WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, I thought we would do some more examples on the derivative.
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Let us jump right on in.
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Typical example, graphs are going to be very huge.
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You are going to be given a function, find the derivative graph.
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Or the derivative graph, find the function, work your way back.
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The following is a graph of f(x), the original function.
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On the same graph, sketch f’(x).
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They just want a sketch of the derivative.
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One of the things that we know -- when you are working with graphs and their derivatives, the most important thing is slope, where slope is 0.
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We know that a slope of 0 means a horizontal tangent line.
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On this graph, we have a horizontal tangent line here, here, and here.
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At these x values, namely, let us say somewhere around there.
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Here it looks like this is 0.
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Here, those are the places where the derivative, because the derivative is the slope of the graph.
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It is where the derivative graph hits 0.
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Now to the left of this point, we notice that the slope is negative.
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It is below the axis.
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As x moves this way, as we hit this point, that is where we are going to hit 0.
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From this point to this point, the slope is positive.
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It changes and it increases a little bit, and it decreases towards 0.
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Basically, it goes like this and then it passes 0 again.
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Now you notice the slope is pass this point, the slope is negative.
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It gets more negative and it starts to go positive again until it hits 0 again.
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It gets more negative, it starts to go positive until it hits 0 again.
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Then, it keeps going up.
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Look for horizontal tangents, when you are working with graphs and their derivatives.
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Whether it is first derivative or second derivative, or whatever.
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There you go, that is the derivative.
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This is black, this is your f’(x).
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Same thing, the following is a graph of f(x), sketch f’(x).
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Here, we notice that f(x) is not differentiable here.
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Anytime you remember when you have a cusp,
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that is one of the ways that a function cannot be differentiable at that point.
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Everything else is fine.
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We look at this and we see that the slope moving from of course left to right,
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we are moving this way from negative values to positive values, the slope is negative.
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It is actually becoming more negative.
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At that point, it is not differentiable.
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The derivative is not defined there.
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It is a negative slope, it means it starts below the x axis, becomes more negative.
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The derivative graph looks like that.
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You are just graphing what the slope does.
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That is what the f’ graph is, that is what the derivative graph is.
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What is the slope doing?
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Starting negative becoming more negative and not touching 0.
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Here, the derivative just passes 0.
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It starts really highly positive, it stays positive but it becomes less positive.
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Highly positive becomes less positive, and that is it.
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This hyperbola looking thing, this is your f’ graph.
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That is all it is, what is the slope doing, that is all you have to ask yourself.
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Find the derivative of the following function using the definition.
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I think I’m going to work in blue here.
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Again, it is always a great idea to write the definition.
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Just get accustomed to writing it in a nice, just to be systematic in your approach.
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We know that the definition f’(x), the derivative = the limit as h approaches 0 of the following quotient, f(x) + h.
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It must be nice if I make my + signs legible.
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x + h - f(x)/ h.
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We form this quotient first, simplify as much as possible, and then we take the limit.
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Let us do it, I'm going to go ahead and just work with the difference quotient first.
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And then, I will do the limit at the end.
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I do not want to keep writing limit as h goes to 0.
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I will just work with the quotient first.
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f(x) + h is going to be 1/ √x + h – f(x).
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f(x) is 1/ √x/ h.
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I take the common denominator for the top.
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It becomes √x – √x + h/ √x × √x + h/ h.
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I’m going to keep simplifying this here.
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Actually you know what, it looks like this particular one going to cause a little bit of a problem
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because we are still going to end up with, if we take h to 0,
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it looks like here the denominator is 0 as h goes to 0, because when we put 0 in there for h.
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I’m going to try an alternate procedure here.
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I’m going to start again.
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I’m going to go 1/ √x + h - 1/ √x/ h.
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This one, the reason I stop is because at this point, there is really not much you can do.
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I mean, this is h/1.
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When you flip it, you are going to end up with an h at the bottom.
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When you take h to 0, you are going to end up with 0 in the denominator.
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That is what is happening here.
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When that happens, because you really cannot simplify this much more than that this.
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Because we have 0 in the denominator, we cannot really do anything with that.
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We have to try an alternate procedure, an alternate manipulation.
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That is why we are going to try an alternate manipulation here.
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I go back to the beginning, this thing.
