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, we are going to continue our practice exam.
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It is going to be Section 1 Part B, where the calculators are actually allowed.
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You are going to be using the calculator regularly in this section.
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Let us go ahead and get started.
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Let me let you know once again where you can find this, in case you already forgot or did not see it.
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Of course, there is a link down below in the quick notes.
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I will go ahead and write it here.
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You can find this at www.online.math.uh.edu/apcalculus/exams.
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When you pull out this page, it is going to be section 1 part B.
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The version that we are going through is version 5.
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There are multiple exams on this page.
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I would recommend just going through all of them as practice.
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But we are going to be doing version 5 for this.
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Let us get started, problem number 1.
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I kind of like this purple, it is nice.
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That is okay, I will go ahead and work in blue, I think.
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It is a nice color that we usually work in.
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Number 1 asks us to give a value of c that satisfies the conclusions of the mean value theorem for a given function.
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Our function is f(x) = x² – x – 1.
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The interval that we are concerned with is 1 to 3.
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The mean value theorem basically says that, if f is continuous on a closed interval, it is differentiable on the open interval.
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Then, there is some number c in that interval such that f’ at c is going to equal the actual slope from point 1 to point 3.
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Let us go ahead and run through this.
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Looking at this, it is definitely continuous.
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Any polynomial, all polynomials are continuous, or the entire real line.
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The first hypothesis is satisfied.
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F is continuous on the closed interval 1,3.
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F is a definitely differentiable because polynomials are differentiable everywhere.
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F is differentiable on the open interval 1,3.
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The conclusion that we can draw, there exists some number c such that c is between 1 and 3,
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such that f’ at that value of c is equal to f(3) – f(1)/ 3 – 1.
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That is what the mean value theorem says.
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Let us go ahead and take the first derivative because we are going to be looking for f’.
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F’ at x is equal to 2x – 1.
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F’ at c is nothing more than 2c – 1.
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We are going to be looking for that value of c.
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Let us go ahead and calculate this value and set this f’(c) equal to this and solve for c.
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That is all you have to do.
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F(3), when I put it in here, I’m going to get a 9 – 3 – 1 is equal to 5.
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f(1) is equal to 1 - 1 - 1 is equal to -1.
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f(3) - f(1)/ 3 - 1 is equal to 5 - a -1/ 3 – 1.
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This is going to be 6/2.
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It is going to equal 3.
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Therefore, we have, f’ at c is going to equal 3 or f’ at c is 2c – 1.
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2c - 1 = 3, then we just solve for c.
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2c = 4 and c = 2.
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Our choice is going to be c, for this one.
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Nice and straightforward, just have to make sure that the hypotheses of the mean value theorem are satisfied.
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It has to be continuous on the closed interval and it has to be differentiable on the open interval.
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If either of those two is not satisfied, you cannot use the mean value theorem.
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Hypothesis is very important.
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Number 2, it asks us to give us a function.
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Let us go ahead and write this function down.
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Our f(x), we have got 4x³ + 2 × e ⁺x.
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It tells us that this function is invertible, it wants us to find the derivative of the inverse at x = 2.
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What it is asking us to find is f inverse' at 2, that is what we want.
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We are going to start off by graphing this.
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Because again, you have this exponential function, you have your calculator, go ahead and graph it to take a look at it.
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The graph is going to look something like this.
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This over here, it is going to go something like that.
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This is f, this is going to be the original function.
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Now the inverse of that, we know what the inverse is.
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The inverse is just a reflection about the line y = x.
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Let me go back to blue here.
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The inverse is going to look something like that.
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This value is 2, f(0) is 2.
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This graph is f inverse, what they want is the slope.
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They want the derivative of the tangent line.
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They want the slope of the curve at that point.
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f inverse' at 2.
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We actually know how to do this.
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It turns out that the derivative of the inverse is nothing more than the reciprocal of the derivative of the original function.
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f inverse', 1/ f’ at a given point.
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We just have to be very careful at which point we are doing it.
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Here is what is going to happen.
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What we are looking for, f inverse' at 2 is actually going to be the inverse of 1/ f’ at 0, not 1/ f’ at 2.
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Here is why, this point is 2,0, it is inverse point, if you will, is 0,2.
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The x value for the inverse function is 2 but this 2, because inverse is switch x and y values,
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when we go up to 2,0, the x value, this point which is the essentially the inverse point of this, its x value is 0.
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Essentially, it is just going to be the reciprocal of that slope, right there.
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That slope is f’ at 0.
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F’ inverse at 2 is not 1/ f’ at 2.
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You have to find the f inverse’ at 2.
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You have to find where, that is going to be the y value of the original function.
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Up here, it is going to be 1/ f’ at 0.
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We need to find f’at (0), I hope that make sense.
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Be very careful, that is why it is a good idea to actually graph this.
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The argument for what we are looking for is going to have to be the y value of the original function,
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where the inverse function is this.
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Let us go ahead and do f’(x).
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F’(x) is going to equal 12x² +,
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I’m sorry, my real function was wrong, e ⁺2x.
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There you go.
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Yes, is it e ⁺2x, now I’m getting confused, 2e ⁺2x.
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What function am I looking at here?
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I think we are good.
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4x³ + 2e ⁺2x, the derivative is going to be 12x² + 2e ⁺2x × 2, 4e ⁺2x.
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We are going to find f’(0), it is going to be 12 × 0 which is 0.
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e⁰ is going to be 1, this is going to be 4.
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f inverse’ at 2 is equal to 1/ f’ at 0.
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It is just equal to ¼, it is going to be c.
