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Multivariable Calculus / LEVEL 1 · DIFFICULTY 4/5

Multivariable Calculus

Read the prerequisite, study a worked example and justify each step in these selected university topics.

3 stages · 24 practice problems · two 6-question assessment forms

STAGE UMV 1.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Partial Derivatives & Directional Change

Useful preparation: Derivatives, Chain Rule & Implicit Change

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Partial derivatives measure changes along coordinate slices. For a differentiable function, the gradient combines these rates into a local linear model for every direction. Directional change is its projection onto a unit direction.

Partial derivative
The instantaneous rate when one input varies and the others remain fixed; subscripts identify the varying coordinate.
Gradient
The vector of first partial derivatives. For a differentiable scalar function it encodes the linear part of all small input changes.
Directional derivative
The rate of change per unit distance in a specified direction. A unit vector sets direction without introducing an extra speed factor.

Conditions before calculation

  • The gradient dot-product formula requires differentiability at the point. Continuous first partial derivatives in a neighborhood are a sufficient condition.
  • Normalize a nonzero direction vector before asking for change per unit distance; its unnormalized version instead includes its magnitude as a speed factor.
  • The gradient identifies a unique steepest-ascent direction only when it is nonzero. Orthogonality to a regular level curve also assumes a nonzero gradient.

Why the method works

1. Differentiate coordinate slices

Vary one input while treating the others as constants. Assemble these coordinate derivatives into a vector at the specified point.

∇f(a,b)=⟨fx(a,b),fy(a,b)⟩\nabla f(a,b)=\langle f_x(a,b),f_y(a,b)\rangle

2. Use the local linear model

Differentiability provides one linear approximation for every small displacement, with error negligible relative to displacement length. Merely existing coordinate partials are insufficient.

f(p+h)=f(p)+∇f(p)⋅h+o(∥h∥),Duf(p)=∇f(p)⋅uf(\mathbf p+\mathbf h)=f(\mathbf p)+\nabla f(\mathbf p)\cdot\mathbf h+o(\|\mathbf h\|),\qquad D_{\mathbf u}f(\mathbf p)=\nabla f(\mathbf p)\cdot\mathbf u

3. Read geometry from the dot product

The projection is largest along a nonzero gradient and vanishes perpendicular to it. Zero first-order change does not imply constancy along a straight path.

∥u∥=1,∇f(p)≠0:umax⁡=∇f(p)∥∇f(p)∥,max⁡Duf=∥∇f(p)∥\|\mathbf u\|=1,\quad \nabla f(\mathbf p)\ne\mathbf0:\qquad \mathbf u_{\max}=\frac{\nabla f(\mathbf p)}{\|\nabla f(\mathbf p)\|},\quad \max D_{\mathbf u}f=\|\nabla f(\mathbf p)\|

Worked examples

EXAMPLE 1

Find the directional derivative at the given point toward the vector.

  1. Compute the coordinate rates before substituting the point. Polynomial smoothness justifies the gradient formula.

    f(x,y)=x2y+3y2,p=(1,2);fx=2xy,fy=x2+6y,∇f(1,2)=⟨4,13⟩f(x,y)=x^2y+3y^2,\quad \mathbf p=(1,2);\qquad f_x=2xy,\quad f_y=x^2+6y,\quad \nabla f(1,2)=\langle4,13\rangle

  2. The direction vector is not a unit vector. Divide by its length to separate direction from speed.

    v=⟨3,4⟩,∥v∥=5,u=⟨35,45⟩\mathbf v=\langle3,4\rangle,\qquad \|\mathbf v\|=5,\qquad \mathbf u=\left\langle\frac35,\frac45\right\rangle

  3. Take the projection. Compare it with the gradient magnitude, the largest possible rate at this point.

    Duf(1,2)=435+1345=645,max⁡Duf(1,2)=185D_{\mathbf u}f(1,2)=4\frac35+13\frac45=\frac{64}{5},\qquad \max D_{\mathbf u}f(1,2)=\sqrt{185}

Conclusion: The requested rate is 12.8 per unit distance, below the steepest possible rate.

