Resit Prep

CalcLin Exam Essentials

This note contains the methods, formulas, decision rules, and common mistakes for the topics that must be mastered for the exam.


1. Stokes’ Theorem

Purpose

Stokes’ theorem converts a line integral around a closed curve into a surface integral of the curl:

∮CF⋅dr=∬S(∇×F)⋅ndS

Here:

When to Use Stokes

Look for:

Work Plan

  1. Compute the curl:
∇×F=|ijk∂x∂y∂zF1F2F3|

Thus,

∇×F=[∂F3∂y−∂F2∂z∂F1∂z−∂F3∂x∂F2∂x−∂F1∂y]
  1. Choose the easiest surface S with boundary C.
  2. Find the correctly oriented normal vector.
  3. Evaluate:
∬S(∇×F)⋅ndS

Surface Written as z=g(x,y)

For upward orientation:

ndS=[−gx−gy1]dxdy

For downward orientation, multiply by −1.

Orientation

Curl the fingers of your right hand in the direction of C:

Important Shortcut

You may replace the original surface with any easier surface having the same boundary, as long as the orientation is preserved.

Common Mistakes


2. Gauss’ Divergence Theorem

Purpose

The divergence theorem converts the outward flux through a closed surface into a triple integral over the enclosed volume:

∬∂VF⋅ndS=∭V∇⋅FdV

The surface ∂V must be closed and n must point outward.

When to Use Gauss

Look for:

Divergence Formula

For

F=[F1F2F3]

the divergence is

∇⋅F=∂F1∂x+∂F2∂y+∂F3∂z

Work Plan

  1. Check that the surface is closed.
  2. Compute ∇⋅F.
  3. Describe the enclosed volume V.
  4. Choose suitable coordinates.
  5. Evaluate:
∭V∇⋅FdV

Open Surfaces

Gauss cannot be applied directly to an open surface. Instead:

  1. Add a cap to close the surface.
  2. Apply Gauss to the entire closed surface.
  3. Subtract the flux through the added cap.
Flux through original surface=total closed flux−cap flux

Common Coordinate Systems

Cylindrical Coordinates

x=rcos⁡θ,y=rsin⁡θ,z=zdV=rdrdθdz

Spherical Coordinates

Using ϕ as the angle from the positive z-axis:

x=ρsin⁡ϕcos⁡θy=ρsin⁡ϕsin⁡θz=ρcos⁡ϕdV=ρ2sin⁡ϕdρdϕdθ

Common Mistakes


3. Stokes or Gauss?

Use Stokes When

Use Gauss When

Memory Rule

Stokes: boundary curve⟷surfaceGauss: closed surface⟷volume

4. Direct Flux Integrals

Definition

The flux of F through an oriented surface S is

∬SF⋅ndS

Flux measures how much of the vector field passes through the surface.

Parametrized Surface

Suppose

r(u,v)=[x(u,v)y(u,v)z(u,v)]

Then

ru=∂r∂u,rv=∂r∂v

The oriented surface element is

ndS=(ru×rv)dudv

Therefore,

∬SF⋅ndS=∬DF(r(u,v))⋅(ru×rv)dudv

Work Plan

  1. Parametrize the surface.
  2. Compute ru and rv.
  3. Compute ru×rv.
  4. Check the orientation.
  5. Substitute the parametrization into F.
  6. Take the dot product.
  7. Integrate over the parameter domain.

Graph z=g(x,y)

Use

r(x,y)=[xyg(x,y)]

Then

rx×ry=[−gx−gy1]

Thus, for upward orientation,

∬SF⋅ndS=∬DF(x,y,g(x,y))⋅[−gx−gy1]dxdy

Orientation Check

If the normal points in the wrong direction, multiply the final normal vector by −1.

For example:

Scalar Surface Integral Versus Flux

Scalar surface integral:

∬SfdS=∬Df(r(u,v))‖ru×rv‖dudv

Flux integral:

∬SF⋅ndS=∬DF(r(u,v))⋅(ru×rv)dudv

For flux, do not take the magnitude of the cross product.


5. Subspaces

Definition

A subset H of a vector space is a subspace when it satisfies all three properties:

  1. The zero vector is in H.
  2. H is closed under addition.
  3. H is closed under scalar multiplication.

Subspace Test

Property 1: Zero Vector

Check:

0∈H

If the zero vector is not in the set, the set is immediately not a subspace.

Property 2: Closure Under Addition

Take arbitrary u,v∈H and show:

u+v∈H

Property 3: Closure Under Scalar Multiplication

Take arbitrary u∈H and c∈R and show:

cu∈H

Fast Recognition

The solution set of a homogeneous linear system

Ax=0

is always a subspace.