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Sorry about the little confusion there.
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I have written two different things on the page and I got a little confused for a moment.
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Let us go on to question number 3.
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What is question number 3 asks us?
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Question 3 is asking us to give a value.
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We actually have a graph, let me see it here.
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They gave us a graph and this graph looks like that as it passes through 0 and it passes through 3.
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1, 2, and 3, it is a little bit of parabola like that.
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It actually passes through that point.
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They tell us that this is actually a graph of f’(x).
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This is the derivative of f.
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What they are asking us to do is they want us to give a value of x where f actually achieves a local minimum.
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A local min, we know that local min or local max, we know that f’(x) has to equal 0.
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Since this is the f’ graph, it is going to be either this point or this point.
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If I look at 3, if I look at the point 3, and the other point is point 0.
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It is either going to be 3 or 0, which one of those is it going to be?
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To the left of 3, the derivative.
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This is the derivative graph, it is negative, it is decreasing.
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To the right of 3, it is positive, it is above the x axis.
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It is increasing.
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Therefore, decreasing then increasing, our local min is at x = 3.
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That is how we do with.
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0, to the left you have a positive, it is increasing and it is decreasing.
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Here it is actually going to be a local max there.
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They want the local min, x = 3.
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Our choice is a.
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Number 4, basically, you are just going to be graphing the function that they gave you.
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We have a graph right here and let us go ahead and say what this graph is.
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I will go ahead and do over here.
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f(x) =, the piece wise graph, we have got - x - 5 whenever x is less than -2.
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We have x² + 1, when x is greater than or equal to -2, less than or equal to 1.
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We have 2x³ -1, when x is greater than 1.
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That is it, 3 piece wise function.
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Now we answer some questions about this question.
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Question number 1 is, is f continuous at x = -2.
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-2, no, it is not continuous there.
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Question 2 is asking us is f differentiable at x = 1?
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At 1, we also have a discontinuity here.
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The left hand limit and the right hand limit are not equal, no it is not differentiable at x = 1.
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3, is x = 0 a local minimum?
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Is x = 0 a local min?
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Yes, it is a local min because to the left and right of that, the values of the function are higher than that.
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Yes, three is definitely true.
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Number 4, is x - 2 an absolute max?
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x - 2 an absolute max, no because there is something higher over here.
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Only 3 is true, our choice is c.
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That is it, graph it, read it right off the graph.
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Let us see what question 5 has for us.
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Question 5, we are given some integrals and we are asked to determine another definite integral.
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They give us that the integral from 0 to 50 of 4 f(x) dx is equal to 3.
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They tell us that the integral from 2 to 50 of f(x) dx is equal to 2.
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They want us to evaluate the integral from 0 to 2 of f(x) dx.
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This is just an application of the properties of the integral.
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We are going to go ahead and just write the integral from 0 to 2 of f(x) dx, which is what we want.
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It is going to equal 1/4 of 4 f(x) dx which is the integral from 0 to 50 +,
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this upper is going to come down here, so that they end up canceling, so to speak.
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50 to 2 of f(x) dx.
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This is equal to this.
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Now I just plug the values in.
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This is equal to ¼, I know that the integral from 0 to 50 is this one, 3.
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This is -2, if I’m not mistaken.
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Yes, that is -2.
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The integral from 2 to 50 is -2.
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The integral from 50 to 2 is going to be the negative of that.
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It is going to be - -2.
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When I add these together, I’m going to end up with 11/4, the choice is d.
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I hope that made sense.
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Just be very careful.
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Here what I have done is, by flipping the integral, I’m basically taking the negative of this one, - -2.
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I hope that made sense.
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Number 6, what is number 6 asks us?
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It wants us to find an approximate location of the local maximum for the particular function.
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The particular function that they gave us is this and I have gone ahead and graphed it.
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F(x) = 4x³ + 3x² - 2x.
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That is it, basically, just go ahead and graph it and use your zoom feature, your trace feature on your calculator,
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whatever your calculator uses or calculate, basically, to find this.
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They want the value of a local maximum.
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That is it, use the zoom feature, in this particular case, it turns out to be this point -0.7287, 1.502.
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In this case, our choice is going to be b.
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That is it, just graph it and read it right off.
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Either zoom or calculate it directly, however is it that you want, go ahead and do it.
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Let us move on to number 7.
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Number 7 asks us to find the approximate average value of function.
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The function that we are concerned with is going to be 4x ln 3x.
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Let me just double check and make sure that it is correct, 4x ln 3x.
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On the interval, of course, average value is always specified on an interval, on the interval 1 to 4.
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Very simple, we know what the average value is, it is equal to 1/ b – a
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where the first number is a and the second number is b × the integral from a to b of the function itself.
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That is it, it is the same as any other average.
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When you take an average, let us say of 10 numbers, you add those 10 numbers and you divide by 10.
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An interval is continuous, it is not 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
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You are essentially adding an infinite number of numbers.
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The sum, if you remember the integral is just the sum.
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Once you add the numbers, that is the integral.
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And then, you divide by how many there are.
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Where, how many there are, is basically the length of the interval.
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This is the same average that you are used to, always.
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This is nice and simple.
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1/ 4 – 1/ the integral 1 to 4, 4x of ln of 3x dx.
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Again, just go ahead and use your calculator to evaluate the integral numerically.
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You are going to get an approximate value of 20.77.
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Your choice is going to be d.
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Number 8, let us see what number 8 has for us.
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We have a region that is enclosed by the graphs of a couple of functions and
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we rotate it around the y axis to generate a solid.
