For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

### Antiderivatives

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Antiderivatives
- Function & Antiderivative
- Example I: Find the Most General Antiderivative for the Following Functions
- Function 1: f(x) = x³ -6x² + 11x - 9
- Function 2: f(x) = 14√(x) - 27 4√x
- Function 3: (fx) = cos x - 14 sinx
- Function 4: f(x) = (x⁵+2√x )/( x^(4/3) )
- Function 5: f(x) = (3e^x) - 2/(1+x²)
- Example II: Given the Following, Find the Original Function f(x)
- Function 1: f'(x) = 5x³ - 14x + 24, f(2) = 40
- Function 2: f'(x) 3 sinx + sec²x, f(π/6) = 5
- Function 3: f''(x) = 8x - cos x, f(1.5) = 12.7, f'(1.5) = 4.2
- Function 4: f''(x) = 5/(√x), f(2) 15, f'(2) = 7
- Example III: Falling Object

- Intro 0:00
- Antiderivatives 0:23
- Definition of an Antiderivative
- Antiderivative Theorem
- Function & Antiderivative 12:10
- x^n
- 1/x
- e^x
- cos x
- sin x
- sec² x
- secxtanx
- 1/√(1-x²)
- 1/(1+x²)
- -1/√(1-x²)
- Example I: Find the Most General Antiderivative for the Following Functions 15:07
- Function 1: f(x) = x³ -6x² + 11x - 9
- Function 2: f(x) = 14√(x) - 27 4√x
- Function 3: (fx) = cos x - 14 sinx
- Function 4: f(x) = (x⁵+2√x )/( x^(4/3) )
- Function 5: f(x) = (3e^x) - 2/(1+x²)
- Example II: Given the Following, Find the Original Function f(x) 26:37
- Function 1: f'(x) = 5x³ - 14x + 24, f(2) = 40
- Function 2: f'(x) 3 sinx + sec²x, f(π/6) = 5
- Function 3: f''(x) = 8x - cos x, f(1.5) = 12.7, f'(1.5) = 4.2
- Function 4: f''(x) = 5/(√x), f(2) 15, f'(2) = 7
- Example III: Falling Object 41:58
- Problem 1: Find an Equation for the Height of the Ball after t Seconds
- Problem 2: How Long Will It Take for the Ball to Strike the Ground?
- Problem 3: What is the Velocity of the Ball as it Hits the Ground?
- Problem 4: Initial Velocity of 6 m/s, How Long Does It Take to Reach the Ground?

### AP Calculus AB Online Prep Course

### Transcription: Antiderivatives

*Hello, and welcome back to www.educator.com, welcome back to AP Calculus.*0000

*Today, we are going to start our discussion of anti-derivatives.*0005

*In a couple of lessons, we are actually going to change the name of that and start calling them integrals.*0009

*This is the second half of calculus.*0013

*The first part is differential calculus, now it is integral calculus.*0015

*As you will see in a minute, it is actually the inverse process.*0018

*Let us jump right on in.*0022

*Up to now, we started with functions and we took derivatives of them.*0026

*Let us go ahead and write this.*0032

*Up to now, we started with some function f(x) and we found its derivative.*0035

*We have found its derivative, in other words, we differentiate it.*0053

*For example, if we had a sin x, we take the derivative of it and we ended up with cos x.*0062

*If we had, let us say, an x³, we take the derivative of it and we end up with something like 3x².*0072

*What if we begin with a function then ask is this the derivative of some function?*0081

*In other words, instead of starting with the function and going to the derivative,*0089

*let us say that this is a function and this the derivative of some previous function.*0093

*That is the question we are going to look at here.*0099

*What if we begin with a function f(x), then ask is f the derivative of some f(x).*0100

*Can I find this F, can I find this other f?*0139

*In other words, we are going to be giving you the cos x for the 3x².*0147

*And we are going to say, how do you recover the sin x?*0151

*How do you recover the x³, if that original function actually exists?*0153

*The answer is yes, fortunately.*0158

*Let us go ahead and start with a definition.*0167

*You know what, I think I will go ahead and put my definition.*0170

*The different color, I will go ahead and put in red.*0175

*The definition F(x) is called an anti-derivative, exactly what it sounds like,*0180

