| Start | One last topic to cover
And then because it's a nice day, when I feel I've covered this sufficiently, I'll stop a little early. I know how sad that makes you.
Those of you who applauded, thank you. Alright, so we really only have one more thing to do. This will be on the final. And it's antiderivatives. That doesn't mean people who hate derivatives, it means going backwards. Ok? |
| 0:31 | And some of you are leaving early.
That's how long my first marriage lasted. Just kidding. That was, the 7th marriage lasted that long. Antiderivatives, antiderivatives, going backwards. So in other words, we give you the derivative. What's the original function? So if I tell you for example |
| 1:02 | If the derivative of something is x^2
What's the original function?
Well, it would be the original function Would be (x^3)/3 Or (1/3)x^3 Why would it be that? Well, take the derivative of this. And you get that. So this is the antiderivative. |
| 1:34 | Okay it's working backwards. How do I know that? Well it gives you that in a second.
Here's one of the problems though. What's the derivative of x cubed over 3 plus 1? It's x squared, what about X cubed over 3 plus 2? It's x squared. So the problem that you get when you're going backwards is, you get the derivative if the constant's 0. |
| 2:02 | There's an infinite number of possibilities from what the original function was.
Ok, so what we do is we write x cubed plus c Where c stands for a constant. We don't know what that constant is, so we just write plus c. Now if we give you more information you can often figure out what c is. So suppose I told you |
| 2:31 | If f(x)
=x^2. And, I'm sorry f'(x)=x^2
And f(0)=4.
And now we say go backwards and figure it out. Say well, I know the original function And that's the big x^3 /3 Plus a constant. And I know ok, when x is 0, y is poor. |
| 3:03 | So 4=0^3 over 3 plus the constant.
The constant must be 4. So now I can say then the real function is x^3 /3 plus 4. Ok, so sometimes I can find the actual function. Sometimes, I can only find the general function. So this is called an indefinite. Ok, and this is definite. |
| 3:31 | Well, not quite. But for now.
Ok? Sometimes you can find the actual value. Sometimes you can't. And how will you know which one? Well, we'll give you the second piece of information. All right, so now let's just take any f(x) f'(x) and say it's x^n. So, if the derivative of a function is x^n What was the original function? Well |
| 4:01 | It's x^(n+1) over n+1, plus some constant.
Take that in your head, bring the n+1 down It's gonna cancel with that. Subtract one and you get x^n. Ok? So notice when we take the derivative Normally, when we take a derivative, what do you do? You bring the power down, so you multiply. And you subtract 1 from the power. So now here we're going backwards. So you add 1 to the power and you divide. Ok, so you're doing the reverse functions, inverses. |
| 4:33 | Ok, and that's how you know you're doing inverse, you're doing derivative and antiderivative.
If I tell you So if you know that the antiderivative of- if you know that the derivative of a function is x^4 + x^3 + x^2 What's the original function? Well, now you're going backwards. |
| 5:01 | You say well, its x^5 /5.
Plus x^4 /4 Plus x^3 /3 Plus the constant. We got that. Muy facil. Si. So that's not very hard, but of course we're gonna make this harder. But let's write that over here on our list of antiderivatives. |
| 5:37 | So one of the ways we write antiderivatives is, you write the derivative as lowercase f,
and the antiderivative as uppercase.
|
| 6:05 | You guys ready for the Roth Pond Regatta on Friday?
It'll be fun. Some of you are gonna get wet. Some of you are gonna get wet. Ok, you know who you are. You get that green stuff all over you. Alright so that was an easy thing. So let's do some more antiderivative stuff. |
| 6:34 | Let's see
What's the derivative of 3x
You got f(x)
Equals 3x, take the derivative
Is 3.
In fact, it's f(x) as a constant. times X The derivative is just a constant. So now let's go the other way. |
| 7:01 | If I know that the function, a function is equal to a constant
then the antiderivative
is kx+c
You always have to write this plus c because it says I don't actually know what the original function is.
All I know is if I take the derivative of this I'll get back to here, but the constant could be anything. Cause if we would have a constant that's always 0. The constants number like 2, or 1/2, or pi. |
| 7:33 | Or pi^10
Ok, or log of 7.
Those are all constants, ok? Alright, now let's do some other fun ones. |
| 8:01 | If f(x)=e^x what's the antiderivative?