The solution set of

Ax=b,b≠0

is generally not a subspace because it usually does not contain the zero vector.

Common Non-Subspace Conditions

Sets involving the following are suspicious:

Disproving a Subspace

You only need one failed property. Useful methods:


6. Linear Dependence and Independence

Definition

Vectors v1,…,vp are linearly independent when

c1v1+⋯+cpvp=0

has only the trivial solution

c1=⋯=cp=0

If a nontrivial solution exists, the vectors are linearly dependent.

Matrix Method

Place the vectors as columns:

A=[v1v2⋯vp]

Solve

Ax=0

Conclusion

Immediate Dependence Rules

A set is automatically linearly dependent if:

Square Matrix Shortcut

For n vectors in Rn:

det(A)≠0⟺vectors are linearly independentdet(A)=0⟺vectors are linearly dependent

Spanning Versus Independence

For vectors in Rm:

For exactly n vectors in Rn, the following are equivalent:


7. Rank Theorem

For an m×n matrix A:

rank(A)+nullity(A)=n

Here:

Therefore,

number of pivot variables+number of free variables=number of variables

Important Dimensions

dim⁡(ColA)=rank(A)dim⁡(RowA)=rank(A)dim⁡(NulA)=n−rank(A)

Basis for Column Space

  1. Row reduce A.
  2. Identify the pivot column numbers.
  3. Take those columns from the original matrix.

Never take the column-space basis from the row-reduced matrix.

Basis for Null Space

  1. Row reduce A.
  2. Identify the free variables.
  3. Express the pivot variables using the free variables.
  4. Write the solution in parametric vector form.
  5. The vectors multiplying the free parameters form a basis for NulA.

Example Structure

Suppose

x=sv1+tv2

Then

NulA=Span{v1,v2}

and

nullity(A)=2

8. Projections

Projection onto One Vector

The projection of y onto a nonzero vector u is

projuy=y⋅uu⋅uu

If u is a unit vector:

projuy=(y⋅u)u

Orthogonal Decomposition

A vector y can be written as

y=y^+z

where:

For projection onto u:

y^=projuyz=y−y^

The error vector satisfies

z⋅u=0

Projection onto an Orthogonal Basis

If {u1,…,up} is an orthogonal basis for W, then

projWy=∑i=1py⋅uiui⋅uiui

Projection onto an Orthonormal Basis

If {q1,…,qp} is orthonormal, then

projWy=∑i=1p(y⋅qi)qi

Projection Matrix

If the columns of Q are orthonormal, then

P=QQT

and

y^=QQTy

For a general full-column-rank matrix A:

P=A(ATA)−1AT

9. Gram–Schmidt

Purpose

Gram–Schmidt converts a linearly independent set

{a1,a2,…,ap}

into an orthogonal or orthonormal basis for the same subspace.

Orthogonal Version

Start with

v1=a1

Then

v2=a2−a2⋅v1v1⋅v1v1

For the third vector:

v3=a3−a3⋅v1v1⋅v1v1−a3⋅v2v2⋅v2v2

In general:

vk=ak−∑j=1k−1projvjak

Normalize

After finding the orthogonal vectors:

qi=vi∥vi∥

The vectors qi form an orthonormal basis.

Work Plan

  1. Set v1=a1.
  2. Remove from a2 its projection onto v1.
  3. Remove from a3 its projections onto both v1 and v2.
  4. Continue similarly.
  5. Normalize only at the end unless unit vectors are specifically needed earlier.

Check

For an orthogonal basis:

vi⋅vj=0(i≠j)

For an orthonormal basis:

qi⋅qj={1,i=j0,i≠j

Common Mistakes


10. Least Squares

Purpose

The system

Ax=b

may have no exact solution when b∉ColA.

The least-squares solution x^ makes

∥Ax−b∥

as small as possible.

Geometric Interpretation

The vector

Ax^

is the orthogonal projection of b onto ColA.

The residual is

r=b−Ax^

The residual is orthogonal to every column of A:

ATr=0

Normal Equations

Starting from

AT(b−Ax^)=0

we obtain

ATAx^=ATb

Work Plan

  1. Compute ATA.
  2. Compute ATb.
  3. Solve:
ATAx^=ATb
  1. Compute the fitted vector:
b^=Ax^
  1. Compute the residual:
r=b−b^
  1. Check:
ATr=0

Explicit Formula

If the columns of A are linearly independent:

x^=(ATA)−1ATb

Do not use this formula if ATA is not invertible.