*it is just the reverse of the derivative, an anti-derivative of f(x).*0194

*If this f(x) happens to be the derivative of the F(x), we will just mark that with an f’, on a specified interval.*0202

*Let us go back to blue here, not necessary but what the heck.*0225

*All we are doing is going backward, that is it.*0232

*That is all this is.*0243

*Now the most difficult part is going to be remembering, am I going forward to differentiating*0245

*or am I going backward and taking the anti-derivative?*0250

*It is a different set of rules, it is a different set of formulas that you use to find that.*0253

*That is going to be probably the most difficult thing that you have to do is remember which direction you are going in.*0258

*If we begin with some f(x), that is the function that is given, we can go ahead and take the derivative which gives us f’.*0265

*Or we can go this way and take the anti-derivative which gives us our F(x).*0276

*Some function is going to be the function that is given to you.*0284

*There are two ways that you can go, depending on what problem that you are trying to solve.*0285

*That is what makes calculus incredibly beautiful, you start here.*0291

*As you will see in a few lessons, there is a relationship between derivative and anti-derivative.*0296

*They call it the fundamental theorem of calculus, it actually connects the two, *0300

*which as we said are inverse processes.*0304

*Let us note the following.*0308

*If we had some f(x) which is equal to sin x + 6 and we had some g(x), *0320

*let us say this is sin x + 4, these are not the same function.*0329

*If you put an x value in, you are going to get two different y values.*0335

*These are not the same function, very important to know that.*0338

*These are not the same function but you can probably see where this is going.*0343

*But f’(x), f’(sin x) + 6 is equal to cos(x).*0357

*Not f, I’m talking about g and g‘(x).*0367

*When I take the derivative of the sin x + 4, it is also going to equal cos(x).*0370

*These are the same derivative.*0378

*You get two different functions that end up going to the same derivative.*0382

*That could be a bit of an issue, but it is not, fortunately.*0387

*If we begin with h(x) = cos x, if we begin with the derivative and ask for its anti-derivative, which one do I choose?*0393

*Do is say that sin x + 6 is the anti-derivative?*0426

*Do I say sin x + 4 is the anti-derivative?*0428

*Is it sin x + 6?*0436

*Is it sin x + 4?*0444

*Or how about just sin x + c, where c can be any constant?*0447

*Because we know that the derivative of a constant = 0.*0465

*I does not matter what that number is, it can be sin x + π, because the derivative of π is 0.*0470

*Here is our theorem, let us go ahead and mark this in red.*0477

*Our theorem says, if f(x) is an anti-derivative of f(x) on an interval that we will just call I,*0483

*then f(x) + c, any other constant is the general solution.*0514

*It is the general solution to this anti-differentiation problem.*0531

*In other words, if I'm given cos(x) and if I take the anti-derivative of that,*0551

*I know that the anti-derivative is going to be sin x + something.*0555

*What is it that I’m going to choose for that something?*0560

*In general, if we are just speaking about general solutions and taking anti-derivatives, you are just going to write cos x + c.*0562

*There is going to be other information in the problem that allows you to find what c is.*0569

*sin x + 6 or sin x +, or sin x + 48, those are specific solutions, particular solutions to a particular problem where certain data is given to you.*0576

*When we speak of the general situation, we put the anti-derivative and you just stick a c right after it,*0589

*to make sure that it is formally correct.*0594

*Are you going to forget the c, yes you are going to forget the c.*0597

*I still forget the c, after all these years.*0601

*Do not worry about it but this is the general solution.*0603

*anti-derivative, without any other information, just add the constant to it.*0605

*I will say, technically, you must always include the c.*0616

*Try your best to remember it.*0628

*Once again, extra information will allow you to find what c is.*0631

*We will see that when we start doing some of the example problems.*0648

*To find what c is, extra information will allow you to find what c is, in a particular case.*0652

*Thus, giving you what we call the specific solution or often called the particular solution.*0677

*Let us do some examples.*0690

*Before we do the examples, I’m going to give you a list of the anti-derivatives that we actually already know.*0703

