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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

Differential Equations - First Order: Modeling - ii




Example 4 A 50 kg object is shot from a cannon straight up with an initial velocity of 10m/s off a bridge that is 100 meters above the ground. If air resistance is given by 5v determine the velocity of the mass when it hits the ground.

First, notice that when we say straight up, we really mean straight up, but in such a way that it will miss the bridge on the way back down. Here is a sketch of the situation.
Notice the conventions that we set up for this problem. Since the vast majority of the motion will be in the downward direction we decided to assume that everything acting in the downward direction should be positive. Note that we also defined the “zero position” as the bridge, which makes the ground have a “position” of 100.
Okay, if you think about it we actually have two situations here. The initial phase in which the mass is rising in the air and the second phase when the mass is on its way down. We will need to examine both situations and set up an IVP for each. We will do this simultaneously. Here are the forces that are acting on the object on the way up and on the way down.
Notice that the air resistance force needs a negative in both cases in order to get the correct “sign” or direction on the force. When the mass is moving upwards the velocity (and hence v) is negative, yet the force must be acting in a downward direction. Therefore, the “-” must be part of the force to make sure that, overall, the force is positive and hence acting in the downward direction. Likewise, when the mass is moving downward the velocity (and so v) is positive. Therefore, the air resistance must also have a “-” in order to make sure that it’s negative and hence acting in the upward direction.
So, the IVP for each of these situations are.
Upmv=mg5vv(0)=10Downmv=mg5vv(t0)=0In the second IVP, the t0 is the time when the object is at the highest point and is ready to start on the way down. Note that at this time the velocity would be zero. Also note that the initial condition of the first differential equation will have to be negative since the initial velocity is upward.
In this case, the differential equation for both of the situations is identical. This won’t always happen, but in those cases where it does, we can ignore the second IVP and just let the first govern the whole process.
So, let’s actually plug in for the mass and gravity (we’ll be using g = 9.8 m/s2 here). We’ll go ahead and divide out the mass while we’re at it since we’ll need to do that eventually anyway.
v=9.85v50=9.8v10v(0)=10This is a simple linear differential equation to solve so we’ll leave the details to you. Upon solving we arrive at the following equation for the velocity of the object at any time t.
v(t)=98108et10Okay, we want the velocity of the ball when it hits the ground. Of course we need to know when it hits the ground before we can ask this. In order to find this we will need to find the position function. This is easy enough to do.
s(t)=v(t)dt=98108et10dt=98t+1080et10+cWe can now use the fact that I took the convention that s(0) = 0 to find that c = -1080. The position at any time is then.
s(t)=98t+1080et101080To determine when the mass hits the ground we just need to solve.
100=98t+1080et101080t=3.32203,5.98147We’ve got two solutions here, but since we are starting things at t = 0, the negative is clearly the incorrect value. Therefore, the mass hits the ground at t = 5.98147. The velocity of the object upon hitting the ground is then.
v(5.98147)=38.61841

This last example gave us an example of a situation where the two differential equations needed for the problem ended up being identical and so we didn’t need the second one after all. Be careful however to not always expect this. We could very easily change this problem so that it required two different differential equations. For instance we could have had a parachute on the mass open at the top of its arc changing its air resistance. This would have completely changed the second differential equation and forced us to use it as well. Or, we could have put a river under the bridge so that before it actually hit the ground it would have first had to go through some water which would have a different “air” resistance for that phase necessitating a new differential equation for that portion.
Or, we could be really crazy and have both the parachute and the river which would then require three IVP’s to be solved before we determined the velocity of the mass before it actually hits the solid ground.

Finally, we could use a completely different type of air resistance that requires us to use a different differential equation for both the upwards and downwards portion of the motion. Let’s take a quick look at an example of this.


Example 5 A 50 kg object is shot from a cannon straight up with an initial velocity of 10m/s off a bridge that is 100 meters above the ground. If air resistance is given by 5v2 determine the velocity of the mass at any time t.

