Problem 5. “At the beginning of 2010, a landfill contained 1,400 tons of solid waste.
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2011 Calculus AB free response #5a AP Calculus AB
The increasing function W–” so I guess W constantly increases– “the increasing function W models the total amount of solid waste stored at the landfill. Planners estimate that W will satisfy the differential equation.” so the derivative of W with respect to time is equal to 1 over 25 times the quantity W minus 300, “for the next 20 years. W is measured in tons.
t is measured in years from the start of 2010.” All right, let’s get into this. So part a. “Use the tangent line to the graph of W at t equals 0 to approximate the amount of solid waste that the landfill contains at the end of the first 3 months of 2010.
” So since time is in years, 3 months would be 1/4 of a year. So t equals 1/4. So it seems at first this is really daunting. They gave us a differential equation.
We don’t even know what the actual function W is. How do we figure out its tangent line? But we just have to kind of think about what they’re asking. So regardless of what W looks like– so let’s think about it a little bit.
So we don’t know yet what W actually does look like. But they tell us it’s an increasing function. So at time 0 it has 1,400 tons in it.
They tell us that right up here. And that it increases. We don’t know what the function actually does look like. But let’s say that that is W.
So what they’re saying in problem a is use a tangent line to the graph at t equals 0. Let me draw a W a little bit differently. So W might look like this.
So what they’re saying is, find the tangent line, find the slope of that tangent line at t equals 0. So there’s some slope right over there. And then we can use this slope to create a linear approximation of where we’re going to be a quarter of a year from then.
So although we don’t know what W is just yet, we could take the slope of this line out– so this is our t axis– to t is equal to 1/4. And wherever that line takes us out after a fourth of a year, that will be at least a decent approximation. We’re extrapolating forward from that first point and the current slope.
So we really just have to figure out the slope of this line, and just see where that line is at t is equal to 1/4. And you say, wait. How do we know what the derivative of W is at 0 without knowing W? And that’s where we can go straight to this differential equation.
We could actually rewrite this differential equation using slightly different notation. This is the derivative of W with respect to t. We could write that as W prime of t is equal to 1 over 25 times the function W, which is a function of t, minus 300. And when you look at it this way, it becomes a little bit clearer on how to figure out what the derivative of W is at 0.
The slope of that line literally is just the derivative of W evaluated at 0. So let’s literally take the derivative of W and evaluate it at 0. So we have W prime of 0 is equal to 1 over 25 times W of 0, which we know.
We know this is 1,400 tons of solid waste. This is how much waste there is at time 0. Minus 300. So this right over here is 1,400.
And so we have W prime at time equals 0. So our slope at time equals 0, or our derivative at time equals 0, is equal to 1 over 25 times 1,400 minus 300 is 1,100. And 25 goes into 1,100 44 times, right? It goes four times into each 100.
We have 11 hundreds here. So it’ll go 44 times. So it goes 44. So the slope of that line is 44.
We could say m for slope. Or actually, let me just write down the word. The slope of this line is 44. And I just ate some peanuts or something.
So my voice is a little dry. So bear with me. But the slope of this line is 24. So how do we use that to find an approximation for the amount of waste that the landfill contains at the end of the first three months? Well, let me zoom in a little bit.
And actually, my fingers are all salty, too. So maybe I don’t have a proper grip on my pen. But I’ll try my best.
Let me just zoom in a little bit more. So we’re starting at 1,400 tons. So this is my W-axis.
This is my time axis. We’re starting at 1,400 tons. And we are increasing from there.
They’re telling us that it is an increasing function. So maybe it looks something like that. And then our slope right over here.
And I don’t know, I haven’t drawn W exactly that accurately. I’m just guessing what it might look like at this point. The tangent line has a slope of 44. So that is our tangent line.
And what that’s saying is if we go out one unit of time, which is one year, then we would have gone up 44 in tonnage. So if we use this line as an approximation, after one year, at this point, would have 1,444 tons. But we’re not trying to approximate a year out.
We’re trying to approximate 1/4 of a year out. So we’re trying to approximate. So this would be half a year out.
This is 1/4 of a year out. We’re trying to approximate that point right over there. So it’s going to be 1,400, this point right over here. And we could write down the equation of this line, if we like.
We could say this line– so we’ll call this the W approximation, because this isn’t exactly our W function– this is equal to the slope of our line, 44, times time, plus our W intercept, we could say, or plus our initial condition, plus 1,400. So if you put time is equal to 1/4 in there, you get it equal to 44. So let me write it this way.
So our approximate W, at time is equal to 1/4 of a year, is equal to 44 times 1/4 plus 1,400. I’m running out of space. And my salty hands are having trouble writing this properly.
And so 44 times 1/4, or 44 divided by 4, is 11 plus 1,400. And so you add them together. You get 1,411 or 1,411 tons. This is our approximation.
We just took the slope from our starting point and used that slope as an approximation. It probably is not the exact amount of tonnage based on the actual function W. But it’s an OK approximation.
But that’s our first answer for Part A. 1,411 tons.
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