# Video: A Related Rates Problem of a Triangle with Two Sides of Constant Length While the Included Angle Increases at a Given Rate

A triangle with sides 𝑎, 𝑏, and contained angle 𝜃 has area 𝐴 = 1/2 𝑎𝑏 sin 𝜃. Suppose that 𝑎 = 4, 𝑏 = 5, and the angle 𝜃 is increasing at 0.6 rad/s. How fast is the area changing when 𝜃 = 𝜋/3?

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### 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.