Finding rate of change: 2015 AP Calculus 2c
About this video
This video solves part (c) of the 2015 AP Calculus free-response question 2. It defines h(x) = f(x) - g(x) as the vertical distance between the graphs in region S, notes that the rate of change of h at x = 1.8 is h'(1.8), and uses a graphing calculator's numerical derivative to get about -3.812.
You'll be able to
- write the vertical distance between two graphs as h(x) = f(x) - g(x) when f is above g.
- explain that the rate at which h changes with respect to x at x = 1.8 is h'(1.8).
- recognise when an AP question invites using a calculator to evaluate a derivative at a point.
- interpret the sign of a rate of change in context.
Key words
- vertical distance
- The gap between two graphs measured straight up and down at the same x, found by top function minus bottom function.
- rate of change
- How fast one quantity changes as another changes; here, the derivative h'(x).
- numerical derivative (nDeriv)
- A calculator feature that estimates the value of a derivative at a specific point.
- region S
- The region from the exam problem where the graph of f lies above the graph of g.
Transcript
- [Voiceover] Let h be the vertical distance between the graphs of f and g and region s. Find the rate at which h changes with respect to x when x is equal to 1.8. We have region s right over here. You can't see it that well since I drew over it. And what you see when we're in region s, the function f is greater than the function g.
It's above the function g. We can write h of x, we could write h of x, as being equal to f of x minus g of x and what we wanna do is we wanna find the rate at which h changes with respect to x. We could write that as h prime of, we could say h prime of x but we want the rate when x is equal to 1.8.
So h prime of 1.8 is what we wanna figure out. We could evaluate f prime of 1.8 and g prime of 1.8 and to do that we would take the derivatives of each of these things and we know how to do that. It's within our capabilities, but it's important to realize when you're taking the AP test that you have a calculator at your disposal.
And a calculator can numerically integrate and can numerically evaluate derivatives. And so when they actually want us to find the area or evaluate an integral, where they give the endpoints, or evaluate a derivative at a point, well that's a pretty good sign that you could probably use your calculator here.
And what's extra good about this is we have already essentially inputted h of x in the previous steps and really in part A. I defined this function here, and this function is essentially h of x. I took the absolute value of it so it's always positive over either region, but I could delete the absolute value if we want.
Delete that absolute value, then I have to get rid of that parentheses at the end. Let me delete that. And so notice this is h of x. We have our f of x, which was 1 plus x plus e to the x squared minus 2x, and then from that we subtract g of x. G of x was a positive x to the fourth but we're subtracting so negative x to the fourth, let me show you g of x right over here.
G of x is right over here, and notice we are subtracting it. So the y1, as I've defined it in my calculator, and I just pressed this y equals button right over here, that is my h of x. Now I can go back to the other screen, and I can evaluate its derivative when x is equal to 1.8.
I go to math. I scroll down. We have n derivative right here and so click Enter there. And then what are we gonna take the derivative of? Well the function y sub 1 that I've already defined in my calculator. I can go to variables, y variables, it's already selected function so I'll just press Enter.
And I'll select the function y sub 1 that I've already defined. So I'm taking the derivative of y sub 1. I'm taking the derivative with respect to x. And I'm going to evaluate that derivative, when x is equal to 1.8. That simple. And then I click Enter. And there you have it.
It's approximately -3.812. So approximately -3.812. And we're done. One thing you might appreciate from this entire question, and even the question 1, is that they really wanted to make sure that you understand the underlying conceptual ideas behind derivatives and integrals.
And if you understand the conceptual ideas, of how do you use them to solve problems, and you have your calculator at disposal, then these are not too hairy, that these can be done fairly quickly.
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