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We want to find the volume of this solid.
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The two functions are y = x³ - 1 and y = x – 1.
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You can go ahead and graph it or if you know what it looks like,
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you can graph it by hand and save the calculator for the actual evaluation of the integral.
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The region is going to look like this.
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1, this is going to be the x³ - 1 graph.
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The x - 1 is going to be the line going this way, something like that.
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Essentially, we are going to be rotating this around the y axis.
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We are going to end up getting that and we are going to end up getting that.
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We want the volume of that solid.
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We can do this with washers, we can do this with shells.
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I think we can just go ahead and do it with shells.
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We are basically going to take a volume element.
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And then, we are going to integrate that volume element.
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Let me go ahead and work in red for this one.
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A volume element, I decided I think I'm going to go ahead and go shells.
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I'm going to take that and that.
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Basically, a little bit of a cylindrical shell.
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And then, I’m going to integrate, add them up, all over from here to here, from 0 to 1.
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That is how I’m going to go ahead and do this.
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Basically, we are going to take the area of a cylindrical shell.
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The area of the cylindrical shell is going to be 2π x × the height.
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This cylindrical shell, if you will, something like that.
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We are just going to add them all up.
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This is going to be a cylindrical shell.
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This thickness is going to be dx.
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What we want is, we want the surface area then we are going to add them up.
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That is going to be our dx.
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2π x × h.
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X is going to be the radius, then 2π r is going to be the circumference.
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And that multiply by h which is that, that is going to give me the area of the outside of the circular shell.
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The dx is going to give me a volume.
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The volume element is 2π x h dx.
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H itself, how do we find this height?
00:23:50.500 --> 00:23:55.900
The height is going to be the difference between that and that.
00:23:55.900 --> 00:23:59.200
The height is going to be the function x – 1.
00:23:59.200 --> 00:24:05.500
For any value x, I have this and I have this.
00:24:05.500 --> 00:24:13.900
I’m going to do x - 1 - x³ – 1.
00:24:13.900 --> 00:24:16.900
It is going to be this function - this function.
00:24:16.900 --> 00:24:20.500
That is going to give me the difference, the height.
00:24:20.500 --> 00:24:22.400
I hope that makes sense.
00:24:22.400 --> 00:24:30.800
If I have a particular x, it is going to be this height - this height.
00:24:30.800 --> 00:24:38.100
It is this height, they give me essentially just the difference between the two functions.
00:24:38.100 --> 00:24:46.000
I get x = x - 1 – x³ + 1
00:24:46.000 --> 00:24:52.300
The 1 cancel, I get a height equal to x - x³.
00:24:52.300 --> 00:25:00.600
Therefore, my volume element is going to be 2π x.
00:25:00.600 --> 00:25:09.300
My h his x - x³ and it is going to be × 2 × dx.
00:25:09.300 --> 00:25:12.800
Now that takes care of this.
00:25:12.800 --> 00:25:25.500
I have to multiply this by 2 because now I have the bottom part.
00:25:25.500 --> 00:25:47.100
Therefore, I have a dv element of 4π × x² – x⁴ dx.
00:25:47.100 --> 00:25:48.300
Now I can integrate.
00:25:48.300 --> 00:25:51.900
Now the volume is going to be the integral from 0 to 1.
00:25:51.900 --> 00:26:00.300
I’m going to add up all the cylindrical shells.
00:26:00.300 --> 00:26:03.300
That is fine, I will just leave it in here, it is not a problem.
00:26:03.300 --> 00:26:13.700
X² - x⁴ dx and when I evaluate that, I get a volume equal to 1.676 cubic units.
00:26:13.700 --> 00:26:16.100
The choice is actually d.
00:26:16.100 --> 00:26:17.900
I hope that made sense.
00:26:17.900 --> 00:26:22.400
You could have done this with washers.
00:26:22.400 --> 00:26:27.500
We take that region and we draw it like that.
00:26:27.500 --> 00:26:32.500
You could have done it this way but then and you have to integrate along y.
00:26:32.500 --> 00:26:36.200
You are going to have to change this, which is why I went with shells.
00:26:36.200 --> 00:26:39.500
It was given in terms of x and y, as a function of x.
00:26:39.500 --> 00:26:41.300
I integrated along x.
00:26:41.300 --> 00:26:48.900
Along x, I decided to just use the shells.
00:26:48.900 --> 00:26:57.900
Number 9, let us see, instantaneous rate of change.
00:26:57.900 --> 00:27:01.800
They want to know the approximate instantaneous rate of change of a given function.
00:27:01.800 --> 00:27:08.100
This function f(t) is actually expressed as an integral.
00:27:08.100 --> 00:27:19.800
0 to 4t of cos(x) dx.
00:27:19.800 --> 00:27:23.700
Fundamental theorem of calculus tell us that the instantaneous rate of change,
00:27:23.700 --> 00:27:26.400
we know that an instantaneous rate of change is the derivative.
00:27:26.400 --> 00:27:39.600
The derivative f’ t is going to be, when I have the function expressed as an integral,
00:27:39.600 --> 00:27:43.200
I go ahead and just drop the integral sign, when I take the derivative.
00:27:43.200 --> 00:27:48.600
It is going to be cos, I put that in here, of 4t.
00:27:48.600 --> 00:27:57.600
But because this is not t, it is a function of t, I have to multiply by the derivative of that.
00:27:57.600 --> 00:28:02.100
They want us to evaluate the instantaneous rate of change at a certain point.
00:28:02.100 --> 00:28:07.900
What they want is f’ at π/4.
