WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, and welcome back to www.educator.com, welcome back to AP Calculus.
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Today, we are going to start our discussion of anti-derivatives.
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In a couple of lessons, we are actually going to change the name of that and start calling them integrals.
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This is the second half of calculus.
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The first part is differential calculus, now it is integral calculus.
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As you will see in a minute, it is actually the inverse process.
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Let us jump right on in.
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Up to now, we started with functions and we took derivatives of them.
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Let us go ahead and write this.
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Up to now, we started with some function f(x) and we found its derivative.
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We have found its derivative, in other words, we differentiate it.
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For example, if we had a sin x, we take the derivative of it and we ended up with cos x.
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If we had, let us say, an x³, we take the derivative of it and we end up with something like 3x².
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What if we begin with a function then ask is this the derivative of some function?
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In other words, instead of starting with the function and going to the derivative,
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let us say that this is a function and this the derivative of some previous function.
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That is the question we are going to look at here.
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What if we begin with a function f(x), then ask is f the derivative of some f(x).
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Can I find this F, can I find this other f?
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In other words, we are going to be giving you the cos x for the 3x².
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And we are going to say, how do you recover the sin x?
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How do you recover the x³, if that original function actually exists?
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The answer is yes, fortunately.
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Let us go ahead and start with a definition.
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You know what, I think I will go ahead and put my definition.
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The different color, I will go ahead and put in red.
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The definition F(x) is called an anti-derivative, exactly what it sounds like,
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it is just the reverse of the derivative, an anti-derivative of f(x).
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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.
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Let us go back to blue here, not necessary but what the heck.
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All we are doing is going backward, that is it.
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That is all this is.
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Now the most difficult part is going to be remembering, am I going forward to differentiating
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or am I going backward and taking the anti-derivative?
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It is a different set of rules, it is a different set of formulas that you use to find that.
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That is going to be probably the most difficult thing that you have to do is remember which direction you are going in.
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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’.
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Or we can go this way and take the anti-derivative which gives us our F(x).
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Some function is going to be the function that is given to you.
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There are two ways that you can go, depending on what problem that you are trying to solve.
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That is what makes calculus incredibly beautiful, you start here.
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As you will see in a few lessons, there is a relationship between derivative and anti-derivative.
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They call it the fundamental theorem of calculus, it actually connects the two,
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which as we said are inverse processes.
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Let us note the following.
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If we had some f(x) which is equal to sin x + 6 and we had some g(x),
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let us say this is sin x + 4, these are not the same function.
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If you put an x value in, you are going to get two different y values.
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These are not the same function, very important to know that.
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These are not the same function but you can probably see where this is going.
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But f’(x), f’(sin x) + 6 is equal to cos(x).
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Not f, I’m talking about g and g‘(x).
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When I take the derivative of the sin x + 4, it is also going to equal cos(x).
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These are the same derivative.
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You get two different functions that end up going to the same derivative.
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That could be a bit of an issue, but it is not, fortunately.
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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?
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Do is say that sin x + 6 is the anti-derivative?
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Do I say sin x + 4 is the anti-derivative?
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Is it sin x + 6?
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Is it sin x + 4?
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Or how about just sin x + c, where c can be any constant?
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Because we know that the derivative of a constant = 0.
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I does not matter what that number is, it can be sin x + π, because the derivative of π is 0.
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Here is our theorem, let us go ahead and mark this in red.
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Our theorem says, if f(x) is an anti-derivative of f(x) on an interval that we will just call I,
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then f(x) + c, any other constant is the general solution.
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It is the general solution to this anti-differentiation problem.
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In other words, if I'm given cos(x) and if I take the anti-derivative of that,
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I know that the anti-derivative is going to be sin x + something.
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What is it that I’m going to choose for that something?
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In general, if we are just speaking about general solutions and taking anti-derivatives, you are just going to write cos x + c.
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There is going to be other information in the problem that allows you to find what c is.
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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.
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When we speak of the general situation, we put the anti-derivative and you just stick a c right after it,
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to make sure that it is formally correct.
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Are you going to forget the c, yes you are going to forget the c.
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I still forget the c, after all these years.
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Do not worry about it but this is the general solution.
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anti-derivative, without any other information, just add the constant to it.
