Good morning. Up to this point we have only worked with definite integrals. Today we introduce the
indefinite integral. A definite integral has defined limits. An indefinite integral does
NOT have defined limits. ♪ Flipping Physics ♪
Is that it? That seems…
Too easy. Yeah.
Right. Well. That’s the basic difference, however, in practice it’s a bit more complicated.
Yea! Yeah. Yep. Let’s start with the derivative definition of
acceleration. Bobby, please rearrange the derivative definition of acceleration
to form an antiderivative or an integral. Uh. Acceleration equals the derivative of velocity
with respect to time. Multiply both sides by dt and we get dv equals acceleration times dt. Take
the integral of both sides and I guess we do not add the limits because we are doing an indefinite
integral and you said indefinite integrals do not have limits. Taking the integral gives us
velocity equals acceleration times time. And this integral assumes the acceleration is constant
and does not change as a function of time. Thank you Bobby, however, I will point out that,
because this is an indefinite integral, what you have said here is not
quite correct. On the left-hand side, we do get velocity, however, notice that velocity
changes as a function of time, so this is velocity as a function of time. Also,
whenever we take an indefinite integral, or an integral without limits, we need to add a constant
of integration or capital C to the equation. What is a constant of integration?
Yeah, what is capital C? Good question. In order to figure that out,
let’s determine the value of the velocity function at the initial time or time equals
zero seconds. Billy, please do that. To determine the value of the velocity
function at time equals zero seconds, we plug zero in for time. When we do that we get,
the velocity at zero seconds equals well, just the constant of integration, capital C. Oh,
that means the constant of integration equals the initial velocity of the function.
In other words, the velocity as a function of time equals acceleration times time plus velocity
initial. And we can rearrange that slightly to get velocity final equals velocity initial plus
acceleration times time. That is one of our uniformly
acceleration motion equations. Oh yeah! Velocity final equals
velocity initial plus acceleration times time is a kinematic equation!
Kinematic equation? The uniformly accelerated
motion equations and kinematic equations are two different names for the same things.
Oh. How did velocity
as a function of time become velocity final? Well, when you plug a time into the velocity
as a function of time equation, you get a velocity at that time,
which is exactly what velocity final is. Oh right. The initial time is zero,
the time final is time t. The initial velocity is at time equals zero and the final velocity is
at time equals t. That makes sense. Thanks. You’re welcome.
But why did we not add a constant of integration to the left-hand side of the equation?
Bo, thanks for helping there. Billy, we certainly could do that. Let’s look at
what happens when we add constants of integration to both sides. Constant 1
on the left side of the equation and Constant 2 to the right side of the equation. We can subtract
Constant 1 from the whole equation and we get Constant 2 minus Constant 1 on the right side.
Oh, I get it. The quantity Constant 2 minus Constant 1 is still just a constant. So,
it does not matter. Adding a single constant C on the right side is equivalent
to adding constants to both sides. Thanks. Thanks for the question. Alright, let’s get
back to the equation we derived, velocity final equals velocity initial plus acceleration times
time. Bo, please substitute the derivative definition of velocity in for final velocity
and rearrange the equation to solve for another one of our uniformly accelerated motion equations.
Sure. Velocity equals the derivative of position with respect to time. Multiply both sides by dt
and take the integral of both sides. Do we do a definite or indefinite integral?
Let’s do an indefinite integral again. Okay. Indefinite integral, so no limits.
The integral of dx is just x but it is position as a function of time.
And that equals well, we are taking the integral with respect to time, so
acceleration times time squared divided by 2 plus uh is it velocity initial squared over 2?
No. Velocity initial is a constant. So, the integral of velocity initial with
respect to time is velocity initial times time. Oh right. And, plus the constant of integration
C. And we need to solve for C. So, we plug in zero for the initial time.
And we get that the constant equals the initial position of the function. Ah, yeah.
We get position final equals position initial plus velocity initial times time plus one half
acceleration times time squared. Which is one of our uniformly accelerated motion equations. Nice.
Correct Bo. And notice the constant of integration, C, for both of the equations
we just solved for equals the value of the function when time equals zero.
The constant of integration is the initial value of the function.
Right. It is the value of the function when time equals zero. Now, let’s go back to the
derivative definition of acceleration and approach it a bit differently. Hopefully
you can see that the derivative of velocity as a function of time is equivalent to the derivative
of velocity with respect to position times the derivative of position with respect to time.
