Using average rate of change and first derivative, 2018 Calculus AB free response 5a,b
About this video
Sal solves 2018 AP Calculus AB 5a and 5b for f(x) = e^x cos(x). For part A he finds the average rate of change on [0, pi] as (f(pi) - f(0))/pi = (-e^pi - 1)/pi. For part B he uses the product rule, f'(x) = e^x cos(x) - e^x sin(x), and the unit circle to find the tangent slope at x = 3pi/2 is e^(3pi/2).
You'll be able to
- find the average rate of change of a function on an interval.
- use the product rule to differentiate e^x cos(x).
- evaluate a derivative at 3pi/2 using the unit circle.
Key words
- Average rate of change
- (f(b) - f(a))/(b - a), the slope between two points on the graph.
- Product rule
- The derivative of f(x)g(x) is f'(x)g(x) + f(x)g'(x).
- Unit circle
- A circle of radius 1 where the x-coordinate is cos of the angle and the y-coordinate is sin.
- Tangent line slope
- The value of the derivative at a point.
Transcript
- [Instructor] We're told let f be the function defined by f of x is equal to e to the X cosine of x. Find the average rate of change of f on the interval from zero to pi. Pause the video, have a go at that, and then we'll do part B. Okay, average rate of change, we essentially just want to find the slope between the points zero, f of zero, and the point pi, f of pi.
And so this is going to be f of pi minus f of zero. And so F of pi is going to be e to the pi times co-sign of pi minus f of zero. F of zero is e to the zero time cosine of zero. All of that over pi minus zero. And so this is going to be equal to cosine of pi is negative one, e of the zero is one, cosine of zero is one.
And so we are left with e to the pi times negative one. So it's negative e to the pi. Did I do that right? Yep. e to the pi, yeah, cosine of pi is negative one, negative e to the pi minus one. All of that over pi. So there we have our average rate of change. And now part B, what is the slope of the line tangent to the graph of f at x is equal to three pi over two?
Pause the video, have a go at that. Well, to figure out the slope of the tangent line, we just have to evaluate the derivative at X equals three pi over two, so f prime of X. We essentially have the product of two functions here. Let me color code this, and we're gonna use the product rule.
So in red I'll have that function, e to the x and it's derivative. And maybe in blue I'll do that for cosine of x. So first, I'm gonna take the derivative of e to the x, which is just e to the x times cosine of x, which is of course just cosine of X. Let me color code it better than that.
Times cosine of x. And then to that I am going to add just e to the x, which happens to be the same as what we had before. Plus e to the x, times the derivative of cosine of x. The derivative of cosine of x is negative sine of x. So sine of x, and I'll turn this one into a negative.
And so f prime of three pi over two. So f prime, let me do that in black in that neutral blue color. So f prime of three pi over two is going to be equal to e to the three pi over two times cosine of three pi over two minus e to the three pi over two times sine of three pi over two.
So let's see, cosine of three pi over two. I always try to remember my unit circle here. And so three pi over two, we're pointing straight down here. So the cosine, which is the exit coordinate, this is just going to be zero. So this whole thing is just going to be to zero, this first term, and sine of three pi over two, that's negative one.
So our final answer, all of this just becomes zero. And so this is going to be equal to negative one times negative e to the three pi over two, which is just e to the three pi over two. And we're done.
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