WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com and AP Calculus.
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Today, we are going to be talking about related rates.
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Related rates, the mathematics is actually quite simple.
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The difficult part with related rates is setting up the problem from the verbal description.
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Let me write this down here.
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Related rates, I’m not going to actually talk about it very much.
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I’m just going to launch right into the examples because the best way to explain it is by doing the examples.
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Related rates is best explained by example.
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The strategy for solving the related rates problems is always going to be the same for every single problem.
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The general outline of the problem is always going to be the same.
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Here is what it looks like.
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The strategy for solving these problems is always the same.
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One, it is always going to be a rate that you are given.
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It may be more than one but generally, it is one.
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But there is a rate or two that you are given.
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There is a rate that you are given.
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Two, there is a rate that you are asked for.
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Three, your job is to find an equation that relates the variable of the rate that they give you,
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to the variable of the rate that they ask for.
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And the variable is going to be your choice.
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In other words, you can label it any way you want.
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It does not have to be x and y.
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It could be whatever it is that you happen to be discussing.
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You will see an example in just a second.
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Find an equation relating the above two variables, whatever the variables you chose.
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This is always the case.
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The 4th and final part is differentiate the entire equation with respect to time t.
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Because again, here we are going to be talking about a real rate, something per second,
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something per minute, something per hour, something like that.
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The denominator, for example, if we have like dy dt, it is going to be with respect to time.
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Once again, let me go ahead and write this down.
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I’m sorry, the last part is solve, solve for the rate requested.
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The rate they ask for.
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That is it, you just rearrange the equation if you need to, and solve for the rate that they ask for.
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Again, the difficulty in these problems is translating the verbal description into viable mathematics.
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I bet that has been the problem with word problems,
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ever since you guys have started doing them in junior high school, something like that.
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Let us launch right into the examples and we will use all of these examples to actually explain what is going on.
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But it is the same basic strategy.
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They are going to ask you, they are going to give you a rate.
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They are going to ask for a rate.
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You have to find an equation that relates those two variables.
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Sometimes it will be direct, sometimes you are going to have to use other information
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that is in the problem to come up with an equation that is a direct.
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In other words, one variable on one side, the other variable on the other.
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And then, you differentiate with respect to t and then you solve.
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Let us see, our first one here.
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Air is being pumped into a spherical balloon at a rate of 15 cm³ /s.
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How fast does the radius of the balloon change, when the volume is 70 cm³?
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Air is being pumped into a spherical balloon.
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For all of these problems, you always have to draw a picture.
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You might get to a point where you do not have to draw a picture,
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when you understand what is happening, especially for the more simple ones.
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But draw a picture, see what is going on, use your physics.
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There is nothing in here that is going to ask you to use something that is counterintuitive.
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You just have to trust that you understand the natural world well enough,
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after being on earth for these many years, that you can actually figure out the rest.
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You can come up with some physical model for what is going on.
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We have a spherical balloon, nice and easy.
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A spherical balloon, air is being pumped into the spherical balloon at a rate of 15 cm³/s.
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This is the rate that they give you.
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This is dv dt that equals 15 cm³/s.
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The reason I chose v for my variable is volume.
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Cubic centimeters is a unit of volume.
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Air is being pumped in, it is a 3 dimensional object, it is a volume.
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This is my rate, a rate of change, a derivative.
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Volume is changing per time, it is 15 cm³/s.
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That is the rate they give you.
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How fast is the radius of the balloon changing?
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The rate that they want is dr.
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Dr dt that is the rate that they want.
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This is r, how fast when the volume is 70 cm³.
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This is another bit of information that we are going to use, when we finally solve.
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This is the rate that they give us, this is the rate that they want.
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My two variables are, let me go ahead and do this in red, are v and r.
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I need to find the relationship that relates v and r.
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Fortunately, I have one.
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Volume of a sphere is 4/3 π r³.
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Now that I have this, I differentiate with respect to t.
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This is going to be dv dt = 4/3 π × 3 r².
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The 3 is canceled, I'm left with 4.
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I’m sorry, I forgot something.
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I’m differentiating with respect to t.
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This is like implicit differentiation except now both the v and r are functions of t.
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It is dv dt = 4/3 π × 3 r² dr dt.
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Dr dt, the 3 is canceled and I'm left with 4π r² dr dt.
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We are looking for dr dt, I’m just going to solve for that.
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Solve for dr dt.
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Let me write this out a little bit better here.
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I have got dv dt = 4π r² dr dt.
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Now I’m going to isolate dr dt by dividing by 4π.