I don’t see it. The dx’s cancel out.
Woah, you can do that? Yep. It’s the chain rule.
The chain rule? Yeah. The derivative of y with respect to x
equals the derivative of y with respect to u times the derivative of u with respect to x.
Cool. Yep.
And the derivative of position with respect to time equals velocity. So, now,
Billy, please rearrange this and use a definite integral to solve for another kinematic equation.
Absolutely. That means we can multiply both sides by dx and get the definite integral of
acceleration with respect to position from position initial to position final equals
the definite integral of velocity with respect to velocity from velocity initial to velocity final.
The integral of acceleration with respect to position equals well, acceleration is a constant
so that integral is just acceleration times position and add the limits.
Substituting in the limits gives us acceleration times position final minus acceleration times
position initial or acceleration times change in position.
The integral of velocity with respect to velocity equals velocity squared over 2 and add the limits.
Substituting in the limits gives us velocity final squared over 2 minus velocity initial
squared over 2. And Oh, I see it! We can multiply the whole equation by 2 and reverse the sides,
and then add velocity initial squared to both sides and we get velocity final squared equals
velocity initial squared plus 2 times acceleration times displacement. And that is a third uniformly
accelerated motion or kinematic equation! Why did we use a definite integral
and not an indefinite integral this time? Thank you Billy. Bo, the reason we used a definite
integral this time and not an indefinite integral is because, in this case, the indefinite integral
solution is a bit more cumbersome. I have included that solution in my lecture notes on my website
if you would like to see it. Now, there is one more uniformly accelerated motion equation we can
solve for. We can use two of the equations we just derived to derive a fourth UAM equation. Starting
with velocity final equals velocity initial plus acceleration times time, we can rearrange that
equation to solve for acceleration. Then we can substitute that equation for acceleration into
the equation position final equals position initial times velocity initial times time
plus one half acceleration times time squared and subtract position initial from both sides.
Time squared over time reduces to just time. And we can factor out time.
Velocity initial minus velocity initial over 2 equals positive velocity initial over 2.
Factoring out one half gives us the fourth, and sadly, often ignored, kinematic equation,
displacement equals one half times the quantity velocity initial plus velocity final
all times time. (mr.p?) Actually,
doesn’t that just equal the average velocity times time which we know equals displacement?
Why yes, Billy, it does equals average velocity times time. I thought it would be more fun to
derive the equation this way. Oh kay.
Uh. I thought we were going to use integrals to solve for all four UAM equations.
Sure. How about y’all do that? Use the fact that uniformly accelerated motion
means the acceleration is constant, displacement equals the definite integral of velocity with
respect to time from time initial to time final, and this example graph
of velocity as a function of time for uniformly accelerated motion.
Uh … I agree.
We know one definition of an integral is that it is the area "under" a function, right?
So? Well, we could pick
a time t and determine the area between the velocity function and the time axis.
Okay. That would be the red area in the graph. We could split that into two different areas,
area 1 and area 2. And the total area would be the sum of those two areas.
Area 2 is a triangle and has an area equal to one half base times height.
Using those variables, area 1 is a rectangle of area equal to base times height. And the base
of both areas has a length equal to just t. The height of area 1 is velocity initial.
Oh, and the height of area 2 is velocity final minus velocity initial!
And that is the same equation we had halfway through the last solution,
so this also gives us the same uniformly accelerated motion equation. Nice.
Yeah. Actually, we could have just used the equation for the area of a trapezoid, right?
What is the area of a trapezoid equation? Area of a trapezoid equals one half times the
quantity base 1 plus base 2 all times height. I’m not seeing it.
The trapezoid is rotated by 90 degrees. Oh. I see it. base 1 is velocity initial,
base 2 is velocity final, and the height is the time. Nice.
Yep. It does work out to give you the fourth uniformly accelerated motion equation.
But what if the velocity function is not a straight line?
Yeah? These equations are for
uniformly accelerated motion. Oh!
and a constant acceleration is a constant slope on a velocity as a function of time graph.
So, the velocity function for uniformly accelerated motion is always a straight line.
Yep. Very nice everybody.
Realize we just used the definitions of velocity and acceleration, and calculus to derive four
uniformly accelerated motion equations, which are also called kinematic equations.
I think that is awesome. (Absolutely! Yeah. Yep.)
Thank you very much for learning with me today, I enjoyed learning with you.