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I think what I’m going to do them is multiply it by the conjugate of the numerator.
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It is going to be × 1/ √x + h + 1/ √x/ 1/ √x + h + 1/ √x.
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This is going to end up equaling 1/ x + h - 1/ x/ h × 1/ √x + h + 1/ √x, when you simplify this.
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Let us see what we have got here.
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I might as well continue on.
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This is going to equal x - x + h/ x × x + h.
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I just found the common denominator for the numerator, and I leave the bottom as is.
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1/ √x + h + 1/ √x.
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This is going to end up equaling, x - x cancels, you are going to end up with -h/ x × x + h/ h × 1/ √x + h -1/ √x.
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Now the h cancel and you are left with,
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Now we have -1/ x × x + h/, do a common denominator here, √x - √x + h/ √ x × √x + h.
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This = -1/ x × x + h × √x × √x + h.
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Again, this is just algebraic manipulation, that is all it is, /√x + √x + h.
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That is fine, I will go ahead and write out the whole thing, not a problem.
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This is going to be -1/ x × x + h ×, this is √ × √, I can combine them.
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It is just going to be x × x + h, under the radical together, / √x + √x + h = -1/,
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This can cancel with that, one of these.
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I’m left with √x × x + h.
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On the bottom, this one, this cancels one of the radicals here.
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This just leaves me with that.
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This is going to be × √x + √x + h.
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Now I take the limit of this.
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I take the limit as h goes to 0.
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This is just the same as f(x), I have just manipulated it.
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The limit as h goes to 0.
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Now I take h to 0 and I end up with -1/ √x² × √x + √x.
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Let me go to the next page, it is not a problem.
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It = -1/ √x × 2√x = -1/ 2, just √x, x².
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x² × x is x³.
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There you go, that is your derivative.
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Or we can write it as 2x³/2, that is another way that we can write it.
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Again, that is one possibility, or a third possibility is you can bring this up to the top.
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You can write -, this x³/2 is going to be x⁻³/2 /2.
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Any one of these is absolutely fine, it is not a problem.
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Notice that f’(x), we have this thing right here.
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If it is 0, if x is 0, we are going to end up with 0 in the denominator.
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The original function f(x) is not differentiable at 0.
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If you try to graph it, you can see that it is not going to be differentiable at 0.
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Analytically, you can see that it is not going to be differentiable at 0.
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Let us try something else here.
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The following is a graph of f(x), f’, and f”.
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Our first and second derivatives along with the original function.
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Decide which is which and be able to explain why you made the choices that you made.
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Again, when you are dealing with situations like this, you work with 0 slopes.
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You look for a function, you look for the graph that has a horizontal tangent.
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You see where it hits 0 on the x axis, that is going to be the derivative of that function.
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Again, work with horizontal tangent which is 0 slopes.
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Let me work in black actually because one of my graphs is blue.
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I have a horizontal tangent here.
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I find that the blue graph is where it hits 0.
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The blue graph is the derivative of the red graph.
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Now the blue graph has a horizontal tangent here.
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As I go straight up from there, I notice the green graph has crosses 0 at the x value
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where the blue graph has a horizontal tangent, which means that the green graph is the derivative of the blue graph.
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What you get is f, the original function is the red graph.
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The blue is f’, the first derivative.
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The green is going to be the f”.
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That is it, just work with horizontal tangents and see where it is.
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The graph has a horizontal tangent, see where the x value crosses 0, that is the derivative of that function.
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Find an equation for the tangent line to the graph of the following function at the given x value.
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A tangent line, we need to find the derivative.
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We need to find the y value, whatever that is going to be, for the original function.
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And then, we are going to do y - y1 = the slope.
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We are going to put the x value into the derivative function to get the slope, × x – x.
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x1 is 4, y1 we are going to find.
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Let us get started.
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We know that f’(x) is equal to the limit as h approaches 0 of f(x) + h - f(x)/ h.
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Again, I'm going to work with just the difference quotient.
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I'm going to get the f(x) + h is 5 × x + h/ x + h + 3² - f(x) which is 5x/ x + 3², and all of that is going to be /h.
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I'm going to end up with, I’m going to take a common denominator here.
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I’m going to get 5 × x + h × x + 3² - 5x × x + h + 3²/ x + h + 3 × x + 3²/ h.
