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!


Implicit Differentiation Practice Problem

Find \(\frac{dy}{dx}\) if \(y=x^x\).

Solution

This is kind of a tricky problem. Obviously, if we need to find \(\frac{dy}{dx}\), we need to take the derivative. And since we are already given an explicit formula for y only in terms of x, it seems like we can just go ahead and take the derivative. But unfortunately, having an x both in the base and the exponent makes it a bit more complicated.

So what can we do then?

Finding the derivative of this function is going to require a little trick that seems a little counter intuitive. What we need to do is actually take the natural log of both sides of this equation. The reason for this is that it will help us get rid of the exponent and put this in a form we can work with more easily.

$$y=x^x$$

$$ln (y) =ln \big( x^x \big)$$

Now that we have done this we can use one of the basic log rules that you will want to remember.

3 log rules to remember

Really quickly I want to list the three main log rules that you will want to remember. These come up frequently, so you will want to remember these.

$$log(ab) = log(a) + log(b)$$ $$log \bigg( \frac{a}{b} \bigg) = lob(a) – log(b)$$ $$log \Big( a^b \Big) = b \cdot log(a)$$

How do we apply this to our problem?

Let’s look back to our equation to see where we were.

$$ln (y) =ln \big( x^x \big)$$

Notice the right side of our equation looks a lot like the third log rule from above. Based on that third log rule, we can move the x in the exponent down in front of the log and multiply rather than having to deal with an exponent.

$$ln(y) = x \cdot ln(x)$$

The reason I think this seems a little counter intuitive is that we no longer have an explicit formula for y. But now the right side of our equation will be easier to take its derivative. So now let’s see what happens if we take the derivative of both sides of the equation with respect to x.

Taking the derivative

The reason we need to take the derivative with respect to x is that the question asked us to find \(\frac{dy}{dx}\). The dx in the denominator is the indicator that tells us that we need to differentiate with respect to x. So let’s do that now.

$$\frac{d}{dx} ln(y) = \frac{d}{dx} \big[x \cdot ln(x) \big]$$

First the left side

Taking the derivative of the left side of the equation will require the use of the chain rule since y is a function of x. This was explained in a bit more detail in my implicit differentiation lesson. You will also use the fact that the derivative of ln(x) is \(\frac{1}{x}\).

$$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx} \big[x \cdot ln(x) \big]$$

Then the right side

In order to take the derivative of the right side of this equation, we will need to use the product rule. As I did in the product rule lesson, we’ll first want to call one part of our function f and the other will be g.

$$f=x$$ $$g=ln(x)$$

Now we need to find f’ and g’ to use the product rule formula.

$$f’=1$$ $$g’=\frac{1}{x}$$

And lastly, we just need to plug these four pieces into the product rule formula.

$$\frac{1}{y} \cdot \frac{dy}{dx} = (x) \bigg( \frac{1}{x} \bigg) + (1) \big( ln(x) \big)$$

Now that we have taken the derivative of both sides of the equation, we just need to simplify the equation and solve for \(\frac{dy}{dx}\).

Solving for dy/dx

$$\frac{1}{y} \cdot \frac{dy}{dx} = (x) \bigg( \frac{1}{x} \bigg) + (1) \big( ln(x) \big)$$

$$\frac{1}{y} \cdot \frac{dy}{dx} = 1 + ln(x)$$

And all we need to do from here is multiply both sides by y to isolate \(\frac{dy}{dx}\).

$$\frac{dy}{dx} = y \big( 1+ln(x) \big)$$

With most implicit differentiation problems this would be a perfectly fine place to stop and say we’ve reached our answer. Finding \(\frac{dy}{dx}\) in terms of x and y is frequently the best we can do. But in this case, we can actually get our answer only in terms of x so that we have an explicit derivative of the original function.

The reason we’re able to do this is that our original function was an explicit formula for y. Since we know \(y=x^x\) we can actually go to our formula for \(\frac{dy}{dx}\) and replace the y with \(x^x\). So,

$$\frac{dy}{dx} = x^x \big( 1+ln(x) \big)$$

And now we can say we have reached our answer!

I just want to circle back to those 3 log rules I discussed above. They can be very useful when taking derivatives of exponential functions, or in some strange cases products and quotients. They can be used to rewrite complex functions in a way that would make their derivative easier to find, so always be sure to be aware of those and know how to use them to manipulate a function when needed.

