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
Example 4 Find the solution and interval of validity to the following IVP.
So, first deal with that minus sign separating the two terms.
Now, find and and check that it’s exact.
So, it’s exact. We’ll integrate the first one in this case.
Differentiate with respect to and compare to .
So, it looks like we’ve got.
This gives us
The implicit solution is then,
Applying the initial condition gives,
The implicit solution is now,
This solution is much easier to solve than the previous ones. No quadratic formula is needed this time, all we need to do is solve for . Here’s what we get for an explicit solution.
Alright, let’s get the interval of validity. The term in the logarithm is always positive so we don’t need to worry about negative numbers in that. We do need to worry about division by zero however. We will need to avoid the following point(s).
We now have three possible intervals of validity.
The last one contains and so is the interval of validity for this problem is . Here’s a graph of the solution.
Now, find and and check that it’s exact.
So, it’s exact. We’ll integrate the first one in this case.
Differentiate with respect to and compare to .
So, it looks like we’ve got.
This gives us
The implicit solution is then,
Applying the initial condition gives,
The implicit solution is now,
This solution is much easier to solve than the previous ones. No quadratic formula is needed this time, all we need to do is solve for . Here’s what we get for an explicit solution.
Alright, let’s get the interval of validity. The term in the logarithm is always positive so we don’t need to worry about negative numbers in that. We do need to worry about division by zero however. We will need to avoid the following point(s).
We now have three possible intervals of validity.
The last one contains and so is the interval of validity for this problem is . Here’s a graph of the solution.
Example 5 Find the solution and interval of validity for the following IVP.
Let’s identify and and check that it’s exact.
So, it’s exact. With the proper simplification integrating the second one isn’t too bad.
However, the first is already set up for easy integration so let’s do that one.
Differentiate with respect to and compare to .
So, it looks like we’ve got
Recall that actually , but we drop the because it will get absorbed in the next step. That gives us . Therefore, we get.
The implicit solution is then
Apply the initial condition.
The implicit solution is then
This is as far as we can go. There is no way to solve this for and get an explicit solution.
So, it’s exact. With the proper simplification integrating the second one isn’t too bad.
However, the first is already set up for easy integration so let’s do that one.
Differentiate with respect to and compare to .
So, it looks like we’ve got
Recall that actually , but we drop the because it will get absorbed in the next step. That gives us . Therefore, we get.
The implicit solution is then
Apply the initial condition.
The implicit solution is then
This is as far as we can go. There is no way to solve this for and get an explicit solution.
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