Finding average velocity 2015 AP Calculus AB BC 3d
About this video
This video solves part (d) of the 2015 AP Calculus AB/BC free response #3: Bob's average velocity for 0 <= t <= 10 using the model B(t) = t^3 - 6t^2 + 300. Average velocity is the area under the velocity curve divided by the time interval, so (1/10) times the integral from 0 to 10 of B(t) dt, which works out to 3500/10 = 350 meters per minute.
You'll be able to
- find average velocity as (1/(b - a)) times the integral of velocity.
- explain average value as area divided by width.
- integrate a polynomial and evaluate it at the bounds.
Key words
- average velocity
- Total displacement divided by the time taken; for a velocity function, its average value.
- average value of a function
- The integral over [a, b] divided by the width b - a: like the average height of the area.
- power rule for integrals
- The integral of t^n is t^(n+1)/(n+1).
Transcript
- [Instructor] All right, let's do the next part. Based on the model B from part C, find Bob's average velocity during the interval from 0 ≤ t ≤ 10. And if the notion of average velocity or average value of a function is completely foreign to you, I encourage you to watch the videos on Khan Academy on find the average of a function.
But just to kind of cut to the chase, the average velocity, the average velocity is going to be the area under the velocity curve ÷ our change in time. So the area under the velocity curve from T = 0 to T = 10 of B(t)dt ÷ our change in time. So it's going to be ÷, well, you're going from 0 to 10, so 10 - 0 = 10.
And if you wanted the intuition here, it's like, "Well, if you know the area of something and if you wanted to find its average height, you could just divide by its width." And that's what we're doing here. If we know the area of something, we want to figure out its average height, and so you divide by its width.
That's, I guess, a very high-level intuition for where this expression came from. And so this is going to be equal to 1/10 times the integral from 0 to 10, and B of t is T^3 - 6T^2 + 300 dT. And so this is going to be equal to 1/10, take the anti-derivative here so this is going to be t^4/4.
And this is going to be, if we increase the exponent here by 1 it's T^3 and then you divide by 3. So it's -6 ÷ 3 is -2t^3. And then + 300t. And I'm going to evaluate it. I am going to evaluate it at 10, and subtract from that and evaluate it at 0. And so this is going to be equal to, it is going to be equal to 1/10th, that same 1/10th there.
And when you evaluate all of this at 10, what are we going to get? Let's see. 10^4 is 10,000 ÷ 4 is 2,500, 2,500, and then - 2 x 10^3, so it's 2 x 1,000. So - 2,000, and then 300 x 10. Well, that's + 3,000. And then you subtract all of this evaluated at 0, which is just going to be 0.
So this is going to be equal to 2,500 - 2,000 is 500 + 3,000, this is all simplifies to 3,500 or 3,500, and then you divide it by 10. This is going to be 350, 350, and it's an average velocity meters per 350 meters per minute. And we are done.
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