00:28:07.900 --> 00:28:18.100
f’ at π/4, when I put π/4 into this expression, I get 1.662.
00:28:18.100 --> 00:28:25.300
That is it, nice and simple, just a straight application of the fundamental theorem of calculus I’m plugging in.
00:28:25.300 --> 00:28:37.100
Number 10, we have a particular integral and they want to know what the area is
00:28:37.100 --> 00:28:44.200
when this integral is evaluated by a trapezoidal rule for n = 3.
00:28:44.200 --> 00:28:52.000
The integral that they gave us is 0 to 1 of sin of π(x) dx.
00:28:52.000 --> 00:28:55.600
Essentially, what we are going to do is we are going to calculate this integral with our calculator.
00:28:55.600 --> 00:28:58.700
We are going to do a trapezoidal rule approximation.
00:28:58.700 --> 00:29:02.200
We are going to subtract 1 for the other and that is going to be our area.
00:29:02.200 --> 00:29:06.100
In this particular case, n = 3.
00:29:06.100 --> 00:29:21.100
When n = 3, the integral from a to b of f(x) dx, the trapezoidal approximation is going to b Δ x/ 2.
00:29:21.100 --> 00:29:23.800
This is just a formula that you have to know.
00:29:23.800 --> 00:29:39.100
It is going to be f(x) is 0 + 2 f(x1) + 2 × f(x2) + f at x3.
00:29:39.100 --> 00:29:42.100
n = 3, you are going to have four terms.
00:29:42.100 --> 00:29:52.000
X0 and x3 and the n, and 2 × down below or Δ x is equal to be - a/ n.
00:29:52.000 --> 00:29:59.200
In this case, b - a/ n.
00:29:59.200 --> 00:30:05.600
b - a is equal to 1 - 0/ 3.
00:30:05.600 --> 00:30:08.400
Δ x = 1/3.
00:30:08.400 --> 00:30:14.900
We just need to find f(x) 0, f(x1), f(x2), f(x3).
00:30:14.900 --> 00:30:28.500
When I plug those in, f(x) is 0, it is going to be this is f right here, sin(π) x.
00:30:28.500 --> 00:30:41.500
It is going to be sin(π) × 0 which is equal to 0.
00:30:41.500 --> 00:30:49.300
Let me write out everything explicitly here, and a little more clear.
00:30:49.300 --> 00:31:10.900
We have x(0) = 0, that is going to give us, f(x0), π × 0 = 0.
00:31:10.900 --> 00:31:20.200
x₁ is equal to 1/3 because Δ x is 1/3, it is going to be 0, 1/3, 2/3, 1.
00:31:20.200 --> 00:31:24.400
We are breaking up this interval into that many parts.
00:31:24.400 --> 00:31:26.500
x1 is equal to 1/3.
00:31:26.500 --> 00:31:44.700
Therefore, f of 1/3 is going to equal the sin(π/3) which is equal to 0.86603.
00:31:44.700 --> 00:31:57.400
x2 is equal to 2/3, therefore, the f(2/3) is going to equal the sin of 2π/ 3.
00:31:57.400 --> 00:32:02.800
It also equals 0.86603.
00:32:02.800 --> 00:32:11.500
x of 3 is equal to 1, f(1) is equal to the sin(π).
00:32:11.500 --> 00:32:16.900
The sin(π) is equal to 0.
00:32:16.900 --> 00:32:24.400
We have the trapezoidal approximation is equal to, we said Δ x/ 2.
00:32:24.400 --> 00:32:28.300
It is going to be 1/3/ 2.
00:32:28.300 --> 00:32:29.500
I will do it in this way.
00:32:29.500 --> 00:32:32.800
I will actually write everything out.
00:32:32.800 --> 00:32:46.600
0 + 2 × 0.86603 + 2 × 0.86603 + 0.
00:32:46.600 --> 00:32:55.100
That is going to give some answer of 0.577353.
00:32:55.100 --> 00:33:11.600
The actual value of the integral itself, when I evaluate this integral, it is going to equal 0.63662.
00:33:11.600 --> 00:33:15.200
I take that number - that number.
00:33:15.200 --> 00:33:30.500
I get an area = 0.63662 - 0.577353.
00:33:30.500 --> 00:33:35.900
It is going to give me an area of 0.05927.
00:33:35.900 --> 00:33:41.000
Our choice of is d, that is it, just got to know this formula.
00:33:41.000 --> 00:33:44.300
That is it, Δ x/ 2, and then, however many you have.
00:33:44.300 --> 00:33:46.400
If n is 6, you are going to have 7 terms.
00:33:46.400 --> 00:33:48.800
If n is 19, you have 20 terms.
00:33:48.800 --> 00:33:52.400
Always n + 1 terms in a trapezoidal approximation.
00:33:52.400 --> 00:34:04.000
The 1 and n stay, f(x) and f(x) ⁺n, the ones in between all are multiplied by a factor of 2.
00:34:04.000 --> 00:34:06.700
Not too bad.
00:34:06.700 --> 00:34:10.300
What does number 11 ask us?
00:34:10.300 --> 00:34:14.800
They are telling us that the amount of money in a bank is increasing at a certain rate.
00:34:14.800 --> 00:34:17.200
The rate is going to be dollars per year.
00:34:17.200 --> 00:34:21.100
They give us a time, a year where t = 0.
00:34:21.100 --> 00:34:24.100
In this case, it turns out to be 2005.
00:34:24.100 --> 00:34:30.900
They want to know, what is the approximate total amount of increase from 2005 to 2007?
