Summary Memory Aid/Cheat Sheet for MATH2310 Calculus for Engineering and Science - DE AND MVP
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Course
MATH2310
Institution
University Of Newcastle (UON
)
Memory aid/cheat sheet for MATH2310, calculus of engineering and science, at the University of Newcastle.
Contains all content covered in both the multivariable calculus and differential equation components of the MATH2310 course.
Function gradient Cartesian: 1st order Linear ODE Tangent line of
is given by:
at a is:
Spherical: Parametric surfaces:
Elliptical paraboloid:
-To find tangent
Solving IVP’s: vector, find
Standard model: parameters and sub
Tangent plane is given by: Elliptical cylinder: into derivative.
-To find normal vector,
Law of cooling: take cross product of
partial derivatives with
Double integrals: subbed parameters.
Resistors:
-Can be permutated:
Arc length is given by:
Capacitors:
Type 1 region:
Inductors:
Surface area is:
Homogenous
solutions:
Type 2 region:
-Distinct root:
Independence:
-Multiple root: -Functions are linearly
independent if they
-Complex root: cannot be expressed
Cartesian function on as a combination of
polar rectangle: each other.
Conversion from -A set of functions are
polar: linearly independent if
their Wronskian is 0.
Conversion to
polar: To change
function
Jacobian = variable,
multiply by For ,
Conversion to Jacobian:
spherical: equilibrium points,
are constant
solutions.
If and are
continuous at ,
Jacobian =
then ,
Triple integral Approximation methods: An nth order IVP has a unique
Jacobian is given by: Euler method: is equivalent to solution.
the 1st order IVP: A direction field is the
Absolute error: plot of the vectors:
Relative error:
Improved Euler method: -A function can be determined from a
Ratio test is: vector field by testing various points.
‘ - defines a gradient vector field.
Runge-Kutta method:
If L>1: divergent. -Summation index of a power series can
If L<1: convergent. be shifted, enabling addition of series’.
If L=1: inconclusive. For a function over a For a change
smooth curve, the line of variables
Solve L<1 for x to integral is: to t, bounds
find radius of are 0 ≤ t ≤ 1.
convergence. If k=0:
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