So, this is basically the same situation as in the previous example. We just changed the air resistance from 5v to 5v2. Also, we are just going to find the velocity at any time t for this problem because, we’ll the solution is really unpleasant and finding the velocity for when the mass hits the ground is simply more work that we want to put into a problem designed to illustrate the fact that we need a separate differential equation for both the upwards and downwards motion of the mass.
As with the previous example we will use the convention that everything downwards is positive. Here are the forces on the mass when the object is on the way and on the way down.
Now, in this case, when the object is moving upwards the velocity is negative. However, because of the v2 in the air resistance we do not need to add in a minus sign this time to make sure the air resistance is positive as it should be given that it is a downwards acting force.
On the downwards phase, however, we still need the minus sign on the air resistance given that it is an upwards force and so should be negative but the v2 is positive.
This leads to the following IVP’s for each case.
Upmv=mg+5v2v=9.8+110v2v(0)=10Downmv=mg5v2v=9.8110v2v(t0)=0
These are clearly different differential equations and so, unlike the previous example, we can’t just use the first for the full problem.
Also, the solution process for these will be a little more involved than the previous example as neither of the differential equations are linear. They are both separable differential equations however.
So, let’s get the solution process started. We will first solve the upwards motion differential equation. Here is the work for solving this differential equation.
19.8+110v2dv=dt10198+v2dv=dt1098tan1(v98)=t+cNote that we did a little rewrite on the integrand to make the process a little easier in the second step.
Now, we have two choices on proceeding from here. Either we can solve for the velocity now, which we will need to do eventually, or we can apply the initial condition at this stage. While, we’ve always solved for the function before applying the initial condition we could just as easily apply it here if we wanted to and, in this case, will probably be a little easier.
So, to apply the initial condition all we need to do is recall that v is really v(t) and then plug in t=0. Doing this gives,
1098tan1(v(0)98)=0+c
Now, all we need to do is plug in the fact that we know v(0)=10 to get.
c=1098tan1(1098)
Messy, but there it is. The velocity for the upward motion of the mass is then,
1098tan1(v98)=t+1098tan1(1098)tan1(v98)=9810t+tan1(1098)v(t)=98tan(9810t+tan1(1098))
Now, we need to determine when the object will reach the apex of its trajectory. To do this all we need to do is set this equal to zero given that the object at the apex will have zero velocity right before it starts the downward motion. We will leave it to you to verify that the velocity is zero at the following values of t.
t=1098[tan1(1098)+πn]n=0,±1,±2,±3,
We clearly do not want all of these. We want the first positive t that will give zero velocity. It doesn’t make sense to take negative t’s given that we are starting the process at t=0 and once it hit’s the apex (i.e. the first positive t for which the velocity is zero) the solution is no longer valid as the object will start to move downwards and this solution is only for upwards motion.
Plugging in a few values of n will quickly show us that the first positive t will occur for n=0 and will be t=0.79847. We reduced the answer down to a decimal to make the rest of the problem a little easier to deal with.
The IVP for the downward motion of the object is then,
v=9.8110v2v(0.79847)=0
Now, this is also a separable differential equation, but it is a little more complicated to solve. First, let’s separate the differential equation (with a little rewrite) and at least put integrals on it.
19.8110v2dv=10198v2dv=dt
The problem here is the minus sign in the denominator. Because of that this is not an inverse tangent as was the first integral. To evaluate this integral we could either do a trig substitution (v=98sinθ) or use partial fractions using the fact that 98v2=(98v)(98+v). You’re probably not used to factoring things like this but the partial fraction work allows us to avoid the trig substitution and it works exactly like it does when everything is an integer and so we’ll do that for this integral.
We’ll leave the details of the partial fractioning to you. Once the partial fractioning has been done the integral becomes,
10(1298)198+v+198vdv=dt598[ln|98+v|ln|98v|]=t+c598ln|98+v98v|=t+c
Again, we will apply the initial condition at this stage to make our life a little easier. Doing this gives,
598ln|98+v(0.79847)98v(0.79847|=0.79847+c598ln|98+0980|=0.79847+c598ln|1|=0.79847+cc=0.79847
The solutions, as we have it written anyway, is then,
598ln|98+v98v|=t0.79847
Okay, we now need to solve for v and to do that we really need the absolute value bars gone and no we can’t just drop them to make our life easier. We need to know that they can be dropped without have any effect on the eventual solution.
To do this let’s do a quick direction field, or more appropriately some sketches of solutions from a direction field. Here is that sketch,
Note that 98=9.89949 and so is slightly above/below the lines for -10 and 10 shown in the sketch. The important thing here is to notice the middle region. If the velocity starts out anywhere in this region, as ours does given that v(0.79847)=0, then the velocity must always be less that 98.
What this means for us is that both 98+v and 98v must be positive and so the quantity in the absolute value bars must also be positive. Therefore, in this case, we can drop the absolute value bars to get,
598ln[98+v98v]=t0.79847
At this point we have some very messy algebra to solve for v. We will leave it to you to verify our algebra work. The solution to the downward motion of the object is,
v(t)=98e1598(t0.79847)1e1598(t0.79847)+1
Putting everything together here is the full (decidedly unpleasant) solution to this problem.
v(t)={98tan(9810t+tan1(1098))0t0.79847(upward motion)98e1598(t0.79847)1e1598(t0.79847)+10.79847ttend(downward motion)
where tend is the time when the object hits the ground. Given the nature of the solution here we will leave it to you to determine that time if you wish to but be forewarned the work is liable to be very unpleasant.

And with this problem you now know why we stick mostly with air resistance in the form cv! Note as well, we are not saying the air resistance in the above example is even realistic. It was simply chosen to illustrate two things. First, sometimes we do need different differential equation for the upwards and downwards portion of the motion. Secondly, do not get used to solutions always being as nice as most of the falling object ones are. Sometimes, as this example has illustrated, they can be very unpleasant and involve a lot of work.

Before leaving this section let’s work a couple examples illustrating the importance of remembering the conventions that you set up for the positive direction in these problems.
Awhile back I gave my students a problem in which a sky diver jumps out of a plane. Most of my students are engineering majors and following the standard convention from most of their engineering classes they defined the positive direction as upward, despite the fact that all the motion in the problem was downward. There is nothing wrong with this assumption, however, because they forgot the convention that up was positive they did not correctly deal with the air resistance which caused them to get the incorrect answer. So, the moral of this story is : be careful with your convention. It doesn’t matter what you set it as but you must always remember what convention to decided to use for any given problem.

So, let’s take a look at the problem and set up the IVP that will give the sky diver’s velocity at any time t.


Example 6 Set up the IVP that will give the velocity of a 60 kg sky diver that jumps out of a plane with no initial velocity and an air resistance of 0.8|v|. For this example assume that the positive direction is upward.

Here are the forces that are acting on the sky diver
Because of the conventions the force due to gravity is negative and the force due to air resistance is positive. As set up, these forces have the correct sign and so the IVP is
mv=mg+0.8|v|v(0)=0The problem arises when you go to remove the absolute value bars. In order to do the problem they do need to be removed. This is where most of the students made their mistake. Because they had forgotten about the convention and the direction of motion they just dropped the absolute value bars to get.
mv=mg+0.8vv(0)=0(incorrect IVP!!)So, why is this incorrect? Well remember that the convention is that positive is upward. However in this case the object is moving downward and so v is negative! Upon dropping the absolute value bars the air resistance became a negative force and hence was acting in the downward direction!
To get the correct IVP recall that because v is negative then |v| = -v. Using this, the air resistance becomes FA = -0.8v and despite appearances this is a positive force since the “-” cancels out against the velocity (which is negative) to get a positive force.
The correct IVP is then
mv=mg0.8vv(0)=0Plugging in the mass gives
v=9.8v75v(0)=0For the sake of completeness the velocity of the sky diver, at least until the parachute opens, which we didn’t include in this problem is.
v(t)=735+735et75

This mistake was made in part because the students were in a hurry and weren’t paying attention, but also because they simply forgot about their convention and the direction of motion! Don’t fall into this mistake. Always pay attention to your conventions and what is happening in the problems.
Just to show you the difference here is the problem worked by assuming that down is positive.


Example 7 Set up the IVP that will give the velocity of a 60 kg sky diver that jumps out of a plane with no initial velocity and an air resistance of 0.8|v|. For this example assume that the positive direction is downward.

Here are the forces that are acting on the sky diver
In this case the force due to gravity is positive since it’s a downward force and air resistance is an upward force and so needs to be negative. In this case since the motion is downward the velocity is positive so |v| = v. The air resistance is then F= -0.8v. The IVP for this case is
mv=mg0.8vv(0)=0Plugging in the mass gives
v=9.8v75v(0)=0Solving this gives
v(t)=735735et75This is the same solution as the previous example, except that it’s got the opposite sign. This is to be expected since the conventions have been switched between the two examples.

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