WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to continue with our AP practice exam.
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We are going to start the free response section.
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Again, this is going to be the same thing.
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For part of it, you are going to be allowed a calculator.
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For part of it, you are not going to be allowed a calculator.
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Let us jump right on in.
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In case you need to pull this up, there is a link down below or let me write it over here.
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It is going to be www.online.math.uh.edu/apcalculus/exams.
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It would be nice if I actually wrote it, /exams.
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You are going to be clicking on where it says section 2, written.
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We are going to be doing version 2.
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Version one is the same thing, it is just different numbers, different arrangement, different order.
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But it is essentially the same exam.
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Let us get started.
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Question number 1, we have a particular region that is going to be bounded by a couple of functions.
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In this particular case, y = -x² and y = 4x², and the y axis.
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The first thing we want to do is go ahead and graph this.
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We see what we are looking at.
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We got y = 4x², it is going to be some parabola, something like that.
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This is our y = 4x².
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Then we have our y = -x² which is going to be a bump function.
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I will go ahead and just mark that there.
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It is going to be something like that.
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This is our function y = e ⁻x², they are all going to look like that.
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These little bump functions.
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It is bounded by the y axis.
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The particular region that we are interested in is that region right there, that little triangular looking thing.
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Part A, find the area of s.
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This is are our region s.
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Find the area of s.
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The first thing we want to do, what we are going to do is we are just going to take the difference in the two graphs.
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We are going to take a little vertical strip and we are going to integrate from 0 to whatever this x value is.
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We need to find where the two graphs meet, so that we have an upper limit for our integral.
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Let us go ahead and do that.
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We set 4x² equal to e ⁻x².
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We get 4x² - e ⁻x² = 0.
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What you are going to do is, again, calculator is allowed here, you are just going to use this to find the roots of this.
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Find the root of this equation, where does it hit the x axis.
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It is going to be this and this, those two.
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We end up with the first root is equal to 0.4515.
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The second root is going to be symmetric.
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On the other side, it is going to be -0.4515.
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Therefore, the area of this region is going to equal the integral.
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S, this region right here.
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That is just going to be from 0 to 0.4515 and this is going to be the length of this strip.
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The length of this strip is going to be, this is some x value,
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it is going to be the height of the top function - the height of the bottom function.
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Height of the top function - the bottom function gives me this height.
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I’m going to add all of those little strips together.
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The top function is e ⁻x².
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The bottom function is 4x² dx.
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I get an answer of 0.300, that is it.
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Nice and straightforward application of an integral to find the area between two curves.
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Let us now move on to part B.
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Part B wants us to find the volume of the solid that is generated when this region is rotated around the x axis.
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Let us go ahead and redraw real quickly, just a small portion of it.
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We have that and we have the bump function.
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This was our region s.
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What we are going to do is we are going to rotate that around the x axis.
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We are going to end up with this region roughly.
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We are taking this region, we are rotating it about the x axis.
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We want to find the volume of that solid.
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I’m going to go ahead and use washers because again, I want to integrate with respect to the x axis.
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Let us recall what our functions were.
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We had y = our outside function, this was our y = e ⁻x².
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Our lower function there which is y = 4x².
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When I take a little strip like a washer, I’m going to take that, I’m going to turn it this way so that I’m looking at it.
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I’m going to be looking at, essentially that.
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I want to take this area and I want to add up all the little volume elements.
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I think I will call this radius 2, I think I will call this outer radius 1.
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The volume is going to equal the integral from a to b, it is going to be from here to here.
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It is going to be the 0.4515 of the area of this circle - the area of this circle which is going to give you this area.
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It is going to give me the area of that × dx.
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It is going to be π × r1² - π × r2² dx.
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r1 = e ⁻x², r2 is equal to 4x².
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r1 is that radius, r2 is that radius.
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Therefore, what we have is the volume equals the integral.
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I pull out the π, 0 to 0.4515 of e ⁻x²² - 4x²² dx.
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When I solve that, I get 1.059, very straightforward.