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I’m not going to do the expansion.
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I’m going to leave the algebra to you.
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When you expand, when you multiply everything out, when you expand, multiply, and simplify, you get the following.
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Once you simplify it, you take the limit as h goes to 0.
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In other words, you put 0 in wherever h shows up in the expression.
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You are going to get your derivative.
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You get f’(x) is equal to -5x + 15/ x + 3³, that is the derivative of the function.
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Now the original f(x), that was equal to 5x/ x + 3².
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We need to find the y value for where the original function = 4.
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This is going to be the hardest part, when dealing with derivatives and functions
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is making sure you keep the original function separate from the derivative function.
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They are not the same.
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Here we want to find f(4), the original function.
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You are going to end up with 20/49, when you put it in.
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The tangent line touches the graph of the original function at the point 4, 20/49.
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Now the slope, that is the derivative.
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We want f’ at 4.
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That is going to equal -5 × 4, we are using this one, + 15/ 4 + 3³.
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It is going to end up equaling -5/ 343, if I done my arithmetic correctly which I often do not.
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Please check that.
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My slope is -5.
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The equation of the tangent line to the original function at x = 4 is, we said it is y - y1 = the slope × x - x1.
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It is going to be y - 20/ 49, that is our y value = our slope -5/ 343 × x – 4.
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That is it, nice and simple.
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You find the derivative, you find the slope.
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You find the x and y values of the point that it passes through.
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Now you got a slope and you got a point, you are done.
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The following graph shows a distance vs. time graph for two runners.
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One runner is the blue.
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One runner is our blue graph and runner two is our red.
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Answer the following questions.
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Describe the race from beginning to end.
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Distance vs. time.
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Distant on the y axis, time on the x axis.
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The slope of the graph is the y axis/ the x axis.
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y/x, Δ y/ Δ x.
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Distance/ time, we already know what distance/ time is.
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Distance/ time = velocity.
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The slope at any point along the graph tells me how fast the runner is going.
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It is always going to be that way.
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The derivative is always going to be the slope, you know that already.
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In this case, it is a physical application.
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Here the y value is distance, the x value is time.
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Distance/ time, velocity.
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In this case, the slope gives me the velocity of the person at any given moment.
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We notice that the red runner, it is constant velocity.
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It is one speed.
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The blue runner starts slowly, his slope is almost horizontal but he speeds up.
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At some point, his slope is actually steeper than the slope of the red.
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It is actually going faster than the other person.
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What is happening here is very simple.
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They start running, the person in red is running faster initially.
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At some point, where the slopes are the same, they are the same speed but the red is further ahead.
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However, as the blue starts to speed up his velocity, eventually, at this value,
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it looks like maybe 14 or 15 seconds, he catches up with the first one.
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They are actually at the same distance, that is what is happening.
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Red, blue catches up eventually, that is what is happening.
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That takes care of part A, when do the runners have the same velocity?
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Velocity is a slope, they have the same velocity when the slope of the blue graph is the same as,
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We look for something like that when they are parallel.
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Maybe somewhere around 8.2 seconds or something like that.
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When is the distance between the runners greatest, the part c.
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The distance is greatest when the gap between them, when that distance is the greatest.
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It looks like it is somewhere around 8 seconds.
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You can argue, maybe it is 7 or 9, something like that.
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But basically you are looking for the gap between the two.
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The distance, that is the y axis, when is that distance the greatest.
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That is it, nice and simple.
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Again, slope is the y axis/ the x axis.
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In any given real world situation, it might be temperature time, it might be distance time, whatever it is.
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In this case, it was distance/ time.
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Distance/ time, we know it is velocity.
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The displacement of a particle from the origin.
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As it moves along the x axis it is given by x(t) = t³ – 7t² + 12t – 3.
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What this means is that as time goes from 0 forward, 1, 2, 3, 4, 5, 6, 7 seconds,
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when I put both t values in here, the number that I get, the x tells me where on the x axis I am.
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It gives the x coordinate, that is what this means.
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Find the displacement, the velocity, and the acceleration of the particle at t = 5 seconds.
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It is asking me, at 5 seconds, where is it, how fast is it going, in what direction, and is it speeding up or slowing down.
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Displacement, where is it.
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Velocity, how fast is it moving and in what direction.
