Calculus, all content (2017 edition)
- Sal interviews the AP Calculus Lead at College Board
- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c
- 2011 Calculus AB free response #1a
- 2011 Calculus AB free response #2 (a & b)
- 2011 Calculus AB free response #2 (c & d)
- 2011 Calculus AB free response #3 (a & b)
- 2011 Calculus AB free response #3 (c)
- 2011 Calculus AB free response #4a
- 2011 Calculus AB free response #4b
- 2011 Calculus AB free response #4c
- 2011 Calculus AB free response #4d
- 2011 Calculus AB free response #5a
- 2011 Calculus AB free response #5b
- 2011 Calculus AB free response #5c
- 2011 Calculus AB free response #6a
- 2011 Calculus AB free response #6b
- 2011 Calculus AB free response #6c
Approximating rate of change and total area under a curve. Trapezoidal sums to approximate integrals. Created by Sal Khan.
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- At4:17: Why is this expression the average temperature?(4 votes)
- Recall the definition of a Definite Integral
You're basically doing a summation of H(t)*dt
which can be read as a summation of Temperature*change in time
You really only need to focus on the units
You're basically adding up a bunch of degrees Celsius*minutes
which then leaves you with [total] degrees Celsius*minutes
But then you divide by 10 [total] [minutes]
What do you get if you divide [total] degrees Celsius*minutes by [total] minutes?
you get [average] degrees Celsius, or in another word: [average] Temperature
The Definite Integral expression yields [average] Temperature
-- Edit to clarify:
The Definite Integral itself yields [total] degrees Celsius*minutes
It is implied that the 1/10 outside the integral is 1/10minutes.
The "minutes" unit cross cancels, so the whole expression leaves just Temperature
And yes.. degrees Celsius*minutes makes no sense
Length*Width can at least be understood as Area
or Area*Height can at least be understood as Volume
degrees Celsius/minutes can be understood as rate of change
but degrees Celsius*minutes..... no intuitive sense(6 votes)
- I am confused when he does the derivative for f(3.5) at2:56. My textbook tells me that you use two secant lines(that are close to the point whose slope you are trying to estimate) to approximate the derivative, is there something I am missing here?(3 votes)
- You can do that if you have a formula, but if you only have a table of values to draw from, you might be limited on how many slopes you can calculate. Sal is using a version of the difference quotient that uses an input value on either side of the input you are looking for. This is sometimes called the "balanced difference quotient" and is a bit more efficient than the standard formulation. The formula for the balanced difference quotient is
lim h->0 of (f(x+h)-f(x-h))/(2h).(3 votes)
- for (a), could you not say that the graph is probably an exponential decay graph? Then you could substitute in a couple of the points and do a little bit of algebra to find the approximate equation of the curve, then take the derivative at that point? I understand it would take longer, but it would give an even better approximation, would it not?(3 votes)
- It would be possible to extrapolate the equation of the curve, but this is one of many questions from a timed test, and the instructions say to use trapezoidal sum. Defining the curve itself algebraically would cost a lot of a limited resource to find a more precise answer than the test is asking for.(4 votes)
- How fast are you expected to do each of these free response questions on the AP test?(3 votes)
- There have been minor changes in recent years. There are two calculator questions in a 30 minute span of time, and then 4 non calculator questions in a 60 minute chunk. However, you can go back and complete the calculator questions (without the use of your calculator) if you finish early in the non calculator section.(2 votes)
- Why was part b more complicated than it needed to be? it would had been easier if we added all the degrees that the sample gave us and divided by five (The answer would then be 53).(2 votes)
- Well, it is true that that approximation is pretty close to the one obtained by integration over the interval. I can think of a couple of reasons not to use that shortcut.
First, the question asks us to use the trapezoidal sum method and to use the defined subintervals that we are given in the table. Usually the rubrics don't take kindly to not following directions on the test, and your method would not demonstrate mastery of the specified technique. Mathematical intuition might get to the ballpark of the required approximation, but probably won't give you points on this part of the test, which is kinda why people take it.
Second, your method would be further off if the curve were different, right? It happens that this curve is not that far off a linear function. But what if the data points were as follow:
You'd start to see more divergence and that is why the technique was invented.(3 votes)
- How do you suggest to complete this problems in a timely manner when stressed and in a hurry during the exam?! Honestly, I would like input.(1 vote)
- First, practice as much as you can. You don't want to be caught having to try to remember how to do integration by parts or a trig substitution or a complicated limit on the exam. You should just practice as much as you can, so that you learn the techniques as well as you can.
Second, memorize as many function/derivative and function/integral pairs as you can. The more you know, the less you have to figure out on the exam.