00:34:30.900 --> 00:34:46.300
The rate at which the money is increasing is equal to 10,000 × e⁰.06 t.
00:34:46.300 --> 00:34:53.300
It is going to be that many dollars per year.
00:34:53.300 --> 00:34:59.000
t is in years.
00:34:59.000 --> 00:35:04.700
They tell us that t = 0, corresponds to the year 2005.
00:35:04.700 --> 00:35:12.800
They want to know the increase, this is the rate of increase,
00:35:12.800 --> 00:35:20.300
they want to know what the increase is from the year 2005 to 2007.
00:35:20.300 --> 00:35:26.900
If t(0) is 2005, then t = 2 is 2007.
00:35:26.900 --> 00:35:40.700
What we are going to do, the increase from 2005 to 2007.
00:35:40.700 --> 00:35:47.000
If the rate of increases this amount per year, I multiply by the number of years.
00:35:47.000 --> 00:35:50.000
But I integrate, because it is a function of t.
00:35:50.000 --> 00:35:56.300
It is the integral from 0 to 2 of r(t) dt.
00:35:56.300 --> 00:36:03.500
That is it, I just calculate that integral with my calculator and it turns out to be $21,250.
00:36:03.500 --> 00:36:06.800
That is it, I hope that makes sense.
00:36:06.800 --> 00:36:11.300
Dollars per year × year.
00:36:11.300 --> 00:36:14.100
Years cancels year, you are left with dollars.
00:36:14.100 --> 00:36:18.000
But because the rate of increase, the dollars per year is not constant.
00:36:18.000 --> 00:36:21.300
I cannot just multiply them, I have it integrate it.
00:36:21.300 --> 00:36:29.100
That is it, very simple.
00:36:29.100 --> 00:36:34.800
Number 12, let us see what is number 12 asking us.
00:36:34.800 --> 00:36:41.700
A particle is moving with a certain acceleration which they give us here and its initial velocity is 0.
00:36:41.700 --> 00:36:46.800
For how many values of t does the particle change the direction?
00:36:46.800 --> 00:36:53.400
A particle changes direction, when velocity goes from positive to negative or negative to positive.
00:36:53.400 --> 00:37:02.400
When the velocity curve actually crosses the x axis, that is when it changes direction.
00:37:02.400 --> 00:37:07.200
The acceleration that they give us here,
00:37:07.200 --> 00:37:14.200
Our acceleration that they give us is 2t² - 4t.
00:37:14.200 --> 00:37:17.200
They tell us that the initial velocity is 0.
00:37:17.200 --> 00:37:24.400
Was that correct , yes, the initial velocity is 0.
00:37:24.400 --> 00:37:35.500
For how many values of t does it change direction?
00:37:35.500 --> 00:37:40.300
For how many values of t does the velocity curve cross the x axis?
00:37:40.300 --> 00:37:43.000
We need to find the function for the velocity.
00:37:43.000 --> 00:37:45.700
We know what the velocity is.
00:37:45.700 --> 00:37:49.600
The velocity is just the integral of the acceleration.
00:37:49.600 --> 00:37:58.900
It is going to be integral of 2t² - 4t dt.
00:37:58.900 --> 00:38:07.900
When we do that, we get 2t³/ 3 - 2t² + c.
00:38:07.900 --> 00:38:12.400
They tell us that the initial velocity is actually equal to 0.
00:38:12.400 --> 00:38:19.000
v(0) is nothing more than, I put 0 into this for t.
00:38:19.000 --> 00:38:25.300
It is going to be 0 + 0 + c which implies that c itself = 0,
00:38:25.300 --> 00:38:36.500
which means now I can go ahead and put that into here to get my velocity function is equal to 2t³/ 3 – 2t².
00:38:36.500 --> 00:38:40.700
Now I graph that function and I see where it crosses the x axis.
00:38:40.700 --> 00:38:43.500
Not hits the x axis only, it have to fully cross.
00:38:43.500 --> 00:38:51.100
It has to go from positive to negative or negative to positive.
00:38:51.100 --> 00:38:56.300
Let us see what we do next.
00:38:56.300 --> 00:39:00.600
This is a graph of the function that we just dealt with.
00:39:00.600 --> 00:39:14.700
This is the velocity function 2t³/ 3 - 2t².
00:39:14.700 --> 00:39:18.900
This is the velocity function, if something changes direction, it is what they want.
00:39:18.900 --> 00:39:20.400
Does the particle change direction?
00:39:20.400 --> 00:39:24.000
It changes direction when the velocity goes from positive to negative or negative to positive.
00:39:24.000 --> 00:39:27.000
Notice here, the velocity goes to 0 but it stays negative.
00:39:27.000 --> 00:39:29.500
It means it is moving to the left.
00:39:29.500 --> 00:39:33.400
It stops but it keeps moving to the left.
00:39:33.400 --> 00:39:39.700
Here it only crosses once, there is only one place where it actually changes direction.
00:39:39.700 --> 00:39:46.600
Our choice is e, hope that makes sense.
00:39:46.600 --> 00:39:54.700
There is only one place where it actually changes direction.
00:39:54.700 --> 00:40:01.300
Number 13, let us see what we have got.
00:40:01.300 --> 00:40:03.100
This is a related rates problem.
00:40:03.100 --> 00:40:13.300
We have a sphere, they want to know fast the volume of the sphere is changing,
00:40:13.300 --> 00:40:26.500
when the surface area is 3m² and the radius is increasing at a rate of ¼ m/ min.