*Some anti-derivative formulas we already know.*0711

*I think I will go ahead and do this in red.*0730

*Here we have the function and here I'm going to put the anti-derivative.*0732

*Once again, we have to make sure that we know what direction we are going in.*0745

*If I have x ⁺n, the way to find the anti-derivative of x ⁺n is you take x ⁺n + 1/ n + 1.*0749

*In other words, if this were the function that we are given,*0761

*you know that what you do is to take the exponent, you bring it down here.*0764

*You subtract one from the exponent.*0769

* Differentiation is this way.*0770

*What we are saying is, if you are starting with a function anti-differentiation, this is the formula that you use.*0773

*Again, you will see in just a minute what we mean.*0779

*Right now, I’m just going to write down some formulas here.*0782

*1/x, the anti-derivative of that is the natlog of the absolute value of x e ⁺x.*0786

*The anti-derivative is e ⁺x.*0795

*If I have cos(x), I know the anti-derivative of that is sin(x).*0798

*If I’m given sin(x), the derivative is cos(x).*0803

*If I’m given cos(x), the anti-derivative is sin(x).*0806

*Direction is very important here.*0810

*Do not worry, you will make a thousand of mistakes, as far as direction is concerned.*0812

*You will be asked to get an anti-derivative and you end up differentiating.*0817

*That is just the process that we go through, do not worry about it.*0819

*One more page here, let us write our function and let us write our anti-derivative.*0825

*If I have a sin x and I want the anti-derivative, it is actually going to be a -cos x.*0841

*Because if I’m given –cos x, the derivative of that is sin x.*0847

*Sec² x, the anti-derivative is the tan(x).*0854

*If I'm given sec x tan x, the anti-derivative sec x.*0859

*If I'm given 1/ 1 - x², all under the radical,*0867

*the anti-derivative of that is the inv sin(x) 1/ 1 + x², that anti-derivative is the inv tan(x).*0872

*One last one, -1/ 1 - x², all under the radical and that is going to be the inv cos(x), that is the anti-derivative.*0886

* If I were given the inv cos, the derivative of that would be -1/ √1 - x². *0896

*Now let us go ahead and jump right into the examples.*0904

*Find the most general anti-derivative for the following functions.*0909

*General means just add c to your answer, that is all that means.*0912

*1, 2, 3, 4, 5, let us go ahead and jump right on in.*0917

*Let me go ahead and make sure that I have everything here.*0924

*I have copied the functions, 14 x⁵, 4/3, and 3 ⁺x.*0928

*Let us start with number 1, I think I will go back to blue here, I hope you do not mind.*0939

*Number 1, our f(x) was x³ – 6x² + 11x – 9.*0947

*We said that the formula for the anti-derivative, when you are given some x ⁺n, *0962

*the anti-derivative of that is x ⁺n + 1/ n + 1.*0968

*Add 1 to the exponent , divide by the number that you get which is now the new exponent, very simple.*0972

*Therefore, our f(x), our anti-derivative is going to be x⁴/ 4.*0979

*How much easier can this possibly be?*0987

*-6, the constant, x³/ 3, we will simplify it, just a minute.*0992

*+ 11 x² because this is 1, add 1 to it and divide by that same number, -9x.*1001

*This is x⁰, x⁰, add 1 to the exponent, it becomes 1, divide by 1, it becomes 9x.*1009

*Now we can simplify, divide where we need to.*1017

*This is perfectly valid, you do not have to take the 6 and the 3, and divide it.*1020

*You can stop there, if you want to.*1023

*It just depends on what you teacher is going to be asking for.*1025

*We have x⁴/ 4, 6/3 is 2.*1028

*It is going to be -2x³, it is going to be +11 x²/ 2 – 9x + c.*1035

*I will go ahead and put that c here.*1045

*Again, we are going to add that c because it is the most general solution.*1046

*There you go, that is your anti-derivative.*1051

*You can always double check by differentiating your F.*1055

*The anti-derivative that you got, just differentiate and see if you get the original function.*1066

*That corroborates the fact that you have done it right, by differentiating f(x).*1070