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I will say, technically, you must always include the c.
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Try your best to remember it.
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Once again, extra information will allow you to find what c is.
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We will see that when we start doing some of the example problems.
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To find what c is, extra information will allow you to find what c is, in a particular case.
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Thus, giving you what we call the specific solution or often called the particular solution.
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Let us do some examples.
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Before we do the examples, I’m going to give you a list of the anti-derivatives that we actually already know.
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Some anti-derivative formulas we already know.
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I think I will go ahead and do this in red.
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Here we have the function and here I'm going to put the anti-derivative.
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Once again, we have to make sure that we know what direction we are going in.
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If I have x ⁺n, the way to find the anti-derivative of x ⁺n is you take x ⁺n + 1/ n + 1.
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In other words, if this were the function that we are given,
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you know that what you do is to take the exponent, you bring it down here.
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You subtract one from the exponent.
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Differentiation is this way.
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What we are saying is, if you are starting with a function anti-differentiation, this is the formula that you use.
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Again, you will see in just a minute what we mean.
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Right now, I’m just going to write down some formulas here.
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1/x, the anti-derivative of that is the natlog of the absolute value of x e ⁺x.
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The anti-derivative is e ⁺x.
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If I have cos(x), I know the anti-derivative of that is sin(x).
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If I’m given sin(x), the derivative is cos(x).
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If I’m given cos(x), the anti-derivative is sin(x).
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Direction is very important here.
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Do not worry, you will make a thousand of mistakes, as far as direction is concerned.
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You will be asked to get an anti-derivative and you end up differentiating.
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That is just the process that we go through, do not worry about it.
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One more page here, let us write our function and let us write our anti-derivative.
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If I have a sin x and I want the anti-derivative, it is actually going to be a -cos x.
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Because if I’m given –cos x, the derivative of that is sin x.
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Sec² x, the anti-derivative is the tan(x).
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If I'm given sec x tan x, the anti-derivative sec x.
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If I'm given 1/ 1 - x², all under the radical,
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the anti-derivative of that is the inv sin(x) 1/ 1 + x², that anti-derivative is the inv tan(x).
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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.
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If I were given the inv cos, the derivative of that would be -1/ √1 - x².
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Now let us go ahead and jump right into the examples.
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Find the most general anti-derivative for the following functions.
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General means just add c to your answer, that is all that means.
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1, 2, 3, 4, 5, let us go ahead and jump right on in.
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Let me go ahead and make sure that I have everything here.
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I have copied the functions, 14 x⁵, 4/3, and 3 ⁺x.
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Let us start with number 1, I think I will go back to blue here, I hope you do not mind.
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Number 1, our f(x) was x³ – 6x² + 11x – 9.
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We said that the formula for the anti-derivative, when you are given some x ⁺n,
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the anti-derivative of that is x ⁺n + 1/ n + 1.
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Add 1 to the exponent , divide by the number that you get which is now the new exponent, very simple.
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Therefore, our f(x), our anti-derivative is going to be x⁴/ 4.
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How much easier can this possibly be?
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-6, the constant, x³/ 3, we will simplify it, just a minute.
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+ 11 x² because this is 1, add 1 to it and divide by that same number, -9x.
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This is x⁰, x⁰, add 1 to the exponent, it becomes 1, divide by 1, it becomes 9x.
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Now we can simplify, divide where we need to.
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This is perfectly valid, you do not have to take the 6 and the 3, and divide it.
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You can stop there, if you want to.
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It just depends on what you teacher is going to be asking for.
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We have x⁴/ 4, 6/3 is 2.
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It is going to be -2x³, it is going to be +11 x²/ 2 – 9x + c.
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I will go ahead and put that c here.
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Again, we are going to add that c because it is the most general solution.
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There you go, that is your anti-derivative.
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You can always double check by differentiating your F.
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The anti-derivative that you got, just differentiate and see if you get the original function.
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That corroborates the fact that you have done it right, by differentiating f(x).
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We have f(x) right here, therefore, f’, let us see what happens when I take the derivative of that.
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It is going to be 4x³/ 4 - 6x² + 22x/ 2 – 9.
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Sure enough, f’(x) is equal to x³ - 6x² + 11x – 9, which is exactly what the original = f(x).