Least-Squares Line

For a line

y=β0+β1x

the design matrix is

A=[1x11x2⋮⋮1xm],x=[β0β1],b=[y1y2⋮ym]

Solve the normal equations for β0 and β1.

Common Mistakes


11. Sequence and Series Convergence

Sequence Convergence

A sequence {an} converges to L if

limn→∞an=L

A necessary condition for a series

∑n=1∞an

to converge is

limn→∞an=0

If

limn→∞an≠0

the series diverges.

However,

limn→∞an=0

does not prove convergence.

Example:

∑n=1∞1n

diverges even though 1/n→0.


12. Standard Series

Geometric Series

∑n=0∞arn

converges when

|r|<1

Its sum is

∑n=0∞arn=a1−r

It diverges when

|r|≥1

p-Series

∑n=1∞1np

converges when

p>1

and diverges when

p≤1

Harmonic Series

∑n=1∞1n

diverges.


13. Choosing a Convergence Test

Step 1: Term Test

Always begin by checking:

limn→∞an

If the limit is not zero, the series diverges immediately.

Step 2: Recognize the Form


14. Integral Test

Conditions

Suppose

an=f(n)

where f is:

  1. Positive.
  2. Continuous.
  3. Decreasing.

Then

∑n=N∞an and ∫N∞f(x)dx either both converge or both diverge.

Work Plan

  1. Identify f(x) such that f(n)=an.
  2. State that f is positive, continuous, and decreasing.
  3. Compute:
∫N∞f(x)dx=limb→∞∫Nbf(x)dx
  1. Conclude:
    • Finite improper integral ⇒ series converges.
    • Infinite improper integral ⇒ series diverges.

Example

For

∑n=1∞1n2

use

f(x)=1x2

Then

∫1∞1x2dx=[−1x]1∞=1

Therefore, the series converges.

Important

The integral test normally determines convergence, not the exact value of the series.


15. Integral-Test Error Bounds

Let

S=∑n=N∞an

and let

Sn=∑k=Nnak

The remainder is

Rn=S−Sn

If the integral-test conditions hold, then

∫n+1∞f(x)dx≤Rn≤∫n∞f(x)dx

Maximum Error

The upper bound gives the maximum possible error:

|Rn|≤∫n∞f(x)dx

Better Estimate for the Sum

Since

S=Sn+Rn

and

Ln≤Rn≤Un

we obtain

Sn+Ln≤S≤Sn+Un

A better estimate is the midpoint:

S≈Sn+Ln+Un2

The maximum error of this midpoint estimate is

maximum error=Un−Ln2

Here,

Ln=∫n+1∞f(x)dxUn=∫n∞f(x)dx

Finding the Required Number of Terms

To guarantee an error smaller than ε, solve

∫n∞f(x)dx<ε

Then round n upward to the next integer.


16. Alternating-Series Test

An alternating series has the form

∑n=1∞(−1)nbnor∑n=1∞(−1)n+1bn

where bn≥0.

The series converges if:

  1. bn+1≤bn eventually.
  2. limn→∞bn=0.

Alternating-Series Error Bound

If the alternating-series test applies, then

|Rn|=|S−Sn|≤bn+1

The error is at most the magnitude of the first omitted term.

Finding the Number of Terms

To guarantee

|Rn|<ε

solve

bn+1<ε

Sign of the Error

The exact sum lies between consecutive partial sums:

SnandSn+1

If the first omitted term is positive, the exact answer lies above Sn.

If the first omitted term is negative, the exact answer lies below Sn.


17. Absolute and Conditional Convergence

A series

∑an

converges absolutely if

∑|an|

converges.

A series converges conditionally if

∑an

converges but

∑|an|

diverges.

Procedure for Alternating Series

  1. Test
∑|an|
  1. If it converges, the original series converges absolutely.
  2. If it diverges, test the original alternating series.
  3. If the original converges, it converges conditionally.
  4. If the original also diverges, the series diverges.

Important Result

Absolute convergence implies convergence.

18. Ratio Test

For

∑an

compute

L=limn→∞|an+1an|

Then:

L<1⇒absolute convergenceL>1⇒divergenceL=1⇒inconclusive

The ratio test is especially useful for:


19. Power Series

A power series centred at a has the form

∑n=0∞cn(x−a)n

It normally has:

Work Plan

  1. Apply the ratio test or root test.
  2. Solve the resulting inequality for |x−a|.
  3. Determine the radius R.
  4. Write the open interval.
  5. Test each endpoint separately.
  6. State the full interval of convergence.