*We have f(x) right here, therefore, f’, let us see what happens when I take the derivative of that.*1085

*It is going to be 4x³/ 4 - 6x² + 22x/ 2 – 9.*1091

*Sure enough, f’(x) is equal to x³ - 6x² + 11x – 9, which is exactly what the original = f(x).*1111

*F’(x) = f(x), that is what our theorem said, that is all we are doing.*1128

*We just have to remember which direction we are going in.*1133

*If we are taking the derivative of x³, it is going to be 3x², the original function.*1137

*If we are taking anti-derivative, it is going to be x⁴/ 4.*1144

*Direction is all that matters.*1148

*Let us go to function number 2, we had f(x) is equal to 14 × √x - 27 × 4√x.*1151

*We are going to write this with rational exponents.*1167

*This is going to be equal to 14 × x ^½ - 27 × x¹/4.*1170

*When we have an x ⁺n, when we take our anti-derivative, our formula is x ⁺n + 1/ n + 1.*1180

*That is it, you just subjected to the same thing.*1188

*It does not matter whether the exponent is rational or not.*1191

*This is going to be 14 × x, ½ + 1 is 3/2 divided by that number 3/2 – 27 × x.*1194

*¼ + 1 is 5/4 divided by 5/4.*1206

*We have to have our + c.*1213

*Therefore, we end up with, our final anti-derivative is going to be 14/ 3/2, that is going to be 28/3 × x³/2 - 4 × 27.*1217

*That is going to be 108 divided by 5 × x 5⁴ + c.*1234

*That is your most general anti-derivative.*1244

*If you took the derivative of this, you would get the original back.*1248

*Example number 3, we have f(x) = cos(x) - 14 × sin(x).*1254

*We are doing anti-derivative.*1267

*We go back to that list where we have the function and its anti derivative, which is also in your book or anywhere on the web.*1270

*You can just look at table of anti-derivatives also called table of integrals.*1278

*The anti-derivative of cos x was sin x - 14 which is the constant.*1284

*The anti-derivative of sin x was -cos x + c.*1296

*When we simplify, we get sin x + 14 cos(x) + c.*1305

*Once again, if you want to go ahead and check, the derivative of sin x is cos x.*1318

*The derivative of 14 cos x is -14 sin x.*1323

*Let us see what we have got, number 4.*1336

*We have x⁵, let me write down f(x).*1341

*F(x) = x⁵ + 2 × √x/ x⁴/3.*1349

*Let us write with rational exponents here.*1364

*We have x⁵ + 2 × x ^½/ x⁴/3.*1366

*I'm going to go ahead and separate this out.*1378

*It is going to be x⁵/ x⁴/3 + 2x ^½ / x⁴/3.*1380

*This is going to be x⁵/ x⁴/3 + 2x ^½/ x⁴/3.*1386

*I do not like writing my fractions that way, sorry about that.*1398

*I’m going to write it as it is supposed to be written, x⁴/3. *1402

*We get this is equal to x ⁺15/3 - 4/3 + 2 × x³/6 – 8/6.*1406

*Our f(x) is actually equal to x ⁺11/3 + 2 × x⁻⁵/6.*1430

*We can go ahead and take the anti-derivative.*1445

*I have just simplified that and made it such that there was an x to some exponent, so that I can use my formula.*1447

*I hope that make sense.*1457

*I cannot do anything with this, I have to convert it to something where I have x ⁺n and x ⁺n.*1458

*Now I can apply the formula.*1463

*The anti-derivative of f(x), now it is equal to, it is going to be x ⁺11/3 + 1 / 11/3 + 1 + 2 × x⁻⁵/6 + 1 all divided by -5/6 + 1 + c.*1465

*We have f(x) = x ⁺14/3/ 14/3 + 2x⁻⁵/6 + 6/6 is x¹/6 divided by 1/6 + c.*1490

*Our final answer is going to be 3/14 x ⁺14/3 + 12x¹/6 + c.*1511

*Our final answer, slightly longer not a problem.*1533

*It was only because of the simplification that we have to do.*1536

*Our number 5, we have our f(x) is equal to 3e ⁺x - 2/1 + x².*1542

*Really simple, this we can just read off.*1555

*The anti-derivative of e ⁺x is e ⁺x.*1558

*This stays 3e ⁺x.*1561

*Hopefully, we recognize that 1/1 + x², the anti-derivative of that is the inv tan.*1564

