WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, welcome back to AP Calculus.
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Today's lesson is going to be a continuation of the Section 1 Part A of the AP practice exam.
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The one with no calculator allowed.
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Let us continue on.
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We finished off with question number 14, in the last lesson.
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Now we are going to start with question number 15.
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Let me go ahead and go to blue.
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Question number 15 is asking us to find the average value of a given function over a particular interval.
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The function at our disposal is 2x + 3³.
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The particular interval over which we want to take the average value is -3 to -1.
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The average value is really easy to calculate.
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It is just, basically, the integral of the function divided by the length of the interval.
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It is usually written as 1/ b – a.
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A is the first number, b is the second number, × the integral from a to b of gx dx.
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It is a simple evaluation of integral.
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This is going to be 1/ 1 – a - 3 × the integral from -3 to -1 of 2x + 3² dx.
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I think it was squared, was not it cubed?
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Yes, squared not cubed.
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I’m just going to go ahead and multiply this out.
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= ½ × the integral from -3 to 1 of, this is going to be 4x² + 12x + 9 dx.
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That is going to equal ½ 4x³/ 3 + 12x²/ 2.
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That is going to be + 6x² + 9x all evaluated from -3 to -1.
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When we do the evaluation, we get ½.
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This is going to end up being.
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I would not go through all the stuffs, I will just do what it is that we have got.
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= ½ - 13/3 + 27/3.
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You are going to end up with a value of 7/3 which is going to be c.
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That is it, just evaluate the definite integral
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Here there is no particular technique to use, just straight application of antiderivatives.
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Number 16, what does number 16 asks us?
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Excuse me, let me move that over.
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Number 16 is asking us to evaluate a limit.
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In this particular case, it is the limit as t goes to 0 of the quotient of tan(π/4) + t – tan(t)/ t.
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The important thing here is instead of actually evaluating limit, you are not going to evaluate the limit.
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You are going to recognize that this limit is the definition of derivative for a particular function.
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This limit is going to be the derivative evaluated at π/4.
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In this particular case, it is going to be y =, this is the derivative of the tangent function evaluated at π/4.
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You just want to recognize that is what it is, instead of actually evaluating this limit.
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Let us write that down.
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Recognize this as the definition of the derivative.
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That is all this is, definition of the derivative of the tan(x) at x = π/4.
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We have y is equal to tan(x).
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We have y’ of tan(x) which is sec² (x).
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y’(π/4) = sec² of π/4 which is equal to √2² which is equal to 2.
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That is choice e.
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Just recognize that that is the definition of derivative.
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Let us see, that is number 16, let us go ahead and go to number 17.
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Let us see what number 17 is asking us.
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17 is giving us a particular function, a function of t.
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They want us to find the instantaneous rate of change at t = 0.
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Nice and straightforward, instantaneous rate of change.
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We know that an instantaneous rate of change is the derivative.
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Our derivative is the slope, it is the rate of change.
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The slope of the secant line is equal to the average rate of change.
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The slope of the tangent line which is the derivative is the instantaneous rate of change.
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Let me write number 17, our function of t is equal to 2t³ - 2t + 4 × √t² + 2t + 4.
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We want to evaluate f’ at 0.
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Let us go ahead and f’.
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f’(t), it is going to be product rule.
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It is going to be a little tedious and long, but that is not too big of a deal.
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We have got 2t³, this × the derivative of that + that × the derivative of this.
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-2t + 4 × ½ t² + 2t + 4⁻¹/2 × 2t + 2 + t² + 2t + 4.
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Instead of running the radical sign, I will go ahead and write it that way.
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× the derivative of what is inside here which is going to be 6t – 2.
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Of course, we just plug in t, we do not have to simplify this.
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We can just plug in our particular 0.
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When you plug 0 in to t for all of this and evaluate that, you are going to end up with 2 - 4 = -2.
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That is our answer and that is choice b.
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Nice and straightforward.
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Let us try number 18.
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Let us see, what is number 18 asking us to do?
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It wants us to calculate, it wants us to do ddx of 11 ⁺cos(x).
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We know that the derivative of some constant e ⁺u is equal to e ⁺u × ln(a) × du.
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If we have 11 ⁺cos(x) and we subject it to the differential operator, in other words, take the derivative of it.
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It is a fancy term for that.
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We subject it to differentiation.
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We get the derivative of 11 ⁺cos(x) is equal to 11 ⁺cos(x) × natlog of 11 × the derivative of the u.