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I get, let me write it over here, dr dt = dv dt divided by 4π r².
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We said dv dt is equal to 15.
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This is going to be 15/ 4π r².
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The question is, what do we put in for r?
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They said the volume was 70cm³.
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I will use the equation that I have.
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Volume = 4/3 π r³.
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They said the volume is 70 4/3 π r³.
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Now I solve for r.
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That will tell me what r is, when the volume happens to be 70.
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I get r³ is equal to 16.71.
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I get r is equal to 2.56.
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Dr dt, we said it is equal to dv dt divided by 4π r², that is going to be 15 divided by 4 × π × 2.56².
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If I did my arithmetic correctly which I often do not, 0.183 and it is going to be cm/s dr.
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R is a length, therefore, it is centimeters per second.
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It is an area which would be square centimeters per second or volume which is cubic centimeters per second.
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In case you are not sure about that, dv dt, the numerator is in cubic centimeters per second.
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Down here, we have r which is in centimeters.
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It is squared so this is going to be cm².
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Cm² cancels this, I’m left with the unit of cm/s.
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If you want to carry the units, that is fine.
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Or you can go back at the end to elucidate what the units are.
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There you go. The radius is growing at 0.183 cm every second, when the volume hits 70.
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Notice, the rate of change actually depends on what the radius is.
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As the radius changes, the rate at which the radius is growing changes.
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They are two different things.
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The radius and this is the rate of change of the radius, per unit time.
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I hope that makes sense.
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Again, related rates, we use the rate of the volume change to find the rate of the radius change.
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Let us try example number 2.
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A ladder 20 ft long rests against the vertical wall.
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If the bottom of the ladder was pulled away at a rate of 1.2 ft/s,
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how fast does the top of the ladder slide down the wall, when the bottom is 7 ft from the wall?
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We go ahead and we draw our picture, always.
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Let me go to black actually.
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I have got a wall here and I have this 20ft ladder, this is 20ft.
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They are pulling this bottom away.
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They are pulling it away.
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This top of the ladder is actually going to be going down.
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This is growing, the bottom is growing.
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I’m going to call it x and I’m going to call this y.
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If the bottom of the ladder is pulled away at a rate of 1.2 ft/s, x is changing at 1.2 ft/s.
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It is growing 1.2 ft/s.
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Dx dt, the rate of change of x per unit time is 1.2 ft/s.
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How fast is the top of the ladder is sliding down the wall?
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They want to know dy dt.
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The rate they give me, rate of change of x, the rate they want is y.
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Now I have to find the relationship between x and y.
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It is a triangle, I have a relationship between x and y.
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I have x² + y² = 20².
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I have x² + y² = 400.
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I differentiate with respect to t.
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This becomes 2x dx dt.
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This becomes 2y dy dt.
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The derivative of 400 is 0.
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I have got 2y dy dt = -2x dx dt.
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The 2 is cancel, I divide by y to get my final dy dt, which is why I’m isolating, = -x/y × dx dt.
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I just plug in my values, I need to plug in an x, plug in a y.
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My dx dt was already given to me, that is the 1.2 ft/s.
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Let me go to the next page here.
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Dy dt we said was equal to –x/y × dx dt, that is equal to,
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They said that x, they wanted to know what it was.
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Sorry about that, this is a ladder, it does not go through the wall or through the floor.
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This was x, they wanted to how fast this is going down, when the ladder is 7 ft from the wall.
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X is 7 and dx dt was 1.2.
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The question is what is y, would we put in for y?
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We have a relationship, we have x² + y² = 400.
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X is 7 so 7², we have 7² + y² = 400.
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We have 49 + y² = 400.
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We have y² = 400 -49 is 351, I think.
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Y ends up being 18.73 ft, there you go.
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Dy dt, the rate of change of y is equal to -7/ 18.73 × 1.2.
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We get that dy dt is equal to -0.448 ft/s, there you go.
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Notice that dy dt is negative.
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Negative means that, this is y, y is getting smaller.
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This ladder is going down.
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This distance right here is actually getting smaller, that is why this is negative.
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This negative tells me that it is decreasing.
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In other words, for every unit change in time 1 second, when the ladder is 7 ft from the wall,
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the vertical distance is changing by 0.448 ft every second.
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Notice, the rate of change depends on what x is and what y is.
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It is going to change depending on what those things are.
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It is not a constant rate, it is a variable rate.
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Notice that dy dt is negative because y is decreasing.
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Example number 3, a water tank has the shape of an inverted circular cone.
