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Calculus III - 3-Dimensional Space: Equations of Lines

In this section we need to take a look at the equation of a line in  R 3 R 3 . As we saw in the previous section the equation  y = m x + b y = m x + b  does not describe a line in  R 3 R 3 , instead it describes a plane. This doesn’t mean however that we can’t write down an equation for a line in 3-D space. We’re just going to need a new way of writing down the equation of a curve. So, before we get into the equations of lines we first need to briefly look at vector functions. We’re going to take a more in depth look at vector functions later. At this point all that we need to worry about is notational issues and how they can be used to give the equation of a curve. The best way to get an idea of what a vector function is and what its graph looks like is to look at an example. So, consider the following vector function. → r ( t ) = ⟨ t , 1 ⟩ r → ( t ) = ⟨ t , 1 ⟩ A vector function is a function that takes one or more variables, one in this case, and returns a...
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Digital Signal Processing - Basic Continuous Time Signals

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Calculus III - 3-Dimensional Space: Equations of Lines

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Differential Equations - Systems: Repeated Eigenvalues - i

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Differential Equations - Systems: Repeated Eigenvalues - ii

Example 3  Solve the following IVP. → x ′ = ( − 1 3 2 − 1 6 − 2 ) → x → x ( 2 ) = ( 1 0 ) x → ′ = ( − 1 3 2 − 1 6 − 2 ) x → x → ( 2 ) = ( 1 0 ) First the eigenvalue for the system. det ( A − λ I ) = ∣ ∣ ∣ ∣ − 1 − λ 3 2 − 1 6 − 2 − λ ∣ ∣ ∣ ∣ = λ 2 + 3 λ + 9 4 = ( λ + 3 2 ) 2 ⇒ λ 1 , 2 = − 3 2 det ( A − λ I ) = | − 1 − λ 3 2 − 1 6 − 2 − λ | = λ 2 + 3 λ + 9 4 = ( λ + 3 2 ) 2 ⇒ λ 1 , 2 = − 3 2 Now let’s get the eigenvector. ( 1 2 3 2 − 1 6 − 1 2 ) ( η 1 η 2 ) = ( 0 0 ) ⇒ 1 2 η 1 + 3 2 η 2 = 0 η 1 = − 3 η 2 ( 1 2 3 2 − 1 6 − 1 2 ) ( η 1 η 2 ) = ( 0 0 ) ⇒ 1 2 η 1 + 3 2 η 2 = 0 η 1 = − 3 η 2 → η = ( − 3 η 2 η 2 ) η 2 ≠ 0 → η ( 1 ) = ( − 3 1 ) η 2 = 1 η → = ( − 3 η 2 η 2 ) η 2 ≠ 0 η → ( 1 ) = ( − 3 1 ) η 2 = 1 Now find  → ρ ρ → , ( 1 2 3 2 − 1 6 − 1 2 ) ( ρ 1 ρ 2 ) = ( − 3 1 ) ⇒ 1 2 ρ 1 + 3 2 ρ 2 = − 3 ρ 1 = − 6 − 3 ρ 2 ( 1 2 3 2 − 1 6 − 1 2 ) ( ρ 1 ρ 2 ) = ( − 3 1 ) ⇒ 1 2 ρ 1 + 3 2 ρ 2 = − 3 ρ 1 = − 6 − 3 ρ 2 → ρ = ( − 6 − 3 ρ 2 ρ 2 ) ⇒ → ρ = ( − 6 0 ) if  ρ 2 = 0 ρ → ...
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Differential Equations - Laplace Transforms: Table

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Differential Equations - Basic Concepts: Definitions

Differential Equation The first definition that we should cover should be that of  differential equation . A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives. There is one differential equation that everybody probably knows, that is Newton’s Second Law of Motion. If an object of mass  m m  is moving with acceleration  a a  and being acted on with force  F F  then Newton’s Second Law tells us. F = m a (1) (1) F = m a To see that this is in fact a differential equation we need to rewrite it a little. First, remember that we can rewrite the acceleration,  a a , in one of two ways. a = d v d t OR a = d 2 u d t 2 (2) (2) a = d v d t OR a = d 2 u d t 2 Where  v v  is the velocity of the object and  u u  is the position function of the object at any time  t t . We should also remember at this point that the force,  F F  may also be a f...
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