*Sorry about that, it is -2 × inv tan(x).*1574

*Of course, we add our c to give us our most general anti-derivative.*1580

*There you go, that takes care of that.*1585

*Hopefully, those examples help.*1589

*Again, it is all based on the basic formulas.*1590

*Let us do example number 2, given the following, find the original function f(x).*1595

*This time, they have given us extra information.*1600

*They have not only given us the f’, we are going to find the anti-derivative which is the f.*1603

*They have given it to us as f’, we just need to find f.*1614

*They also gave us other information, they said that the original f at 2 is equal to 40.*1618

*This extra information now is going to allow us to find what c is, in a particular case.*1623

*We are going to find the general solution.*1629

*And then, we are going to use this extra information to find the particular constant.*1630

*Let us get started here.*1637

*I think I have the wrong number here.*1646

*F(2) = 40, let me double check and make sure that my numbers are correct here.*1652

*I think I ended up actually using a different number when I solve this.*1661

*I had f(2) = 47, sorry about that, slight little correction.*1665

*Number 1, we have that f’(x) is equal to 5x³ - 14x + 24.*1671

*They tell us that f(2) is equal to 47.*1684

*F(x), notice that if I’m using prime notation, f is the anti-derivative of f’.*1691

*This just becomes 5x⁴/4 - 14x²/ 2 + 24x + c.*1701

*x ⁺n, just add 1 to the exponent, put that new exponent also in the denominator.*1716

*Let us go ahead and simplify a little bit.*1722

*f(x) = 5/4 x⁴ - 7x² + 24x + c.*1724

*This is our f, they tell me f(2) is equal to 47.*1740

*I put 2 wherever I have an x, I set it equal to 47, and I solve for c.*1744

*They tell me that f(2) which is 5/4⁴ - 7 × 2² + 24 × 2 + c.*1752

*They are telling you that all of that actually = 47.*1765

*I hope that I have done my arithmetic correctly.*1775

*We have got 20 - 28 + 48 + c = 47 and that gives me a final c = 7.*1777

*I’m hoping that you will confirm.*1790

*Now that I have c which is equal to 7, I can go ahead and put it back in to my equation that I have got.*1793

*My anti-derivative, my specific, my particular solution is going to be 5/4 x⁴ - 7x² + 24x + 7.*1801

*I found my constant and I have a particular solution, a specific solution, that is all I'm doing.*1818

*Do the anti-derivative and then use the information that is given to you to find the rest.*1825

*Number 2, we have an f’(x) is equal to 3 × sin(x) + sec² (x).*1834

*They are telling me that the f(π/6) happens to equal 5.*1847

*If this is f’, I take the anti-derivative.*1856

*This is going to be my f without the prime symbol.*1858

*It is going to be -3 × cos(x) because the anti-derivative of sin x is -cos(x).*1861

*The anti-derivative of sec² is tan(x).*1868

*This is going to be + c.*1874

*Now I use my information, f(π/6) is equal to -3 × cos(π/6) + tan(π/6) + c.*1875

*They are telling me that all of that is equal to 5.*1892

*Here we have cos(π/6) is going to be √3/2, -3 √3/2.*1896

*Tan(π/6) is going to be 1/ √3 + c is equal to 5.*1905

*Therefore, my c is going to equal, I’m not going to solve for of them, I’m just going to write it out straight.*1913

*It is going to be 5 – 1/ √3 + 3 √3/2, that is my c.*1919

*Therefore, I stick my c there and I get f(x) is equal to -3 × cos(x).*1928

*I will make my o's a little closer here, × cos(x) + tan(x) + whatever c I got which is 5 – 1/ √3 + 3 √3/ 2.*1941

*There you go, nice and simple.*1957

*Your teacher can tell you about the extent to which they want this simplified, put together, however they want to see it.*1961