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The base has a diameter of 5ft while the height is 7 ft.
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It is being emptied at a rate of 1.8 m³/min.
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The water is leaving this tank at 1.8 m³/min.
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How fast does the water level dropping, when the water level is 3.5 ft?
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We are looking at something like this.
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We have an inverted cone, something like that, and it is full of water.
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This is the water.
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Let us go ahead and do just a full side view of this.
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It looks something like this.
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This is the water level.
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They said the base has a diameter of 5 ft.
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This is going to be 5 ft, the height is 7 ft.
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It is being emptied at a rate of 1.82 m³/min.
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This is the rate that they give us.
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This is dv dt, it is the water is being emptied.
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It is -1.8 m³/min.
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In other words, this is how much is leaving, it is decreasing.
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How fast does the water level dropping?
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The water level is this height right here, from here to here.
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Let us go ahead and call it h.
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What they want is dh dt.
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The rate that they give us is dv dt.
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The rate that they want is dh dt.
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We need to find a relationship between the volume and the h.
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Let me write this out.
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We do not have a direct relation between volume and height.
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But we do have a relation among three variables, among volume, height, and radius.
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That equation is the volume of a right circular cone is equal to 1/3 π r² × h.
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Let me go to the next page here.
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I have volume = 1/3 π r² h.
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We want only h on the right side of the equality.
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This is single variable calculus.
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I’m relating two variables.
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Volume, I have two variables on the right, r² and h.
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I need just h, I need to find a relation between r and h, that I can substitute in for r.
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We want only h on the right side.
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In other words, we want volume to be only a function of h.
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We must find a relation between r and h, that is some r = some function of h,
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and substitute in for r in our equation.
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I’m going to take half of my cone.
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We said that this height is 7.
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The diameter was 5, the radius is 2.5.
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The height of the water level is this thing right here, that is h.
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This is r, notice how as the water level drops, r actually gets smaller.
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It goes toward the tip of the cone.
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These are similar triangles, this one and that one.
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I have a relationship here.
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I can write 7 is to 2.5.
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Let me go back to blue here.
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I have 7 is to 2.5, as h is to r.
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This gives me 7r = 2.5 h.
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I solve for r, r = 2.5 h/7.
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I stick this value into here and I will get an equation only with h on the right side.
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V = 1/3 π r² h, becomes v = 1/3 × π × 2.5 h/7² × h.
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I do all of my math here and I end up with v = 0.1336 h³.
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I have my equation that relates the two variables, v and h.
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Now I differentiate with respect to t.
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I have got dv dt = 0.1336 × 3 h² dh dt.
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I get dv dt = 0.4007 dh dt.
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I end up solving for dh dt.
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Therefore, dh dt is equal to the dv dt divided by 0.4007 h².
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I think I forgot the h² on the previous page.
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They want this for a water level that is 3.5.
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H = 3.5, I stick that in there.
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Therefore, I get dh dt is equal to, the dv dt that was -1.8.
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And then, we have 0.4007 × 3.5².
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That leaves me with a dh dt is equal to -0.3667.
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I think it was m/s because the height is a distance.
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Notice that it is negative, that means the water level is decreasing,
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which is what you expect when a tank has been emptied with some water.
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The negative sign means the height is decreasing, which makes physical sense.
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Again, if you ended up with something like a positive here, hopefully, you will stop and say that does this make physical sense?
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No, it does not.
00:28:43.700 --> 00:28:46.800
If water is being emptied, water level is not dropping, it is rising.
00:28:46.800 --> 00:28:49.800
Somewhere along the way, there was a minor arithmetic mistake.
00:28:49.800 --> 00:28:55.900
You can use your final answer to make sure you ended up in the right place.
00:28:55.900 --> 00:29:05.600
It is decreasing, in other words, water level is dropping.
00:29:05.600 --> 00:29:11.700
Clearly, the most difficult aspect of these is just arranging it, drawing the picture, seeing what is going on, picking the variables.
00:29:11.700 --> 00:29:16.200
Finding with is changing, what relation do I use?
00:29:16.200 --> 00:29:17.800
Sometimes, it will be obvious.
00:29:17.800 --> 00:29:26.000
When you clearly have a triangle, sometimes it is not going to be so obvious.
00:29:26.000 --> 00:29:30.200
Let us see, example number 4.
00:29:30.200 --> 00:29:41.000
One car is traveling east of 40 mph toward a certain intersection, while another is traveling north at 50 mph towards the same intersection.