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F’(x) = f(x), that is what our theorem said, that is all we are doing.
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We just have to remember which direction we are going in.
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If we are taking the derivative of x³, it is going to be 3x², the original function.
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If we are taking anti-derivative, it is going to be x⁴/ 4.
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Direction is all that matters.
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Let us go to function number 2, we had f(x) is equal to 14 × √x - 27 × 4√x.
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We are going to write this with rational exponents.
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This is going to be equal to 14 × x ^½ - 27 × x¹/4.
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When we have an x ⁺n, when we take our anti-derivative, our formula is x ⁺n + 1/ n + 1.
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That is it, you just subjected to the same thing.
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It does not matter whether the exponent is rational or not.
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This is going to be 14 × x, ½ + 1 is 3/2 divided by that number 3/2 – 27 × x.
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¼ + 1 is 5/4 divided by 5/4.
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We have to have our + c.
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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.
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That is going to be 108 divided by 5 × x 5⁴ + c.
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That is your most general anti-derivative.
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If you took the derivative of this, you would get the original back.
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Example number 3, we have f(x) = cos(x) - 14 × sin(x).
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We are doing anti-derivative.
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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.
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You can just look at table of anti-derivatives also called table of integrals.
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The anti-derivative of cos x was sin x - 14 which is the constant.
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The anti-derivative of sin x was -cos x + c.
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When we simplify, we get sin x + 14 cos(x) + c.
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Once again, if you want to go ahead and check, the derivative of sin x is cos x.
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The derivative of 14 cos x is -14 sin x.
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Let us see what we have got, number 4.
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We have x⁵, let me write down f(x).
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F(x) = x⁵ + 2 × √x/ x⁴/3.
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Let us write with rational exponents here.
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We have x⁵ + 2 × x ^½/ x⁴/3.
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I'm going to go ahead and separate this out.
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It is going to be x⁵/ x⁴/3 + 2x ^½ / x⁴/3.
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This is going to be x⁵/ x⁴/3 + 2x ^½/ x⁴/3.
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I do not like writing my fractions that way, sorry about that.
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I’m going to write it as it is supposed to be written, x⁴/3.
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We get this is equal to x ⁺15/3 - 4/3 + 2 × x³/6 – 8/6.
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Our f(x) is actually equal to x ⁺11/3 + 2 × x⁻⁵/6.
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We can go ahead and take the anti-derivative.
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I have just simplified that and made it such that there was an x to some exponent, so that I can use my formula.
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I hope that make sense.
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I cannot do anything with this, I have to convert it to something where I have x ⁺n and x ⁺n.
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Now I can apply the formula.
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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.
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We have f(x) = x ⁺14/3/ 14/3 + 2x⁻⁵/6 + 6/6 is x¹/6 divided by 1/6 + c.
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Our final answer is going to be 3/14 x ⁺14/3 + 12x¹/6 + c.
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Our final answer, slightly longer not a problem.
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It was only because of the simplification that we have to do.
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Our number 5, we have our f(x) is equal to 3e ⁺x - 2/1 + x².
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Really simple, this we can just read off.
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The anti-derivative of e ⁺x is e ⁺x.
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This stays 3e ⁺x.
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Hopefully, we recognize that 1/1 + x², the anti-derivative of that is the inv tan.
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Sorry about that, it is -2 × inv tan(x).
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Of course, we add our c to give us our most general anti-derivative.
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There you go, that takes care of that.
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Hopefully, those examples help.
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Again, it is all based on the basic formulas.
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Let us do example number 2, given the following, find the original function f(x).
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This time, they have given us extra information.
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They have not only given us the f’, we are going to find the anti-derivative which is the f.
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They have given it to us as f’, we just need to find f.
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They also gave us other information, they said that the original f at 2 is equal to 40.
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This extra information now is going to allow us to find what c is, in a particular case.
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We are going to find the general solution.
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And then, we are going to use this extra information to find the particular constant.
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Let us get started here.
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I think I have the wrong number here.
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F(2) = 40, let me double check and make sure that my numbers are correct here.
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I think I ended up actually using a different number when I solve this.
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I had f(2) = 47, sorry about that, slight little correction.