*Number 3, let us see what we have got here.*1970

*Number 3, this one involves taking the anti-derivative twice.*1977

*They are telling me that f”(x) is equal to 8x - cos x.*1982

*They gave me two bits of information.*1990

*They are giving me f(1.5) is equal to 12.7.*1992

*They are telling me that f’(1.5) is equal to 4.2.*1998

*I have to take two anti-derivatives.*2006

*Therefore, I'm going to have two initial conditions.*2009

*One is going to be for f, one is going to be for f’.*2011

*I’m going to take the anti-derivative once, find f’.*2014

*Use this information, the f’(1.5) = 4.2, to find that constant.*2017

*I’m going to take the anti-derivative again, we will call it integration later, it is not a problem.*2023

*We are going to take the anti-derivative again, of the f’ to get our original function f.*2030

*We are going to use this first bit of information to find that constant.*2034

*Each step has a constant in it.*2039

*From f”, we are going to take the anti-derivative which means find f’.*2042

*This is going to be 8x²/ 2 - the anti-derivative of cos x which is sin x.*2049

*I will call this constant 1.*2057

*This is f’(x) = simplify a little bit, we have got 4x² – sin x + the constant of 1.*2061

*Now I use this information right here, the f(1) f’.*2072

*F' of 1.5 is equal to 4 × 1.5² – sin(1.5) + c1, they are telling me that it = 4.2.*2076

*I will write it all, that is not a problem.*2096

*When I solve this, I get 9 - 0.997 + c1 = 4.2.*2097

*I get that my c1 is equal to -3.803.*2106

*I found my first c1, that is the one that I’m going to plug in to here.*2113

*Therefore, my f’(x) is going to equal 4x² – sin x - 3.803.*2119

*Now that I have my f’, I want my original function f.*2134

*I’m going to take the anti-derivative again.*2136

*F(x) = 4x³/ 3 + cos(x) because the anti-derivative of sin x is -cos(x).*2140

*It is going to be -3.803x.*2154

*This is x⁰, it becomes x¹/1, and then now, + c2, always add that constant.*2158

*I know, I always forget.*2165

*I think the only reason that I actually remember is because I'm doing a lesson now, I’m trying hard to remember this, to put that c there.*2169

*Now we use this bit of information.*2177

*They are telling me that f(1.5) which is equal to 4/3 × 1.5³ + cos(1.5) - 3.803 × 1.5 + c2.*2180

*They are telling me that it = 12.7.*2200

*When I do that, I have got f(1.5) is equal to, it is going to be 4.5 + 0.0707 - 5.303 + c2 = 12.7.*2207

*When I solve, I get 13.432, that is my c2.*2228

*I put it back to my original and I end up with f(x) = 4/3 x³ + cos x - 3.803x + 13.432.*2236

*This is my particular solution to this particular anti-differentiation problem, given those two initial conditions.*2256

*Let us try another one of those.*2270

*This time we have, sorry about that, this is a double prime.*2274

*f”(x) is equal to 5/ √x.*2283

*We have f(2) is equal to 15 and we also have f’(2) is equal to 7.*2289

*Two initial conditions.*2299

*Let us write this in a way that we can manipulate.*2301

*f”(x) is equal to 5 × x⁻¹/2.*2307

*I take the anti-derivative so this is now going to become f’(x) and *2314

*this is going to be 5 × x⁻¹/2 + 1/ -1/2 + 1 which = 5x ^½/ ½, which is equal to 10x ^½ + c1.*2318

*There you go, that is my f’.*2342

*They tell me that f’(2) which is going to be 10 ^½ + c1 is equal to 7.*2348

*Therefore, my c1 is going to equal -7.14, when I do the calculation.*2364

*Therefore, I put this 7.14 into there and I get my f’, my specific solution f’(x) is equal to 10 x ^½ - 7.14.*2371

*f(x), I will do my f(x), I take the anti-derivative of this.*2392

*This is going to be 10x ½ + 1 is 3/2/ 3/2 - 7.14 × x + c2.*2398

*Let me see what I have got here, let me go to the next page.*2416

*I have got, when I simplify this, I have got f(x) = 20/3 x³/2 - 7.14 × x + c2.*2429

