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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 vector. Note as we
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Differential Equations - Systems: Repeated Eigenvalues - ii



Example 3 Solve the following IVP.x→′=(−132−16−2)x→x→(2)=(10)

First the eigenvalue for the system.
det(A−λI)=|−1−λ32−16−2−λ|=λ2+3λ+94=(λ+32)2⇒λ1,2=−32Now let’s get the eigenvector.
(1232−16−12)(η1η2)=(00)⇒12η1+32η2=0η1=−3η2η→=(−3η2η2)η2≠0η→(1)=(−31)η2=1Now find ρ→,
(1232−16−12)(ρ1ρ2)=(−31)⇒12ρ1+32ρ2=−3ρ1=−6−3ρ2ρ→=(−6−3ρ2ρ2)⇒ρ→=(−60)if ρ2=0The general solution for the system is then,
x→(t)=c1e−3t2(−31)+c2(te−3t2(−31)+e−3t2(−60))Applying the initial condition gives,
(10)=x→(2)=c1e−3(−31)+c2(2e−3(−31)+e−3(−60))Note that we didn’t use t=0 this time! We now need to solve the following system,
−3e−3c1−12e−3c2=1e−3c1+2e−3c2=0}⇒c1=e33,c2=−e36The actual solution is then,
x→(t)=e33e−3t2(−31)−e36(te−3t2(−31)+e−3t2(−60))=e−3t2+3(013)+te−3t2+3(12−16)

And just to be consistent with all the other problems that we’ve done let’s sketch the phase portrait.



Example 4 Sketch the phase portrait for the system.x→′=(−132−16−2)x→

Let’s first notice that since the eigenvalue is negative in this case the trajectories should all move in towards the origin. Let’s check the direction of the trajectories at (1,0).
(−132−16−2)(10)=(−1−16)So, it looks like the trajectories should be pointing into the third quadrant at (1,0). This gives the following phase portrait.

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