Almost Plus c. The TA's will take off if they don't see the plus c. Ok? So far so good? |
| 8:30 | What's the derivative of the log of x?
f(x)=ln(x) inverse and the antiderivative is 1/x +c. So let's add that to our list. I wrote that backwards, hang on. |
| 9:04 | Okay, basically the derivative of this is this.
So the antiderivative of this, is that. Ok? Really all you're doing is just going backwards. Alright, what else do we have to learn the antiderivative of? Oh Some trig functions. You never escape trigonometry folks. Never. You're gonna be there studying for your MCATs |
| 9:30 | We're gonna ask you trig. You're gonna be in med school.
And you're gonna say what do I do? And they're gonna say oh, patient me the vital sines. Just had to get that one in there. I went off on a tangent there. You know, I can keep going. Alright. If you have a function, what's the derivative of sine? Cosine. So the antiderivative of sine Is negative cosine. |
| 10:03 | Because, take the derivative of this.
You get that. Ok? So the antiderivative of sine we'll make another column. |
| 10:54 | Ok?
You ready to try some? |
| 11:10 | Ok, find the antiderivative.
If you start with that function, what's the antiderivative? |
| 11:41 | I'll give you a clue. You might want to multiply that out first.
I'm just copying the problems in the book, hold on. I didn't write this book. I'd be wealthy if I wrote this book. |
| 12:05 | The guy who wrote this book is probably a millionaire
You get a lot of money from books.
|
| 12:40 | The c at the end.
The whole thing, plus c. That you know, specifically. If you have x, it has to start with x^2 /2. If you have 3x, then you have 3x^2. |
| 13:05 | Write it down, take the derivative.
That's long enough. |
| 13:33 | Long enough?
Alright. So what would we do if we wanna find the antiderivative of this? Well first, I would multiply it out. And get 2x^2 Plus x-1. And now we get the antiderivative 2x^3 /3 plus x^2 /2 minus x Plus a constant. See, if you're not sure |
| 14:00 | Take the derivative, go backwards.
How do I know that the antiderivative of 1 is x? Because the derivative of x is 1. 1 is constant, right? If you have 1 as a constant The antiderivative is a constant times x. So get 1, and the antiderivative is 1x. Ok, let's do another one. Make sure you guys can do this, this is gonna be on the final. |
| 14:38 | You have 5x^(1/4) - 7x^(3/4).
|
| 16:23 | Alright.
Let's do this one. So this is the annoying part, is you have to add 1 to 1/4 and 1 to 3/4. |
| 16:33 | So, we have 5x^(1/4), 1/4 + 1 is 5/4.
Divided by 5/4. Minus 7x^(7/4) Divided by 7/4. Let's see. Now, of course, you should be able to simplify that. You flip the fraction. |
| 17:10 | You flip the fractions and multiply, and you get
4x^(5/4) minus 4x^(7/4)
Plus c. But this is perfectly fine.
Ok, you would want to turn it into this if you needed to do something else. |
| 17:31 | Ok?
What if you had What if you had f= 3e^x + 7sec^2 x? |
| 18:08 | Well, derivative of what is sec^2?
Tangent. So look at that. |
| 18:34 | So if we wanted to do the antiderivative of this,
it's 3e^x + 7tanx
plus c.
When you're not sure, take the derivative of the other direction. Okay, let's try something slightly different. |
| 19:25 | Okay, now let's suppose I tell you |
| 19:30 | that the derivative is 5x^4 - 2x^5, and that the original function at the value 0 equals 4?
That's gonna help you figure out what c is. |
| 20:01 | You good?
|
| 20:54 | You figure this one out?
|
| 21:11 | Alright
So, we do the antiderivative.
You get 5x^5 over 5 Minus 2x^6 /6 plus c And I know that that f(0) equals 4 |
| 21:31 | So 4 is, you cross out those 5's. Is 0-0+c.
so c=4. Therefore, the antiderivative is x^5 -2x^6 /6 which you certainly could reduce plus 4. So this information helps us find the exact function. |
| 22:01 | Suppose I had |
| 25:20 | Alright, so antiderivative I have capital F(x) 3x + (5x^2)/2 - (2x^3)/3 |
| 25:34 | plus a constant
Okay, now we know that F(1) is 2
So 2=3(1) + 5(1/2)
Well, that's a fairly sophisticated fraction but I think we can do that.
