### Video Transcript

In this video, weβll learn how to
find the average rate of change of a function between two π₯-values and use limits
to find the instantaneous rate of change. Weβll 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 π of π₯ over an interval between two points given by π, π of
π and π, π of π 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 π¦ divided by
the change in π₯. Well, in this case, the change in
π¦ would be given by the difference between the value of the function at π and the
value of the function at π. Thatβs π of π minus π of π,
whereas the change in π₯ is simply π minus π. And so, the slope of our secant
line and therefore the average rate of change of our function is given by π of π
minus π of π over π minus π. But letβs just say we actually want
to define the second point, thatβs π, π of π, by its relationship to the first
point.

Letβs instead say that the
horizontal distance between these two points is β. Then, we define π as being equal
to π plus β, and π of π, we can say, is equal to π of π plus β. The average rate of change can now
be written as π of π plus β minus π of π over β. The average rate of change function
is sometimes called π΄ of β. This latter formula is the one that
weβre primarily going to look at in this video. And now, letβs have a look at how
we might apply it to a simple average rate of change problem.

Determine the average rate of
change function π΄ of β for π of π₯ equals four π₯ squared plus three π₯ plus
two at π₯ equals one.

Remember, the average rate of
change of a function π of π₯ between two points defined by π, π of π and π
plus β, π of π plus β is π of π plus β minus π of π all over β. We can see in this question
that π of π₯ has been defined for us. Itβs four π₯ squared plus three
π₯ plus two. We want to find the average
rate of change function for π of π₯ at π₯ equals one. So, we let π be equal to
one. We donβt actually know what β
is, but thatβs 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 β with this function. Letβs break this down and begin
by working out what π of π plus β is.

We said π is equal to one, so
weβre actually looking to find π of one plus β. We go back to our function π
of π₯, and each time we see an π₯, we replace it with one plus β. So, π of one plus β is four
times one plus β squared plus three times one plus β plus two. Letβs distribute our
parentheses. One plus β all squared is one
plus two β plus β squared, and three times one plus β is three plus three β. We can then distribute these
parentheses and we get four plus eight β plus four β squared. Finally, we collect like terms
and we get four β squared plus 11β plus nine.

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

Now, what does this actually tell
us? Well, letβs 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 π₯
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 π of π₯ as being π
of π₯ plus β minus π of π₯ all over β. And if we do use that, it gives us
a function that we can use for any π₯ and β. Letβs 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 π of π₯ equals the square root of two π₯ minus one when π₯ varies
from five to 5.62.

Remember, the average rate of
change of a function π of π₯ as it varies from π₯ equals π to π₯ equals π
plus β is π of π plus β minus π of π all over β. In this question, weβre told
that π of π₯ is the square root of two π₯ minus one, and that π₯ is varying
from five to 5.62. So, letβs let π be equal to
five, and β is the amount that π₯ varies from. So, itβs 5.62 minus five, which
is 0.62. Once weβve defined all of
these, all thatβs left is to substitute each value into our formula. We need π΄ of β, thatβs π΄ of
0.62. Itβs the average rate of change
of our function as π₯ varies by 0.62. And this is, of course, equal
to π of five plus 0.62 minus π of five all over 0.62. Letβs simplify this to π of
5.62 minus π of five over 0.62.

Weβre quite clearly going to
need to evaluate π of 5.62 and π of five. So, π of 5.62 is found by
replacing π₯ with 5.62. So, itβs the square root of two
times 5.62 minus one. Thatβs the square root of
10.24, which is equal to 3.2. π 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. π΄ of 0.62 is therefore 3.2
minus three over 0.62, which is 10 over 31. As π₯ varies from five to 5.62,
the average rate of change of the function itself, the function π of π₯ equals
the square root of two π₯ minus one, is 10 over 31.