*They are telling me that f(2) which is equal to 20/3 × 2³/2 - 7.14 × 2 + c2.*2445

*They are telling me that that = 15.*2459

*When I solve for this, I get c2 is equal to 10.42.*2461

*I have my final f(x), my original function is 20/3 x³/2 - 7.14 × x + 10.42.*2468

*I think I did that right, I hope I did that right.*2491

*That is it, just anti-derivative, anti-derivative.*2501

*With each anti-derivative that you take, you want to go ahead and make sure to put the c *2504

*and then use the other information for wherever you are to find that c, and then take the next step.*2509

*Let us do a practical problem here.*2518

*A steel ball was dropped from rest from a tower 500 ft high, answer the following questions.*2521

*Take the acceleration of gravity to be 9.8 m/s2.*2527

*The first thing we want to do is find an equation for the height of the ball, after t seconds.*2533

*After I have dropped it, how long will it take for the ball to hit the ground?*2537

*What is the velocity of the ball as it hits the ground?*2543

*Part 4, if the stone is not dropped from rest, but if the stone is actually thrown downward with an initial velocity of 6 m/s,*2546

*how long does it take to reach the ground?*2555

*Let us see what we have got.*2561

*We have this tower, let us go ahead and draw this out.*2568

*This is the ground level, I’m just going to make this tower like that.*2571

*They tell us that this is 500 ft high.*2575

*I'm going to go ahead and take that as ground 0.*2580

*This is the 500, right.*2584

*Number 1 wanted the equation for the height of the ball after t seconds.*2590

*After a certain number of seconds, the ball is going to be like right there.*2610

*The height is going to be 500 - the distance that it actually traveled.*2614

*What I'm going to do is I'm going to find an equation for the distance that it actually traveled, and then take 500 – that.*2621

*That will give us our equation.*2627

*We will let s(t) be the distance function, how long it travels?*2630

*Be the distance function also called the position function.*2638

*You remember we called the position before.*2644

*That is the actual function that I’m looking for.*2647

*I’m looking for s(t).*2649

*S’(t), I know that the derivative of the position function is my velocity function.*2651

*It is my velocity and I know that if I take the derivative again, in other words, s”(t),*2661

*that is my acceleration, that is my acceleration function.*2670

*What do we know, we know that the acceleration of gravity is 9.8.*2682

*It is just a constant, that is it, you are just dropping it from rest.*2686

*The only force that is acting on this is the acceleration of gravity.*2689

*Therefore, our acceleration function s”(t) is actually just equal to 9.8, it is a constant.*2693

*Therefore, s’(t), when I take the first anti-derivative, that is just going to equal 9.8 t + c1.*2704

*What do I know, I know that s’(t) which is the velocity, *2720

*let us try it again, I know that s’ is the velocity.*2731

*I know that the velocity at time 0 which is the s’ at time 0 was starting from rest.*2736

*I’m just dropping it, it is 0.*2742

*S’(0) is 0, let us plug it in here.*2746

*That means that s’(0) is equal to 9.8 × 0 + c1.*2749

*I know that that = 0, that implies that c1 is actually equal to 0.*2762

*When I plug that in to my original equation, to my s’, I get s’(t) = 9.8 t.*2768

*I found an equation for the velocity at time t, it is 9.8 t.*2777

*Now I'm looking for s(t), now I’m going to take the anti-derivative of that.*2786

*Now I have got s’(t) = 9.8.*2794

*Therefore, s(t) is going to equal 9.8 t²/ 2 + c2.*2797

*What else do I know, now I need to find c2.*2812

*I know that s(t) or s(0), in other words the position at time 0 is 0.*2817

*I take that as my 0 position, that also = 0.*2828

*I’m going to put that in here, s(0) = 4.9 × 0² + c2 is equal to 0, that implies that c2 is equal to 0.*2833

*Therefore, my position function s(t) is equal to 4.9 t².*2849

*That means after a certain number of seconds, t seconds, I have actually traveled 4.9 × t² ft, m, whatever the length is.*2857