EXAMPLE 2

Can a direction with zero directional derivative still change the function value?

  1. At the specified point, compute the gradient and choose a perpendicular unit direction.

    T(x,y)=10−x2−4y2,p=(1,1),∇T(p)=⟨−2,−8⟩,u=⟨4,−1⟩17T(x,y)=10-x^2-4y^2,\quad \mathbf p=(1,1),\quad \nabla T(\mathbf p)=\langle-2,-8\rangle,\quad \mathbf u=\frac{\langle4,-1\rangle}{\sqrt{17}}

  2. The dot product vanishes, so the initial first-order rate along the straight path is zero.

    DuT(p)=−8+817=0,r(s)=p+suD_{\mathbf u}T(\mathbf p)=\frac{-8+8}{\sqrt{17}}=0,\qquad \mathbf r(s)=\mathbf p+s\mathbf u

  3. Substitute the entire path into the function. Linear terms cancel, but the quadratic change remains for every nonzero step.

    T(r(s))=10−(1+4s17)2−4(1−s17)2=5−2017s2T(\mathbf r(s))=10-\left(1+\frac{4s}{\sqrt{17}}\right)^2-4\left(1-\frac{s}{\sqrt{17}}\right)^2=5-\frac{20}{17}s^2

Conclusion: The value decreases for every nonzero step, despite a zero initial directional derivative.

Pitfalls & counterexamples

Tempting claim: Existing partial derivatives guarantee differentiability.

Correction: Coordinate slices alone do not control all approaches to a point.

Test the claim: Away from the origin, divide the coordinate product by the sum of coordinate squares; assign zero at the origin. Both origin partials vanish, but diagonal values equal one half, contradicting continuity.

Check your understanding

What happens to the gradient dot product if a direction vector is doubled without normalization?

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Compare your reasoning

It doubles.

The dot product is linear in the vector; the result now measures change along a path moving twice as fast.

Connect this idea

Gradients supply tangent planes, optimization conditions, and conservative vector fields. The local linear model generalizes to Jacobian matrices for vector-valued functions.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    Let f (x,y)=2 x2 y\displaystyle f\, \left(x, y\right) = 2\, x^{2}\, y. Find fx (1,3)\displaystyle f_{x}\, \left(1, 3\right).

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  2. Problem 2

    Let f (x,y)=x2+2 y2\displaystyle f\, \left(x, y\right) = x^{2} + 2\, y^{2}. Find the y-component of its gradient at (1,2).

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  3. Problem 3

    A gradient at a point is (2, 4). Find the directional derivative in the unit direction (35,45)\displaystyle \left(\frac{3}{5}, \frac{4}{5}\right).

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  4. Problem 4

    A surface has f (0,0)=4\displaystyle f\, \left(0, 0\right) = 4, fx (0,0)=2\displaystyle f_{x}\, \left(0, 0\right) = 2 and fy (0,0)=3\displaystyle f_{y}\, \left(0, 0\right) = 3. What does its tangent-plane approximation predict at (0.1,0.2)?

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  5. Problem 5New context

    A height surface is f (x,y)=3 x2 y\displaystyle f\, \left(x, y\right) = 3\, x^{2}\, y. Find fx (1,3)\displaystyle f_{x}\, \left(1, 3\right).

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  6. Problem 6New context

    A field has potential f (x,y)=x2+3 y2\displaystyle f\, \left(x, y\right) = x^{2} + 3\, y^{2}. Find the y-component of its gradient at (1,2).

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  7. Problem 7New context

    A temperature gradient at a point is (3, 6). Find the directional derivative in the unit direction (35,45)\displaystyle \left(\frac{3}{5}, \frac{4}{5}\right).

