Analyzing related rates problems: expressions | AP Calculus AB | Khan Academy

The base ( b ) of the triangle is decreasing at a rate of 13 meters per hour, and the height ( h ) of the triangle is increasing at a rate of 6 meters per hour. At a certain instant ( t_0 ), the base is 5 meters and the height is 1 meter.

What is the rate of change of the area ( A ) of the triangle at that instant?

So, our area is going to be a function of ( t ). What is the rate of change of the area of the triangle at that instant? Instead of going straight and trying to solve it, what we need to do is identify the various units of different expressions. Then, we’ll think about what information is given and what's not, which will equip us to solve this rate of change problem.

Let's do this first part: let’s match each expression with its units. Like always, pause the video and see if you can do it on your own.

All right, so the first one is ( b' ) of ( t ). This is the rate of change of the base with respect to time. If we think about it, ( b(t) ) is the base that is going to be in meters.

So, this is going to be in meters. If we say ( b' ) of ( t ), this will indicate how much our base is changing with respect to time. So, this is going to be meters per. They give us right over here that it’s decreasing at a rate of 13 meters per hour, so the units here are meters per hour. Therefore, ( b' ) of ( t ) is going to be in meters per hour.

Now, ( A ) at time ( t_0 ): remember, ( A ) is the area of our triangle and we're measuring everything in meters. From the information given, we can tell that the area is going to be in square units; hence, it's going to be in square meters.

Now, the height at time ( t_0 ): both the base and the height are lengths that are going to be measured in meters. Thus, our height at time ( t_0 ) is going to be in meters.

Then, we have the rate of change of our area with respect to time. We already established that the area is in square meters. However, we want to find the rate of change of this area with respect to time. So, this will be an amount of area per unit time. Here, we’re using hours as you can see from the information provided. This will be area per unit time or meters squared per hour.

Now they ask us to match each expression with its given value.

So, what is the base of the triangle at time ( t_0 )? Do they give that to us? Well, let's see. They say at a certain instant ( t_0 ), the base is 5 meters.

So they say the base at time ( t_0 ) is a function of time, but they tell us that it is 5 meters.

Now, what about the rate of change of the base with respect to time? Do they tell us that?

Look right over here; that is actually the first piece of information they gave us. The base ( b(t) ) of the triangle is decreasing at a rate of 13 meters per hour.

So, the rate of change of the base, that is ( b' ) of ( t ), which is equal to ( \frac{db}{dt} ), is decreasing at a rate of 13 meters per hour. Therefore, that would be negative 13 meters per hour.

The rate of change of the base with respect to time is going to be negative 13; they provided that.

Now, ( A' ) of ( t ) is the rate of change of the area at time ( t_0 ). Did they give us this? Well, they ask us: "What is the rate of change of the area ( A ) of the triangle at that instant?"

This is what we actually need to figure out, but they haven't provided it to us. Otherwise, there would be no problem to solve. So, this one right over here is not given; in fact, this is what we are trying to solve for.

Finally, we have the change, the first derivative of the height with respect to time. You could view this as ( \frac{dh}{dt} ). What is this going to be?

Do they give it to us? Well, look right over here; they say the height of the triangle is increasing at a rate of 6 meters per hour.

So if they say ( h(t) ) is increasing, they're giving us the rate of change of ( h(t) ) with respect to time. That’s ( h' ) of ( t ), and they tell us that it is increasing at 6 meters per hour.

So it's going to be positive 6 meters per hour. They did indeed provide that.

Now, why is all of this a useful exercise? Well, now we are really ready to solve the question. Because in general, if we’re talking about any triangle, we know that area ( A ) is equal to one-half base times height.

In this situation, area, base, and height are all going to be functions of ( t ). So we could write ( A(t) = \frac{1}{2} b(t) h(t) ).

If we want to find the rate of change of our area at that instant, the instant they’re referring to is at time ( t_0 ). What we would want to do is take the derivative of both sides with respect to ( t ).

The derivative on the left-hand side with respect to ( t ) would be ( A' ) of ( t ). On the right-hand side, it would be one-half times, and we would actually use a combination of the product rule right over here. The derivative of the first function with respect to ( t ) is ( b' ) of ( t ) times the second function plus the first function ( b(t) ) times the derivative of the second function with respect to time.

We want to determine ( A' ) at time ( t_0 ). That will be equal to one-half times ( b' ) of ( t_0 ) times ( h(t_0) ) plus ( b(t_0) ) times ( h' ) of ( t_0 ).

Now this might seem daunting, except they have given us a lot of this information.

What is ( b' ) of ( t_0 )? They tell us the rate of change of ( b ) with respect to time is negative 13 meters per hour, so we got this.

What is the height at time ( t_0 )? They tell us right over here that at a certain instant, the base is 5 meters and the height is 1 meter.

So they provided both ( b ) and ( h ) at ( t_0 ). What is the rate of change of the height at time ( t_0 )? They tell us the height of the triangle is increasing at a rate of 6 meters per hour.

They provided all that information, so you just have to plug it in to figure out what is the rate of change of the area at ( t_0 ) at that instant.