*Therefore, that means that is this distance.*2869

*After a certain number of seconds, I traveled this distance.*2876

*Therefore, my height above the ground is going to be 500 - this distance, *2879

*my height function is going to be 500 - 4.9 t².*2884

* Again, this is only based from the fact that I chose this as my 0.*2892

*You could have chosen this as your 0, just a different frame of reference.*2896

*I hope that make sense.*2902

*Let us go back to blue here.*2910

*How long before the ball strikes the ground, in other words, how many seconds go by before it?*2915

*How long before the ball strikes the ground?*2930

*Let us see, we came up with our s(t) which was going to be 4.9 t².*2937

*That was our position function.*2948

*We are falling 500 ft, our tower was this one.*2953

*We need to find out, we are going to set s(t) which is 4.9 t², we are going to set it to 500.*2959

*In other words, how many seconds does it take to go 500 ft?*2969

*When I solve this, I get t = 10.10 s.*2973

*Nice and simple, I have my position function.*2982

*How long does it take to go the 500 ft?*2986

*Number 3, velocity, as the ball hits the ground.*2990

*We said that our velocity function which was our first anti-derivative, s’(t), we said that that = 9.8 × t.*3008

*After 10.10 s which is when the ball is hitting the ground, I get the velocity at 10.10 s = 9.8 × 10.10 s.*3019

*It is going to be 99 m/s.*3033

*That is it, very straight forward.*3038

*Let us do the last one, number 4, if our initial velocity is 6 m/s downward, what is the velocity of the ball as it hits the ground?*3044

*Let us do this again, let us start from the beginning?*3083

*We have s”(t) which is our acceleration function, we know that that = 9.8.*3085

*When I take the anti-derivative of that, that is going to give my velocity function.*3092

*That is what I’m interested in.*3095

*Again, I have s’(t) which is my velocity function, that is going to equal 9.8 t + c1.*3096

*Standard anti-differentiation but now it is slightly different.*3109

*Now my initial velocity, in other words, my s’(0) which is my v(0) is now 6, it is not 0.*3113

*My velocity at time 0 which is 9.8 × 0 for t + c1 is equal to 6 m/s.*3126

*This implies that my c1 is actually equal to 6, that goes in here.*3137

*Therefore, my s’(t) function is actually equal to 9.8 t + 6.*3143

*When I find my s(t), I take my anti-derivative again,*3156

*I’m going to get my 9.8 t²/ 2 + 60 + c(2) which is equal to 4.9 t² + 60 + c/2.*3160

*I have a different function now and I also know that s(0) is still 0.*3182

*My 0 point is my starting point.*3189

*Therefore, I have got s(0) is equal to 4.9 × 0² + 6 × 0 + c2 = 0, which implies that our c2 is equal to 0.*3191

*Therefore, I get s(t) is now equal to 4.9 t² + 6t, that is my equation.*3207

*I need to find out how many seconds it takes, now that I have thrown it with an initial velocity which introduces the second term, *3222

*which was not there before, now I need to set this equal to, s(t) = 4.9 t² + 60.*3227

*I need to set that equal to 500, when I do that, I get t is equal to 9.51 s.*3242

*Exactly, what I expect, I threw it down with initial velocity instead of dropping from it rest.*3251

*It is going to take less time for it to get to the ground.*3256

*It took 10.1 seconds, now it is only taking 9.51 seconds.*3258

*This 9.51 is now, what I actually am going to put into my velocity function.*3263

*I knew velocity function which includes this extra term for the initial velocity.*3275

*S’(t) which is my velocity function is equal to 9.8 t + 6.*3286

*Therefore, the velocity of 9.51 = 9.8 × 9.51 + 6.*3295

*I get my velocity 1.51 = 99.2 m/s.*3306

*Not a lot faster but certainly faster.*3314

*There you go, that takes care of anti-derivatives.*3320

*Thank you so much for joining us here at www.educator.com.*3323

*We will see you next time, bye.*3325

1 answer

Last reply by: Professor Hovasapian

Thu Aug 25, 2016 5:38 PM

Post by Isaac Martinez on August 25 at 12:44:06 PM

Hello Professor Hovasapian,

I was wondering how you got 13.432 as an answer for the second derivative of your example II, Function 3.

Thank you,

Isaac