00:29:41.000 --> 00:29:43.500
How fast does the distance between the cars changing,
00:29:43.500 --> 00:29:50.900
when car 1 is 0.5 miles from the intersection and car 2 is 0.6 miles from the intersection?
00:29:50.900 --> 00:29:53.300
Let us draw this out.
00:29:53.300 --> 00:29:57.500
Here is how our intersection, I will just put it right over here.
00:29:57.500 --> 00:30:08.000
We have one car over here, that is actually moving at 40 mph.
00:30:08.000 --> 00:30:15.700
I’m going to draw a little line towards the intersection.
00:30:15.700 --> 00:30:19.100
That is one rate.
00:30:19.100 --> 00:30:22.300
Another is traveling north at 50 mph.
00:30:22.300 --> 00:30:33.100
I put the other car here and it is traveling this way at 50 mph.
00:30:33.100 --> 00:30:36.100
How fast does the distance between the cars changing?
00:30:36.100 --> 00:30:41.400
They want this.
00:30:41.400 --> 00:30:43.200
Let us assign some variables.
00:30:43.200 --> 00:30:55.300
I’m going to call this x, I'm going to call this y, and I’m going to call this z.
00:30:55.300 --> 00:30:59.600
The rate they gave me is, they gave me dx dt and they gave me dy dt.
00:30:59.600 --> 00:31:03.800
They told us it is travelling east of 40 mph towards a certain intersection.
00:31:03.800 --> 00:31:07.300
Dx, how fast is x changing?
00:31:07.300 --> 00:31:12.000
It is getting smaller at a rate of 40 mph.
00:31:12.000 --> 00:31:25.800
Therefore, dx dt = -40 mph.
00:31:25.800 --> 00:31:28.700
Why, this car is traveling 50 mph this way.
00:31:28.700 --> 00:31:32.600
Therefore, y is getting smaller.
00:31:32.600 --> 00:31:40.500
Dy dt = -50 mph.
00:31:40.500 --> 00:31:43.300
I will do it as mph.
00:31:43.300 --> 00:31:46.900
What they want is how fast does the distance between them is changing?
00:31:46.900 --> 00:31:51.400
They want dz dt.
00:31:51.400 --> 00:31:55.300
Dz dt equals what?
00:31:55.300 --> 00:32:03.400
I need a relation between x, y, z.
00:32:03.400 --> 00:32:07.900
Dx dt, dy dt, what they give me, what they want is dz dt.
00:32:07.900 --> 00:32:11.900
These are the variables involved, I need relation between those variables.
00:32:11.900 --> 00:32:20.300
I have one, I have x² + y² = z².
00:32:20.300 --> 00:32:25.300
Let us differentiate.
00:32:25.300 --> 00:32:28.700
Once I have the equation, I can go ahead and differentiate.
00:32:28.700 --> 00:32:35.600
I have got x² + y² = z².
00:32:35.600 --> 00:32:51.600
This is going to be 2x dx dt because we are differentiating with respect to time, + 2y dy dt = 2z dz dt.
00:32:51.600 --> 00:32:56.400
Let us go ahead and cancel the 2.
00:32:56.400 --> 00:33:26.700
I end up with dz dt is equal to x dx dt + y dy dt/ z.
00:33:26.700 --> 00:33:30.900
Let us plug in what we know.
00:33:30.900 --> 00:33:40.700
Dz dt =, they said when x is 0.5 miles away from the intersection, that is going to be 0.5.
00:33:40.700 --> 00:33:45.200
Dx dt that is the -40.
00:33:45.200 --> 00:33:49.600
They said when y is 0.6 miles from the intersection, that is 0.6.
00:33:49.600 --> 00:33:58.700
The rate of change of y is -50/ z.
00:33:58.700 --> 00:34:05.000
What is z?
00:34:05.000 --> 00:34:07.100
We have a relation for z.
00:34:07.100 --> 00:34:12.600
We have z² = x² + y².
00:34:12.600 --> 00:34:25.300
Z² = 0.5², I will do it this way, + 0.6².
00:34:25.300 --> 00:34:36.900
I get z² = 0.61 which means that z is equal to 0.781.
00:34:36.900 --> 00:34:43.200
That number, I put in there, and then I solve this.
00:34:43.200 --> 00:35:01.400
I get dz dt = 0.5 × -40 + 0.6 × -50 all divided by 0.781.
00:35:01.400 --> 00:35:21.600
I get that my rate of change of distance between them, per unit time, is equal to -64 mph distance over time.
00:35:21.600 --> 00:35:30.100
The distance between those cars.