So you get 2=3+(11/6) |
| 26:01 | plus c.
So you get -17/6 Is that right? Ah, hang in there for one more day folks. You know how I do that? I'm old, we didn't have a calculator growing up. We just discovered fire, you don't think we had electricity back then, do you? |
| 26:35 | Okay, so that's your original function.
Here's something else we could ask you. Suppose the second derivative of x is |
| 27:03 | 6x + 12x^2
Find the original function, you knew this was too easy.
Alright, well the first derivative would be 6x^2 over 2 plus 12x^3 over 3 plus a constant Which we can simplify to 3x^2 +4x^3. |
| 27:30 | Plus some c. So that's the first derivative
So now we just have to do it a second time.
So that means the original function is 3x^3 over 3 plus 4x^4 over 4. Plus some constant times x, plus another constant, we'll call that c1. So you'd probably have to solve information twice to get to the constants. |
| 28:02 | We have to give you information about the derivatives and about the first function.
And of course you can simplify that Where would that show up? Oh, physics. Just pick a topic at random. |
| 28:41 | Suppose we had this.
|
| 29:03 | Why do I have the c and the c1? They're not the same constant.
Okay, so when you do that you have two different constants. So, you have constant c and you have constant c1. You can call them A and B. You could call them velocity and displacement So, the second derivative is 2+x+x^2 And I know that the first derivative is equal to 4 at 0 and the original function, sorry that should be an F |
| 29:35 | Is equal to 10 at 0, now I can solve this.
Say well the first derivative must've been 2x plus x^2 over 2 plus x^3 over 3 plus c. And I know when x is 0 this was equal to 4 So 4 equals 0+0+0+c |
| 30:02 | The original derivative, the 1st derivative
is 2x
plus x^2 over 2
plus x^3 over 3
plus 4.
And now find the original function. So, thats gonna be 2x^2 over 2 + x^3 over 3(2) which is 6. |
| 30:40 | Plus x^4 over 3(4) which is 12.
Plus 4x plus c. Or I'll call it c1. But you can call it c, yeah, just don't confuse yourself. And I know that this time, when I plug in 0 I get 10. So 10 |
| 31:01 | equals 0 plus 0
plus 0 plus 0 plus c.
So c equals 10. Well c1. So my original function must have been x^2 + x^3 over 6 plus x^4 over 12 plus 4x plus 10. |
| 31:30 | And if you get a problem like this, check. It's very easy to do. Take the derivative, take the derivative again.
How'd I get that 6? I had x^2 over 2. I raised the power to get x^3, x^2 over 2. |
| 32:02 | I get x^3 over 2.
But I'm straight. Because this is x^3 over 3. Ok? So the 2(3) becomes 6. Ok? Alright, we've got a couple others to try. We've got one more thing to learn. |
| 34:30 | What's the first derivative gonna be?
Well, the antiderivative of sine is -cosine. So here's something that people will mess up on the final. Ok? The derivative of sine is +cosine, the antiderivative of sine is -cosine. So you have to remember whether you're doing the derivative or the antiderivative. You're gonna get the minus sign wrong, ok? Very easy to do. |
| 35:01 | The antiderivative of cosine is sine.
Just check. The derivative of sine is cosine so the antiderivative of cosine is sine. Plus a constant. And now I know f'(0) is 1. So 1=-cos(0)+sin(0) Plus c. Sin(0) is 0 and cos(0) is 1. So c=2. |
| 35:31 | Alright?
Now I gotta get my original function. Well f(x) is -sinx-cosx+2x plus c. Check it, the derivative of sine is cosine, the derivative of -sine, -cosine. The derivative of cosine is -sine, the derivative of -cosine is +sine. Then I have, c came out 2 so, the derivative of 2x is 2. |
| 36:02 | Plus another constant.
And f(0) is 1, so 1 is -sin(0)-cos(0) plus 0 plus c. Sin(0) is 0 and cos(0) is 1 So you get c is 2 again. So your function is F(x)= -sinx-cosx+2x+2. |
| 36:42 | Where does this show up?
