Original public practice · MV-1 · October 2026

Multivariable calculus: cumulative practice

Connect spatial geometry, local change, coordinate integration and oriented boundary arguments, then defend changed cases. This form contains 12 multipart questions and 120 practice points.

This is calculus-based undergraduate practice. Review single-variable derivatives and integrals, vectors, coordinate maps and the linked multivariable lessons. Every lesson remains directly accessible. These original problems sample the published material; they are not an official university paper, a calibrated grade prediction or an examination of every course objective.

Attempt, review and revise

Write on paper or in your own document before revealing solutions. This page does not save responses or award completion. Record any help or solution exposure; closing a solution does not make the next attempt unseen.

For an optional rehearsal, allow 240 minutes across the form, or divide it into separate sessions with breaks and accommodations. This is an author estimate, not a measured learner timing. There is no enforced timer or pass threshold.

State the domain, units, coordinate conventions and theorem hypotheses. Draw the region or describe its boundaries, check orientation and distinguish an exact equality from a local approximation. You may cite results taught in the linked lessons unless a part asks for their proof. Supply intermediate reasoning and the requested calculations; a theorem name or numerical match alone is not a proof. A calculator or local numerical code may help check arithmetic; no paid tool is required. Supplied theorem statements may be used only under their stated assumptions. Alternative valid methods can earn the same points.

Print this page for questions only. Solutions, review advice and navigation are excluded from print even when their disclosures are open. Keep a separate answer sheet.

1. Geometry and local change

40 practice points.

Question 1: Recover a plane from spatial constraints

10 points

The line L is (x,y,z)=(1,0,1)+t(1,2,−1), t real. A second point is B=(0,1,1). Coordinates use one fixed length unit.

(a) · 5 points

Construct the plane containing L and B. Verify the entire line and the point satisfy it. Find the distance from the origin to this plane.

(b) · 5 points

Find the closest point on the plane to the origin and verify perpendicularity. If B is replaced by (2,2,0), does the same construction specify a unique plane? Explain.

Review question 1: worked argument and scoring criteria

(a)

A direction from (1,0,1) to B is (−1,1,0). The line direction crossed with this displacement gives (1,1,3), so the plane is x+y+3z=4. Substitution gives 1+t+2t+3(1−t)=4 on L and 0+1+3=4 at B. The origin distance is |−4|/√11=4/√11 length units.

2 points: Constructs a nonzero normal from two independent in-plane directions.
1 point: Obtains and checks the plane on the entire line and point.
2 points: Normalizes the signed plane expression to obtain distance.

(b)

The closest point lies along n=(1,1,3): αn satisfies 11α=4, giving (4/11,4/11,12/11). It lies on the plane; its displacement from the origin is normal to every in-plane direction. The replacement point is on L at t=1. The two directions become dependent and the cross product vanishes. Infinitely many planes contain L, so a unique normal or plane cannot be inferred from those data.

2 points: Finds and checks the perpendicular foot.
1 point: Identifies the replacement point on the line.
2 points: Explains loss of uniqueness without interpreting a zero cross product as a plane normal.

Targeted lesson returns:

Construct and verify spatial normals

Question 2: Change speed without changing the track

10 points

A particle follows r(t)=(t,t²,0), t>0. A second traversal is q(u)=r(2−u), 0<u<2. Curvature is κ=|r′×r″|/|r′|³ for a regular twice differentiable curve.

(a) · 5 points

At t=1 find velocity, acceleration, speed, unit tangent, curvature and the rate of change of speed. Explain why curvature is not acceleration.

(b) · 5 points

Compare q at u=1 with r at t=1, including tangent, acceleration and curvature. Resolve acceleration into tangential and normal components at this point for r, and reconcile their magnitudes with |r″|.

Review question 2: worked argument and scoring criteria

(a)

The vectors are r′=(1,2,0) and r″=(0,2,0); speed is √5 and unit tangent (1,2,0)/√5. The cross product is (0,0,2), hence κ=2/(5√5). Differentiating √(1+4t²) gives speed derivative 4t/√(1+4t²), or 4/√5 at 1. Curvature is turning per length; acceleration also includes changing speed and has different units.

