Lesson Video: Average and Instantaneous Rates of Change Mathematics

Video Transcript

In this video, we’ll learn how to find the average rate of change of a function between two 𝑥-values and use limits to find the instantaneous rate of change. We’ll learn how the average rate of change links to the slope of a line and use this to derive a formula to find the average rate of change of a function before considering the applications of this formula. We say that the average rate of change of a function 𝑓 of 𝑥 over an interval between two points given by 𝑎, 𝑓 of 𝑎 and 𝑏, 𝑓 of 𝑏 is the slope of the secant line connecting those two points.

We also might recall that the formula to help us find the slope of a line is given by the change in 𝑦 divided by the change in 𝑥. Well, in this case, the change in 𝑦 would be given by the difference between the value of the function at 𝑏 and the value of the function at 𝑎. That’s 𝑓 of 𝑏 minus 𝑓 of 𝑎, whereas the change in 𝑥 is simply 𝑏 minus 𝑎. And so, the slope of our secant line and therefore the average rate of change of our function is given by 𝑓 of 𝑏 minus 𝑓 of 𝑎 over 𝑏 minus 𝑎. But let’s just say we actually want to define the second point, that’s 𝑏, 𝑓 of 𝑏, by its relationship to the first point.

Let’s instead say that the horizontal distance between these two points is ℎ. Then, we define 𝑏 as being equal to 𝑎 plus ℎ, and 𝑓 of 𝑏, we can say, is equal to 𝑓 of 𝑎 plus ℎ. The average rate of change can now be written as 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 over ℎ. The average rate of change function is sometimes called 𝐴 of ℎ. This latter formula is the one that we’re primarily going to look at in this video. And now, let’s have a look at how we might apply it to a simple average rate of change problem.

Determine the average rate of change function 𝐴 of ℎ for 𝑓 of 𝑥 equals four 𝑥 squared plus three 𝑥 plus two at 𝑥 equals one.

Remember, the average rate of change of a function 𝑓 of 𝑥 between two points defined by 𝑎, 𝑓 of 𝑎 and 𝑎 plus ℎ, 𝑓 of 𝑎 plus ℎ is 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 all over ℎ. We can see in this question that 𝑓 of 𝑥 has been defined for us. It’s four 𝑥 squared plus three 𝑥 plus two. We want to find the average rate of change function for 𝑓 of 𝑥 at 𝑥 equals one. So, we let 𝑎 be equal to one. We don’t actually know what ℎ is, but that’s fine. This question is essentially asking us to derive a function that will allow us to find the average rate of change for any value of ℎ with this function. Let’s break this down and begin by working out what 𝑓 of 𝑎 plus ℎ is.

We said 𝑎 is equal to one, so we’re actually looking to find 𝑓 of one plus ℎ. We go back to our function 𝑓 of 𝑥, and each time we see an 𝑥, we replace it with one plus ℎ. So, 𝑓 of one plus ℎ is four times one plus ℎ squared plus three times one plus ℎ plus two. Let’s distribute our parentheses. One plus ℎ all squared is one plus two ℎ plus ℎ squared, and three times one plus ℎ is three plus three ℎ. We can then distribute these parentheses and we get four plus eight ℎ plus four ℎ squared. Finally, we collect like terms and we get four ℎ squared plus 11ℎ plus nine.

Next, we work out 𝑓 of 𝑎. Well, of course, we know that that’s 𝑓 of one. This one is slightly more straightforward than 𝑓 of one plus ℎ. We simply replace 𝑥 with one. And we get four times one squared plus three times one plus two. And that’s equal to nine. We’re now ready to substitute everything into the average rate of change formula. We have 𝑓 of one plus ℎ minus 𝑓 of one and that’s all over ℎ. Well, nine take away nine is zero, and then we can divide through by ℎ. And it simplifies really nicely to four ℎ plus 11. And so, the average rate of change function 𝐴 of ℎ for 𝑓 of 𝑥 equals four 𝑥 squared plus three 𝑥 plus two at 𝑥 equals one is four ℎ plus 11.

Now, what does this actually tell us? Well, let’s go back to our graph. Given any other point on the graph, we can work out the slope of the secant line between this point and the point at 𝑥 equals one. And this in turn tells us the average rate of change of the function. Now, whilst we did use one formula, we could define the average rate of change function for a given 𝑓 of 𝑥 as being 𝑓 of 𝑥 plus ℎ minus 𝑓 of 𝑥 all over ℎ. And if we do use that, it gives us a function that we can use for any 𝑥 and ℎ. Let’s now consider an example which will lead us to computing the average rate of change over a given interval.

Evaluate the average rate of change of 𝑓 of 𝑥 equals the square root of two 𝑥 minus one when 𝑥 varies from five to 5.62.

