Instantaneous Rate of Change - Problem 2 3,688 views
Another way of calculating the instantaneous rate of change at a certain time is to draw a tangent line at that point on a given graph. The slope of the tangent line will be slope of the curve at that point, so it is the instantaneous rate of change. To calculate the slope of the tangent line, you can use any two points on the tangent line, and find the slope as per usual (change in y divided by the change in x). The rate of change is decreasing if it is greater at one time than it is at a later time. Or, it is increasing if it is less at one time than it is at a later time.
A really important idea about instantaneous rate of change, is that, you can measure it by looking at the slope of the graph of a function. Let’s take a look at an example; Estimate the instantaneous rate of change of a function f at x equals 1, by measuring the slope of the tangent line.
Here, I’ve a function y equals f(x) graphed, and I want to find the instantaneous rate of change at x equals 1. That’s at this point right here. So I want to really carefully draw a tangent line and measure its slope. So remember that the tangent line has got to go in the same direction as the curve. So right about like that. I’ll extend it backwards and extend it forward. And I like to extend these tangent lines pretty far because I think it’s best to pick points that are far apart when you’re measuring slope. I’m going to use this point right here and I’m going to use, how about this point right here?
I try to find the points where the grid lines intersect because those are going to be nice points. Here 1 unit is 4 squares, so each square is going to be a quarter units. And that means that this point has coordinates (-.75, 1). That’s one point. Now I need a second point. This point here is at (2.5, 4.75).
Now all I have to do is measure the slope. So it’s the change in y over the change in x. 4.75 minus 1 over 2.5 minus -.75. This is going to be 3.75 over this is 2.5 plus .75, 3.25. I want to go ahead and use my calculator for this. So 3.75 divided by 3.25, and I get about 1.15. A little more than 1. We don’t have units here so we don’t have to worry about that.
Now let’s do the same thing for x equals 4. I want to draw a tangent line and measure its slope and that will give me the instantaneous rate of change of this function at x equal s4. So that’s the point x equals 4. First I want to line up my ruler here so it’s going in the same direction as the curve and extend my tangent line to the right and to the left. You want to extend it pretty far again and let’s pick two points that are far apart. This looks like a good point here.
Let’s take this point right there. So this left point has x coordinates zero, y coordinate 3.25. So I’ll write that down, (0, 3.25). I need to get my second point. Way up here we are at x equals 5 and y equals 5.5. So (5, 5.5). Again change in y over change in x. I have 5.5 minus 3.25 over the change in x, 5 minus 0. This is going to be 2.25 over 5. Again my calculator, 2.25/5, .45. Again no units.
But you can see that the instantaneous rate of change is less. When you’re looking at the graph of the function, the slope of the curve tells you the instantaneous rate of change and here the slope is less than it was here. The slope is decreasing and that helps us answer the last part; does the rate of change appear to b increasing or decreasing?
The rate of change is given by the slope of the curve. The slope of the tangent line and it does appear to be decreasing. The slope is very high at the beginning, smaller at the end so it appears to be decreasing as x increases.