### Video Transcript

A triangle with sides π, π, and
contained angle π has area π΄ equals a half ππ sin π. Suppose that π equals four, π
equals five, and the angle π is increasing at 0.6 radians per second. How fast is the area changing when
π is π by three.

Okay, so letβs try to picture
whatβs happening in this scenario. This scenario involves a triangle
with sides of length π and π and contained angle π. So that means that the angle
between the two sides π and π has measure π. Weβre told that the area of this
triangle is a half ππ sin π.

And you might recognize this as the
formula for the area of a triangle given that the lengths of two sides and the
measure of the included angle. We are additionally told that the
values of the side lengths π and π are four and five, respectively. And so we can update our diagram
and also substitute these values into the general formula for the area of the
triangle. So the area of our triangle is a
half times four times five times sin π which is 10 sin π.

Continuing to read the question, we
see that π is increasing at 0.6 radians per second. What does that mean? Well, we can draw another picture
of our triangle after some time has passed and the angle π has increased. And this helps us to understand
what it means for the angle π to increase as the side length stays the same. But we would like to find some
mathematical way of stating that the angle π is increasing at 0.6 radians per
second.

When we see words like increasing
or decreasing and units of something over seconds, then we think about rates of
change. The variable π is changing at 0.6
radians per second. The derivative of the measure of
the angle π with respect to time π‘ is 0.6.

The last sentence of the question
tells us what weβre looking for. Weβre looking for how fast the area
is changing when π is π by three. The natural way to express how the
area is changing is with a derivative as before. Weβre looking for the rate of
change of area with respect to time β ππ΄ by ππ‘. In particular, weβre looking for
the instantaneous rate of change of the area with respect to time when π is π by
three.

Okay, so now that weβve read
through the question and translated all the statements into mathematical notation,
letβs recap. We were given in the question that
the rate of change of the measure of one of the angles of the triangle with respect
to time ππ by ππ‘ is 0.6. We deduced from the information in
the question that the area π΄ of the triangle is 10 sin π. And we are required to find the
rate of change of the area of the triangle with respect to time at π equals π by
three.

Weβre looking for ππ΄ by ππ‘ and
we have the value of ππ by ππ‘. So if we had a relation between
ππ΄ by ππ‘ and ππ by ππ‘, then weβd be pretty much done. Okay, but we donβt have a relation
between ππ΄ by ππ‘ and ππ by ππ‘, which would allow us to find ππ΄ by ππ‘ in
terms of ππ by ππ‘. But we do have a relation between
π΄ and π. Now, how does this help us find a
relation between ππ΄ by ππ‘ and ππ by ππ‘? Well, we can differentiate this
relation implicitly with respect to π‘. So letβs do that.

On the left-hand side, we get ππ΄
by ππ‘ which is after all what weβre looking for. So this is looking promising. On the right-hand side, weβve got
π by ππ‘ of something involving π. So weβre going to want to use the
chain rule. π by ππ‘ can be replaced by ππ
by ππ‘ times π by ππ. So letβs make this change.

We can differentiate 10 sin π with
respect to π quite easily. The derivative of sin is cos and so
this is 10 cos π. How about ππ by ππ‘? Well, if you remember we were given
its value in the question, ππ by ππ‘ is 0.6. Multiplying these together, we get
that ππ΄ by ππ‘ is six cos π. Weβre looking for the value of ππ΄
by ππ‘ when π is π by three. And so we have to substitute π by
three for π. We make this substitution. And as we know that cos π by three
is 0.5, π by three being a special angle, we see that ππ΄ by ππ‘ is six times 0.5
which is three.

Letβs interpret this result in
context then. In a triangle with sides of lengths
four and five, where the included angle π is increasing at a rate of 0.6 radians
per second, then at the instant when π is π by three, the instantaneous rate of
change of the area with respect to time is three square units per second.

This question is a related rates
question and we solved it using the standard methods for solving such questions. We use derivatives to express the
scenario described in the question mathematically. For example, we saw that we were
required to find the value of ππ΄ by ππ‘. And we were given the value of ππ
by ππ‘. We then want to relate to the rate
ππ΄ by ππ‘ to the rate ππ by ππ‘ which is what makes it a related rates
problem. And we did this by first finding a
relation between π΄ and π, before what differentiating implicitly to turn this into
a relation between ππ΄ by ππ‘ and ππ by ππ‘. Having done this, we just
substituted the values of ππ by ππ‘ and π to find our answer.