Remember, the average rate of change of a function 𝑓 of 𝑥 as it varies from 𝑥 equals 𝑎 to 𝑥 equals 𝑎 plus ℎ is 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 all over ℎ. In this question, we’re told that 𝑓 of 𝑥 is the square root of two 𝑥 minus one, and that 𝑥 is varying from five to 5.62. So, let’s let 𝑎 be equal to five, and ℎ is the amount that 𝑥 varies from. So, it’s 5.62 minus five, which is 0.62. Once we’ve defined all of these, all that’s left is to substitute each value into our formula. We need 𝐴 of ℎ, that’s 𝐴 of 0.62. It’s the average rate of change of our function as 𝑥 varies by 0.62. And this is, of course, equal to 𝑓 of five plus 0.62 minus 𝑓 of five all over 0.62. Let’s simplify this to 𝑓 of 5.62 minus 𝑓 of five over 0.62.

We’re quite clearly going to need to evaluate 𝑓 of 5.62 and 𝑓 of five. So, 𝑓 of 5.62 is found by replacing 𝑥 with 5.62. So, it’s the square root of two times 5.62 minus one. That’s the square root of 10.24, which is equal to 3.2. 𝑓 of five is the square root of two times five minus one, which is the square root of nine, which we know to be equal to three. 𝐴 of 0.62 is therefore 3.2 minus three over 0.62, which is 10 over 31. As 𝑥 varies from five to 5.62, the average rate of change of the function itself, the function 𝑓 of 𝑥 equals the square root of two 𝑥 minus one, is 10 over 31.

Now, what’s great about this formula is that it works really well for real-life applications too, in particular, within physics and specifically motion. We can apply the formula to a function for displacement to help us find the average rate of change of displacement, for example, which ends up giving us the function for velocity. We can also use it in geometrical problems as we’ll now see.

A lamina in the shape of an equilateral triangle expands whilst maintaining its shape. Find the average rate of change of its area when its sides change from 12 centimetres to 14 centimetres.

In this question, we’re looking to find the average rate of change of the area of the equilateral triangle. Now, for a function 𝑓 of 𝑥 that varies from 𝑥 equals 𝑎 to 𝑥 equals 𝑎 plus ℎ, the average rate of change is given by 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 over ℎ. But what is 𝑓 of 𝑥 here? Well, remember, we’re interested in the rate of change of the area. So, we need a function that describes the area of our triangle. So, let’s sketch the triangle out. We can define the side length to be 𝑥 or 𝑥 centimetres. This is our variable. We know the triangle is equilateral, so its interior angles are all 60 degrees. And then, we can use the formula the area of a triangle is a half 𝑎𝑏 sin 𝐶. And then, in this case, the area function will be a half times 𝑥 times 𝑥 times sin 60.

Well, we know that sin of 60 degrees is equal to the square root of three over two. So, this becomes the square root of three over four times 𝑥 squared. We’re told that the side length changes from 12 centimetres to 14 centimetres. So, we let 𝑎 be equal to 12, and then ℎ is the amount 𝑥 varies by; it’s 14 minus 12, which is equal to two. And so, now we can substitute everything we have into our formula for the rate of change. It’s 𝐴 of ℎ, so here that’s 𝐴 of two, and of course this is going to be equal to 𝑓 of 12 plus two minus 𝑓 of 12 all over two. We simplify 𝑓 of 12 plus two to 𝑓 of 14.

And now we need to work out 𝑓 of 14 minus 𝑓 of 12. It’s the square root of three over four times 14 squared minus the square root of three over four times 12 squared. Those values are obtained simply by substituting 𝑥 equals 14 and 𝑥 equals 12 into our function. We factor root three over four and then divide that by two to get root three over eight. 14 squared is 196, and 12 squared is 144. And so, this becomes root three over eight times 52. And then, we simplify by dividing through by four to give us 13 root three over two. And so, the average rate of change of its area is 13 root three over two. And we might say that’s 13 root three over two centimetres squared per centimetre.

We’re now going to look at the reverse of this process.

The average rate of change of 𝑓 as 𝑥 varies from two to 2.6 is negative 1.67. If 𝑓 of two is equal to negative 13, what is 𝑓 of 2.6?

Remember, the average rate of change of a function 𝑓 as 𝑥 varies from 𝑎 to 𝑎 plus ℎ is given by 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 over ℎ. Now, in this question, we don’t actually know what 𝑓 of 𝑥 is. But we do see that 𝑥 varies from two to 2.6. So, we let 𝑎 be equal to two. And then, ℎ is the amount 𝑥 varies by. It’s 2.6 minus two, which is 0.6. We want to find the average rate of change function, so that’s 𝐴 of ℎ, which is 𝐴 of 0.6. And so, according to our formula, that’s 𝑓 of two plus 0.6 minus 𝑓 of two over 0.6. This simplifies to 𝑓 of 2.6 minus 𝑓 of two over 0.6.