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Positive velocity to the right.
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Negative velocity to the left.
00:24:31.000 --> 00:24:35.500
We are on the x axis, we are moving this way and this way.
00:24:35.500 --> 00:24:38.800
Positive velocity is that way, negative velocity is that way.
00:24:38.800 --> 00:24:43.600
Acceleration, if it is positive, it is speeding up at that point.
00:24:43.600 --> 00:24:47.000
If it is negative, it is slowing down.
00:24:47.000 --> 00:24:50.600
You can have a particle with a positive velocity and a negative acceleration.
00:24:50.600 --> 00:24:54.200
It is moving to the right but it is slowing down.
00:24:54.200 --> 00:24:59.900
It is getting ready to stop.
00:24:59.900 --> 00:25:04.100
f(t), displacement.
00:25:04.100 --> 00:25:11.900
It is t³ – 7t² + 12t – 3.
00:25:11.900 --> 00:25:14.000
We want to know what x of 5 is.
00:25:14.000 --> 00:25:17.300
When we put it in, we get 7.
00:25:17.300 --> 00:25:27.600
At 5 seconds, on the x axis, that point, my point is there.
00:25:27.600 --> 00:25:43.900
Now velocity, the velocity is a function of time, it is equal to the first derivative of the displacement function.
00:25:43.900 --> 00:25:51.000
The derivative of this function, you can either do it by doing the definition of derivative
00:25:51.000 --> 00:25:54.600
or you can do that thing that we told you about.
00:25:54.600 --> 00:26:04.500
3t² - 14t + 12, remember, I think we discussed that real quickly at the end of one lesson
00:26:04.500 --> 00:26:06.400
or at the end of the problem lessons.
00:26:06.400 --> 00:26:08.500
You take this number, you bring it down here.
00:26:08.500 --> 00:26:12.100
You multiply it by that and you reduce the degree by 1.
00:26:12.100 --> 00:26:13.600
This is the velocity function.
00:26:13.600 --> 00:26:18.800
At any time t, this function tells me how fast I’m moving.
00:26:18.800 --> 00:26:23.600
The velocity at 5 = 17.
00:26:23.600 --> 00:26:36.200
This is positive which means that I'm here and I’m actually moving to the right at 17 m/s, ft/s, mph, whatever my unit is.
00:26:36.200 --> 00:26:38.600
Now the acceleration function.
00:26:38.600 --> 00:26:44.600
The acceleration at any given time t is equal to the derivative of the velocity,
00:26:44.600 --> 00:26:49.700
which is equal to the second derivative of the displacement.
00:26:49.700 --> 00:26:52.100
Now I take the derivative of that.
00:26:52.100 --> 00:26:57.200
2 × 3 is 6, 6t, drop the degree by 1.
00:26:57.200 --> 00:27:06.500
1 × 14 is 14, -14, drop it by 1, it is just t⁰ which is 1.
00:27:06.500 --> 00:27:11.000
It just stays 14, that is my acceleration function.
00:27:11.000 --> 00:27:18.500
My acceleration at 5 is equal to 16.
00:27:18.500 --> 00:27:21.800
This is positive.
00:27:21.800 --> 00:27:26.000
I’m at 7, I'm traveling at 7, let us say m/s.
00:27:26.000 --> 00:27:30.200
I’m accelerating, I’m speeding up going to the right.
00:27:30.200 --> 00:27:39.300
Not only am I going to the right, but I’m actually getting faster going to the right, which means 16 m/s² acceleration.
00:27:39.300 --> 00:27:42.600
At t = 5, I’m at 17.
00:27:42.600 --> 00:27:46.200
At 6 seconds, 1 second later, I’m at 17 + 16.
00:27:46.200 --> 00:27:48.900
I’m 33 m/s.
00:27:48.900 --> 00:27:55.200
One second after that, I’m at 33 + 16, I’m 49 m/s.
00:27:55.200 --> 00:28:03.600
I'm here, I’m traveling this fast, and I’m speeding up that fast per second.
00:28:03.600 --> 00:28:11.400
Let us say this is meters, this is going to be meters per second, this is going to be meters per square second.
00:28:11.400 --> 00:28:14.400
Displacement, when you take the first derivative, you get the velocity.
00:28:14.400 --> 00:28:21.600
When you take the derivative of velocity, you get the acceleration.
