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Introductory Calculus - Lecture 1
Jul 22, 2024
Introductory Calculus - Lecture 1
Practical Information
Lectures
: 16 lectures total
Notes available online (written by Cath Wilkins)
Lecturer: Dan Ciubotaru
Schedule: Mondays and Wednesdays at 10 am
8 problem sheets, first 2 available online
4 tutorials in college (1 hour each)
Reading List
: Available online
Recommended book:
Mathematical Methods in Physical Sciences
by Mary Boas
Syllabus Overview
First Half
: Differential Equations (7-8 lectures)
Ordinary Differential Equations (ODEs)
Partial Differential Equations (PDEs)
Second Half
: Line and Double Integrals (3 lectures)
Compute arc lengths & areas
Final Lectures
: Calculus of functions in two variables
Introduction to multivariable calculus
Topics: Surfaces, gradients, normal vectors, Taylor's theorem in two variables, critical points, Lagrange multipliers
Course Utility
Interaction with other preliminary courses (e.g., Multivariable Calculus, Dynamics, PDEs)
Useful for applied mathematics options in Part A (e.g., differential equations, fluid and waves)
Mandatory course
Differential Equations (DEs)
Ordinary Differential Equations (ODEs)
Involves independent variable ( x ) and function ( y(x) )
Example: ( \frac{dy}{dx} = f(x) )
Solved via integration: ( y = \int f(x) dx )
Example in Mechanics
: Newton's second law
Force = mass ( \times ) acceleration (( \frac{d^2r}{dt^2} ))
Example in Electrical Circuits
: RLC circuit
Components: Resistor (R), Inductor (L), Capacitor (C), Voltage source (V)
Kirchhoff’s Law: ( V(t) = RI(t) + L \frac{dI}{dt} + \frac{1}{C}Q(t) )
Differential equation in ( Q ): ( L \frac{d^2Q}{dt^2} + R \frac{dQ}{dt} + \frac{1}{C}Q = V(t) )
Integration Techniques
Integration by Parts
: Originates from the product rule
Formula: ( \int f(x)g'(x) dx = f(x)g(x) - \int f'(x)g(x) dx )
Example
: ( \int x^2 \sin(x) dx )
Recursive Integration
: Involves reduction formulas
Example
: ( \int \cos^n(x) dx )
Derive a recurrence relation to solve higher integrals
Separable Differential Equations
Form: ( \frac{dy}{dx} = A(x)B(y) )
Solution: ( \int \frac{1}{B(y)} dy = \int A(x) dx )
Example
: Solve ( x(y^2 - 1) + y(x^2 - 1) \frac{dy}{dx} = 0 )
Steps:
Rearrange: ( y(x^2 - 1) \frac{dy}{dx} = -x(y^2 - 1) )
Separate variables: ( \frac{y}{y^2 - 1} dy = - \frac{x}{x^2 - 1} dx )
Integrate both sides: ( \int \frac{y}{y^2 - 1} dy = - \int \frac{x}{x^2 - 1} dx )
Solve the integrals to find the solution implicitly
Conclusion
Review integration techniques if necessary
Next lecture: More on differential equations
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