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Because u is a function of x, it is not just x.
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The derivative of cos(x) is - sin(x).
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That is it, basically, you just end up with -sin x 11 cos(x) ln(11) which is d.
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Again, if you want to separate them out, so you can actually not get confused as to which is which with the symbols,
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just put some parentheses around it, not a problem.
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Great, nice and straightforward.
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Those are the ones that we love.
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Number 19, let us see what we have got here.
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Number 19, we have a solid which is generated by rotating a particular region that is enclosed by these graphs and these lines.
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We are rotating about the x axis and they want to know which one of these integrals
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actually gives you the volume of the particular solid that is generated.
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Let us see what we got.
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We have y = √x and we have x = 1, x = 2, and y = 1.
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Let us go ahead and draw this out.
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Something like this, we know what the √x function looks like.
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It looks something like that.
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X = 1, let us just put it here.
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x = 2, let us go ahead and put it here.
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y = 1, that is this line right here.
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This is our origin, y = 1.
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I will just go ahead and put it right there.
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It is the region that is bounded by all of these.
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Bounded by x = 1, x = 2, the function y = √x, and the line y is 1.
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This is the region that we are looking for.
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That is the region that we are going to rotate around the x axis.
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Basically, we want to come down here.
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Now we have that one, we are going to have this.
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This is the cell that is going to be generated.
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We are going to have this solid, we want to find the volume of that solid or the integral representing that volume.
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I think what I’m going to do here is I’m going to go ahead and use washers.
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I'm going to integrate from 1 to 2.
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I’m going to integrate in the horizontal direction.
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I’m definitely going to be using dx.
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My upper limit and lower limits of integration, my lower to upper is actually I’m going to go from 1 to 2.
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Essentially, what I'm doing is I’m taking a little bit of a washer here.
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It looks like this.
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It is going to be something like this.
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This is the region.
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We are going to take the area of that washer and we are going to add up all the areas along the x axis.
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Let us go ahead and call this radius the outer radius.
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We will call this one the inner radius.
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The volume is going to be the integral from a to b of an area element × dx.
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We said we are going to integrate from 1 to 2, that is 1 to 2.
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The area of this region is going to be π × the outer radius² - π × inner radius².
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In other words, the big circle - the inner circle.
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That is just π × the outer radius² - the inner radius² × dx.
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Now we just need to know what the outer and inner radiuses are, as functions of x.
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This is going to equal, I’m going to pull the π out of the integral sign, it is 1 to 2
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The outer radius is, if this is my x value, the outer radius is √x.
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The inner radius is 1.
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It is going to be the outer radius, it is going to be √x²,
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that is the outer radius² - the inner radius which is from here to here, from here to here, which is 1² dx.
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We have π × the integral 1 to 2 of x - 1 dx.
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That is our answer, it did not ask us to evaluate it.
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This is going to be choice d.
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In this particular case, we decided to use washers.
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I’m going to stick with red, I guess it is kind of nice.
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This is going to be question number 20.
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There is no calculator involved in this particular section, the first section.
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The first section is no calculator, the second section is there are going to be calculator.
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When you do the free response questions, one of the sections is going to be no calculator,
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the other one is going to be with calculator.
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Let us see what 20 is asking us.
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It is asking is to calculate a limit.
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Let us see what we have got.
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They want us to evaluate the limit as x goes to 0 of 4x/ sin(3x) + x/ cos(3x).
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The limit of the sum is the sum of the limits.
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Let me actually separate this out.
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This is equal to the limit as x approaches 0 of 4x/ sin(3x) + the limit as x approaches 0 of x/ cos(3x).
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This one right here, when I put 0 in, I get 0 sin(3x), sin(3) × 0, sin(0) is 0.
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What I ended up with is 0/0.
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This is an indeterminate form.
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We have something great for this, L’Hospitals rule.
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I just take the derivative of that or the derivative of that, and I take the limit again.
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This limit is actually equal to the limit as x approaches 0.
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The derivative of 4x is equal to 4.
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The derivative of sin of 3x is 3 cos of 3x.
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Now when I take x to 0, the cos(3) × 0 is cos(0).
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The cos(0) is 1.
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The limit of 4/3, this is actually equal to 4/3.
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This first limit right here, that is equal to 4/3.
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This one, when I put 0 in, I end up with 0/1 which is 0.
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My final answer is 4/ 3 + 0 = 4/3.
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My answer is b.
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Again, just a straight application of l’Hospitals rule, any of the indeterminate forms.