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Number 1, we have that f’(x) is equal to 5x³ - 14x + 24.
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They tell us that f(2) is equal to 47.
00:28:11.400 --> 00:28:20.700
F(x), notice that if I’m using prime notation, f is the anti-derivative of f’.
00:28:20.700 --> 00:28:36.000
This just becomes 5x⁴/4 - 14x²/ 2 + 24x + c.
00:28:36.000 --> 00:28:42.000
x ⁺n, just add 1 to the exponent, put that new exponent also in the denominator.
00:28:42.000 --> 00:28:44.300
Let us go ahead and simplify a little bit.
00:28:44.300 --> 00:29:00.300
f(x) = 5/4 x⁴ - 7x² + 24x + c.
00:29:00.300 --> 00:29:04.500
This is our f, they tell me f(2) is equal to 47.
00:29:04.500 --> 00:29:11.700
I put 2 wherever I have an x, I set it equal to 47, and I solve for c.
00:29:11.700 --> 00:29:24.900
They tell me that f(2) which is 5/4⁴ - 7 × 2² + 24 × 2 + c.
00:29:24.900 --> 00:29:34.900
They are telling you that all of that actually = 47.
00:29:34.900 --> 00:29:37.600
I hope that I have done my arithmetic correctly.
00:29:37.600 --> 00:29:50.500
We have got 20 - 28 + 48 + c = 47 and that gives me a final c = 7.
00:29:50.500 --> 00:29:53.300
I’m hoping that you will confirm.
00:29:53.300 --> 00:30:00.800
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.
00:30:00.800 --> 00:30:17.900
My anti-derivative, my specific, my particular solution is going to be 5/4 x⁴ - 7x² + 24x + 7.
00:30:17.900 --> 00:30:24.800
I found my constant and I have a particular solution, a specific solution, that is all I'm doing.
00:30:24.800 --> 00:30:33.800
Do the anti-derivative and then use the information that is given to you to find the rest.
00:30:33.800 --> 00:30:47.300
Number 2, we have an f’(x) is equal to 3 × sin(x) + sec² (x).
00:30:47.300 --> 00:30:55.700
They are telling me that the f(π/6) happens to equal 5.
00:30:55.700 --> 00:30:58.400
If this is f’, I take the anti-derivative.
00:30:58.400 --> 00:31:01.400
This is going to be my f without the prime symbol.
00:31:01.400 --> 00:31:08.600
It is going to be -3 × cos(x) because the anti-derivative of sin x is -cos(x).
00:31:08.600 --> 00:31:13.700
The anti-derivative of sec² is tan(x).
00:31:13.700 --> 00:31:15.500
This is going to be + c.
00:31:15.500 --> 00:31:32.000
Now I use my information, f(π/6) is equal to -3 × cos(π/6) + tan(π/6) + c.
00:31:32.000 --> 00:31:35.900
They are telling me that all of that is equal to 5.
00:31:35.900 --> 00:31:45.500
Here we have cos(π/6) is going to be √3/2, -3 √3/2.
00:31:45.500 --> 00:31:53.600
Tan(π/6) is going to be 1/ √3 + c is equal to 5.
00:31:53.600 --> 00:31:59.200
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.
00:31:59.200 --> 00:32:08.400
It is going to be 5 – 1/ √3 + 3 √3/2, that is my c.
00:32:08.400 --> 00:32:21.300
Therefore, I stick my c there and I get f(x) is equal to -3 × cos(x).
00:32:21.300 --> 00:32:37.300
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.
00:32:37.300 --> 00:32:41.600
There you go, nice and simple.
00:32:41.600 --> 00:32:50.600
Your teacher can tell you about the extent to which they want this simplified, put together, however they want to see it.
00:32:50.600 --> 00:32:57.200
Number 3, let us see what we have got here.
00:32:57.200 --> 00:33:02.000
Number 3, this one involves taking the anti-derivative twice.
00:33:02.000 --> 00:33:09.800
They are telling me that f”(x) is equal to 8x - cos x.
00:33:09.800 --> 00:33:11.900
They gave me two bits of information.
00:33:11.900 --> 00:33:18.200
They are giving me f(1.5) is equal to 12.7.
00:33:18.200 --> 00:33:26.600
They are telling me that f’(1.5) is equal to 4.2.