1 point: Computes velocity, acceleration, speed and unit tangent.
2 points: Computes curvature with the cubic speed denominator.
2 points: Derives speed change and explains the distinction.

(b)

At u=1, q′=−r′, q″=r″ and the unit tangent reverses. Speed and curvature are unchanged; the chain rule changes the sign of the cross product but not its length. For r the tangential acceleration magnitude is 4/√5 and the normal magnitude is κv²=2/√5. Their orthogonal squared magnitudes sum to 16/5+4/5=4=|r″|². Tangential acceleration vector is (4/5,8/5,0); normal component is (−4/5,2/5,0).

2 points: Applies the reparametrization derivatives and compares the same point.
1 point: Shows curvature invariance and tangent reversal.
2 points: Obtains both acceleration components and their orthogonal norm check.

Targeted lesson returns:

Speed, curvature and regular reparametrization

Question 3: Decide whether a local model is justified

10 points

Define g(x,y)=x²y/(x⁴+y²) off (0,0), and g(0,0)=0. Separately define f(x,y)=exp(x) cos y. For the second calculation you may use the following Hessian remainder bound: if the operator norm of the Hessian is at most M on the segment from a to a+h, then |f(a+h)−f(a)−∇f(a)·h|≤M|h|²/2.

(a) · 5 points

Find the coordinate partial derivatives of g at the origin and decide continuity and differentiability there. Give an explicit path argument; do not infer continuity from the two partials.

(b) · 5 points

Linearize f at (0,0) and estimate f(0.02,0.01). Justify an absolute error bound, using that its Hessian has operator norm exp(x). State which conclusion remains approximate.

Review question 3: worked argument and scoring criteria

(a)

Both axis restrictions are identically zero, so the two partial derivatives at zero are zero by their one-variable difference quotients. On y=x² with x≠0, g=x⁴/(2x⁴)=1/2. Along either axis the limit is zero. There is no two-variable limit, so g is neither continuous nor differentiable at the origin.

2 points: Computes the partials from axis difference quotients.
2 points: Exhibits conflicting paths with their values.
1 point: Concludes both failures using differentiability implies continuity.

(b)

At zero f=1 and gradient=(1,0), giving L=1+x and estimate 1.02. The segment has 0≤x≤0.02, so M=e^0.02 works. The squared displacement is 0.0005, yielding error at most 0.00025e^0.02, approximately 0.000255051. This is a rigorous bound under the supplied theorem; 1.02 is still an approximation, not an exact value. The Hessian is e^x times [[cos y,−sin y],[−sin y,−cos y]], whose square is the identity after removing e^x, confirming its norm.

1 point: Obtains the tangent approximation.
2 points: Checks the segment bound and applies the squared displacement.
2 points: Interprets the bound and distinguishes it from the actual signed error.

Targeted lesson returns:

Question 4: A coordinate derivative is not a unit-distance rate

10 points

A smooth scalar field F has gradient (2,−1,3) at P=(1,1,2). Define H(r,θ,s)=F(r cos θ,r sin θ,sr). Consider (r,θ,s)=(√2,π/4,√2), which maps to P. Angles are in radians.

(a) · 5 points

Compute ∂H/∂θ and ∂H/∂r at the given coordinate point, holding the other coordinates fixed each time. Explain which physical displacement each derivative uses.

(b) · 5 points

Find the derivative of F per unit distance in the increasing-θ direction. Find its maximum directional derivative at P and a unit direction attaining it. Can the given gradient determine the exact change for a finite angular change of 0.1?

Review question 4: worked argument and scoring criteria

(a)

The θ tangent is (−r sin θ,r cos θ,0)=(−1,1,0), so Hθ=(2,−1,3)·(−1,1,0)=−3. The r tangent is (cos θ,sin θ,s)=(1/√2,1/√2,√2), giving Hr=7/√2. Varying r at fixed s changes z as well as horizontal radius; varying θ at fixed r,s follows a horizontal circle.

2 points: Constructs the θ tangent and derivative.
2 points: Includes the z change in the r tangent and derivative.
1 point: Interprets the two distinct coordinate paths.

(b)

The θ tangent has norm √2, so the unit-distance derivative is −3/√2. The maximum is |∇F|=√14, attained in direction (2,−1,3)/√14 by Cauchy–Schwarz. An angular change 0.1 gives first-order prediction −0.3 for H, but an exact finite change cannot be recovered from one gradient. A remainder estimate requires additional control of derivatives along the path.

2 points: Normalizes the coordinate tangent.
2 points: Gives the maximizing unit direction and magnitude with justification.
1 point: Separates the first-order prediction from an exact finite change.

Targeted lesson returns:

Coordinate chain rule and normalized directional derivatives

2. Regions and coordinate integration

40 practice points.

Question 5: Interior, boundary and singular constraints

10 points

Let D={(x,y):x≥0,y≥0,x+y≤4}. Investigate f(x,y)=xy on this closed triangle. Separately consider h(x,y)=x constrained by x²+y²=0.

(a) · 5 points

Find all global maxima and minima of f on D. Show that your candidate search includes interior points, edges and vertices, and justify attainment.

(b) · 5 points

Apply the regular Lagrange-multiplier equations on the sloping edge and compare with part (a). For the separate constraint on h, explain why an extremum can exist even though no multiplier solves ∇h=λ∇(x²+y²).

Review question 5: worked argument and scoring criteria

(a)

Continuity and compactness guarantee extrema. The interior gradient (y,x) cannot vanish with x,y>0. On the two coordinate edges f=0. On x+y=4, f=x(4−x)=4−(x−2)², so the maximum is 4 at (2,2). The minimum is 0 at every point of the two coordinate edges, including all relevant vertices; elsewhere f>0. Endpoints of the sloping edge have already been included.

1 point: Invokes continuity and compactness.
2 points: Checks interior and all edge/endpoint cases.
2 points: States the maximum and the complete minimum set.

(b)

On x+y=4 the constraint gradient is (1,1), so (y,x)=λ(1,1), giving x=y=2. This supplies an edge stationary candidate; endpoints and other edges still need checking. The separate feasible set is the singleton (0,0), hence h has both global minimum and maximum 0. At that point ∇h=(1,0) and the constraint gradient is zero; no λ works. The regularity hypothesis fails, so there is no contradiction.

2 points: Solves the regular edge equations and retains boundary checks.
1 point: Identifies the singleton extrema.
2 points: Shows the multiplier failure and identifies its missing hypothesis.

Targeted lesson returns:

Question 6: Reverse a region whose description changes

10 points

A lamina occupies 0≤x≤2 and x²≤y≤x+2. Here x,y are dimensionless numerical coordinates: physical coordinates are (ℓx,ℓy), with fixed length ℓ>0. Its surface density is ρ=kx, where k>0 has units mass per area.

(a) · 5 points

Describe the boundaries and intersections, and express its mass in both integration orders. Explain why the reversed order must split.

(b) · 5 points

Evaluate the mass and both center-of-mass coordinates. Show the moment integrals and check whether the result is geometrically plausible.

Review question 6: worked argument and scoring criteria

(a)

The lower parabola and upper line intersect at x=2 (the other intersection x=−1 is outside the specified strip). At x=0 the vertical boundary runs from y=0 to 2. Mass is kℓ²∫₀²∫ₓ²^(x+2) x dy dx. Reversing gives kℓ²[∫₀²∫₀^√y x dx dy+∫₂⁴∫_(y−2)^√y x dx dy]. The line imposes x≥y−2, which overtakes the lower bound zero only above y=2.

2 points: Identifies the actual bounded region and relevant intersections.
2 points: Gives both orders with the split at y=2.
1 point: Explains the changing lower bound geometrically.

(b)

In normalized coordinate arithmetic, M/(kℓ²)=∫₀²x(x+2−x²)dx=8/3. The physical x-weighted moment divided by kℓ³ is ∫₀²x²(x+2−x²)dx=44/15. The physical y-weighted moment divided by kℓ³ is (1/2)∫₀²x[(x+2)²−x⁴]dx=6. Thus the physical coordinates are x̄=11ℓ/10 and ȳ=9ℓ/4. At numerical x=1.1 the lower and upper boundaries are 1.21 and 3.1, so this center lies inside D. Mass has units mass, moments mass·length, and their ratios length; k cancels from the coordinates.