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If you have 0/0, if you have infinity/ infinity, if you have infinity × 0, things like that.
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Let us try number 21.
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What is 21 asking us?
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For 21, it says that y is greater than 0 and it is telling us that dy dx is equal to 3x² + 4x/ y.
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They are telling us that the point 1√10 is on the graph.
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What is 0y?
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In other words, find the y value.
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They are telling us the y is greater than 0.
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They are telling us that the derivative is equal to 3x² + 4x/ y.
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They are telling us that the graph itself contains the point 1 √10.
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They we want to know what the y value is, when x = 0 of the function.
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Let us what we have got.
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We know y’, that is dy dx.
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y’ is equal to 3x² + 4x/ y.
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I'm going to go backwards.
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This is essentially an implicit differentiation.
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Somebody who has done implicit differentiation and they come up with this.
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Just go backwards.
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What you end up with is yy’, just multiply through by y = 3x² + 4x.
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What you will end up, when you rearrange this, you get -3x² - 4x + y, y’ is equal to 0.
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Essentially, now you are just going to integrate this function one at a time.
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The integral of -3x² is x³.
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We are just going backwards to what y is equal.
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This is -2x², the integral of -4x is -2x² because the derivative of -2x² is -4x.
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Here, this derivative is ½ y², the antiderivative.
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Because if I take ½ y², take the derivative of it, implicitly, it is going to be 2 × ½ y × y’.
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The 2 and ½ cancel, you are left with y and y’.
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That is equal to some constant c.
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We do not know what c is.
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Let us go ahead and work it out.
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We figured out what our original function is, in implicit form.
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That is this thing.
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We also know that we have an x and y value.
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We know that 1√10 is on this graph.
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This is –x³, sorry.
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-1³ - 2 × 1² + 1/2 y which is √10² is equal to c.
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Here what we end up with is c is equal to 2.
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If c is equal to 2, let us plug that back in.
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Now we know that –x³ - 2x² + ½ y² is equal to 2.
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We want to know what y is, when x is 0, just plug this in.
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I get 0 - 0 + ½ y² is equal to 2.
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y² is equal to 4.
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y is equal to + or -2.
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But they said early on, y is greater than 0 so that means that y = 2.
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I took the implicit derivative formula and I worked my way backwards integrating to the original function.
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I hope that made sense, this choice was a.
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Let us see, that was number 21.
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We are almost done with this section.
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We are actually moving along pretty well.
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This is 22, what is 22 asking us to do?
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It looks like it is asking us to evaluate,
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Excuse me, let me turn the page here.
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I have difficulty turning these things.
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It wants us to evaluate an integral.
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Let us evaluate the integral from 1 to 2 of 1/ 4 - t², all under the radical.
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Again, this is just a question of recognizing antiderivatives and doing a little manipulation.
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I’m going to rewrite this as.
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I think I want to go back to blue, I like it better.
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The integral from 1 to 2, I’m going to factor out a √4 in the bottom, 1 - t²/ 4.
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Let me write this a little bit better.
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It is going to be 1/ √4 × √1 - t²/ 4.
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I hope that makes sense that this is this.
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I factored out a √4 under the radical sign.
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I just pulled it out of the radical sign, because a radical × a radical is equal to a radical.
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All I’m doing is a little bit of mathematical manipulation.
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This 1/ √4 comes out as ½ × the integral 1 to 2, 1/ 1 -, t²/ 4 is the same as t/2² under the radical dt.
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Now I do a u substitution, u = t/2, du = ½ dt, dt = 2 du.
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Therefore, this integral is equal to ½ × the integral from 1 to 2 of 1/ 1 - u² × 2 du which is equal to the integral from 1 to 2.
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This is 2/2 which turns into 1, 1/ 1 - u² du.
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I recognize this as the inv sin(u), from 1 to 2.
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I plug u back in as the inv sin of t/2, from 1 to 2 = the inv sin of, when I put 2 in there,
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it is going to be the inv sin of 1 - the inverse sin of ½.
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The inverse sin of 1 is π/2, the inverse sign of 1/2 is π/6.
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I'm left with π/3 which is choice a.
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That is it, just a little bit of manipulation to turn it into an antiderivative that we recognize.
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In this case, the inverse sin function.
00:25:48.200 --> 00:25:58.300
That is 22, let us try number 23.
00:25:58.300 --> 00:26:02.100
23 is asking us to evaluate an integral.
00:26:02.100 --> 00:26:06.000
Again, this time it looks like an indefinite integral.
