Video Transcript
A triangle with sides 𝑎, 𝑏, and
contained angle 𝜃 has area 𝐴 equals a half 𝑎𝑏 sin 𝜃. Suppose that 𝑎 equals four, 𝑏
equals five, and the angle 𝜃 is increasing at 0.6 radians per second. How fast is the area changing when
𝜃 is 𝜋 by three.
Okay, so let’s try to picture
what’s happening in this scenario. This scenario involves a triangle
with sides of length 𝑎 and 𝑏 and contained angle 𝜃. So that means that the angle
between the two sides 𝑎 and 𝑏 has measure 𝜃. We’re told that the area of this
triangle is a half 𝑎𝑏 sin 𝜃.
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 𝑎 and 𝑏 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 𝜃 which is 10 sin 𝜃.
Continuing to read the question, we
see that 𝜃 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 𝜃 has increased. And this helps us to understand
what it means for the angle 𝜃 to increase as the side length stays the same. But we would like to find some
mathematical way of stating that the angle 𝜃 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 𝜃 is changing at 0.6
radians per second. The derivative of the measure of
the angle 𝜃 with respect to time 𝑡 is 0.6.
The last sentence of the question
tells us what we’re looking for. We’re looking for how fast the area
is changing when 𝜃 is 𝜋 by three. The natural way to express how the
area is changing is with a derivative as before. We’re looking for the rate of
change of area with respect to time — 𝑑𝐴 by 𝑑𝑡. In particular, we’re looking for
the instantaneous rate of change of the area with respect to time when 𝜃 is 𝜋 by
three.
Okay, so now that we’ve read
through the question and translated all the statements into mathematical notation,
let’s 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 𝑑𝜃 by 𝑑𝑡 is 0.6. We deduced from the information in
the question that the area 𝐴 of the triangle is 10 sin 𝜃. And we are required to find the
rate of change of the area of the triangle with respect to time at 𝜃 equals 𝜋 by
three.
We’re looking for 𝑑𝐴 by 𝑑𝑡 and
we have the value of 𝑑𝜃 by 𝑑𝑡. So if we had a relation between
𝑑𝐴 by 𝑑𝑡 and 𝑑𝜃 by 𝑑𝑡, then we’d be pretty much done. Okay, but we don’t have a relation
between 𝑑𝐴 by 𝑑𝑡 and 𝑑𝜃 by 𝑑𝑡, which would allow us to find 𝑑𝐴 by 𝑑𝑡 in
terms of 𝑑𝜃 by 𝑑𝑡. But we do have a relation between
𝐴 and 𝜃. Now, how does this help us find a
relation between 𝑑𝐴 by 𝑑𝑡 and 𝑑𝜃 by 𝑑𝑡? Well, we can differentiate this
relation implicitly with respect to 𝑡. So let’s do that.
On the left-hand side, we get 𝑑𝐴
by 𝑑𝑡 which is after all what we’re looking for. So this is looking promising. On the right-hand side, we’ve got
𝑑 by 𝑑𝑡 of something involving 𝜃. So we’re going to want to use the
chain rule. 𝑑 by 𝑑𝑡 can be replaced by 𝑑𝜃
by 𝑑𝑡 times 𝑑 by 𝑑𝜃. So let’s make this change.
We can differentiate 10 sin 𝜃 with
respect to 𝜃 quite easily. The derivative of sin is cos and so
this is 10 cos 𝜃. How about 𝑑𝜃 by 𝑑𝑡? Well, if you remember we were given
its value in the question, 𝑑𝜃 by 𝑑𝑡 is 0.6. Multiplying these together, we get
that 𝑑𝐴 by 𝑑𝑡 is six cos 𝜃. We’re looking for the value of 𝑑𝐴
by 𝑑𝑡 when 𝜃 is 𝜋 by three. And so we have to substitute 𝜋 by
three for 𝜃. 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 𝑑𝐴 by 𝑑𝑡 is six times 0.5
which is three.
Let’s interpret this result in
context then. In a triangle with sides of lengths
four and five, where the included angle 𝜃 is increasing at a rate of 0.6 radians
per second, then at the instant when 𝜃 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 𝑑𝐴 by 𝑑𝑡. And we were given the value of 𝑑𝜃
by 𝑑𝑡. We then want to relate to the rate
𝑑𝐴 by 𝑑𝑡 to the rate 𝑑𝜃 by 𝑑𝑡 which is what makes it a related rates
problem. And we did this by first finding a
relation between 𝐴 and 𝜃, before what differentiating implicitly to turn this into
a relation between 𝑑𝐴 by 𝑑𝑡 and 𝑑𝜃 by 𝑑𝑡. Having done this, we just
substituted the values of 𝑑𝜃 by 𝑑𝑡 and 𝜃 to find our answer.