We’re told, however, that this is equal to negative 1.67, and also that 𝑓 of two is equal to negative 13. So, we find that negative 1.67 must be equal to 𝑓 of 2.6 minus negative 13 over 0.6. To find 𝑓 of 2.6 as this question is asking us, we need to solve this equation for 𝑓 of 2.6. We’ll begin by multiplying through by 0.6. And that gives us negative 1.002 on the left. And then on the right, we’re left with 𝑓 of 2.6 minus negative 13, which is, of course, 𝑓 of 2.6 plus 13. Next, we subtract 13 from both sides, and we find that 𝑓 of 2.6 is negative 14.002. Correct to the nearest whole number, 𝑓 of 2.6 is negative 14.

Now, I’d like to go right back to our original diagram. I’d like us to think about what happens as ℎ grows smaller. As ℎ grows smaller and smaller, the slope of the secant line approaches the slope of the curve at the point 𝑎, 𝑓 of 𝑎. This means that rather than finding the average rate of change over a given interval, we’re actually finding the rate of change at that exact point. We call this the instantaneous rate of change of the function. And since this is found by letting ℎ get smaller, we define it as being equal to the limit as ℎ approaches zero of the average rate of change. We say that the instantaneous rate of change of a function 𝑓 of 𝑥 at a point 𝑥 equals 𝑎 is the limit as ℎ approaches zero of 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 all over ℎ. Let’s now look at how this might work.

Find the instantaneous rate of change of 𝑓 of 𝑥 is equal to the square root of 𝑥 at 𝑥 equals 𝑥 one, which is greater than zero.

Remember, the instantaneous rate of change of a function 𝑓 of 𝑥 at a point 𝑥 equals 𝑎 is found by taking the limit as ℎ approaches zero of the average rate of change function. That’s the limit as ℎ approaches zero of 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 all over ℎ. In this case, we know that 𝑓 of 𝑥 is equal to the square root of 𝑥, and we want to find the instantaneous rate of change at 𝑥 equals 𝑥 one. So, we’ll let 𝑎 be equal to 𝑥 one. Let’s substitute what we know into our formula. We want to compute the limit as ℎ approaches zero of 𝑓 of 𝑥 one plus ℎ minus 𝑓 of 𝑥 one all over ℎ. We need to find the limit as ℎ approaches zero of the square root of 𝑥 one plus ℎ minus the square root of 𝑥 one all over ℎ.

Now, we can’t do this with direct substitution. If we do, we end up dividing by zero and we know that to be undefined. And so instead, we multiply the numerator and denominator of the function by the conjugate of the numerator, by the square root of 𝑥 plus one plus ℎ plus the square root of 𝑥 one. On the denominator, we simply have ℎ times the square root of 𝑥 one plus ℎ plus the square root of 𝑥 one. Then on the numerator, we have the square root of 𝑥 one plus ℎ times the square root of 𝑥 one plus ℎ, which is simply 𝑥 one plus ℎ. Then, we multiply the square root of 𝑥 one plus ℎ by the square root of 𝑥, and negative the square root of 𝑥 one times the square root of 𝑥 one plus ℎ. When we find their sum, we get zero.

So, all that’s left to do is to multiply negative the square root of 𝑥 one by the square root of 𝑥 one. And we simply get negative 𝑥 one. 𝑥 one minus 𝑥 one is zero. And then, we divide through by ℎ. And so, this becomes the limit as ℎ approaches zero of one over the square root of 𝑥 one plus ℎ plus the square root of 𝑥 one. And we can now evaluate this as ℎ approaches zero. We’re left with one over the square root of 𝑥 one plus the square root of 𝑥 one, which is one over two times the square root of 𝑥 one. The instantaneous rate of change function of 𝑓 of 𝑥 is equal to the square root of 𝑥 is therefore one over two times the square root of 𝑥 one.

In this video, we’ve learned that the average rate of change of a function 𝑓 of 𝑥 over an interval between two points given by 𝑎, 𝑓 of 𝑎 and 𝑎 plus ℎ, 𝑓 of 𝑎 plus ℎ is the slope of the secant line that connects those two points. We often define this as 𝐴 of ℎ, and it’s given by 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 all over ℎ. We also saw that the instantaneous rate of change of a function is found by letting ℎ approach zero. It’s the limit as ℎ approaches zero of 𝐴 of ℎ, of 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 all over ℎ.