RELATED RATES – Triangle Problem (changing angle)

A plane flies horizontally at an altitude of 5 km and passes directly over a tracking telescope on the ground. When the angle of elevation is \(\pi\)/3, this angle is decreasing at a rate of \(\pi\)/6 rad/min. How fast is the plane traveling at this time?

1. Draw a sketch

We are going to go ahead and proceed with the 4 steps that I use for all related rates problems. You can check those out in my related rates lesson. As with any related rates problem, the first thing we should do is draw a sketch of the situation being described in this problem.

related rates triangle problem plane
Figure 1

In this drawing the plane is moving from left to right. Keep in mind that the problem also said that the angle \(\theta\) is \(\frac{\pi}{3}\) radians and is decreasing at a rate of \(\frac{\pi}{6} \ \frac{rad}{min}\). I did not label these facts, but they are important and we should remember them.

2. Come up with your equation

Now that we have made our sketch, we need to use this to come up with an equation relating the necessary quantities and measurements. When doing this we need to consider what the question is asking us to find and what information was given.

However, this may be difficult with the drawing we have so far. First we will need to create and label some other measurements based on some geometric shape we already know something about. Looking back at the initial sketch, it seems like our situation can be described with a triangle. So let’s consider the following drawing instead.

labeled triangle for related rates
Figure 2

Notice that I have added a few things that weren’t in the first drawing. First of all, it is much more clear that we can look at this situation as a triangle. Also I have added labels to the bottom side and hypotenuse of this triangle.

Keep in mind that the side labeled as 5 km will measure the height of the plane as it moves to the right. Therefore, it will always maintain a right angle with the ground. It will also always be 5 km because the altitude of the plane is not changing as it flies horizontally.

Now let’s consider how to make an equation out of this triangle.

What are we looking for?

The question that this problem asks us is “how fast is the plane traveling at this time?” Another way to think about how fast the plane is traveling is how quickly its position is changing, or the rate of change of the plane’s position.

In order to measure the plane’s position at different times, we need to measure how far away it is from some stationary point. This is exactly what the tracking telescope is (shown as the black semi-circle on the ground in our drawing). Since the plane is not moving exactly in the opposite direction of the tracking telescope, we can’t simply use the side of the triangle that measures the distance between these two things (the side labeled z).

However, since the plane is moving horizontally, there is another distance we have labeled that can be used. Since it’s moving horizontally, we will need to use a horizontal distance. In our drawing this is exactly what side x represents! It is the horizontal distance between the telescope and the point directly beneath the plane!

So the rate of change of the plane’s position is represented by the rate of change of the horizontal distance between the plane and the telescope. Therefore, the rate of change of x will give us our answer. Since we will introduce the rate of change when we take the derivative of our equation, we just need to be sure to include x in our initial equation.

What do we know about?

The problem gives us a few pieces of information that we may need to know.

  • Plane’s altitude is always 5 km (it won’t change because it says the plane flies horizontally)
  • Angle of elevation is \(\frac{\pi}{3}\) radians
  • Angle of elevation is decreasing at a rate of \(\frac{\pi}{6} \ \frac{rad}{min}\)

In general, the angle of elevation is by definition the angle between the object being measured and the horizontal. In this case we made the ground perfectly horizontal to make this easier to visualize, but as long as we measure against some horizontal line the ground doesn’t have to be flat.

So to summarize, we know that the right side of the triangle will be a constant 5 km. We also have some information about our angle \(\mathbf{\theta}\) and about its rate of change.

I also want to point out what we can figure out if needed. Since we know two of the three angles, we could use this to find the third. Knowing all three angles and the length of a side, we can use these to find either of the other two sides. This means we can find x or z at this instant if we need to, so they are fine to use in our equation.

Putting it into an equation.

Up to this point we have sorted through what we need to find, which told us that we need to use x in our equation. We also know that we can use \(\theta\), z, or the constant 5 km side. So let’s think about what we can do with this.