00:26:06.000 --> 00:26:20.300
We have the integral e ⁺2x × √e ⁺x + 1 dx.
00:26:20.300 --> 00:26:22.500
This is going to be a little strange.
00:26:22.500 --> 00:26:30.300
I’m going to start with u = e ⁺x + 1.
00:26:30.300 --> 00:26:36.000
Du = e ⁺x dx.
00:26:36.000 --> 00:26:43.800
I’m also going to write dx is equal to du/ e ⁺x.
00:26:43.800 --> 00:27:01.200
What this gives me is, when I plug these into that, I get the integral of e ⁺2x × √e ⁺x + 1, that is just u and dx.
00:27:01.200 --> 00:27:07.700
dx is equal to du/ e ⁺x.
00:27:07.700 --> 00:27:14.100
This is equal to e ⁺2x divided by e ⁺x.
00:27:14.100 --> 00:27:22.200
I get the integral of e ⁺x × that u du.
00:27:22.200 --> 00:27:31.900
I have the u, I have the du, I just need to take care what this e ⁺x is.
00:27:31.900 --> 00:27:35.200
Let me write that down.
00:27:35.200 --> 00:27:39.400
I need to get everything in terms of u, one variable.
00:27:39.400 --> 00:27:44.200
I have taken care of that, I just need to take care of e ⁺x.
00:27:44.200 --> 00:27:50.500
We said that u is equal to e ⁺x + 1.
00:27:50.500 --> 00:28:04.000
U - 1 = e ⁺x which means x is equal to natlog of u – 1.
00:28:04.000 --> 00:28:14.200
Therefore, this integral now becomes the integral,
00:28:14.200 --> 00:28:17.500
I do not really need to do that.
00:28:17.500 --> 00:28:23.200
Should I leave it as u⁻¹?
00:28:23.200 --> 00:28:30.400
I do not need to do this part, that is not even necessary.
00:28:30.400 --> 00:28:33.200
Again, I have got e ⁺x, now it is u⁻¹.
00:28:33.200 --> 00:28:40.100
Now I have the integral of u⁻¹ × √u × du.
00:28:40.100 --> 00:28:45.000
Perfect, this is great, this is equal to the integral of, this distribute.
00:28:45.000 --> 00:28:51.700
This says u³/2 - u ^½.
00:28:51.700 --> 00:29:14.100
du = 2/5 u⁵/2 - 2/3 u³/2 + c.
00:29:14.100 --> 00:29:32.100
When we substitute back in what u is, we get that this integral is equal to 2/5 × e ⁺x + 1⁵/2 - 2/3
00:29:32.100 --> 00:29:39.000
× e ⁺x + 1, because u was e ⁺x + 1³/2 + c.
00:29:39.000 --> 00:29:40.000
This is choice e.
00:29:40.000 --> 00:29:42.700
That is it, nice application.
00:29:42.700 --> 00:29:45.500
We set the mess around with that e ⁺x.
00:29:45.500 --> 00:29:47.000
We can work one function at a time.
00:29:47.000 --> 00:29:50.900
Whenever you are doing the u substitution, everything has to be changed into u.
00:29:50.900 --> 00:29:57.400
You may not be able do it in one step, you may have to take that extra step.
00:29:57.400 --> 00:30:05.900
Number 24, let us see what does number 24 asks us.
00:30:05.900 --> 00:30:09.200
Let us come over here.
00:30:09.200 --> 00:30:17.900
Number 24, we are given an acceleration and we are given an initial position of velocity.
00:30:17.900 --> 00:30:23.900
What is the position of the particle at another time?
00:30:23.900 --> 00:30:31.100
I’m given acceleration a ⁺t is equal to 12t + 4.
00:30:31.100 --> 00:30:35.600
They are telling me that my initial position x(0) is equal to 2.
00:30:35.600 --> 00:30:41.600
They are telling me that my velocity at 1 is equal to 5.
00:30:41.600 --> 00:30:49.800
They want to know what is my position at time t = 2.
00:30:49.800 --> 00:30:55.200
We know that if you are given that position graph, the first derivative is velocity, the second derivative is acceleration.
00:30:55.200 --> 00:30:59.100
If you are given an acceleration, integrate once, you get velocity.
00:30:59.100 --> 00:31:02.400
Integrate twice, you get position.
00:31:02.400 --> 00:31:31.200
v(t) or velocity of t = the integral of a(t) dt which is equal to the integral of 12t + 4 dt, which is equal to 6t² + 4t + c.
