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Current time:0:00Total duration:9:16

Video transcript

as a pot of tea cools the temperature of the tea is modeled by a differentiable function H for zero is less than or equal to T is less than or equal to ten where time T is measured in minutes and temperature H of T is measured in degrees Celsius values of H of T at selected taught at selected values of time T are shown in the table above so that's right over there use the data in the table to approximate the rate at which the temperature of the tea is changing at time T equals three point five show the computations that lead to your answer so they give us a bunch of points so let's just graph this just so we can visualize what this data is telling us a little bit and I suspect that this might be useful so this right over here is our temperature axis this is our temperature at any moment in degrees Celsius and then this is our time axis and they gave us the time zero to five nine and ten or they give us a temperature at those times so this is zero to five is a little bit further five nine and then 10 and at time zero the temperature is 66 degrees Celsius 66 so it's right over there at time two it is sixty degrees Celsius at time two it is 60 degrees Celsius at time five it is 52 degrees Celsius so time five maybe it's right over here this is 52 degrees Celsius at time nine an hour after nine minutes I should say it's at 44 degrees Celsius so this is 44 degrees Celsius and after 10 minutes it's at 43 degrees Celsius so this is really just a graph showing how it cools off over those 10 minutes over those 10 minutes that's what the data is telling us it's kind of a curve like this and we've sampled it at these points so going back to Part A use the data in the table to approximate the rate at which the temperature of the tea is changing at time T equals 3.5 so the rate of change is really just the slope of this curve at time T is equal to 3.5 so really we just want to find the slope at that point right over there we want to find the slope and we don't know the actual function so the best way to approximate the slope at that point is really just find the slope between time between minute five and minute two so the rate of change so we could say we could say that the rate the rate is going to be approximately our change in temperature over that over those three minutes so H of five minus H of two and this is obviously going to be in degrees Celsius over the number of minutes that changed so we from five minutes or from we end at five minutes and we started at two minutes and so the rate the rate is going to be approximately at five minutes our temperature is 52 degrees 52 degrees at two minutes our temperature is 60 degrees 60 degrees and then this is over a change in three minutes 5 minus 2 is 3 so this gets us let me scroll to the right a little bit this gives us negative 8 negative 8 degrees Celsius over 3 minutes or the rate at the rate at taught at 3 and a half minutes is going to be approximately negative 8/3 degrees Celsius per minute so that is Part A Part A and then Part B using correct units explaining the meaning of 1 over 10 1 over 10 times the definite integral from 0 to 10 of H of T DT in the context of this problem use a trapezoidal sum with the four subintervals indicated by the table to estimate this thing so the integral from zero to 10 of H of T is really the area of this entire under this curve right over here that's the integral that's this part right over here and then we're going to divide that we're going to divide that by we're going to divide that by 10 so what this is really giving us this is this expression right over here is the average this is the average temperature this is the average temperature over that time period the integral of the temperature function divided by the total amount of time that has that has elapsed and then in using correct units well it's the average temperature the average temperature is once again going to be in degrees degrees Celsius and it makes sense because this right over here is in degree Celsius you're multiplying it by time right over here but then you could divide by you're dividing I'm assuming by time right over here so this is times minutes divided by minutes so then you just get degrees Celsius again so this is this expression right over here is just the average temperature over those 10 minutes then they say use a trapezoidal sum with the four subintervals indicated by the table to estimate this so the trapezoidal sum is we don't know the exact function here so we won't be able to analytically evaluate this definite integral but what we can do is divide is divide this area into four sections they tell us to use this four subintervals and we'll essentially divided into four trapezoids so this is one trapezoid right over here so this is one trapezoid right over there that's my first trapezoid from zero to two on the base and you see on the left end of that trapezoid is high 266 the right end is 60 and then the next trapezoid will go from 2 to 5 let me do that in a more a color that contrasts a little bit better the next trapezoid will go from 2 to 5 next trapezoid goes from 2 to 5 and then the third trapezoid goes from goes from 5 to 9 the third trapezoid goes from 5 to 9 and then the fourth trapezoid goes from 9 to 10 goes from 9 to 10 and so if I want to if I want to approximate the definite integral part right here before we divide by 10 I just need to find the area of these four trapezoids so let's do that the area of this first trapezoid is going to be the base which is 2 times the average height the height on the left side of this trapezoid is 66 the height on the right side of this trapezoid is 60 the average height right over here is going to be 63 just the average of 60 and 66 so this area right over here is this is 126 that's the area of the green part right over there by the same logic the area of this orange part I want to do that in the orange color the area of this orange part the base right over here is three base is three and then the average height the height over here is sixty the height over here is 52 this is there's either eight apart so the average is going to be far away from each of is going to be 56 so this is the area here is going to be 3 times 3 times 56 which is 150 plus 18 150 plus 18 is 168 that's the area of this orange trapezoid and then the area of this blue trapezoid the base right over here is 4 and then the average height the height here is 52 the height here is 44 the average of 52 and 44 let's see there 10 they are sorry they are 8 apart 8 apart so then it's just going to be 4 from each of these so it's 48 so the average height here is 48 and so 4 times 48 is 160 plus 32 so it's 192 and then finally this last trapezoid its base is only 1 its base right over here is only 1 and then it's height is it's 43 at the left side 43 at the right side so it's average height is forty three point five forty three point five so it's forty three point five times one so that's just going to be forty three point five and so if we add up the areas under or the areas of all these trapezoids we have a pretty good approximation for the definite integral I'll just use do the use the calculator for this part right over here so we have 126 plus 168 plus 192 plus forty three point five forty three point five gives us five hundred and twenty nine point five so this is equal to five hundred twenty nine point five and so that's our approximation for the definite integral part but then we also have to divide it by ten or we have to multiply it times 1/10 so we've evaluated so this part right over here we got five hundred twenty nine point five as our Meishan it's not going to be exact but using the using the trapezoidal sum the sum of the areas of these trapezoids and now we have to multiply it by one-tenth so let's do that so or we can just divide by ten which actually we don't need a calculator for that 52.95 so this whole thing this whole thing evaluates to 52.95 and i'll do the next two parts in the next video
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