Your class textbook should provide you with a list of common integral/derivatives. I strongly suggest you memorize as many of them as possible.
Here are some review/cheat sheets I'd recommend memorizing:
- Why does the area of a trapezoid A = 1/2H(B1 +B2) not used?(1 vote)
- It is used.
He is saying (1/2)(b1+b2)*h
Where h is the length of the bottom, and the (1/2)(b1+b2) he calls the "average of the heights"
So when he finds the average of 66 and 60, he is finding (66+60)/2, then he multiplies that by the "height" which is 2(2 votes)
- For problem b part 1. Is it ok if you say: The average temperature in °C/min? that was my guess. Thank you(1 vote)
The average temperature is expressed (in this case) in °C and is a scalar, and would not be per unit time. The time units cancel: [H(t) (°C) x dt (min)] / [10 (min)]
I hope this helps.(2 votes)
- What about the rest of number 1?(1 vote)
- Sal did the rest of #1 in the "2011 Calculus AB Free Response #1 (b, c, & d)" video. https://www.khanacademy.org/math/ap-calculus-ab/ab-solved-exams-new/ab-solved-exams-2011-new/v/2011-calculus-ab-free-response-1-parts-b-c-d(1 vote)
- For B, could we just leave the answer in the sub-interval form as 1/10(1/2(2)(66+60)+1/2(3)(60+52)+1/2(4)(52+44)+1/2(1)(44+43)) degrees Celsius? (as to save time on the test)(1 vote)
- I assume you mean the AP Test. On the AP test, that would be totally fine! Make sure you show your work, though. (Mention the function) Anything that can be typed into a calculator and give a number is considered a numeric answer on the test and can be left in that form.(1 vote)
As a pot of tea cools, the temperature the tea is modeled by a differentiable function h. For 0 is less than or equal to, t is less than or equal to 10. Where time t is measured in minutes. And temperature h of t is measured in degrees Celsius. Values of h of t, at selected time, 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 t is changing at time t equals 3.5. 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's 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 of 0 2 5 9 and 10. They give us a temperature of those times. So, this is 0. 2, 5's A little bit further. 5, 9, And then 10. And at time zero, the temperature's 66 degrees, Celsius. 66, So it's right over there. At time 2, it is 60 degrees Celsius. At time two, it is 60 degrees celsius. At time 5, it is 52 degrees Celsius. So time 5, maybe it's right over here, this is 52 degrees Celsius. At time 9, or after 9 minutes, I should say, it's at 44 degrees Celsius. So this 44 degrees Celsius, and after 10 minutes it's 43 degrees Celsius. So this is really just a graph showing how it cools off over those ten minutes, over those ten 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 a slope of this curve at time t is equal to 3.5. So really, we just wanna find the slope at that point right over there. We wanna 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 5 minus h of 2, and this is obviously going to be in degrees Celsius, over the number of minutes that changed, so we went from 5 minutes or from, we end at 5 minutes, and we started at 2 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 change in three minutes. 5 Minus 2 is 3. So, this gets us, let me just pull to the right a little bit,. This gives us negative 8. Negative 8 degrees Celsius over three minutes. Or the rate at, the rate at, at three and a half minutes is going to be approximately negative 8 3rds, degrees Celsius per minute. So that is part a, part a. And then part b, using correct units, explain 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 4 sub-intervals indicated by the table to estimate this thing. So, the integral from 0 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 and the average temperature is once again going to be in degrees. Degrees Celsius. And it makes sense, cuz this right over here is in degrees 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 as, 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 ten minutes. They they say use a trapezoidal sum with the four sub intervals indicated by the table to estimate this. So, a 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 divide it 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 height of 66 the right end is 60. And the the next trapezoid will go from two to five. Let me do that in a more. Color that contrast a little bit better. The next trapezoid will go from 2 to 5. The 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 if the left side for this trapped 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 3. Base is 3. And then the average height. The height over here is 60. The height over here is 52. This is, they're seated 8 apart, so the average is going to be 4 away from each 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 then average height, the height here is 52, the height here is 44, the average of 52 and 44, let see they're 10, they are, sorry, they are 8 apart. So then it's just gonna 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, it's base is only 1. Its base right over here is only 1, and then its height is 43 at the left side, 43 at the right side. So its average height is 43.5, 43.5. So it's 43.5 times 1. So that's just going to be 43.5, 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 43.5, 43.5. Gives us 529.5. So this is equal to 529.5. And so that's our approximation for the definite integral part. But then we also have to divide it by 10, or we have to multiply time 1 10th. So we've evaluated, so this part right over here, we got 529.5 as our approximation. 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 by 1 10th, so lets do that. So, or we could just divide by 10, 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.