00:31:31.200 --> 00:31:33.600
That gives us our velocity, we have to find c.
00:31:33.600 --> 00:31:38.700
We have an initial value.
00:31:38.700 --> 00:31:42.600
V(1) which is equal to putting 1 in here.
00:31:42.600 --> 00:31:46.200
You get 6 + 4 + c.
00:31:46.200 --> 00:31:49.200
They tell me that v(1) is actually equal to 5.
00:31:49.200 --> 00:31:52.600
I get c is equal to -5.
00:31:52.600 --> 00:31:58.000
My velocity function v(t), I just put this -5 back into here.
00:31:58.000 --> 00:32:04.000
I get 6t² + 4t – 5.
00:32:04.000 --> 00:32:25.300
For f(t), f(t) is equal to the integral of v(t) which is equal to the integral of 6t² + 4t - 5 dt is equal to 2t³, not²,
00:32:25.300 --> 00:32:35.800
2t³ + 2t² - 5t, + this time I will just call it d.
00:32:35.800 --> 00:32:38.500
They are telling me that x(0) is equal to 2.
00:32:38.500 --> 00:32:48.400
X(0) which is equal to 0 + 0 - 0 + d is equal to 2.
00:32:48.400 --> 00:32:55.900
I get d is equal to 2.
00:32:55.900 --> 00:33:07.400
Therefore, my function, my x(t) is equal to 2t³ + 2t² - 5t + 2.
00:33:07.400 --> 00:33:23.000
Therefore, my function at t = 2, just plug the 2 in, is equal to 2 × 2³ + 2 × 2² - 5 × 2 + 2.
00:33:23.000 --> 00:33:28.100
You end up with 16, e is our choice.
00:33:28.100 --> 00:33:37.100
Very nice and straightforward.
00:33:37.100 --> 00:33:40.100
What is number 25 asking us?
00:33:40.100 --> 00:33:45.800
They ask us to determine an integral, combination of integrals here.
00:33:45.800 --> 00:34:00.600
We have the integral from 0 to π/2 of the sin(3x) dx + the integral from 0 to π/6.
00:34:00.600 --> 00:34:08.100
Is that correct, yes, π/6 of the cos(3x) dx.
00:34:08.100 --> 00:34:11.700
Just straight evaluation of an integral.
00:34:11.700 --> 00:34:15.000
Sin(3x) dx, this is going to be a u substitution.
00:34:15.000 --> 00:34:18.000
Hopefully, you have done enough of these to recognize that,
00:34:18.000 --> 00:34:24.300
when you are taking the trig of some constant × x, that constant comes out as fraction.
00:34:24.300 --> 00:34:27.700
This is going to equal, the integral of sin x - cos x.
00:34:27.700 --> 00:34:38.400
It is going to be – 1/3 of cos(3x), evaluated from 0 to π/2 + the integral of cos.
00:34:38.400 --> 00:34:48.000
It is going to be + 1/3 × sin(3x), evaluated from 0 to π/6.
00:34:48.000 --> 00:34:52.200
You are going to end up this, plug in these, this – that, this – that.
00:34:52.200 --> 00:35:11.700
You are going to end up with 0 - a - 1/3 + 1/3 – 0.
00:35:11.700 --> 00:35:14.700
You are going to end up with 2/3.
00:35:14.700 --> 00:35:17.400
This is choice c.
00:35:17.400 --> 00:35:18.900
That is it, straight application.
00:35:18.900 --> 00:35:21.100
All you have to worry about is that.
00:35:21.100 --> 00:35:32.800
The integral of sin of a(x) is equal to -1/ a cos of a(x).
00:35:32.800 --> 00:35:45.700
The integral of cos of a(x) is equal to 1/a × sin of a(x).
00:35:45.700 --> 00:35:49.000
When we do a u substitution, it will work out.
00:35:49.000 --> 00:35:55.100
It will make itself clear.
00:35:55.100 --> 00:35:57.500
Let us see number 25.
00:35:57.500 --> 00:36:01.100
Let us see number 26, we are almost there.
00:36:01.100 --> 00:36:07.400
Number 26, it is asking us to determine the derivative of this particular function at π/2.
00:36:07.400 --> 00:36:18.800
We got f(x) is equal to cos(2x) - 2².
00:36:18.800 --> 00:36:25.500
They want us to find f’ at π/2.
00:36:25.500 --> 00:36:29.400
F’(x), this is just an application, a very careful application of the chain rule.