00:33:26.600 --> 00:33:29.000
I have to take two anti-derivatives.
00:33:29.000 --> 00:33:31.100
Therefore, I'm going to have two initial conditions.
00:33:31.100 --> 00:33:34.500
One is going to be for f, one is going to be for f’.
00:33:34.500 --> 00:33:37.500
I’m going to take the anti-derivative once, find f’.
00:33:37.500 --> 00:33:42.900
Use this information, the f’(1.5) = 4.2, to find that constant.
00:33:42.900 --> 00:33:50.100
I’m going to take the anti-derivative again, we will call it integration later, it is not a problem.
00:33:50.100 --> 00:33:54.600
We are going to take the anti-derivative again, of the f’ to get our original function f.
00:33:54.600 --> 00:33:58.800
We are going to use this first bit of information to find that constant.
00:33:58.800 --> 00:34:01.800
Each step has a constant in it.
00:34:01.800 --> 00:34:09.300
From f”, we are going to take the anti-derivative which means find f’.
00:34:09.300 --> 00:34:16.800
This is going to be 8x²/ 2 - the anti-derivative of cos x which is sin x.
00:34:16.800 --> 00:34:20.800
I will call this constant 1.
00:34:20.800 --> 00:34:31.900
This is f’(x) = simplify a little bit, we have got 4x² – sin x + the constant of 1.
00:34:31.900 --> 00:34:36.400
Now I use this information right here, the f(1) f’.
00:34:36.400 --> 00:34:56.200
F' of 1.5 is equal to 4 × 1.5² – sin(1.5) + c1, they are telling me that it = 4.2.
00:34:56.200 --> 00:34:57.400
I will write it all, that is not a problem.
00:34:57.400 --> 00:35:06.300
When I solve this, I get 9 - 0.997 + c1 = 4.2.
00:35:06.300 --> 00:35:13.200
I get that my c1 is equal to -3.803.
00:35:13.200 --> 00:35:19.200
I found my first c1, that is the one that I’m going to plug in to here.
00:35:19.200 --> 00:35:33.700
Therefore, my f’(x) is going to equal 4x² – sin x - 3.803.
00:35:33.700 --> 00:35:36.600
Now that I have my f’, I want my original function f.
00:35:36.600 --> 00:35:39.900
I’m going to take the anti-derivative again.
00:35:39.900 --> 00:35:53.700
F(x) = 4x³/ 3 + cos(x) because the anti-derivative of sin x is -cos(x).
00:35:53.700 --> 00:35:57.900
It is going to be -3.803x.
00:35:57.900 --> 00:36:05.400
This is x⁰, it becomes x¹/1, and then now, + c2, always add that constant.
00:36:05.400 --> 00:36:09.000
I know, I always forget.
00:36:09.000 --> 00:36:16.800
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.
00:36:16.800 --> 00:36:19.900
Now we use this bit of information.
00:36:19.900 --> 00:36:39.700
They are telling me that f(1.5) which is equal to 4/3 × 1.5³ + cos(1.5) - 3.803 × 1.5 + c2.
00:36:39.700 --> 00:36:47.200
They are telling me that it = 12.7.
00:36:47.200 --> 00:37:08.500
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.
00:37:08.500 --> 00:37:16.600
When I solve, I get 13.432, that is my c2.
00:37:16.600 --> 00:37:36.500
I put it back to my original and I end up with f(x) = 4/3 x³ + cos x - 3.803x + 13.432.
00:37:36.500 --> 00:37:50.300
This is my particular solution to this particular anti-differentiation problem, given those two initial conditions.
00:37:50.300 --> 00:37:54.200
Let us try another one of those.
00:37:54.200 --> 00:38:02.700
This time we have, sorry about that, this is a double prime.
00:38:02.700 --> 00:38:09.600
f”(x) is equal to 5/ √x.
00:38:09.600 --> 00:38:19.200
We have f(2) is equal to 15 and we also have f’(2) is equal to 7.
00:38:19.200 --> 00:38:21.600
Two initial conditions.
00:38:21.600 --> 00:38:26.700
Let us write this in a way that we can manipulate.
00:38:26.700 --> 00:38:34.200
f”(x) is equal to 5 × x⁻¹/2.