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  8. Problem 8New context

    A local response surface has f (0,0)=5\displaystyle f\, \left(0, 0\right) = 5, fx (0,0)=2\displaystyle f_{x}\, \left(0, 0\right) = 2 and fy (0,0)=3\displaystyle f_{y}\, \left(0, 0\right) = 3. What does its tangent-plane approximation predict at (0.1,0.2)?

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Open stage UMV 1.1 in the student workspace →

STAGE UMV 1.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Hessians & Multivariable Extrema

Useful preparation: Partial Derivatives & Directional Change

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

At a smooth interior unconstrained optimum, every first-order directional change vanishes. The Hessian distinguishes second-order shapes. Constraints permit only feasible directions, so a constrained optimum can have a nonzero ordinary gradient.

Stationary point and saddle
A stationary point has zero gradient. A saddle is a stationary point with arbitrarily nearby larger and smaller function values, preventing either type of local extremum.
Hessian
The matrix of second partial derivatives. For a twice continuously differentiable function it is symmetric and describes the quadratic part of local change.
Lagrange multiplier
A scalar relating objective and constraint gradients at a regular constrained stationary point; the constraint must also hold.

Conditions before calculation

  • The Hessian test requires continuous second partials near an interior stationary point. A zero determinant makes it inconclusive.
  • A continuous function on a nonempty compact feasible set attains its absolute extrema; interior tests alone do not check the boundary.
  • For one smooth equality constraint, the standard Lagrange necessary condition requires a nonzero constraint gradient. Inspect singular constraint points separately.

Why the method works

1. Find first-order candidates

Every directional rate must vanish at an interior differentiable extremum. Solve all first-partial equations simultaneously.

∇f(p)=0\nabla f(\mathbf p)=\mathbf0

2. Classify the quadratic shape

For two variables, at a stationary point, a positive Hessian determinant gives a definite quadratic form, classified by the first diagonal entry. A negative determinant gives opposite directional signs and a saddle.

D=fxxfyy−fxy2;D>0, fxx>0⇒local minimum;D>0, fxx<0⇒local maximum;D<0⇒saddleD=f_{xx}f_{yy}-f_{xy}^2;\qquad D>0,\ f_{xx}>0\Rightarrow\text{local minimum};\quad D>0,\ f_{xx}<0\Rightarrow\text{local maximum};\quad D<0\Rightarrow\text{saddle}

3. Restrict to feasible motion

At a regular constrained extremum, both gradients are normal to the constraint tangent and hence parallel. Solve this condition together with the constraint.

∇f=λ∇g,g(x,y)=c\nabla f=\lambda\nabla g,\qquad g(x,y)=c

Worked examples

EXAMPLE 1

Locate the quadratic's stationary point, classify it, and determine whether the result is global.

  1. Set both coordinate derivatives to zero and solve the linear system.

    f(x,y)=x2+xy+y2−4x−2y;2x+y−4=0,x+2y−2=0 ⟹ (x,y)=(2,0)f(x,y)=x^2+xy+y^2-4x-2y;\qquad 2x+y-4=0,\quad x+2y-2=0\ \Longrightarrow\ (x,y)=(2,0)

  2. The Hessian has positive determinant and positive first diagonal entry, giving a strict local minimum.

    H=(2112),D=3>0,f(2,0)=−4H=\begin{pmatrix}2&1\\1&2\end{pmatrix},\qquad D=3>0,\qquad f(2,0)=-4

  3. Complete squares to compare with every point in the plane. Both squared terms are nonnegative and vanish together only at the candidate.

    f(x,y)=(x−2+y2)2+34y2−4≥−4f(x,y)=\left(x-2+\frac y2\right)^2+\frac34y^2-4\ge-4

Conclusion: The point with coordinates 2 and 0 is the unique global minimum; there is no global maximum.

EXAMPLE 2

Find the largest and smallest coordinate products on the circle.

