, giving trivial estimate.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Tangent plane exists if f is differentiable at (a,b), regardless of second derivative behavior nearby. However, unbounded second partials mean the remainder term in Taylor's theorem cannot be bounded uniformly, so approximation deteriorates rapidly near the point. This distinguishes pointwise differentiability from uniform approximability—a nuanced concept critical in numerical analysis and sensitivity modeling where local linearity assumptions break down.

Q11. A climate model uses sea surface temperature T(x,y,t). At fixed time, researchers compute normal vector to isotherm surface T=constant. If ocean currents advect warm water such that $\partial T/\partial t > 0$ while spatial gradient remains unchanged, how does this affect the instantaneous tangent plane to the isotherm?

A.The tangent plane rotates because temporal change alters surface orientation.

B.The tangent plane remains identical since it depends only on spatial gradients at that instant. ✅

C.Isotherms cease to be smooth surfaces when temperature changes temporally.

D.Normal vector gains a time component, making it four-dimensional.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Tangent plane to level surface T(x,y,z)=c at fixed time depends solely on ∇T evaluated at that moment. Temporal derivative affects evolution of the surface but not its instantaneous geometry. This separates spatiotemporal dynamics from static differential geometry, testing whether students conflate rate of change with shape. Crucial in fluid dynamics where Eulerian vs Lagrangian perspectives differ.

Q12. In optimizing $f(x,y)$ subject to constraint $g(x,y)=k$ , Lagrange multipliers require $\nabla f = \lambda \nabla g$ . Geometrically, this means level curves share tangent lines. Extending to 3D: for extremizing $f(x,y,z)$ on surface $g(x,y,z)=k$ , why must $\nabla f$ be parallel to surface normal $\nabla g$ ?

A.Because extrema occur where directional derivative of f along every tangent direction vanishes. ✅

B.Because ∇f must be perpendicular to ∇g to maximize projection.

C.Because surface normals always point toward increasing f values.

D.Because Lagrange method assumes surfaces are spheres.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: At constrained extremum, f cannot increase/decrease along any direction tangent to constraint surface. Thus $\nabla f \cdot \mathbf{t} = 0$ for all tangent vectors t, meaning ∇f is orthogonal to tangent space—hence parallel to normal ∇g. This synthesizes optimization, geometry, and linear algebra. Distractors exploit confusion between orthogonality conditions and misremembered multiplier rules, requiring deep conceptual integration beyond algorithmic application.

Q13. A student claims that if $f_x(a,b) = f_y(a,b) = 0$ , then the tangent plane at (a,b,f(a,b)) must be horizontal AND the point is a local extremum. Which part of this statement is incorrect and why?

A.The tangent plane is never horizontal when partials vanish.

B.Vanishing partials guarantee saddle points, not extrema.

C.Horizontal tangent plane does not imply extremum; could be saddle or inflection-like behavior. ✅

D.Partials being zero contradicts differentiability.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Critical points (zero gradient) yield horizontal tangent planes, but classification requires second derivative test or further analysis. Saddle points like $z=x^2-y^2$ at origin have horizontal tangent yet are not extrema. This error analysis targets pervasive misconception equating stationary points with optima. Understanding this distinction is vital in machine learning loss landscapes and physics potential energy surfaces where saddles are common.

Q14. Consider surface $z = x^3 - 3xy^2$ (monkey saddle). At origin, all first partials vanish. Unlike standard saddle, this surface has threefold symmetry. What unique property does its tangent plane exhibit regarding higher-order contact?

A.The tangent plane intersects the surface along three lines through origin, unlike quadratic saddles with two. ✅

B.The tangent plane coincides with surface to third order in all directions.

C.No tangent plane exists due to cubic singularity.

D.Contact order is identical to elliptic paraboloid despite different symmetry.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Standard saddle $z=x^2-y^2$ has tangent plane intersecting along two lines (asymptotes). Monkey saddle’s cubic homogeneity means $z=0$ when $x^3=3xy^2$ , i.e., $x(x^2-3y^2)=0$ , giving three lines: x=0 and $y=\pm x/\sqrt{3}$ . Tangent plane (z=0) thus has triple-line intersection. This Olympiad-style problem connects algebraic geometry to differential concepts, testing ability to generalize beyond textbook quadratics and recognize how degree affects local structure.

🔗 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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