00:38:34.200 --> 00:38:38.200
I take the anti-derivative so this is now going to become f’(x) and
00:38:38.200 --> 00:39:02.000
this is going to be 5 × x⁻¹/2 + 1/ -1/2 + 1 which = 5x ^½/ ½, which is equal to 10x ^½ + c1.
00:39:02.000 --> 00:39:08.300
There you go, that is my f’.
00:39:08.300 --> 00:39:23.900
They tell me that f’(2) which is going to be 10 ^½ + c1 is equal to 7.
00:39:23.900 --> 00:39:30.800
Therefore, my c1 is going to equal -7.14, when I do the calculation.
00:39:30.800 --> 00:39:52.400
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.
00:39:52.400 --> 00:39:57.800
f(x), I will do my f(x), I take the anti-derivative of this.
00:39:57.800 --> 00:40:15.800
This is going to be 10x ½ + 1 is 3/2/ 3/2 - 7.14 × x + c2.
00:40:15.800 --> 00:40:28.700
Let me see what I have got here, let me go to the next page.
00:40:28.700 --> 00:40:45.200
I have got, when I simplify this, I have got f(x) = 20/3 x³/2 - 7.14 × x + c2.
00:40:45.200 --> 00:40:58.700
They are telling me that f(2) which is equal to 20/3 × 2³/2 - 7.14 × 2 + c2.
00:40:58.700 --> 00:41:01.100
They are telling me that that = 15.
00:41:01.100 --> 00:41:08.400
When I solve for this, I get c2 is equal to 10.42.
00:41:08.400 --> 00:41:30.900
I have my final f(x), my original function is 20/3 x³/2 - 7.14 × x + 10.42.
00:41:30.900 --> 00:41:41.400
I think I did that right, I hope I did that right.
00:41:41.400 --> 00:41:43.800
That is it, just anti-derivative, anti-derivative.
00:41:43.800 --> 00:41:49.500
With each anti-derivative that you take, you want to go ahead and make sure to put the c
00:41:49.500 --> 00:41:57.700
and then use the other information for wherever you are to find that c, and then take the next step.
00:41:57.700 --> 00:42:00.700
Let us do a practical problem here.
00:42:00.700 --> 00:42:07.000
A steel ball was dropped from rest from a tower 500 ft high, answer the following questions.
00:42:07.000 --> 00:42:12.700
Take the acceleration of gravity to be 9.8 m/s2.
00:42:12.700 --> 00:42:17.400
The first thing we want to do is find an equation for the height of the ball, after t seconds.
00:42:17.400 --> 00:42:22.800
After I have dropped it, how long will it take for the ball to hit the ground?
00:42:22.800 --> 00:42:26.400
What is the velocity of the ball as it hits the ground?
00:42:26.400 --> 00:42:35.100
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,
00:42:35.100 --> 00:42:41.500
how long does it take to reach the ground?
00:42:41.500 --> 00:42:47.800
Let us see what we have got.
00:42:47.800 --> 00:42:51.400
We have this tower, let us go ahead and draw this out.
00:42:51.400 --> 00:42:55.300
This is the ground level, I’m just going to make this tower like that.
00:42:55.300 --> 00:43:00.400
They tell us that this is 500 ft high.
00:43:00.400 --> 00:43:04.600
I'm going to go ahead and take that as ground 0.
00:43:04.600 --> 00:43:10.300
This is the 500, right.
00:43:10.300 --> 00:43:30.100
Number 1 wanted the equation for the height of the ball after t seconds.
00:43:30.100 --> 00:43:34.600
After a certain number of seconds, the ball is going to be like right there.
00:43:34.600 --> 00:43:40.900
The height is going to be 500 - the distance that it actually traveled.
00:43:40.900 --> 00:43:47.000
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.
00:43:47.000 --> 00:43:50.600
That will give us our equation.
00:43:50.600 --> 00:43:57.800
We will let s(t) be the distance function, how long it travels?
00:43:57.800 --> 00:44:03.900
Be the distance function also called the position function.
00:44:03.900 --> 00:44:06.900
You remember we called the position before.
00:44:06.900 --> 00:44:08.800
That is the actual function that I’m looking for.