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Let us go ahead and go on to part C.
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Now what they want us to do is, now what they say is, this region is the base of a solid,
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where the cross section that is perpendicular to the x axis is an equilateral triangle.
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Find the volume of the solid.
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Let us go ahead and draw it again.
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I have got a 4x² part, I have the x² part.
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Basically, what they are telling me is this.
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This is our region s that we have.
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This function right here, this function, this was our y = e ⁻x².
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This function right here was y = our 4x².
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They are telling me that this is the base of a solid.
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If I take a vertical slice perpendicular to the x axis, the cross section, if I were to take this and flip it,
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in other words look at it from that direction, it is going to be an equilateral triangle.
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What they are saying is this.
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I’m going to take this and I’m going to turn it this way.
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What I’m looking at is an equilateral triangle.
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If I’m looking at it from that direction, I see a bunch of equilateral triangles getting bigger.
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That is what is happening.
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They want to know what the volume of the solid is.
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This region is the base of the solid.
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The cross section is going to be an equilateral triangle.
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Let me draw my triangle a little bit better here.
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In fact, I can draw my triangle over here.
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I’m going to move over here, this, that, and that.
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This right here is this right there.
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That height is equal to e ⁻x² - 4x².
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I’m going to draw out a perpendicular and I’m going to call that h.
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H because this is an equilateral triangle, all of the angles are 60.
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When I draw out the perpendicular, I get a 30, 60, 90.
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This is a perpendicular, therefore h is equal to half of this thing × √3.
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It is equal to half of e ⁻x² - 4x, √3.
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The area of that triangle is equal to ½ the base × the height that = ½ the base is that thing,
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e ⁻x² – 4x² × the height × √3/2 × e ⁻x² - 4x.
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Volume = the area of this thing integrated from here to 0.4515.
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I’m going to add up now all of these triangles.
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I’m going to add up all the triangles.
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The volume = the integral from a to b of the area × dx,
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the integral from 0 to 0.4515 of the area which is now √3/4 × e ⁻x² - 4x²² dx.
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I get a volume equal to 0.1037.
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I hope that makes sense.
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Let us go ahead and go on to problem number 2.
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We are given, we have f(x) is equal to the integral from 0 to x of cos².
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A function that is expressed in terms of an integral cos².
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I’m changing my variables dt, on the interval from 0 to 3π/ 2.
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What do we want here?
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Number 2, they want us to approximate f(π) using a trapezoidal rule with n equal to 4.
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Let us see what we have got.
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f(π) = the integral from 0 to π cos² t dt.
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The trapezoidal rule where n is equal to 4, that is going to equal Δ x/ 2 × f(x0) + 2 × f(x1) + 2 × f(x2).
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You definitely want to write this out on your exam so they know what you are using.
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+ 2 × f(x3) + f(x4), you see that you have five terms.
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It is equal to 4, you have 1, 2, 3, 4, 5 terms, 0 to 4.
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Δ x here, Δ x is equal to b - a/ 4.
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It is going to be π - 0/4 = π/ 4.
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Therefore your x0, x1, x2, x3, x4, are going to be 0, π/4, 2π/ 4, 3π/ 4, 1.
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That is what is happening.
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At x = 0, x(0) = 0, we have f(0) is equal to,
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f is this, small f, this is small f.
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We are finding F(π) which is an integral.
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We are evaluating and approximating that integral with trapezoids.
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F is that.
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f(0) = 1² = 1, x1 = π/4, f(π/4) = 1/ √2² because the cos of π/4 is 1/ √2.
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1/ √2² is ½, x2 = π/2, f(π/2) = 0² = 0, x3 = 3π/ 4, f(3π/4) = -1/ √2² which is ½.
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x4 is equal to π, f(π) = -1, -1² is equal to 1.
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Therefore, our approximation t4 is equal to π/4/ 2 × 1 + twice this, twice this, twice this, + that.
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1 + 2 × ½ + 2 × 0 + 2 × ½ + 1 + π/8 × 4 = π/2.
