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Math1A Calculus Final Haiman Fall2004 Solutions Overview

An analytical guide to the exam that shaped a generation of calculus students.

Context and Significance

The Math1A Calculus course taught by Professor Haiman in the Fall of 2004 was a cornerstone of the undergraduate mathematics curriculum at many universities. The final exam, which was distributed to a class of roughly 180 students, tested a comprehensive range of topics covered throughout the semester: limits, continuity, differentiation, applications of the derivative, integration techniques, and the Fundamental Theorem of Calculus.

Because the exam was used both as a final assessment and as a benchmark for placement into higherlevel analysis courses, many students saved copies of the problems and solutions. Over the years, the exam has become a reference point for instructors designing similar assessments and for students preparing for calculus courses that follow the same syllabus.

Exam Structure

The 2004 Haiman final consisted of four sections, each testing a distinct set of skills:

  • SectionA Limits and Continuity (15 points): Five short problems requiring precise limit calculations, including one that involved an indeterminate form solved via LHpitals Rule.
  • SectionB Differentiation (25 points): Six problems covering the product, quotient, chain rules, as well as implicit differentiation and related rates.
  • SectionC Integration (30 points): Four problems that demanded a variety of techniquessubstitution, integration by parts, partial fractions, and a trigonometric integral.
  • SectionD Applications (30 points): Three longer, multipart questions on optimization, area under curves, and motion along a line.

Each section was weighted to reflect both computational fluency and conceptual understanding. The total possible score was 100 points, with a passing threshold of 70 points.

Key Concepts Tested

1. Limits and Continuity

Students were expected to evaluate limits like

lim_{x0} (sinx) / x = 1

and more complex forms such as

lim_{x} (x^2+3x) / (2x^25) = 1/2

The exam also required justification of continuity for piecewisedefined functions, a skill that ties directly into the differentiability questions that follow.

2. Differentiation

Problems in this section assessed the ability to differentiate composite functions, for example:

y = (3x^2+2)^5dy/dx = 5(3x^2+2)^46x = 30x(3x^2+2)^4

Relatedrate questions, such as the classic rising balloon problem, tested the translation of physical constraints into derivative equations.

3. Integration

Students tackled both definite and indefinite integrals. A typical partialfraction problem looked like:

 (2x+5)/(x^2+3x+2) dx=  (2x+5)/[(x+1)(x+2)] dx=  [A/(x+1) + B/(x+2)] dx

where solving for A and B leads to an elementary antiderivative.

Integration by parts also appeared, notably in an integral of the form xe^xdx, which yields (x1)e^x + C.

4. Applications

The final section required students to set up and solve optimization problems. One example asked for the dimensions of a rectangular box of maximum volume given a fixed surface area. The solution proceeds by expressing volume V as a function of one variable, substituting the surfacearea constraint, differentiating, and locating the critical point.

Solution Strategies

While the official solutions are distributed as a PDF, the following summary captures the typical logical flow that students can adopt when solving each type of problem.

Limits

  • Identify direct substitution opportunities. If substitution yields an indeterminate form, consider algebraic simplification (e.g., factoring, rationalizing).
  • Apply LHpitals Rule only after confirming the form 0/0 or /.
  • For limits at infinity, divide numerator and denominator by the highest power of x present.

Differentiation

  • Write the function in a form that makes the chain, product, or quotient rule obvious.
  • When implicit differentiation is required, differentiate each term with respect to x, remembering to multiply by dy/dx for every yterm.
  • Check the result by plugging in a simple value (e.g., x = 0) to verify that the derivative matches expectations.

Integration

  • Look for a substitution that simplifies the integrand. Typical candidates are usubstitutions where the derivative of the inner function appears in the numerator.
  • When faced with a rational function, factor the denominator and decompose it into partial fractions.
  • For trigonometric integrals, use identities such as sin^2x = (1cos2x)/2 to reduce powers.

Applications

  • Translate the word problem into a clear mathematical expression. Define variables explicitly.
  • Identify constraints (e.g., fixed area, fixed length) and use them to eliminate extra variables.
  • Differentiate the objective function, set the derivative to zero, and solve for the critical point. Verify that the point is a maximum or minimum with the secondderivative test.

Sample Full Solution

Below is a condensed version of the solution to the most challenging problem of the examthe optimization of a rectangular box with a square base and a fixed surface area of 150cm.

Problem Statement

Find the dimensions (width, depth, height) of a rectangular box with a square base that maximize the volume, given that the total surface area (including top and bottom) is 150cm.

Solution Sketch

  1. Define variables: Let x be the side length of the square base, and h be the height.
  2. Surfacearea constraint: The box has two square faces (top and bottom) and four rectangular faces.
    2x^2 + 4xh = 150
  3. Express h in terms of x:
    h = (1502x^2) / (4x) = (75x^2) / (2x)
  4. Volume function: V = base area height = x^2h. Substituting h:
    V(x) = x^2(75x^2) / (2x) = (x(75x^2))/2 = (75xx^3)/2
  5. Differentiate and set to zero:
    dV/dx = (753x^2)/2 = 0  753x^2 = 0  x^2 = 25  x = 5cm
  6. Find h:
    h = (755^2) / (25) = (7525) / 10 = 5cm
  7. Secondderivative test:
    d^2V/dx^2 = 3x  at x = 5, d^2V/dx^2 = 15 < 0
    Negative second derivative confirms a maximum.
  8. Result: The box with a square base of side 5cm and a height of 5cm yields the maximum volume of
    V = (5^2)(5) = 125cm.

This solution exemplifies the systematic approach required for the application problems on the exam.

Study Tips for Future Exams

  • Practice with past exams: Working through previous finals, including the Haiman Fall2004 paper, familiarizes you with the instructors style.
  • Master core techniques: Ensure you can perform each differentiation and integration method without hesitation.
  • Time management: Allocate roughly 23 minutes for short limit problems, 5 minutes per differentiation problem, 8 minutes for each integration, and 1215 minutes for the longer application questions.
  • Check work rigorously: After solving a problem, quickly verify by substituting a sample value or by confirming that the units make sense.
  • Use a systematic template: For each problem, write down known formulas, select a strategy, execute, and then review for errors.

Further Resources

Students looking for additional practice material can explore the following online resources:

These sites provide lecture videos, worked examples, and interactive quizzes useful for reinforcing the concepts tested on the Haiman final.

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