00:44:08.800 --> 00:44:11.200
I’m looking for s(t).
00:44:11.200 --> 00:44:21.500
S’(t), I know that the derivative of the position function is my velocity function.
00:44:21.500 --> 00:44:30.200
It is my velocity and I know that if I take the derivative again, in other words, s”(t),
00:44:30.200 --> 00:44:41.700
that is my acceleration, that is my acceleration function.
00:44:41.700 --> 00:44:46.300
What do we know, we know that the acceleration of gravity is 9.8.
00:44:46.300 --> 00:44:49.100
It is just a constant, that is it, you are just dropping it from rest.
00:44:49.100 --> 00:44:53.600
The only force that is acting on this is the acceleration of gravity.
00:44:53.600 --> 00:45:04.100
Therefore, our acceleration function s”(t) is actually just equal to 9.8, it is a constant.
00:45:04.100 --> 00:45:20.600
Therefore, s’(t), when I take the first anti-derivative, that is just going to equal 9.8 t + c1.
00:45:20.600 --> 00:45:31.200
What do I know, I know that s’(t) which is the velocity,
00:45:31.200 --> 00:45:36.300
let us try it again, I know that s’ is the velocity.
00:45:36.300 --> 00:45:42.400
I know that the velocity at time 0 which is the s’ at time 0 was starting from rest.
00:45:42.400 --> 00:45:46.300
I’m just dropping it, it is 0.
00:45:46.300 --> 00:45:49.000
S’(0) is 0, let us plug it in here.
00:45:49.000 --> 00:46:02.500
That means that s’(0) is equal to 9.8 × 0 + c1.
00:46:02.500 --> 00:46:08.100
I know that that = 0, that implies that c1 is actually equal to 0.
00:46:08.100 --> 00:46:16.900
When I plug that in to my original equation, to my s’, I get s’(t) = 9.8 t.
00:46:16.900 --> 00:46:25.900
I found an equation for the velocity at time t, it is 9.8 t.
00:46:25.900 --> 00:46:33.700
Now I'm looking for s(t), now I’m going to take the anti-derivative of that.
00:46:33.700 --> 00:46:37.000
Now I have got s’(t) = 9.8.
00:46:37.000 --> 00:46:52.400
Therefore, s(t) is going to equal 9.8 t²/ 2 + c2.
00:46:52.400 --> 00:46:57.200
What else do I know, now I need to find c2.
00:46:57.200 --> 00:47:08.600
I know that s(t) or s(0), in other words the position at time 0 is 0.
00:47:08.600 --> 00:47:12.800
I take that as my 0 position, that also = 0.
00:47:12.800 --> 00:47:28.700
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.
00:47:28.700 --> 00:47:37.100
Therefore, my position function s(t) is equal to 4.9 t².
00:47:37.100 --> 00:47:49.600
That means after a certain number of seconds, t seconds, I have actually traveled 4.9 × t² ft, m, whatever the length is.
00:47:49.600 --> 00:47:55.900
Therefore, that means that is this distance.
00:47:55.900 --> 00:47:59.000
After a certain number of seconds, I traveled this distance.
00:47:59.000 --> 00:48:04.400
Therefore, my height above the ground is going to be 500 - this distance,
00:48:04.400 --> 00:48:12.500
my height function is going to be 500 - 4.9 t².
00:48:12.500 --> 00:48:16.400
Again, this is only based from the fact that I chose this as my 0.
00:48:16.400 --> 00:48:22.400
You could have chosen this as your 0, just a different frame of reference.
00:48:22.400 --> 00:48:30.200
I hope that make sense.
00:48:30.200 --> 00:48:35.000
Let us go back to blue here.
00:48:35.000 --> 00:48:50.600
How long before the ball strikes the ground, in other words, how many seconds go by before it?
00:48:50.600 --> 00:48:57.500
How long before the ball strikes the ground?
00:48:57.500 --> 00:49:08.100
Let us see, we came up with our s(t) which was going to be 4.9 t².
00:49:08.100 --> 00:49:13.200
That was our position function.
00:49:13.200 --> 00:49:19.200
We are falling 500 ft, our tower was this one.
00:49:19.200 --> 00:49:29.100
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.