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Just a nice application of the trapezoidal rule, which I actually personally dislike very much.
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I do not know why.
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Part B, find f’(x).
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f(x) = the integral from 0 to x of cos² t dt.
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f’(x) is nothing more than cos² (x).
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Straight application of the fundamental theorem of calculus.
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C, average value of f’ on the interval.
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They want the average value of f’ on the interval.
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The average value = 1/ b – a, integral from a to b, f’ is cos² x dx 1/,
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What was the particular interval, over the interval of 0 to π.
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It is going to be π – 0, the integral from 0 to π, cos² x dx and at 0.5.
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That is it, straightforward, nothing strange happening here.
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Let us do question number 3 and we will done with the first half, very nice.
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We have a particle that is moving along the x axis and its acceleration is given by, our acceleration function is 24t – 12.
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Let me write this a little bit better.
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It equals 24t – 12.
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They are telling me that the velocity at 1 is equal to 8.
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They are telling me that the position at 1 is equal to 10.
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Part A wants us to find an expression for the velocity of this.
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We know what the velocity is, the velocity of t is nothing more than the integral of the acceleration
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which is the integral of 24t - 12 dt, which is going to give us 12t² - 12t + c.
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They gave us a value of v1.
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v1 which is going to equal 12 × 1², just plug in 1 for t - 12 + c.
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They tell us that this equals 8.
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When we solve this, we get c = 8.
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12 – 12, we get c = 8.
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Therefore, we plug this back into there.
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Our velocity function which is what we wanted is going to be 12t² - 12t + 8.
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There we go, that is part A.
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Part B, at what values of t does the particle change direction?
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The particle changes direction when the velocity goes from positive to negative.
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The particle changes direction when v(t) changes sign, positive to negative or negative to positive,
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where it crosses the x axis completely, not touches it.
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Find the root of v(t).
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When I plug this in my calculator and I check the roots, there are no real roots.
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Does not change direction, the particle does not change direction.
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It might slow down, then speed up again, but it does not change direction.
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Now part C, now they want us to find an expression for the position of the particle.
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We know what that is.
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We know that the position of the particle is nothing more that the integral of the velocity function.
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It is going to be the integral of 12t² - 12t + 8 dt.
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We are going to end up with 4t³ – 6t² + 8t, again + another constant.
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They told us what x₁ is, what the position at t = 1 is.
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It is going to be 4 × 1³ - 6 × 1² + 8 × 1 + c.
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They tell me that that = 10.
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When I solve this, I get c is equal to 4.
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I plug that back in, therefore my position function = 4t³ – 6t² + 8t + 4.
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That is what I want.
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D, the total distance traveled from t = 1 to t = 4.
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The particle does not change direction.
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I do not know what is the best way to do this?
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Total distance traveled.
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It does not change direction.
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I did it one way but now that I'm sort of sitting here and thinking what is the best way to do this.
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The total distance traveled by the particle.
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The velocity × time gives me total distance.
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That is fine, I can just go ahead and take,
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Our total distance traveled, because it did not change direction, I do not have to split up the,
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From t = 1 to t = 4 because it does not change direction,
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I do not have to break it up into moving to the left, moving to the right, things like that.
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I can just go ahead and I can find the distance at t = 4.
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I can find the distance at = 1 which is equal to 196, which is equal to 10,
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and I can just take the difference which is going to be essentially the same as the total distance traveled.
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If you want, you could have done it this way, the total distance.
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Distance = rate × time, rate is velocity, velocity × time.
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Because velocity is not constant, the total distance = the integral from 1 of 4 of the velocity function dt.
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You can go ahead and put the velocity function which is this thing into here and integrate that,
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which is the same essentially as this.
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Because when you integrate the velocity function, you are going to get the distance function.
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When you evaluate the distance function at 4, this is function at 1, you are going to subtract.
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It is the same thing, however you want to look at it.
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That is it for this part, where the calculator was allowed.
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The next part is going to be another three questions where no calculator is allowed.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.