1 point: Gets mass 8kℓ²/3.
2 points: Computes the two coordinate-weighted moment integrals.
2 points: Normalizes to (11ℓ/10,9ℓ/4) and checks geometry and units.

Targeted lesson returns:

Question 7: Branch choice and multiplicity in a substitution

10 points

In the positive quadrant define u=xy and v=y/x. Let D consist of points satisfying 1≤xy≤4 and 1≤y/x≤9.

(a) · 5 points

Find the inverse map on D, the absolute inverse Jacobian and its parameter rectangle. Evaluate the area of D and ∫∫D xy dA.

(b) · 5 points

Remove the positive-quadrant restriction, retaining the same two inequalities. Explain what fails in using the same inverse as a global parameterization, and compute the new area and weighted integral. Would a negative determinant justify negative area?

Review question 7: worked argument and scoring criteria

(a)

For positive coordinates, x=√(u/v), y=√(uv), with (u,v)∈[1,4]×[1,9]. Directly det ∂(u,v)/∂(x,y)=y/x+y/x=2v>0, so |det ∂(x,y)/∂(u,v)|=1/(2v). Area=(3/2)ln9. The weighted integral is ∫₁⁴∫₁⁹ u/(2v)dv du=(15/4)ln9. Positive bounded u,v make the inverse smooth and one-to-one on the interior.

2 points: Finds the positive inverse and rectangle.
1 point: Derives the absolute Jacobian.
2 points: Obtains both integral values with the measure factor.

(b)

The inequalities imply x and y have the same sign. There are two components, with inverse branches ±(√(u/v),√(uv)); each parameter pair has two preimages. The positive branch alone omits the negative component. Splitting by sign gives area 3ln9 and weighted integral (15/2)ln9, since xy=u on both. Each branch uses an absolute determinant. Orientation signs never turn geometric area negative; they also do not automatically account for multiplicity.

2 points: Identifies the two preimages and missing global injectivity.
2 points: Splits branches and doubles both values.
1 point: Separates orientation from positive measure and multiplicity.

Targeted lesson returns:

Inverse maps, absolute determinants and coverage

Question 8: One weighted solid in two coordinate systems

10 points

A solid has 0≤z≤2 and x²+y²≤z². Here physical coordinates are (ℓx,ℓy,ℓz), where ℓ>0 is a fixed length and x,y,z are dimensionless. Its volume density is ρ=kz, where k>0 has units mass per volume. Use cylindrical radius r and spherical coordinates (R,φ,θ), where φ is measured from the positive z-axis.

(a) · 5 points

Set up and evaluate mass and the vertical first moment in cylindrical coordinates. Find the center of mass, explaining the horizontal coordinates by symmetry.

(b) · 5 points

Set up both integrals in spherical coordinates and verify the mass by evaluation. Explain why using R≤2 would describe a different solid.

Review question 8: worked argument and scoring criteria

(a)

Bounds are 0≤θ≤2π, 0≤z≤2, 0≤r≤z, with physical volume element ℓ³r dr dz dθ. Mass is kℓ³∫₀²z·πz² dz=4πkℓ³. The physical vertical first moment is kℓ⁴∫₀²z²·πz² dz=32πkℓ⁴/5. Therefore z̄=8ℓ/5, while x̄=ȳ=0 by rotation/reflection symmetry. Mass, moment and centroid have units mass, mass·length and length respectively.

2 points: Derives bounds and includes the cylindrical measure factor.
2 points: Computes mass, vertical moment and z̄.
1 point: Justifies horizontal coordinates and units.

(b)

The cone gives 0≤φ≤π/4 and the top plane gives 0≤R≤2/cosφ; θ spans 0 to 2π. Density is kR cosφ and dV=ℓ³R²sinφ dR dφ dθ. Mass is kℓ³∫₀²π∫₀^(π/4)∫₀^(2/cosφ)R³cosφ sinφ dR dφ dθ=8πkℓ³∫₀^(π/4)sinφ/cos³φ dφ=4πkℓ³. The physical vertical moment has integrand kℓ⁴R⁴cos²φ sinφ with the same bounds. R≤2 would use a spherical cap boundary rather than the plane z=2. Coordinate degeneracy at the axis/apex affects only measure-zero sets here.