This shows up in calculus. I'm sorry, in physics. Using calculus. Ok, something called rectilinear motion, motion in a line. Essentially, it shows up when you're doing velocity and acceleration, position, velocity and acceleration. Have any of you taken physics yet? |
| 37:04 | Some of you, physics in high school?
If you know the position of an object, the derivative of that position gives you velocity and the derivative of velocity gives you acceleration. And, going backwards, you know the acceleration, you can find the velocity. And if you know the velocity you can find the position. So let's write that down someplace clever. |
| 37:46 | So if you take the derivative, usually with respect to time with position, you get velocity And if you take the derivative of velocity |
| 38:00 | you get acceleration.
For those of you who don't know what this means, velocity is how fast something is moving and what direction it's moving. Ok? Kind of like speed, but the direction also matters. So, you can go 50 miles an hour east, you can go 50 miles an hour west. Ok? That's velocity. Actually, yeah that's east and that's west. Whatever. Ok, um |
| 38:31 | So, if you have your location, how fast your location is changing with respect to time gives you velocity.
How fast your velocity is changing is your acceleration. That's why when you step on the accelerator, also known as the gas pedal When you step on the accelerator, you accelerate and you go faster. Ok, so acceleration is the change in velocity. So if we tell you how fast an object is accelerating you can go backwards and figure out some stuff about its location. |
| 39:04 | For example A particle moves with the acceleration 5+4t-2t^2 |
| 39:35 | And we know its initial velocity, 3 meters/second.
And its initial position is 10 meters. Ok, find the formula for its position with respect to time. This looks really scary but it's just a calculus problem. |
| 40:02 | So how do we do this?
This is just what we've been doing. Acceleration is the second derivative. So we're gonna do the antiderivative to get velocity and antiderivative again to get position. So, if we know that the acceleration is 5+4t-2t^2 then the velocity is the antiderivative of that. Which is 5t+ 4t^2 over 2 - 2t^3 over 3 + a constant. |
| 40:44 | And we know that the velocity at time 0 was 3 meters/second
So 3=0+0+0+c.
The constant's 3 so the velocity was 5t + 2t^2 -2t^3 over 3 +3. |
| 41:05 | So I'm gonna rewrite that up here.
The velocity is 5t + 2t^2 -2t^3 over 3 + a constant. So now I wanna find the position. You use s for position. I don't know why either. s(t) is 5t^2 over 2 + 2t^3 over 3 - 2t^4 over 3(4) is 12 |
| 41:47 | Oh, I'm sorry, I knew what c was. c was 3.
Plus 3t plus a constant |
| 42:00 | And I know that position 0 is 10.
10, when you plug in 0 these terms are all 0 so it's just c=10. So your position function 5t^2 over 2 + 2t^3 over 3 - t^4 over 6 + 3t +10. How'd we do? You get that one? |
| 42:47 | Huh, I don't know if we're gonna actually ask about that. But maybe.
Ok, if we throw objects. If you throw them up and you drop them you need to know what the acceleration is automatically. |
| 43:06 | I can't believe we're doing this, but
I can't believe I have to tell you this, but, if you have a problem that involves gravity, like if I throw something up or if I drop it
ok, then the acceleration is either -9.8 meters/second^2 or -32 feet/second^2.
|
| 43:40 | Ok, so for a sample problem
When you take physics, you do a problem, you do a lab called Atwood's machine. You come up with that number.
|
| 44:04 | Kind of a nice experiment.
Ok, a ball is thrown upwards with a speed of 48 feet/second, 48 meters. Ok, a ball is thrown upward from a cliff with a speed of 48 feet/second. |
| 44:31 | Ok, so you're tossing it up.
And the cliff is 432 feet above the ground. So you've got a cliff, and you toss the ball up. And this is 432 feet. So, that means your initial position is 432 and your initial velocity is 48. Ok? |
| 45:02 | What time does it hit the ground?
Well, it's time for physics class. I know. But, here we go. The acceleration is -32. So, if we wanna find the velocity, you just take the antiderivative. Say well the velocity is -32t + a constant. And I know that the initial velocity is 48. |
| 45:33 | So 48=-32(0)+c
So c is 48.
For the initial velocity it's -32t+48. Now, we're gonna find the initial position. |
| 46:02 | I take the antiderivative of that.