  1. The circle is compact and its constraint gradient never vanishes on it, so extrema exist and satisfy the regular multiplier equations.

    f(x,y)=xy,g(x,y)=x2+y2=2;⟨y,x⟩=λ⟨2x,2y⟩f(x,y)=xy,\quad g(x,y)=x^2+y^2=2;\qquad \langle y,x\rangle=\lambda\langle2x,2y\rangle

  2. Multiply the first component equation by y and the second by x. This avoids dividing by a possibly zero coordinate and forces equal squares.

    y2=2λxy=x2,x2+y2=2 ⟹ x2=y2=1y^2=2\lambda xy=x^2,\qquad x^2+y^2=2\ \Longrightarrow\ x^2=y^2=1

  3. Compare all four feasible candidates. Equal signs yield the largest product; opposite signs yield the smallest.

    f(1,1)=f(−1,−1)=1,f(1,−1)=f(−1,1)=−1f(1,1)=f(-1,-1)=1,\qquad f(1,-1)=f(-1,1)=-1

Conclusion: The absolute maximum is 1 and the absolute minimum is negative 1, each attained at two points.

Pitfalls & counterexamples

Tempting claim: Zero gradient guarantees a local minimum.

Correction: It eliminates first-order change but leaves higher-order shape unresolved.

Test the claim: The difference of the squared coordinates has zero gradient at the origin, yet increases along the horizontal axis and decreases along the vertical axis.

Check your understanding

If the Hessian determinant is zero at a stationary point, does that prove a saddle?

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Compare your reasoning

No.

The test is inconclusive: the sum of fourth powers has a strict minimum at the origin, while the difference of fourth powers has a saddle there.

Connect this idea

The Hessian extends the one-variable second-derivative test. Constraint gradients explain why optimization algorithms project proposed changes onto feasible directions.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    For f (x,y)=2 x2+3 y2\displaystyle f\, \left(x, y\right) = 2\, x^{2} + 3\, y^{2}, find the Hessian determinant at (0,0).

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  2. Problem 2

    For f (x,y)=x2−2 y2\displaystyle f\, \left(x, y\right) = x^{2} - 2\, y^{2}, classify the origin: 1 minimum, 2 maximum, 3 saddle.

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  3. Problem 3

    Minimize x2+y2\displaystyle x^{2} + y^{2} subject to x+y=6\displaystyle x + y = 6. Find the minimizing x.

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  4. Problem 4

    For f (x,y)=(x−1)2+(y−2)2+7\displaystyle f\, \left(x, y\right) = \left(x - 1\right)^{2} + \left(y - 2\right)^{2} + 7, what is the minimum value?

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  5. Problem 5New context

    A quadratic cost is f (x,y)=3 x2+3 y2\displaystyle f\, \left(x, y\right) = 3\, x^{2} + 3\, y^{2}, find the Hessian determinant at (0,0).

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  6. Problem 6New context

    A surface is f (x,y)=x2−3 y2\displaystyle f\, \left(x, y\right) = x^{2} - 3\, y^{2}, classify the origin: 1 minimum, 2 maximum, 3 saddle.

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  7. Problem 7New context

    Two allocations minimize x2+y2\displaystyle x^{2} + y^{2} subject to x+y=8\displaystyle x + y = 8. Find the minimizing x.

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  8. Problem 8New context

    A cost surface is f (x,y)=(x−2)2+(y−3)2+7\displaystyle f\, \left(x, y\right) = \left(x - 2\right)^{2} + \left(y - 3\right)^{2} + 7, what is the minimum value?

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STAGE UMV 1.3 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Multiple, Line & Surface Integrals

Useful preparation: Hessians & Multivariable Extrema

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Integration needs a quantity and a geometric measure: area for a sheet, arc length for a wire, and surface area for a surface. Vector fields additionally introduce directional work and flux.

Multiple integral
Accumulation using area or volume. Iterated integration accumulates one coordinate at a time, with bounds describing the region.
Scalar and vector line integrals
Scalar line integrals weight arc length; work integrals project vector fields onto directed displacement.
Surface integral and flux
Scalar surface integrals weight area; flux first projects a vector field onto an oriented unit normal.