Endpoint Warning

The ratio test usually gives no conclusion at the endpoints. Each endpoint may:


20. Taylor and Maclaurin Series

Taylor Series

The Taylor series of f about x=a is

f(x)=∑n=0∞f(n)(a)n!(x−a)n

Maclaurin Series

For a=0:

f(x)=∑n=0∞f(n)(0)n!xn

Important Standard Series

Geometric

11−x=∑n=0∞xn,|x|<1

Exponential

ex=∑n=0∞xnn!

Sine

sin⁡x=∑n=0∞(−1)nx2n+1(2n+1)!

Cosine

cos⁡x=∑n=0∞(−1)nx2n(2n)!

Natural Logarithm

ln⁡(1+x)=∑n=1∞(−1)n+1xnn,−1<x≤1

Arctangent

arctan⁡x=∑n=0∞(−1)nx2n+12n+1,|x|≤1

Transforming Known Series

Starting from

11−x=∑n=0∞xn

you may:

Always update the interval of convergence.


21. Taylor Error Bound

The degree-n Taylor polynomial is

Tn(x)=∑k=0nf(k)(a)k!(x−a)k

The remainder is

Rn(x)=f(x)−Tn(x)

Lagrange Error Bound

If

|f(n+1)(t)|≤M

for all t between a and x, then

|Rn(x)|≤M|x−a|n+1(n+1)!

Work Plan

  1. Identify the degree n.
  2. Find f(n+1).
  3. Find a maximum value M on the interval between a and x.
  4. Substitute into:
|Rn(x)|≤M|x−a|n+1(n+1)!

Choosing the Degree

To guarantee an error below ε, find the smallest n for which

M|x−a|n+1(n+1)!<ε

22. Error-Bound Summary

Alternating Series

|Rn|≤bn+1

Use the first omitted term.

Integral Test

∫n+1∞f(x)dx≤Rn≤∫n∞f(x)dx

Maximum error using Sn:

|Rn|≤∫n∞f(x)dx

Midpoint estimate:

S≈Sn+12(∫n+1∞f(x)dx+∫n∞f(x)dx)$Template

Midpoint maximum error:

12(∫n∞f(x)dx−∫n+1∞f(x)dx)

Taylor Polynomial

|Rn(x)|≤M|x−a|n+1(n+1)!

23. Final Exam Decision Guide

Closed Line Integral

∮CF⋅dr

Use Stokes’ theorem. Compute:

∇×F

Outward Flux Through a Closed Surface

∬∂VF⋅ndS

Use Gauss’ theorem. Compute:

∇⋅F

Flux Through an Open Surface

Parametrize the surface and use

∬DF(r(u,v))⋅(ru×rv)dudv

Alternatively, close the surface and use Gauss, then subtract the added part.


Is a Set a Subspace?

Check:

  1. Zero vector.
  2. Closure under addition.
  3. Closure under scalar multiplication.

A homogeneous solution set is automatically a subspace.


Are Vectors Linearly Independent?

Put the vectors in the columns of a matrix and row reduce.


Need Rank or Nullity?

Use

rank(A)+nullity(A)=n

Need an Orthogonal Basis?

Use Gram–Schmidt:

vk=ak−∑j=1k−1projvjak

Inconsistent System but Need Best Approximation?

Use least squares:

ATAx^=ATb

Series with Alternating Signs?

Check:

  1. bn decreases.
  2. bn→0.

Error:

|Rn|≤bn+1

Use the integral test.

Error:

∫n+1∞f(x)dx≤Rn≤∫n∞f(x)dx

Factorials or Exponentials?

Use the ratio test.


Taylor Approximation Error?

Use

|Rn(x)|≤M|x−a|n+1(n+1)!

24. Formula Sheet

Vector Calculus

∇⋅F=F1x+F2y+F3z∇×F=[F3y−F2zF1z−F3xF2x−F1y]∮CF⋅dr=∬S(∇×F)⋅ndS∬∂VF⋅ndS=∭V∇⋅FdVndS=(ru×rv)dudv

Linear Algebra

projuy=y⋅uu⋅uurank(A)+nullity(A)=nATAx^=ATbx^=(ATA)−1ATbr=b−Ax^ATr=0

Series

∑n=0∞arn=a1−r,|r|<1∑n=1∞1np converges iff p>1|Rn|alternating≤bn+1∫n+1∞f(x)dx≤Rn≤∫n∞f(x)dx|Rn(x)|Taylor≤M|x−a|n+1(n+1)!

25. Last-Minute Checklist

Before the exam, make sure you can do the following without notes: