- [Narrator] Welcome to an example of
an application of first order differential equations
involving a falling object.
In this video we'll model a falling object with
a differential equation
and then solve a falling object problem.
For this example we'll consider a falling object
influenced by gravity
and air resistance that is proportional to
the velocity of the object.
There are two forces acting on the object.
The force due to gravity given by the weight
which is going to be equal to m times g
or mass times the acceleration due to gravity
and the force due to air resistance
is proportional to the velocity
so it's going to be equal to negative k times v
where k is the constant of proportionality
and it's going to be negative because
it's going to be in the opposite direction
of the force due to gravity.
The net force, or big F,
is going to be equal to m times g minus k times v.
Let's take a look at a diagram
to get a better feel for what's happening here.
If this circle represents our falling object
the force downward again is going to be m times g.
So in this case downward will be
the positive direction and therefore the force of
the air resistance we're going to label negative k times v.
because it's going to be in the opposite direction
of the force downward.
This horizontal line here would be the ground.
Now let's see how we can come up with our
differential equation to model this falling object.
Using Newton's second law of motion
we know force is equal to mass times acceleration
or F equals m times a.
Well the first thing we can do is
replace acceleration with dv/dt
with a change in velocity with respects to time.
Next we now know that our net force or big F
is equal to mg minus k times v.
So here's a differential equation
that we can use to model the falling object
but we do want to change the form of
this differential equation
so the next step we're going to
add kv to both sides of the equation
which would give us this equation here
and because this is a linear first order
differential equation we want to write this
in standard form so we'll divide
both sides of the equation by m.
Notice how this first term would be dv/dt,
the second term would be k divided by m times v
then on the right side the m would simplify to one
leaving us with g.
So the differential equation that we're going to use
to solve our problem is dv/dt plus k divided by m
times v equals g.
It's important to recognize this only works
when air resistance is proportional to the velocity.
Now let's take a look at our example.
A steel ball weighing one pound is dropped
from 2500 feet with no velocity.
As it falls air resistance is equal to v divided by eight
or 1/8 times v in pounds
where v is the velocity of the ball
in feet per second.
We want to find the limiting velocity
and the time it takes for the ball to hit the ground.
So to set up our differential equation
we want to find k, m, and g.
Let's start with g which is
the acceleration due to gravity.
Because in this problem we're dealing with feet
the acceleration due to gravity
is going to be 32 feet per second squared
or g equals 32.
Again we're using a positive value here
because we're assuming downward
is the positive direction.
Now we need to find k and v.
Let's talk about k which relates
the air resistance to the velocity.
We're told that the air resistance
is v divided by eight pounds or 1/8 times v.
That means k is equal to 1/8
and then for the mass we need to use
the equation w equals m times g.
We're told the weight is one pound
so w is one.
We don't know the mass
but we know g is equal to 32, so we have 32m.
Divide both sides by 32.
We have m is equal to one over 32 slugs
or just one over 32.
This gives us all the information we need
to write our differential equation.
We have dv/dt
plus this would be 1/8 over one over 32
times v must equal positive 32.
Well 1/8 divided by one over 32 is going to be four
so we have dv/dt plus 4v equals 32.
Again this is a linear first order
differential equation so we're going to
solve this using integrating factor.
Let's go ahead and do that on the next slide.
For integrating factor you have t
is going to be equal to e raised to
the power of the integral of the function
being multiplied by v or in this case
just the constant 4dt.
So the integrating factor here
is going to be e to the power of 4t.
Now we're going to multiply everything in this
equation here by our integrating factor.
Now when using integrating factor to
solve this linear first order differential equation
the left side of this equation or this side here
is always equal to the derivative of the product
of the integrating factor and our function, v.
So if the left side is equal to this derivative
this derivative must also be equal to
the right side, or 32e to the 4t
and now we can solve this by integrating
both sides of the equation with respects to t.
On the left side the integral and derivative undo each other
so we're left with e to the 4t times v must equal
here we do have to perform a u substitution
where u is going to be equal to 4t
so differential u is equal to 4dt.
Dividing both sides by four.
We write this in terms of u we'll have
an extra factor of 1/4 so we're going to have
32 divided by four times e to the 4t plus c.
Now we'll solve this equation for v
by dividing everything by e to the 4t.
So this simplifies to v, or v of t.
This simplifies to one and 32 divided by four
is eight, so eight plus we'll write this as
c times e to the negative 4t power.
Now we need to find our constant c.
To do this we know that v of zero is
equal to zero.
The initial velocity is equal to zero
so that means that eight plus c times
this would be e to the zero or one
must equal zero, therefore c is equal to
negative eight so our velocity function is
v of t equals eight minus eight times
e to the negative 4t.
So to find our limiting velocity
we need to find the limit of this function
as t approaches infinity.
Let's go ahead and do that on the next slide.
To do this it might be helpful to
rewrite this second term as a fraction.
So we would have eight minus eight
all over e to the power of 4t.
In this form we can see the second term here
is going to approach zero
because the denominator is decreasing
without bound as t approaches infinity.
Therefore this limit is just equal to eight
which means our limiting velocity
is equal to eight feet per second.
That's the fastest this object will fall
until it hits the ground, due to the air resistance.
Now for the next part
we're asked to determine when
the object is going to hit the ground.
To do that we want to find the position function.
Remember the velocity function or v of t
is the derivative of the position function
which means if we integrate the velocity function
with respects to t we will be able to
find the position function.
We'll have s of t equals the integral of
eight minus eight times e raised to
the power of negative 4t.
Notice again, integrate the second term
we have to perform a u substitution
where u is going to be equal to negative 4t
so differential u is going to be negative 4dt.
Divide both sides by negative four
so if we're performing this substitution
we'll have an extra factor of negative 1/4.
So s of t is going to be equal to 8t minus 8 times
negative 1/4, e to the negative 4t plus c.
So simplifying here we'll have s of t equals
8t, it's going to be plus two
times e to the negative 4t plus c.
Now before we find the constant of integration
let's go back to our diagram just for a moment.
We know this is 2500 feet above the ground.
We're actually going to let this height here
be zero and let the ground be 2500 feet
so the object starts here and then when it
hits the ground it will travel 2500 feet.
So we'll let s of zero equal zero
so this would give us eight times zero, that's zero
plus two times e to the zero which is one
plus c equals zero
so we know c must equal negative two
so our position function s of t is equal to
8t plus two times e to the negative 4t minus two.
So to determine how long it's going to take
for this object to hit the ground
we're going to set our position function
equal to 2500 and then solve for t.
If we want to solve the equation 8t
plus two times e to the negative 4t minus two
equals 2500.
Now solving this equation
would not be easy.
It would actually take trial and error by hand.
What we're going to do is solve this graphically.
We'll graph the left side of the equation as
let's say Y1 and we'll graph the right side
as let's say Y2.
So if we graph this, the blue graph
would be the graph of the left side of
the equation or Y1, and the graph of
the right side of the constant
would be the red graph or Y2.
So this point of intersection represents
our solution, because we're looking for
a time that's going to be the x coordinate
so it's going to take approximately
312.75 seconds for this object to hit the ground.
Okay, we'll go ahead and stop here for this video.
I hope you found this helpful.