Analyzing related rates problems: equations (trig) | AP Calculus AB | Khan Academy

A 20 meter ladder is leaning against a wall. The distance ( x(t) ) between the bottom of the ladder and the wall is increasing at a rate of 3 meters per minute. At a certain instant ( t_0 ), the top of the ladder is a distance ( y(t_0) ) of 15 meters from the ground. What is the rate of change of the angle ( \theta(t) ) between the ground and the ladder at that instant?

So, what I'm going to do is draw this out. Really, the first step is to think about, well, what equation will be helpful for us to solve this problem? Then we might just go ahead and actually solve the problem.

A 20 meter ladder is leaning against a wall, so let me draw ourselves a wall here. That is my wall. Now, let me draw our 20 meter ladder. So maybe it looks something like that. So, that is 20 meters. They say the distance ( x(t) ) between the bottom of the ladder and the wall. So, it's this distance right over here. This distance right over here is ( x(t) ). They say it's increasing at a rate of 3 meters per minute.

So we know that we could either say ( x'(t) ), which is the same thing as ( \frac{dx}{dt} ), is equal to 3 meters. I'll write it out because it's hard if I just said "m per m"; it might not be that clear. Meters per minute, so they give us that piece of information.

The rate of change of ( x ) with respect to time is given to us. They also give us at a certain instant ( t_0 ) that the top of the ladder is a distance of 15 meters. So, the top of the ladder—let's make this very clear—so this distance right over here is ( y(t) ). They say at time ( t_0 ), ( y(t) ) is 15 meters. Let me just write over here ( y(t_0) ) is equal to 15 meters.

So, maybe let me write this right over here. This is ( y(t_0) ). Let's just assume that we're drawing it at that moment ( t_0 ) because I think that's going to be important. ( y ) at ( t_0 ) is equal to 15 meters.

So, they want to know what is the rate of change of the angle ( \theta(t) ) between the ground and the ladder. This is the same. ( \theta ) is also going to change with respect to time. It's going to be a function of time between the ground and the ladder at that instant. So, ( \theta )—let me get a new color here—( \theta ) is this angle right over here; this is ( \theta ) and it's also going to be a function of time.

So, what we'll always want to do in these related rates problems is we want to set up an equation. Really, an algebraic equation, maybe a little bit of trigonometry involved, that relates the things that we care about. Then we're likely to have to take the derivative of both sides of that in order to relate the related rates.

So, we want to know the rate of change of the angle between the ground and the ladder at that instant. We need to figure out ( \theta'(t_0) ). This is what we want to figure out.

Now, they've given us some interesting things. They've given us, I guess, our rate of change of ( x ) with respect to time is constant at 3 meters per minute. We know what ( y ) is at that moment, so let’s see—we can create a relationship because they gave us ( \frac{dx}{dt} ).

Actually, it will be more useful to find a relationship between ( x ) and ( \theta ), and then take the derivative of both sides. Then we could use this information possibly to figure out what the appropriate value of ( x ) or ( \theta ) is at that moment.

So, how does ( x ) relate to ( \theta )? Well, we use a little bit of trigonometry right over here. If you took the hypotenuse times the cosine of ( \theta ), you would get ( x ). So, let me write this right over here: ( x(t) = 20 \times \cos(\theta(t)) ). I could say the cosine of ( \theta(t) ) just to make it clear that this is a function of time.

This comes straight out of trigonometry, actually, our basic trigonometric function definitions. Now, why is this useful? Why do I think this is useful?

Well, let's think about what happens when I take the derivative of both sides. Using the chain rule on the left-hand side, I am going to have ( x'(t) ), and then that's going to be equal to what I end up with on the right-hand side.

Well, using the chain rule, first I’ll take the derivative with respect to ( \theta ), and so that's just going to be ( -20 \sin(\theta(t)) ). Then, I need to multiply that by ( \theta'(t) ).

So, what I could do is say, "Hey, look at ( t_0 ). I know what x' of t is." I could try to figure out what ( \sin(\theta(t)) ) is, and then I’ll just solve for this right over there.

So, let’s do that. At ( t_0 ), so at ( t = t_0 ), what we're going to have ( x'(t) )—well that's at every time—it's 3 meters per minute. We'll assume that our rates are in meters per minute, and just our values are in meters when we're talking about distance, and our angles are in radians.

So, this is going to be equal to 3, is equal to ( -20 \times \sin(\theta(t)) \times \frac{d\theta}{dt} ). So how do we figure out what ( \sin(\theta(t)) ) is going to be?

Well, let's just use that other information they gave us, and I'm going to scroll down a little bit, to get a little bit more real estate. So, ( \sin(\theta) )—let me write it over here: ( \sin(\theta(t_0)) )—that's what we care about. When ( t = t_0 ), what's that going to be?

Well, sine is opposite over hypotenuse, so that's going to be ( y(t_0) ) over our hypotenuse of 20 meters. Well, that's going to be equal to—they tell us ( y(t_0) = 15 ) meters over 20 meters, which is the same thing as ( \frac{3}{4} ).

So, by this yellow information, they actually told us that this right over here is going to be equal to ( \frac{3}{4} ). So, this times ( \frac{3}{4} ) times the rate of change of ( \theta ) with respect to ( t ).

Now we just solve for this, and we're done. So, what is ( -20 \times \frac{3}{4} )? That is ( -15 ). That is ( -15 ).

If we divide both sides by ( -15 ), we get ( \theta'(t) = \frac{3}{-15} ), which is the same thing as being equal to ( -1 ).

And this, the units here would be in radians per minute, because our rates are all in per minute. So, if I wanted to, I could write "radians per minute." Ideally, I would write it right over here.

But there you go; we were able to figure out this is an interesting one because they give you that information on ( y ), but really use that information of ( y ) to figure out what ( \sin(\theta(t)) ) is. But the equation you set up involves ( x ) and ( \theta ).