You may be thinking that we should use Pythagorean Theorem at this point, so let’s think about this. If we do use Pythagorean Theorem our equation will use all three side lengths and will not use our angle \(\theta\). We don’t necessarily have to use our angle, so there’s nothing wrong with this.

However, think about what will happen when we take the derivative of this equation with respect to time. At this point we would introduce \(\frac{dz}{dt}\) and \(\frac{dx}{dt}\)! Since we don’t know either of these values, this would leave us with two unknowns in only one equation, which isn’t very helpful because it doesn’t allow us to solve for the one variable we want.

But what else can we do?

Well the last thing we tried didn’t include our angle \(\theta\), so let’s think about what we can do with that. What function do you know about that deals with an angle of a right triangle and any number of its sides?

Sine, cosine, and tangent!

Remember soh, cah, toa? We need to use one of these here. Keep in mind that each of these uses one angle and two of the sides of the triangle. Since we already know some information about \(\mathbf{\theta}\) and its rate of change, we will use that as our angle. But how do you decide which sides to use?

We already know that we need to use x. We decided that several paragraphs ago (click here to go back up there). So this leaves the side z and the constant 5 km side to chose from for our other side. If we chose z, we will run into the same problem as if we use Pythagorean Theorem, we don’t know anything about it’s rate of change!

Notice since the 5 km side is constant, we don’t need to worry about the rate of change. This side is a constant so its rate of change would be 0! So we will use x and the constant 5 km side in our equation. Now there’s just one last thing to figure out.

Do we use sine, cosine, or tangent?

When deciding between sine, cosine, and tangent we will want to use soh, cah, toa:

Sine: Opposite / Hypotenuse
Cosine: Adjacent / Hypotenuse
Tangent: Opposite / Adjacent

We know that we are using the angle \(\theta\), and the x side and the constant 5 km side. Let’s think for a second about where these sides lie in relation to the angle we’re looking at.

First of all, neither of these two sides are the hypotenuse. The hypotenuse of a right triangle is the longest side which is opposite from the right angle. Therefore, one will be the side adjacent to \(\theta\) and the other will be opposite to \(\theta\). In general, the adjacent side would be the one that’s touching \(\theta\). In this case, the adjacent side would be x. The opposite side would be the one side that is not touching our angle, making the 5 km side opposite.

So we need to chose between sine, cosine, and tangent so that we incorporate the adjacent side and the opposite side. Using soh, cah, toa, we can see that tangent uses the opposite and adjacent sides. This tells us that taking the tangent of our angle gives us the opposite side divided by the adjacent side. $$tan(\theta) = \frac{\textrm{opposite}}{\textrm{adjacent}}$$ $$tan(\theta) = \frac{5}{x}$$

3. Implicit differentiation

Now that we have come up with our equation, we need to take its derivative with respect to time. This will require the use of the chain rule since we are differentiating with respect to time. The reason for this is that we need to treat \(\theta\) and x as functions of time.

Notice that our equation has a fraction. As a result we would need to use the quotient rule to find the derivative of the right side of our equation. However, we can rewrite this in another form to make the derivative a bit simpler.

Remember, we can always move a term from the bottom of a fraction up to the top by making its power negative. Therefore, we can write our equation as $$tan(\theta) = 5x^{-1}.$$

Now that we just have a constant multiplied by an x term to a constant power, we can use the power rule instead of the quotient rule on the right side of the equation. We do still have to use the chain rule, but at least we made things a bit simpler. In order to find the derivative of \(tan(\theta)\) we would need to use the fact that \(tan(x)=\frac{sin(x)}{cos(x)}\) then use the quotient rule. I’m not going to show this process, but instead I’ll use Wolfram Alpha to find that $$\frac{d}{dx} tan(x) = sec^2(x) = \frac{1}{cos^2(x)}.$$

Applying this to our problem we can implicitly differentiate our equation. $$\frac{d}{dt} \big[ tan(\theta) \big]= \frac{d}{dt} \Big[ 5x^{-1} \Big]$$ $$sec^2(\theta) \cdot \frac{d \theta}{dt} = -5x^{-2} \cdot \frac{dx}{dt}$$