That's (-32t^2)/2 + 48t + a constant. And the initial position is 432. When I plug in 0 I'm gonna get 0 and 0, so c is gonna come out 432. Everybody starting to see that pattern? Ok, so the position is -15t^2 + 48t +432. |
| 46:35 | So when does it hit the ground? What's the height when it hits the ground?
Sure. How do you know? Well, you plug here. Zero. Ok, that was your physics experiment for the day. So, we wanna know when s is 0. So you set it equal to 0. I bet they're all divisible by 16, what do you think? I think they're all divisible by -16. You get t^2 - 3t - I'm sorry, 27. |
| 47:12 | Ok, that factors quite nicely.
No it doesn't. That doesn't factor nicely. So we have to do the quadratic formula. |
| 47:46 | Ok?
It can't be 3 - sqrt(117) over 2, that's a negative number. So the time it hits the ground has to be 3 + sqrt(117) over 2 seconds. |
| 48:03 | If you take the time to make 432, then you should take the time to make it factorable.
I don't think that's that much to ask, right Mr. Stewart? I would've made it factorable. It's my punishment for reading a problem from the book. Ok, let's try another one of these. We're not gonna do too many more of these, I promise. |
| 48:41 | I'm sorry.
Really? Oh yeah. Well we'll do another problem. Ok. |
| 49:05 | We'll have you guys work one out.
|
| 50:30 | Ok.
I told you, we'll stop early. I know it's a nice day and I know you guys have the exam at 5:30. A ball is thrown upward from the roof of a 128 foot tall building with an initial velocity of 32 feet/second. When does it hit the ground? So this is just like the last one. Should we do it as a team or should I have you guys try it first? Who wants to try it? |
| 51:00 | Wow, one person.
Wow, I think we need to do it as a team then. I'll go slowly. I'll go really slowly. Alright So you got the roof of the building, and you throw the ball and it goes like that. And that building is 128 feet tall. And we know the acceleration is what? What's the acceleration? |
| 51:30 | Feet.
Ok, so we start with the acceleration of -32. Good, and I know that the initial velocity is also positive 32. And the initial position is 128. So what do I do first? We're doing it as a team. So first, I have to do the antiderivative. What do I get? You said you wanna do it by yourself, so what do I do first? |
| 52:02 | Ah ok, we'll do it as a team with me being in charge.
Alright, so the velocity is -32t because the derivative of 32t is 32. So -32t is -32. Plus a constant. That constant is gonna turn out to be 32. Ok, because we're gonna plug in 0 and get 32= -32(0)+c. c is 32 |
| 52:31 | So the velocity is now -32t + 32.
Alrighty, now let's find the position. So you do this again. I do the antiderivative of 32t which is -32t^2 /2 Plus 32t + a constant. |
| 53:02 | This becomes, with a little simplifying, -16t^2 + 32t +128.
Where'd the 128 come from? Because when t is 0, 0 and 0 these go away. And you get 128 equals c. Alright, now I wanna set, find when that is equal to 0. So I take -16t^2 + 32t + 128 = 0. |
| 53:36 | Divide through by -16.
Which we don't have to, but it makes life easier. You get t^2 - 2t - 8 = 0. Does that factor? Minus 8. Could you factor by now? Yes it does. That wasn't so hard, Mr. Stewart. |
| 54:03 | I don't understand what his problem was.
So you get t equals 4 seconds, that's when it hits the ground. Or t equals -2 seconds. It can't be -2, -2 seconds would be going back in time. Now one other thing we could ask you, we could ask you what's the maximum height? How high does it get? So you would find the maximum height. What happens when you throw something in the air? At the top of its height the velocity is 0. |
| 54:34 | Right, go up, stop, back down.
So we would put the first derivative and set it equal to 0. Does that make sense? Because the velocity would be 0 at the top of its arc. So the first derivative is gonna equal 0 at t=1. You should be able to do that one in your head. Ok, and then if you take 1 and you put it in the position equation you know how high it got. |
| 55:02 | Ok, again, if you wanna know the maximum height you take the velocity and you find where its 0
That'll give you the time that its at the maximum height. Then you take that time and you plug it in position and you get the height.
You guys all look very restless. So, we'll stop And I'll see everybody on Monday. Have fun on Friday at the Regatta. |