00:36:29.400 --> 00:36:33.600
Easy, just you got to be careful, that is all.
00:36:33.600 --> 00:36:51.900
We are going to get 3 × cos(2x) - 2² × the derivative of what is inside which is - sin(2x) – 2
00:36:51.900 --> 00:37:16.000
× the derivative of what is inside, × 2 = -6 × cos(2x) - 2² × sin(2x) – 2.
00:37:16.000 --> 00:37:18.400
Now you just basically just plug π/2 in.
00:37:18.400 --> 00:37:51.400
F’(π/2), here is -6 × cos(2) × π/2 - 2² × sin(2) × π/2 – 2.
00:37:51.400 --> 00:37:57.700
I think if I'm not mistaken, the choice is c.
00:37:57.700 --> 00:38:03.400
Number 27, we have come to the end of this first part.
00:38:03.400 --> 00:38:10.300
What is number 27 asks you to do?
00:38:10.300 --> 00:38:16.300
We are going to compute the derivative of something that is given to us, in terms of an integral.
00:38:16.300 --> 00:38:31.900
f(x) is defined as the integral from 0 to x² of the natlog of t² + 1 dt.
00:38:31.900 --> 00:38:39.400
We have learned from the fundamental theorem of calculus that whenever f(x) is defined as an integral from some constant to x,
00:38:39.400 --> 00:38:42.900
really all you do is you are taking the derivative of an integral.
00:38:42.900 --> 00:38:47.200
Since, the derivative of the integrals are inverse processes, all you are doing is getting rid of the integral sign.
00:38:47.200 --> 00:38:52.600
Your answer would be have been ln(t)² + 1 or ln(x)² + 1.
00:38:52.600 --> 00:38:57.100
However, this upper limit is not x, it is x².
00:38:57.100 --> 00:39:04.700
Really, all you have to do is, remember, f’(x), everything is exactly the same.
00:39:04.700 --> 00:39:14.300
It is going to be the ln of,
00:39:14.300 --> 00:39:17.600
You are going to put this into here.
00:39:17.600 --> 00:39:21.500
It is going to be x⁴ + 1.
00:39:21.500 --> 00:39:28.000
You just have to remember, because this is an x, you have to multiply by the derivative of this thing.
00:39:28.000 --> 00:39:31.300
The derivative of x is just 1. It would be just 1, normally.
00:39:31.300 --> 00:39:35.500
But the derivative of x² is 2x.
00:39:35.500 --> 00:39:40.000
It is literally that simple.
00:39:40.000 --> 00:39:45.800
This is choice e, the only difference is that they put the x and the 2x in front.
00:39:45.800 --> 00:39:54.400
Once again, here, just plug in this into here, that will give you the function.
00:39:54.400 --> 00:40:03.500
But remember that, if this is not x, then you need to just multiply by the derivative of this thing.
00:40:03.500 --> 00:40:12.500
It looks like we have one more, I apologize, there are 28 questions not 27.
00:40:12.500 --> 00:40:29.000
Number 28, here it wants us to evaluate the derivative with respect to x of the ln of 2 - cos x.
00:40:29.000 --> 00:40:31.700
That is one parenthesis, that is two parentheses, and a bracket.
00:40:31.700 --> 00:40:35.900
This is just an application of chain rule, it is not a not a big deal.
00:40:35.900 --> 00:40:48.800
f’(x) which is ddx is equal to the derivative of natlog is 1/ the argument.
00:40:48.800 --> 00:40:57.500
This is the argument, it is going to be 1/ derivative of ln(u), it is 1/ u du dx.
00:40:57.500 --> 00:41:23.700
1/ ln(2) - cos x × the derivative of the argument ×, the derivative of ln(2) - cos x is 1/ 2 - cos x
00:41:23.700 --> 00:41:32.700
× the derivative of the argument which is the derivative of -cos x is sin x.
00:41:32.700 --> 00:41:35.400
That is it, it is choice c.
00:41:35.400 --> 00:41:39.300
Just have to watch the differentiation.
00:41:39.300 --> 00:41:45.000
That takes care of Section 1, Part A, no calculator allowed.
00:41:45.000 --> 00:41:52.200
Next lesson, we will start with section 1 part B, where the calculator is allowed on the multiple choice sections.
00:41:52.200 --> 00:41:54.600
Thank you so much for joining us here at www.educator.com.
00:41:54.600 --> 00:41:55.000
We will see you next time, bye.