4. Solve for desired rate of change

Remember the thing we need to find in this problem is the rate of change of x. So all we need to do now is isolate \(\frac{dx}{dt}\) because that’s exactly what represents the rate of change of x. $$sec^2(\theta) \cdot \frac{d \theta}{dt} = -5x^{-2} \cdot \frac{dx}{dt}$$ $$-\frac{1}{5} sec^2(\theta) \cdot \frac{d \theta}{dt} = x^{-2} \cdot \frac{dx}{dt}$$ $$-\frac{1}{5} x^2 \cdot sec^2(\theta) \cdot \frac{d \theta}{dt} = \frac{dx}{dt}$$

And once we’ve done that, we just need to plug in all of the other variables to find \(\frac{dx}{dt}\)! The other variables that we’ll need to plug in are x, \(\theta\), and \(\frac{d \theta}{dt}\).

We already know from the actual problem that “the angle of elevation is \(\pi\)/3,” and “this angle is decreasing at a rate of \(\pi\)/6 rad/min.” This directly tells us that $$\theta = \frac{\pi}{3}.$$ And since this angle is decreasing we know its rate of change will be negative, so $$\frac{d \theta}{dt} = -\frac{\pi}{6}.$$

Now we just need to find x. Since it wasn’t directly given, finding x will require a little work but it’s not too complicated.

We will actually need to use the equation we created before taking its derivative. $$tan(\theta) = \frac{5}{x}$$ Since we need to find x at the moment when \(\theta = \frac{\pi}{3}\) we can plug this in here to find x. $$tan \bigg( \frac{\pi}{3} \bigg) = \frac{5}{x}$$ $$\frac{sin(\pi/3)}{cos(\pi/3)} = \frac{5}{x}$$

And now using the unit circle. $$\frac{\sqrt{3}/2}{1/2} = \frac{5}{x}$$ $$\sqrt{3} = \frac{5}{x}$$ $$\sqrt{3} \cdot x = 5$$ $$x = \frac{5}{\sqrt{3}}$$

Plugging them all in

$$\frac{dx}{dt} = -\frac{1}{5} x^2 \cdot sec^2(\theta) \cdot \frac{d \theta}{dt}$$ $$\frac{dx}{dt} = -\frac{1}{5} \bigg( \frac{5}{\sqrt{3}} \bigg)^2 \cdot sec^2\bigg( \frac{\pi}{3} \bigg) \cdot \bigg( -\frac{\pi}{6} \bigg)$$ $$\frac{dx}{dt} = -\frac{1}{5} \bigg( \frac{25}{3} \bigg) \cdot \frac{1}{cos^2\big( \frac{\pi}{3} \big)} \cdot \bigg( -\frac{\pi}{6} \bigg)$$ $$\frac{dx}{dt} = -\frac{1}{5} \bigg( \frac{25}{3} \bigg) \cdot \frac{1}{1/4} \cdot \bigg( -\frac{\pi}{6} \bigg)$$ $$\frac{dx}{dt} = -\frac{1}{5} \bigg( \frac{25}{3} \bigg) \cdot 4 \cdot \bigg( -\frac{\pi}{6} \bigg)$$ $$\frac{dx}{dt} = \frac{100\pi}{90} = \frac{10\pi}{9}$$

So we know that the plane is traveling at a speed of \(\mathbf{\frac{10\pi}{9} \ \frac{km}{min}}\)!

Hopefully that all helped this problem make a little more sense. If you still want some more practice with triangle related rates problems, check some of these out:

Solution – At noon, ship A is 150 km west of ship B.  Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h.  How fast is the distance between the ships changing at 4:00 PM?

Solution – A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?

Or you can check out my related rates lesson where I have a list of other related rates problems with solutions if you want more practice with these.

If you have any questions about this problem, please let me know! I’d be happy to help. You can email any questions to me at jakesmathlessons@gmail.com or use the form below to join my email list and I’ll send you my calc 1 study guide as a welcome gift!


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