00:49:29.100 --> 00:49:33.400
In other words, how many seconds does it take to go 500 ft?
00:49:33.400 --> 00:49:41.900
When I solve this, I get t = 10.10 s.
00:49:41.900 --> 00:49:45.800
Nice and simple, I have my position function.
00:49:45.800 --> 00:49:50.300
How long does it take to go the 500 ft?
00:49:50.300 --> 00:50:07.700
Number 3, velocity, as the ball hits the ground.
00:50:07.700 --> 00:50:19.100
We said that our velocity function which was our first anti-derivative, s’(t), we said that that = 9.8 × t.
00:50:19.100 --> 00:50:33.200
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.
00:50:33.200 --> 00:50:37.700
It is going to be 99 m/s.
00:50:37.700 --> 00:50:43.700
That is it, very straight forward.
00:50:43.700 --> 00:51:23.400
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?
00:51:23.400 --> 00:51:25.300
Let us do this again, let us start from the beginning?
00:51:25.300 --> 00:51:31.900
We have s”(t) which is our acceleration function, we know that that = 9.8.
00:51:31.900 --> 00:51:35.000
When I take the anti-derivative of that, that is going to give my velocity function.
00:51:35.000 --> 00:51:36.500
That is what I’m interested in.
00:51:36.500 --> 00:51:48.800
Again, I have s’(t) which is my velocity function, that is going to equal 9.8 t + c1.
00:51:48.800 --> 00:51:52.700
Standard anti-differentiation but now it is slightly different.
00:51:52.700 --> 00:52:05.900
Now my initial velocity, in other words, my s’(0) which is my v(0) is now 6, it is not 0.
00:52:05.900 --> 00:52:16.700
My velocity at time 0 which is 9.8 × 0 for t + c1 is equal to 6 m/s.
00:52:16.700 --> 00:52:23.300
This implies that my c1 is actually equal to 6, that goes in here.
00:52:23.300 --> 00:52:35.900
Therefore, my s’(t) function is actually equal to 9.8 t + 6.
00:52:35.900 --> 00:52:39.800
When I find my s(t), I take my anti-derivative again,
00:52:39.800 --> 00:53:02.000
I’m going to get my 9.8 t²/ 2 + 60 + c(2) which is equal to 4.9 t² + 60 + c/2.
00:53:02.000 --> 00:53:09.300
I have a different function now and I also know that s(0) is still 0.
00:53:09.300 --> 00:53:10.900
My 0 point is my starting point.
00:53:10.900 --> 00:53:26.800
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.
00:53:26.800 --> 00:53:41.900
Therefore, I get s(t) is now equal to 4.9 t² + 6t, that is my equation.
00:53:41.900 --> 00:53:47.600
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,
00:53:47.600 --> 00:54:01.800
which was not there before, now I need to set this equal to, s(t) = 4.9 t² + 60.
00:54:01.800 --> 00:54:10.800
I need to set that equal to 500, when I do that, I get t is equal to 9.51 s.
00:54:10.800 --> 00:54:15.900
Exactly, what I expect, I threw it down with initial velocity instead of dropping from it rest.
00:54:15.900 --> 00:54:18.300
It is going to take less time for it to get to the ground.
00:54:18.300 --> 00:54:23.400
It took 10.1 seconds, now it is only taking 9.51 seconds.
00:54:23.400 --> 00:54:34.900
This 9.51 is now, what I actually am going to put into my velocity function.
00:54:34.900 --> 00:54:45.700
I knew velocity function which includes this extra term for the initial velocity.
00:54:45.700 --> 00:54:55.000
S’(t) which is my velocity function is equal to 9.8 t + 6.
00:54:55.000 --> 00:55:06.200
Therefore, the velocity of 9.51 = 9.8 × 9.51 + 6.
00:55:06.200 --> 00:55:14.600
I get my velocity 1.51 = 99.2 m/s.
00:55:14.600 --> 00:55:19.700
Not a lot faster but certainly faster.
00:55:19.700 --> 00:55:23.100
There you go, that takes care of anti-derivatives.
00:55:23.100 --> 00:55:25.300
Thank you so much for joining us here at www.educator.com.
00:55:25.300 --> 00:55:26.000
We will see you next time, bye.