Now, whatβs 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 weβll 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, weβre looking
to find the average rate of change of the area of the equilateral triangle. Now, for a function π of π₯
that varies from π₯ equals π to π₯ equals π plus β, the average rate of change
is given by π of π plus β minus π of π over β. But what is π of π₯ here? Well, remember, weβre
interested in the rate of change of the area. So, we need a function that
describes the area of our triangle. So, letβs sketch the triangle
out. We can define the side length
to be π₯ or π₯ 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 ππ sin πΆ. And then, in this case, the
area function will be a half times π₯ times π₯ 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 π₯ squared. Weβre told that the side length
changes from 12 centimetres to 14 centimetres. So, we let π be equal to 12,
and then β is the amount π₯ varies by; itβs 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. Itβs π΄ of β, so here thatβs π΄
of two, and of course this is going to be equal to π of 12 plus two minus π of
12 all over two. We simplify π of 12 plus two
to π of 14.

And now we need to work out π
of 14 minus π of 12. Itβs 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 π₯ equals 14 and π₯ 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 thatβs 13 root
three over two centimetres squared per centimetre.

Weβre now going to look at the
reverse of this process.

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

Remember, the average rate of
change of a function π as π₯ varies from π to π plus β is given by π of π
plus β minus π of π over β. Now, in this question, we donβt
actually know what π of π₯ is. But we do see that π₯ varies
from two to 2.6. So, we let π be equal to
two. And then, β is the amount π₯
varies by. Itβs 2.6 minus two, which is
0.6. We want to find the average
rate of change function, so thatβs π΄ of β, which is π΄ of 0.6. And so, according to our
formula, thatβs π of two plus 0.6 minus π of two over 0.6. This simplifies to π of 2.6
minus π of two over 0.6.

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

Now, Iβd like to go right back to
our original diagram. Iβd like us to think about what
happens as β grows smaller. As β grows smaller and smaller, the
slope of the secant line approaches the slope of the curve at the point π, π of
π. This means that rather than finding
the average rate of change over a given interval, weβre 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
β get smaller, we define it as being equal to the limit as β approaches zero of the
average rate of change. We say that the instantaneous rate
of change of a function π of π₯ at a point π₯ equals π is the limit as β
approaches zero of π of π plus β minus π of π all over β. Letβs now look at how this might
work.

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

Remember, the instantaneous
rate of change of a function π of π₯ at a point π₯ equals π is found by taking
the limit as β approaches zero of the average rate of change function. Thatβs the limit as β
approaches zero of π of π plus β minus π of π all over β. In this case, we know that π
of π₯ is equal to the square root of π₯, and we want to find the instantaneous
rate of change at π₯ equals π₯ one. So, weβll let π be equal to π₯
one. Letβs substitute what we know
into our formula. We want to compute the limit as
β approaches zero of π of π₯ one plus β minus π of π₯ one all over β. We need to find the limit as β
approaches zero of the square root of π₯ one plus β minus the square root of π₯
one all over β.

Now, we canβt 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 π₯ plus one plus β plus the square root of π₯ one. On the denominator, we simply
have β times the square root of π₯ one plus β plus the square root of π₯
one. Then on the numerator, we have
the square root of π₯ one plus β times the square root of π₯ one plus β, which
is simply π₯ one plus β. Then, we multiply the square
root of π₯ one plus β by the square root of π₯, and negative the square root of
π₯ one times the square root of π₯ one plus β. When we find their sum, we get
zero.

So, all thatβs left to do is to
multiply negative the square root of π₯ one by the square root of π₯ one. And we simply get negative π₯
one. π₯ one minus π₯ one is
zero. And then, we divide through by
β. And so, this becomes the limit
as β approaches zero of one over the square root of π₯ one plus β plus the
square root of π₯ one. And we can now evaluate this as
β approaches zero. Weβre left with one over the
square root of π₯ one plus the square root of π₯ one, which is one over two
times the square root of π₯ one. The instantaneous rate of
change function of π of π₯ is equal to the square root of π₯ is therefore one
over two times the square root of π₯ one.

In this video, weβve learned that
the average rate of change of a function π of π₯ over an interval between two
points given by π, π of π and π plus β, π of π plus β is the slope of the
secant line that connects those two points. We often define this as π΄ of β,
and itβs given by π of π plus β minus π of π all over β. We also saw that the instantaneous
rate of change of a function is found by letting β approach zero. Itβs the limit as β approaches zero
of π΄ of β, of π of π plus β minus π of π all over β.