Conditions before calculation

  • Continuity on the bounded elementary regions here permits iterated integration. Changing order requires new bounds describing the same region.
  • Coordinate changes require the absolute Jacobian determinant and a valid one-to-one description apart from boundary overlaps. Here polar coordinates use r≥0\displaystyle r \ge 0, so the area factor is r; a signed-radius parameterization requires ∣r∣\displaystyle \left|r\right| and a check for repeated coverage.
  • Use piecewise smooth curves and regular surfaces. Reversing orientation negates work and flux, but preserves scalar integrals.

Why the method works

1. Describe the domain before integrating

Changing slice widths require variable inner bounds. Polar angle increments produce longer arcs farther from the origin, giving the radial area factor.

∬Df dA=∫ab∫ℓ(x)u(x)f(x,y) dy dx,dA=r dr dθ,r≥0\iint_D f\,dA=\int_a^b\int_{\ell(x)}^{u(x)}f(x,y)\,dy\,dx,\qquad dA=r\,dr\,d\theta,\quad r\ge0

2. Convert a curve to a parameter interval

Speed converts parameter width into length. Work instead uses the velocity vector itself, preserving direction.

∫Cρ ds=∫abρ(r(t))∥r′(t)∥ dt,∫CF⋅dr=∫abF(r(t))⋅r′(t) dt\int_C\rho\,ds=\int_a^b\rho(\mathbf r(t))\|\mathbf r'(t)\|\,dt,\qquad \int_C\mathbf F\cdot d\mathbf r=\int_a^b\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt

3. Convert a surface patch to parameter area

Parameter tangents span a small parallelogram. Their cross-product magnitude scales area, while its direction chooses the flux normal.

dS=∥Ru×Rv∥ du dv,∬SF⋅n dS=∬UF(R)⋅(Ru×Rv) du dvdS=\|\mathbf R_u\times\mathbf R_v\|\,du\,dv,\qquad \iint_S\mathbf F\cdot\mathbf n\,dS=\iint_U\mathbf F(\mathbf R)\cdot(\mathbf R_u\times\mathbf R_v)\,du\,dv

Worked examples

EXAMPLE 1

Find the mass of a triangular sheet with the displayed density.

  1. Vertical slices run from the horizontal axis to the sloping edge. The density is nonnegative throughout this triangle.

    D={(x,y):x≥0, y≥0, x+y≤2},ρ(x,y)=x+yD=\{(x,y):x\ge0,\ y\ge0,\ x+y\le2\},\qquad \rho(x,y)=x+y

  2. Integrate with respect to y while holding x fixed. The upper bound depends on x, reflecting the tapering region.

    M=∫02∫02−x(x+y) dy dx=∫02[x(2−x)+(2−x)22]dxM=\int_0^2\int_0^{2-x}(x+y)\,dy\,dx=\int_0^2\left[x(2-x)+\frac{(2-x)^2}{2}\right]dx

  3. Simplify and perform the outer integral. The triangle's area is 2, so the resulting average density provides a consistency check.

    M=∫02(2−x22)dx=83,ρavg=M2=43M=\int_0^2\left(2-\frac{x^2}{2}\right)dx=\frac83,\qquad \rho_{\mathrm{avg}}=\frac{M}{2}=\frac43

Conclusion: Mass is eight thirds; average density lies within the density range from zero to two.

EXAMPLE 2

Compare accumulation on a sloping wire and tilted surface, explaining why parameter size alone is insufficient.