00:35:30.100 --> 00:35:34.300
We have this going this way.
00:35:34.300 --> 00:35:43.000
This distance right here is changing when this guy is 0.5 miles away and when this guy is 0.6 miles away.
00:35:43.000 --> 00:35:47.600
This distance is changing at 64 mph.
00:35:47.600 --> 00:35:55.500
It is negative, it is decreasing because they are getting closer and closer, therefore, this is getting shorter.
00:35:55.500 --> 00:36:00.200
The negative sign tells me that the distance between them is decreasing, and that is the rate.
00:36:00.200 --> 00:36:10.200
Again, notice it is not constant, it depends on x, it depends on y, and it depends on z.
00:36:10.200 --> 00:36:22.900
The rate at which z is changing depends on x and y and z.
00:36:22.900 --> 00:36:31.800
A man 6ft tall on the ground watches a bird flying horizontally at a speed of 7 m/s and an altitude of 200ft above the ground.
00:36:31.800 --> 00:36:38.900
What is the rate of change of the angle the man’s line of sight with the bird makes with the horizontal,
00:36:38.900 --> 00:36:42.400
when the bird is 300ft from the man?
00:36:42.400 --> 00:36:44.500
Let us go ahead and draw this out.
00:36:44.500 --> 00:36:46.800
I can go back to blue here.
00:36:46.800 --> 00:36:57.100
Here is the ground and I have a man who is 6 ft tall.
00:36:57.100 --> 00:37:01.100
They say he is watching a bird flying horizontally.
00:37:01.100 --> 00:37:08.200
Let us put the bird over here, he is flying at 7 m/s.
00:37:08.200 --> 00:37:10.800
The bird is over here.
00:37:10.800 --> 00:37:19.000
They tell me that its altitude is 200ft off the ground.
00:37:19.000 --> 00:37:24.600
We said that this guy is 6ft tall.
00:37:24.600 --> 00:37:28.300
What is the rate of change of the angle the man’s line of sight with the bird?
00:37:28.300 --> 00:37:35.400
The man’s line of sight with the bird is this thing right here.
00:37:35.400 --> 00:37:40.400
What is the rate of change the angle of the man’s line of sight the bird makes with the horizontal?
00:37:40.400 --> 00:37:47.200
The horizontal is right here.
00:37:47.200 --> 00:37:50.600
The line of sight, the horizontal, this is the angle.
00:37:50.600 --> 00:37:59.400
We will call that θ, when the bird is 300ft from the man.
00:37:59.400 --> 00:38:07.600
This distance is, I’m just going to go ahead and call that z.
00:38:07.600 --> 00:38:13.200
Let us see, I have got myself a little triangle here actually.
00:38:13.200 --> 00:38:22.100
This is a fixed distance, this is going to be 200ft - the height of the man.
00:38:22.100 --> 00:38:35.300
This height right here is going to be 194ft.
00:38:35.300 --> 00:38:38.100
Let me stick with the variable that I use.
00:38:38.100 --> 00:38:42.300
Let us call this x.
00:38:42.300 --> 00:38:50.000
Here, I have got this triangle and what they want to know is dθ dt.
00:38:50.000 --> 00:38:54.400
That is what they want.
00:38:54.400 --> 00:38:58.400
What rate do they actually give us?
00:38:58.400 --> 00:39:02.500
I’m going to leave that as z.
00:39:02.500 --> 00:39:09.300
The rate that they give us is horizontally at a speed of 7 m/s.
00:39:09.300 --> 00:39:13.800
Therefore, this distance right here, we will call that x.
00:39:13.800 --> 00:39:17.000
It is x because the bird is flying horizontally.
00:39:17.000 --> 00:39:29.600
It is flying this way, that means this distance between the man directly below the bird, that changes at 7 m/s.
00:39:29.600 --> 00:39:38.300
That is going to be dx dt, that is 7 m/s.
00:39:38.300 --> 00:39:43.100
This is the rate that they give us, this is the rate that they want.
00:39:43.100 --> 00:39:53.900
Therefore, I need to find a relation between x and θ.
00:39:53.900 --> 00:39:59.500
What relation exists between x and θ?
00:39:59.500 --> 00:40:04.900
This triangle, with other information that I have, I know this height.
00:40:04.900 --> 00:40:07.700
Actually, what I can do is I can write the following.
00:40:07.700 --> 00:40:09.600
I can write tan θ.
00:40:09.600 --> 00:40:19.100
I can use tan θ = 194 divided by x.