00:40:26.500 --> 00:40:28.000
Let us see what we have got.
00:40:28.000 --> 00:40:30.100
They tell me that the radius is changing.
00:40:30.100 --> 00:40:33.100
This is a basic related rates problem.
00:40:33.100 --> 00:40:37.900
You may be given at least one rate, at least one, you might be given more.
00:40:37.900 --> 00:40:40.600
You are going to asked to find another rate.
00:40:40.600 --> 00:40:46.400
Your goal, your task is only to find the relationship between the two variables that you choose.
00:40:46.400 --> 00:40:50.900
And then, differentiate implicitly and just plug your values in.
00:40:50.900 --> 00:41:09.300
They tell me that the radius is changing at ¼ m/ min.
00:41:09.300 --> 00:41:12.900
I’m going to skip the units here.
00:41:12.900 --> 00:41:16.200
The rate of change of the radius is ¼.
00:41:16.200 --> 00:41:19.800
It is increasing at ¼ m / min.
00:41:19.800 --> 00:41:22.500
What they want is the rate of change of volume.
00:41:22.500 --> 00:41:24.600
They want dv dt, that is what they want.
00:41:24.600 --> 00:41:30.300
Our task is to find the relationship between that and that.
00:41:30.300 --> 00:41:39.300
We know what that is, the volume of the sphere is 4/3 π r³.
00:41:39.300 --> 00:41:55.800
Now we differentiate, dv dt is equal to 4 π r² dr dt.
00:41:55.800 --> 00:42:06.900
They are telling me, they want to know what this value is when the surface area is equal to 3.
00:42:06.900 --> 00:42:12.600
We know the formula for surface area, it is 4 π r².
00:42:12.600 --> 00:42:17.500
They want to know when that is equal to 3.
00:42:17.500 --> 00:42:24.800
It is 4 π r², the rate of change of the volume, when the surface area is 3,
00:42:24.800 --> 00:42:32.600
I just put in the 3 where I see 4 π r², and I multiply by dr dt which I have.
00:42:32.600 --> 00:42:44.300
3/4 m³/ min is my answer, which is going to be choice e because that is in decimal form, 0.75.
00:42:44.300 --> 00:42:45.500
I hope that makes sense.
00:42:45.500 --> 00:42:47.900
All related rates problems are the same.
00:42:47.900 --> 00:42:50.900
They will give you a rate, they will ask for a rate.
00:42:50.900 --> 00:42:56.600
Your job is to find an explicit expression between this variable and that variable.
00:42:56.600 --> 00:43:02.300
Using whatever other information is given to you, and then differentiate that expression.
00:43:02.300 --> 00:43:04.700
Here is the expression, differentiate the expression.
00:43:04.700 --> 00:43:05.900
This is what we want.
00:43:05.900 --> 00:43:16.900
We arrange it for whatever they ask, that is all.
00:43:16.900 --> 00:43:24.100
Let us see what do we got number 14.
00:43:24.100 --> 00:43:31.600
Rectangle, upper two vertexes, the graph, give the decimal approximation of the maximum possible area.
00:43:31.600 --> 00:43:38.800
What we have is this situation.
00:43:38.800 --> 00:43:51.500
We have a function which is e ⁻9x².
00:43:51.500 --> 00:43:59.600
When I graph this function, I get something like this.
00:43:59.600 --> 00:44:02.800
It is called a dump function, because it looks like a little dump.
00:44:02.800 --> 00:44:09.100
They are telling me that there is this rectangle that is in here.
00:44:09.100 --> 00:44:15.700
They want to know what the maximum area of the rectangle can be.
00:44:15.700 --> 00:44:19.400
Let us take a look at what we have got.
00:44:19.400 --> 00:44:26.600
This is going to be x, this is going to be f(x).
00:44:26.600 --> 00:44:34.700
Therefore, the area of this rectangle is going to be 2x × f(x).
00:44:34.700 --> 00:44:45.900
2x × f(x) which is going to equal 2x × e ⁻9x².
00:44:45.900 --> 00:44:49.500
What we are going to do, we are going to graph the area.
00:44:49.500 --> 00:44:55.800
We are going to graph the derivative because the area is what we want to maximize.
00:44:55.800 --> 00:45:00.000
When you maximize something, we maximize the function like this one.
00:45:00.000 --> 00:45:03.000
You take the derivative and you set it equal to 0.
00:45:03.000 --> 00:45:06.900
Since we have our calculator, we are just going to graph the function, graph the derivative,
00:45:06.900 --> 00:45:10.500
and we are going to see where the derivative crosses the x axis.
00:45:10.500 --> 00:45:14.400
We are going to find where the derivative = 0.
00:45:14.400 --> 00:45:21.400
That is the x value that we are going to use to put in our particular function.
00:45:21.400 --> 00:45:22.900
Let us go ahead and do that.
00:45:22.900 --> 00:45:28.900
Once again, maximization means this.
00:45:28.900 --> 00:45:49.900
Find the x values where a’(x) = 0.
00:45:49.900 --> 00:45:53.800
This green right here, this is our original function.
00:45:53.800 --> 00:45:55.300
This is our area function.
00:45:55.300 --> 00:45:58.000
This is the function that we want to maximize.
00:45:58.000 --> 00:46:03.500
This purple function, this is a’(x).
00:46:03.500 --> 00:46:05.600
That is the derivative of that.
00:46:05.600 --> 00:46:11.000
Our a(x), remember we said this is 2x e ⁻9x².
00:46:11.000 --> 00:46:12.200
It does not matter what this is.
00:46:12.200 --> 00:46:13.400
It is the derivative of this.
00:46:13.400 --> 00:46:18.300
I need to know where the derivative crosses the x axis, where the derivative is equal to 0.
00:46:18.300 --> 00:46:21.300
It is equal to 0 to places here and here.
00:46:21.300 --> 00:46:27.600
This is a negative value, I cannot have a negative length because this is a box that we are talking about.
00:46:27.600 --> 00:46:31.800
A box whose length is this part is x.
00:46:31.800 --> 00:46:34.800
That is the value that I'm interested in.
00:46:34.800 --> 00:46:40.300
That value, I read it off the graph, zoom in, use my calculator, however it is that I want to use it.
00:46:40.300 --> 00:46:47.800
I end up with this value is 0.236.
00:46:47.800 --> 00:46:54.700
a(0.236), you can see that the area function is maximized here.
00:46:54.700 --> 00:46:58.100
The x value is here, the derivative confirms that.
00:46:58.100 --> 00:46:59.300
That is all we are doing.
00:46:59.300 --> 00:47:11.600
It is a 0.236, the area of 0.236, when I put 0.236 into here, I get an area of 0.2859.
00:47:11.600 --> 00:47:18.800
My choice is a, I hope that made sense.
00:47:18.800 --> 00:47:27.800
Number 15, let us see what number 15 is asking us.
00:47:27.800 --> 00:47:34.700
They are saying that a rough approximation from the natlog of 5 is going to be 1.609.
00:47:34.700 --> 00:47:44.600
They want us to use this approximation and differentials or linear approximations to approximate.
00:47:44.600 --> 00:47:56.600
They tell us that the natlog of 5 is equal to 1.609.
00:47:56.600 --> 00:48:04.400
They want us to find an approximate value for the natlog of 521/ 100.
00:48:04.400 --> 00:48:09.700
That is just the natlog of 5.21.
00:48:09.700 --> 00:48:12.400
We are going to use a linear approximations and differentials here.
00:48:12.400 --> 00:48:17.800
Actually linear approximations not differentials.
00:48:17.800 --> 00:48:33.400
The linear approximation of a value is equal to that function at that value + the derivative at that value × x - x0.
00:48:33.400 --> 00:48:39.700
In this particular case, we want to find the value at 5.21.
00:48:39.700 --> 00:48:53.700
It is going to be the f(5) + f’ at 5 × 5.21 – 5.
00:48:53.700 --> 00:48:59.400
f(x) is just equal to the natlog of x.
00:48:59.400 --> 00:49:09.000
f’(x) = 1/ x, f’ at 5 = 1/5.
00:49:09.000 --> 00:49:18.000
Our answer, the approximation of the natlog of 5.21 is equal to f(5),
00:49:18.000 --> 00:49:36.600
which is f(5) which is ln of 5 which is 1.609 + f'(5) which is 1/5 × 5.21 – 5.
00:49:36.600 --> 00:49:40.300
When I calculate this, I end up with 1.651.
00:49:40.300 --> 00:49:46.100
My answer is c, linear approximation, that is it.
00:49:46.100 --> 00:49:47.900
That is all you are doing.
00:49:47.900 --> 00:49:52.800
This is going to be the difference between what you know and what you are looking for.
00:49:52.800 --> 00:49:57.900
That is all that is happening here.
00:49:57.900 --> 00:50:00.900
Just remember, linear approximation.
00:50:00.900 --> 00:50:05.400
Let us say this is a function, the whole idea of the derivative is 2.
00:50:05.400 --> 00:50:13.800
As long as you are staying pretty close to the curve, the line itself is actually pretty good approximation.
00:50:13.800 --> 00:50:19.200
This point was our ln 5.
00:50:19.200 --> 00:50:27.000
This point was going to be our ln 5.21.
00:50:27.000 --> 00:50:36.500
These are linear approximation, we used the derivative instead of actually using the logarithm function because it is very close to it.
00:50:36.500 --> 00:50:42.800
That is all we have done here.
00:50:42.800 --> 00:50:50.600
Let us take a look at number 16, let us see what we have got.
00:50:50.600 --> 00:50:56.400
What is 16 asking us, it is giving us a function and it is telling us that it is differentiable everywhere.
00:50:56.400 --> 00:51:00.400
They want to know what n is.
00:51:00.400 --> 00:51:06.700
f(x) is equal to, it is a piece wise function here.
00:51:06.700 --> 00:51:14.200
We have got nx³ - x for x less than or equal to 1.
00:51:14.200 --> 00:51:22.600
We have got nx² + 5 for x greater than 1.
00:51:22.600 --> 00:51:30.800
Once again, they tell us that it is differentiable everywhere.
00:51:30.800 --> 00:51:35.600
They want to know what is n.
00:51:35.600 --> 00:51:37.700
We want to know what n is.
00:51:37.700 --> 00:51:41.600
Let us go ahead and work in red here.
00:51:41.600 --> 00:51:47.600
Differentiable means it is continuous.
00:51:47.600 --> 00:51:54.800
Because differentiability implies continuity, not the other way around, continuous.
00:51:54.800 --> 00:52:02.600
We know that this function is continuous everywhere.
00:52:02.600 --> 00:52:18.800
The differentiability also means that the left hand derivative = the right hand derivative.
00:52:18.800 --> 00:52:27.800
The left hand derivative, I take the derivative of this, the derivative of that.
00:52:27.800 --> 00:52:30.200
The left hand derivative, as I approach it from the left.
00:52:30.200 --> 00:52:40.100
I have got 3nx² - 1 is equal to 2nx.