  1. The wire parameter runs from zero to one, but the spatial displacement is longer. Compute its speed before integrating.

    r(t)=⟨t,2t,0⟩,0≤t≤1;∥r′(t)∥=5,∫C1 ds=5\mathbf r(t)=\langle t,2t,0\rangle,\quad 0\le t\le1;\qquad \|\mathbf r'(t)\|=\sqrt5,\quad \int_C1\,ds=\sqrt5

  2. The surface is a tilted unit parameter square. Its two tangent vectors determine the actual area multiplier.

    R(u,v)=⟨u,v,u⟩,0≤u,v≤1;Ru=⟨1,0,1⟩,Rv=⟨0,1,0⟩\mathbf R(u,v)=\langle u,v,u\rangle,\quad 0\le u,v\le1;\qquad \mathbf R_u=\langle1,0,1\rangle,\quad \mathbf R_v=\langle0,1,0\rangle

  3. Use the cross-product magnitude for scalar area. For upward flux of a unit vertical field, use the oriented cross product instead.

    Ru×Rv=⟨−1,0,1⟩,∬S1 dS=2,∬S⟨0,0,1⟩⋅n dS=1\mathbf R_u\times\mathbf R_v=\langle-1,0,1\rangle,\quad \iint_S1\,dS=\sqrt2,\qquad \iint_S\langle0,0,1\rangle\cdot\mathbf n\,dS=1

Conclusion: Wire length, surface area, and vertical flux respectively equal square root of 5, square root of 2, and 1.

Pitfalls & counterexamples

Tempting claim: The parameter differential alone is always the geometric measure.

Correction: Parameter intervals and patches must be scaled into actual length or area; the needed factor depends on the geometry.

Test the claim: The wire parameter interval has length one, but the physical wire has length square root of 5.

Check your understanding

What changes when the chosen surface normal is reversed?

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Compare your reasoning

Flux changes sign; a scalar surface-area integral does not.

Flux contains the signed normal direction, while scalar accumulation uses a nonnegative area magnitude.

Connect this idea

These measures support mass, work, and transport models. Green's, Stokes', and the divergence theorems later relate certain interior accumulations to boundary accumulations under their respective hypotheses.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    Evaluate ∫\displaystyle \int from y=0\displaystyle y = 0 to 2 ∫\displaystyle 2\, \int from x=0\displaystyle x = 0 to 2 of x dx dy.

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  2. Problem 2

    Integrate the constant density 3 over 0≤x≤2\displaystyle 0 \le x \le 2, 0≤y≤2\displaystyle 0 \le y \le 2, 0≤z≤4\displaystyle 0 \le z \le 4. Find the triple integral.

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  3. Problem 3

    Find the line integral F=(2,3)\displaystyle F = \left(2, 3\right) along r (t)=(t,2 t)\displaystyle r\, \left(t\right) = \left(t, 2\, t\right), from t=0\displaystyle t = 0 to t=2\displaystyle t = 2. Find ∫ F⋅dr\displaystyle \int\, F \cdot dr.

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  4. Problem 4

    A surface patch has area 3, upward unit normal (0,0,1), and constant field F=(1,2,4)\displaystyle F = \left(1, 2, 4\right). Find the upward flux.

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  5. Problem 5New context

    A rectangular lamina has mass given by ∫\displaystyle \int from y=0\displaystyle y = 0 to 2 ∫\displaystyle 2\, \int from x=0\displaystyle x = 0 to 3 of x dx dy.

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  6. Problem 6New context

    A box has density 3 over 0≤x≤3\displaystyle 0 \le x \le 3, 0≤y≤2\displaystyle 0 \le y \le 2, 0≤z≤4\displaystyle 0 \le z \le 4. Find the triple integral.

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  7. Problem 7New context

    A force field is F=(2,3)\displaystyle F = \left(2, 3\right) along r (t)=(t,2 t)\displaystyle r\, \left(t\right) = \left(t, 2\, t\right), from t=0\displaystyle t = 0 to t=3\displaystyle t = 3. Find ∫ F⋅dr\displaystyle \int\, F \cdot dr.

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  8. Problem 8New context

    A horizontal sensor patch has area 4, upward unit normal (0,0,1), and constant field F=(1,2,4)\displaystyle F = \left(1, 2, 4\right). Find the upward flux.

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