2 points: Gives the angular and variable radial bounds.
1 point: Gives both correct integrands with density and volume factors.
2 points: Verifies mass and distinguishes a plane from a spherical radius bound.

Targeted lesson returns:

Coordinate descriptions of weighted solids

3. Fields, surfaces and theorem choice

40 practice points.

Question 9: A local derivative check does not settle a loop

10 points

On the punctured plane define F=(−y/(x²+y²)+y, x/(x²+y²)+x). Let C be the circle of radius 2 traversed once counterclockwise.

(a) · 5 points

Compute the circulation along C directly, separating any gradient contribution. Decide whether F can have a global scalar potential on the punctured plane.

(b) · 5 points

Apply Green’s theorem correctly on 1≤x²+y²≤4. Give both boundary orientations and integrals. Explain precisely why applying Green to the disk bounded by C would be invalid, and what happens when the entire boundary orientation is reversed.

Review question 9: worked argument and scoring criteria

(a)

The added field (y,x)=∇(xy), so its loop contribution is zero. On r(t)=(2cos t,2sin t), 0≤t≤2π, the singular field dotted with r′ is 1. Thus circulation is 2π. A global differentiable potential would give zero around every closed curve, so none exists on this domain despite the vanishing scalar curl away from zero.

2 points: Identifies and verifies the gradient part.
2 points: Computes the singular-field circulation with orientation.
1 point: Uses the nonzero loop to exclude a global potential.

(b)

The field is C¹ near the closed annulus and its scalar curl is zero. Its positive boundary has outer radius 2 counterclockwise, integral 2π, and inner radius 1 clockwise, integral −2π; the sum is zero. On the disk the field is undefined at the origin, so the neighborhood regularity hypothesis fails. Reversing the entire annular boundary negates both individual integrals but leaves their sum zero. Reversing only the inner boundary would no longer be the induced boundary orientation.

2 points: Checks valid annular hypotheses and both orientations.
1 point: Computes the cancelling integrals.
2 points: Identifies the disk failure and explains reversal consistently.

Targeted lesson returns:

Question 10: Surface area, surface mass and signed transport

10 points

S is the graph z=x²+y² above the unit disk x²+y²≤1, oriented upward. For flux use F=(x,y,z). For a separate surface-mass calculation use surface density σ=kz with k>0, with units chosen so σ is mass per area.

(a) · 5 points

Parameterize S using x,y. Derive its oriented area vector and scalar area element, then compute its area. Explain why projected disk area is insufficient.

(b) · 5 points

Compute upward flux and surface mass. Explain their different integrands and the effect of reversing the orientation.

Review question 10: worked argument and scoring criteria

(a)

Use R(x,y)=(x,y,x²+y²). Rx×Ry=(−2x,−2y,1) points upward, while its length is √(1+4x²+4y²). Area is 2π∫₀¹r√(1+4r²)dr=π(5√5−1)/6. The projected disk has area π; surface slopes increase the local scalar area factor. A unit normal alone is not the vector area unless multiplied by dS.

2 points: Derives vector and scalar area elements with upward orientation.
2 points: Evaluates surface area.
1 point: Explains projection versus surface measure.

(b)

Flux integrand F(R)·(−2x,−2y,1)=−x²−y², so upward flux is −2π∫₀¹r³dr=−π/2. Surface mass is 2πk∫₀¹r³√(1+4r²)dr=kπ(25√5+1)/60, by u=1+4r². Flux uses the directional projection and can be negative; mass uses nonnegative density times scalar area. Orientation reversal changes flux to π/2 but leaves area and mass unchanged.

2 points: Computes oriented flux including its sign.
2 points: Computes the density-weighted scalar surface integral.
1 point: Distinguishes orientation effects on flux, area and mass.

Targeted lesson returns:

Question 11: Verify a piecewise boundary and a changed spanning surface

10 points

Let F=(−y/2,x/2,xy). The triangle S is z=1−x−y above x≥0,y≥0,x+y≤1, with upward normal. Its vertices are A=(0,0,1), B=(1,0,0), C=(0,1,0).

(a) · 5 points

Use Stokes’ theorem to compute circulation around the induced boundary. Specify the vertex order and verify the result by parameterizing all three segments.