00:52:40.100 --> 00:52:41.600
I got myself one equation there.
00:52:41.600 --> 00:52:44.600
Because I know that the left hand derivative is equal to the right hand derivative.
00:52:44.600 --> 00:52:50.400
That has to be true.
00:52:50.400 --> 00:53:21.300
The limit as x approaches 1 from the left of f(x) = the limit as x approaches 1 from the left of nx³ - x is equal to, I plug 1 in for x.
00:53:21.300 --> 00:53:35.900
It is going to be n - 1 and the limit as x approaches 1 from above, they are differentiable, left hand and right hand derivative are equal.
00:53:35.900 --> 00:53:38.900
What I’m saying is that it is continuous which means that
00:53:38.900 --> 00:53:46.400
the left hand limit and the right hand limit are going to be the same because it is continuous everywhere.
00:53:46.400 --> 00:53:52.400
Therefore, those two things have to be the same.
00:53:52.400 --> 00:54:12.000
If f(x) = the limit as x approaches 1 from above of that function mx² + 5 = m + 5.
00:54:12.000 --> 00:54:18.900
These are equal, precisely because f is continuous everywhere.
00:54:18.900 --> 00:54:23.100
It is continuous everywhere because it is differentiable everywhere.
00:54:23.100 --> 00:54:34.500
Now I have got, we have got n - 1 is equal to m + 5.
00:54:34.500 --> 00:54:43.500
Let us go ahead and write this as n - 6 is equal to m.
00:54:43.500 --> 00:54:50.100
I’m going to go ahead and plug n – 6, wherever I see that.
00:54:50.100 --> 00:55:03.600
I got 3nx² - 1 = 2 × n - 6 × x.
00:55:03.600 --> 00:55:10.900
I get 3nx² – 1 = 2nx - 12x.
00:55:10.900 --> 00:55:24.400
If I did my math correctly and I end up with x = 1.
00:55:24.400 --> 00:55:29.200
I know the x = 1 because I'm talking about 1,
00:55:29.200 --> 00:55:34.300
because 1 is the point at which it is going to be both continuous and differentiable.
00:55:34.300 --> 00:55:40.400
X = 1, I get 3n.
00:55:40.400 --> 00:55:50.900
When I plug in x = 1 to there and to there, I get 3n - 1 = 2n – 12.
00:55:50.900 --> 00:55:55.400
I get n = -11, my choice is e.
00:55:55.400 --> 00:56:00.600
In this particular case, when they said it is differentiable everywhere, it implies that it is continuous everywhere.
00:56:00.600 --> 00:56:06.300
Because it is differentiable, the left hand and right hand derivative have to be the same.
00:56:06.300 --> 00:56:10.200
I take the derivative of both functions, to the left of 1 or the right of 1
00:56:10.200 --> 00:56:15.300
because it is differentiable at 1 and I set them equal to each other, that is one of my equations.
00:56:15.300 --> 00:56:21.600
Because it is continuous, I know that the left hand limit, the original function = the right hand limit of the original function.
00:56:21.600 --> 00:56:23.400
I set those two things equal to each other.
00:56:23.400 --> 00:56:26.700
That gives me another equation, an m and n.
00:56:26.700 --> 00:56:27.900
Now I just solve the two equations.
00:56:27.900 --> 00:56:30.300
In this case, they just wanted n.
00:56:30.300 --> 00:56:35.800
I could have done both, not a problem.
00:56:35.800 --> 00:56:43.600
Let us see number 17, which I think is our last problem here.
00:56:43.600 --> 00:56:52.900
They want to know which of the following functions has a vertical asymptote at -1 and a horizontal asymptote at y = 2.
00:56:52.900 --> 00:56:54.400
They gave us some choices.
00:56:54.400 --> 00:57:15.100
Vertical asymptote at x = -1 and they want a horizontal asymptote at y = 2.
00:57:15.100 --> 00:57:18.700
This is what they are asking.
00:57:18.700 --> 00:57:22.600
Basically, you are just going to check each function, that is it.
00:57:22.600 --> 00:57:30.100
Denominator, where the denominator is 0, that is going to be your vertical asymptote.
00:57:30.100 --> 00:57:36.700
Horizontal asymptote, it depends on the nature of the rational function, what the top is and what the bottom is.
00:57:36.700 --> 00:57:51.400
I notice that the 2x² + 1/ x² – 1, it goes to 2 as x goes to infinity.
00:57:51.400 --> 00:57:55.300
Horizontal asymptote means taking x to infinity and see what happens.
00:57:55.300 --> 00:57:59.300
Essentially, as x goes to infinity, these do not really matter.
00:57:59.300 --> 00:58:00.900
x²/ x², they cancel.
00:58:00.900 --> 00:58:04.500
You are just left with 2/1 which is 2.
00:58:04.500 --> 00:58:20.800
Sure enough, y = 2 is a horizontal asymptote and x² – 1, this is x – 1 × x + 1.
00:58:20.800 --> 00:58:27.400
x cannot be -1 because it is actually going to end up making the denominator 0.
00:58:27.400 --> 00:58:33.100
Our choice is b, nice and straightforward.
00:58:33.100 --> 00:58:40.600
That takes care of Section 1, Part B, where the calculator was allowed for the exam.
00:58:40.600 --> 00:58:44.600
Next time, we are going to start with our free response question.
00:58:44.600 --> 00:58:46.500
Thank you so much for joining us here at www.educator.com.
00:58:46.500 --> 00:58:47.000
We will see you next time, bye.