00:28:21.600 --> 00:28:25.000
Now graph the displacement function from the previous problem.
00:28:25.000 --> 00:28:30.900
Then use the graph to describe the motion of the particle during the first 5 seconds.
00:28:30.900 --> 00:28:36.100
The function was x(t) = t³ – 7t² + 12t – 3.
00:28:36.100 --> 00:28:38.200
Here is the graph of the function.
00:28:38.200 --> 00:28:42.400
We want you to describe what is happening to this particle.
00:28:42.400 --> 00:28:47.200
Describe its motion during the first 5 seconds.
00:28:47.200 --> 00:28:49.000
5 seconds is here.
00:28:49.000 --> 00:28:55.000
Between 0 seconds and 5 seconds, what is the particle doing?
00:28:55.000 --> 00:29:02.500
At t = 0, I'm at, it looks like -3.
00:29:02.500 --> 00:29:06.100
Let me draw it up here.
00:29:06.100 --> 00:29:13.100
Here is my 0, let us say this is -3.
00:29:13.100 --> 00:29:16.400
I start at -3, that is what this graph is telling time.
00:29:16.400 --> 00:29:18.500
At time = 0, it is this way.
00:29:18.500 --> 00:29:29.600
Now my slope is positive which means that the particle is actually going to be moving to the right.
00:29:29.600 --> 00:29:33.200
It is positive but it is slowing down.
00:29:33.200 --> 00:29:36.500
At 1 second, the velocity hits 0.
00:29:36.500 --> 00:29:42.500
At 1 second, the particle actually stops momentarily.
00:29:42.500 --> 00:29:46.700
The velocity which is the slope becomes negative and becomes more negative.
00:29:46.700 --> 00:29:51.900
Now the particle is moving this way.
00:29:51.900 --> 00:30:00.100
Again, but it starts to slow down and at just about 3.5 seconds, it hits 0 again.
00:30:00.100 --> 00:30:02.800
The slope is 0, the velocity is 0 again.
00:30:02.800 --> 00:30:05.200
Now the velocity becomes positive.
00:30:05.200 --> 00:30:08.800
It turns around again and starts moving to the right.
00:30:08.800 --> 00:30:12.500
This time the slope starts accelerating to the right.
00:30:12.500 --> 00:30:13.700
That is what is happening.
00:30:13.700 --> 00:30:20.300
At 5 seconds, it actually hits 7.
00:30:20.300 --> 00:30:28.700
It starts at -3, it starts moving to the right.
00:30:28.700 --> 00:30:32.300
Stops momentarily then starts moving to the left.
00:30:32.300 --> 00:30:35.300
Stops again and starts moving to the right.
00:30:35.300 --> 00:30:40.400
At 5 seconds, I’m at 7, that is what is happening.
00:30:40.400 --> 00:30:45.200
Once again, if this is my 0, let us say this is my 3.
00:30:45.200 --> 00:30:47.600
Let us say this is my 7.
00:30:47.600 --> 00:30:53.700
My particle starts here, moves to the right, moves to the left, moves to the right.
00:30:53.700 --> 00:30:57.000
At 5 seconds, I'm there, that is what is happening.
00:30:57.000 --> 00:30:58.500
That is what the graph is telling me.
00:30:58.500 --> 00:31:01.200
Again, distance vs. time.
00:31:01.200 --> 00:31:04.500
Time is on this axis, x is the displacement.
00:31:04.500 --> 00:31:08.400
It is the distance, it is how far I am along the x axis.
00:31:08.400 --> 00:31:15.000
The graph is not describing the path that the particle is following.
00:31:15.000 --> 00:31:16.500
It is not moving in two dimensions.
00:31:16.500 --> 00:31:18.300
It is moving in one dimension.
00:31:18.300 --> 00:31:21.900
The path is telling you what is happening in that one dimension.
00:31:21.900 --> 00:31:23.700
The slope is the velocity.
00:31:23.700 --> 00:31:27.400
Positive velocity moving to the right.
00:31:27.400 --> 00:31:29.300
0 velocity, it stops momentarily.
00:31:29.300 --> 00:31:35.500
Negative velocity, it is heading the other direction, that is what is happening.
00:31:35.500 --> 00:31:37.300
Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.