00:40:19.100 --> 00:40:25.300
That is a relationship between θ and x, two variables, perfect.
00:40:25.300 --> 00:40:29.300
We have a relation, time to go ahead and go to the next page here.
00:40:29.300 --> 00:40:30.500
Let us write our relation again.
00:40:30.500 --> 00:40:35.600
We have the tan θ = 194/x.
00:40:35.600 --> 00:40:39.300
Now that we have a relation, we can go ahead differentiate.
00:40:39.300 --> 00:40:49.300
The derivative of tan θ is sec² θ dθ dt.
00:40:49.300 --> 00:41:00.400
The derivative of this is -194/ x² dx dt.
00:41:00.400 --> 00:41:02.700
I solve for dθ dt because that is what I want.
00:41:02.700 --> 00:41:21.400
Dθ dt = -194 divided by x² sec² θ × dx dt.
00:41:21.400 --> 00:41:27.900
I have the dx dt, they gave us dx dt.
00:41:27.900 --> 00:41:32.200
They said that that was 7 m/s.
00:41:32.200 --> 00:41:48.200
Therefore, this is going to be -194 × 7/ x² sec² θ.
00:41:48.200 --> 00:41:51.000
Now we just need to find x².
00:41:51.000 --> 00:41:54.200
Let me go back to blue here, actually, let me try black.
00:41:54.200 --> 00:42:03.100
We need to find what x is and we need to find what sec² θ is, so that we can plug it in and solve for that.
00:42:03.100 --> 00:42:14.700
Our questions now are, what is x and what is θ?
00:42:14.700 --> 00:42:25.100
Rather, what is sec² θ?
00:42:25.100 --> 00:42:30.900
Let us take care of the x first.
00:42:30.900 --> 00:42:39.100
We have this triangle, remember, this was 194.
00:42:39.100 --> 00:42:43.500
Here was the man was right here and the bird was right here.
00:42:43.500 --> 00:42:53.600
They said they wanted this rate, how fast is this θ changing when the bird is 300ft away?
00:42:53.600 --> 00:42:55.600
This is z, I have a relationship.
00:42:55.600 --> 00:43:00.600
I have z² = x² + y².
00:43:00.600 --> 00:43:07.200
Z is 300 = x² + 194².
00:43:07.200 --> 00:43:08.500
I can actually find this.
00:43:08.500 --> 00:43:21.700
X² is going to equal 52,364, which means x is actually going to equal to 228.8.
00:43:21.700 --> 00:43:26.900
That takes care of x, that we can plug in here.
00:43:26.900 --> 00:43:33.800
Or better yet, we already have x², we can just plug that in there.
00:43:33.800 --> 00:43:36.700
I want sec² θ.
00:43:36.700 --> 00:43:43.000
I finished my triangle, this is 228.8.
00:43:43.000 --> 00:43:54.400
Here, the sec θ is equal to 300/ 228.8.
00:43:54.400 --> 00:43:59.900
Therefore, the sec θ = 1.31.
00:43:59.900 --> 00:44:02.500
Perfect, now I can go ahead and plug everything in.
00:44:02.500 --> 00:44:19.900
I have got dθ dt is equal to -194 × 7 which was the dx dt.
00:44:19.900 --> 00:44:23.600
We are going to divide that by x².
00:44:23.600 --> 00:44:36.500
This is going to be the 228.8² × sec² θ, 1.31².
00:44:36.500 --> 00:44:44.400
When I solve that, I end up with -0.015.
00:44:44.400 --> 00:44:51.500
This is an angle, it is in radians per second.
00:44:51.500 --> 00:44:55.800
I hope that made sense.
00:44:55.800 --> 00:45:00.000
Again, there is a rate that you are given, a rate that you want.
00:45:00.000 --> 00:45:04.200
Identify those and try to figure out some relationship between them.
00:45:04.200 --> 00:45:13.400
Once you have the relation between them, exclusively with those whatever is given, whatever is asked for, then differentiate.
00:45:13.400 --> 00:45:19.900
Rearrange the equation and then use the other information that you have, in order to fill in the rest.
00:45:19.900 --> 00:45:23.100
In this case, we needed the x and we needed the sec θ.
00:45:23.100 --> 00:45:27.300
We found the x first and then we found the sec θ.
00:45:27.300 --> 00:45:29.400
And then, we plug in and we solve.
00:45:29.400 --> 00:45:31.800
Thank you so much for joining us here at www.educator.com.
00:45:31.800 --> 00:45:33.000
We will see you next time, bye.