(b) · 5 points

Replace the graph by z=1−x−y+a xy(1−x−y) over the same triangle, for any fixed real a. Without expanding another flux integral, determine the upward flux of curl F through it. Explain the hypotheses and the effect of taking the downward orientation.

Review question 11: worked argument and scoring criteria

(a)

Curl F=(x,−y,1). The upward vector area is (1,1,1)dxdy, so the curl flux is ∫₀¹∫₀^(1−x)(1+x−y)dy dx=1/2, the x and y contributions cancelling by symmetry. The induced order A→B→C→A is counterclockwise in the xy projection. On AB use (t,0,1−t), giving F·r′=0. On BC use (1−t,t,0), giving F·r′=1/2. On CA use (0,1−t,t), giving zero. Integrating each for 0≤t≤1 confirms 1/2.

2 points: Computes curl flux and checks the upward boundary order.
2 points: Parameterizes and evaluates all three segment contributions.
1 point: Matches the two independent calculations.

(b)

The added term vanishes on each edge, so the spatial boundary is unchanged. Every fixed a gives a smooth graph; its graph vector area has positive z component and therefore induces the same boundary traversal. F is polynomial and C¹ on a neighborhood of either surface. Stokes gives the same upward curl flux 1/2. Taking the downward orientation reverses the induced boundary and gives −1/2. This does not assert that flux of F itself is invariant under changing surfaces.

2 points: Verifies identical boundary and compatible orientation.
2 points: Checks smoothness and uses Stokes to obtain the invariant curl flux.
1 point: Handles reversed orientation and distinguishes flux of F from flux of its curl.

Targeted lesson returns:

Question 12: Close an open surface and account for every face

10 points

Let E={(x,y,z):0≤z≤4−x²−y²}. Its curved top S is oriented outward. Use F=(x,y,z+1).

(a) · 5 points

Compute outward flux through S by closing the boundary, then verify it directly. Identify the bottom-cap contribution explicitly.

(b) · 5 points

Change the lower plane to z=1, retaining z≤4−x²−y². Find the new curved-surface flux. Explain why subtracting the old cap flux, or using only the new volume flux, fails.

Review question 12: worked argument and scoring criteria

(a)

Divergence is 3. The solid volume is 2π∫₀²(4−r²)r dr=8π, so total closed flux is 24π. The bottom disk z=0 has outward normal −ez, so flux is −1 times its area 4π, or −4π. Thus S flux is 28π. Directly the upward vector area is (2x,2y,1)dxdy and F on S is (x,y,5−r²). Dot product is 5+r², integrating over the radius-2 disk to 20π+8π=28π.

2 points: Computes divergence and closed volume flux.
1 point: Includes the nonzero outward bottom flux.
2 points: Computes the curved-surface flux independently and reconciles it.

(b)

The new projection has radius √3. Volume is ∫₁⁴π(4−z)dz=9π/2 and closed flux is 27π/2. On the new bottom F·(−ez)=−2, so cap flux is −6π. Curved flux is 27π/2+6π=39π/2. Direct integration of 5+r² over r≤√3 gives 15π+9π/2, confirming it. The cap changes both area and field value; the divergence theorem applies to the complete closed boundary, not the curved piece alone.

2 points: Updates the solid and computes its volume flux.
1 point: Updates cap normal, area and field value.
2 points: Gets 39π/2 and explains both invalid shortcuts.

Targeted lesson returns:

Turn the review into a new argument

Award points for reasoning actually present. Allow valid later reasoning based on a carried-forward value unless its criterion requires the correct value; do not award correct-result points for an incorrect result. Record missing hypotheses, incorrect bounds, lost Jacobian factors, orientation and arithmetic errors separately.

The total is out of 120, with no university-grade conversion. Preserve your first attempt, consult the targeted lesson, explain the repair and attempt a fresh case. A teacher or peer may review the complete argument; if nobody did, label it self-review. Public worked solutions cannot certify independent proficiency.

Return to Multivariable calculus: cumulative practice for focused practice. The sustained spatial modeling investigation supplies a separate report and changed-case task. Independent assessment remains a separate requirement.

Return to the mathematics program