As I always say, the best way to learn this stuff is practice, practice, practice! So check out some of my other lessons and problem solutions on derivatives. The more you work with these concepts, the better you’ll start to understand them.

If you can’t find the answer to your question or the topic you want to read about on my site, send me an email at jakesmathlessons@gmail.com and I’ll get back to you as soon as I can. You can also use the form below to join my email list and I’ll send you my calculus 1 study guide!


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?

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?

Jake’s Math Lessons Complete Calculus 1 Package

 

1. Draw a sketch

Here we have a related rates problem.  As I said when I discussed related rates problems initially, the first thing I like to do with these problems is draw a sketch of the scene that is being described.  If you want to refer back to that, you can click here.  Otherwise, let’s sketch the problem described here.

A kite 100ft above the ground moves horizontally at a speed of 8 ft/s.

2. Come up with your equation

The next thing we need to do is set up our equation which will relate our different quantities.  To do this, we will want to consider what value the question is asking us to find.

What are we looking for?

It asks “at what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?”  Therefore, the value we are looking for is the “rate of change of the angle between the string and the horizontal.”  This just means we will need to consider the angle between the string and the ground (the ground is the horizontal in this case).  If you look back at our drawing, you will see that this angle is represented by \(\theta\).  Since our goal is to find how fast \(\theta\) is changing, we need \(\theta\) to be in our original equation.

What do we know about?

Now we need to consider what other quantities or variables we know something about.  Clearly we know something about the two sides of the triangle that are labeled as being 100 ft and 200 ft.  And we can use these two sides to figure out the length of the third side, which is not labeled in our drawing.

Although we could simply call one of those sides \(a\) and the other one \(b\) and proceed from there, there is another option that may simplify our problem.

Consider the fact that the kite is moving horizontally.  This means that the kite is not getting any further from or closer to the ground as it moves.  Therefore, the side that is labeled 100 ft will actually be 100 ft at any point in this kite’s flight.  Because of this we actually don’t need to designate a variable to this side of the triangle.  Instead this side is simply a constant 100 ft.

Now we just need to use one of the other two sides of the triangle.  We could technically use either one, but one will be a lot easier than the other.  It looks like the hypotenuse would be the easier of the two, because we know it’s 200 ft at this moment.  However, we don’t know exactly how fast it’s changing.  We can figure that out but it wouldn’t be easy.

We do know exactly how fast the unlabeled side is changing.  The question states that the kite is moving horizontally at a speed of 8 \(\frac{ft}{s}\).  Since this unlabeled side is exactly horizontal, we know its rate of change is also 8 \(\frac{ft}{s}\).  We can figure out its length using Pythagorean Theorem later, but this would certainly be easier than finding the rate of change of the hypotenuse.  Therefore, I will go ahead and use the unlabeled side.

Since this unlabeled side is going to be changing we will need to designate a variable to this side of the triangle.  As the kite moves away from the person flying it, the person holding the string has to let more string out and allow it to become longer.  This means that this unlabeled side in our drawing will need to be described with a variable.  We will call it side \(a\).

Putting it into an equation.

Now we have three different quantities we need to relate somehow:

  1. Angle \(\theta\) (this will be changing as the kite moves).
  2. Side \(a\) (this will be changing as the kite moves and the string is let out).
  3. Side labeled 100 ft (this will not change and can be treated as a constant).

So we have two sides and an angle that we need to make an equation with.  To do this, think about where these sides are in relation to the angle \(\theta\).  The side labeled 100 ft is the side opposite to the angle \(\theta\) and the side we’re calling \(a\) is adjacent to the angel \(\theta\).

Usually when dealing with two sides and one angle of a triangle, you will want to use either sine, cosine, or tangent to relate the three.  So which one should be used when we know the opposite side and the adjacent side to the angle in question?

Remember soh, cah, toa?

  • Sine Opposite Hypotenuse
  • Cosine Adjacent Hypotenuse
  • Tangent Opposite Adjacent

Since we have the opposite side and the adjacent side, we want to use tangent.  Therefore we can say:

$$tan(\theta) = \frac{100}{a}$$

Since it will make finding the derivative easier, I am going to rewrite this as

$$tan(\theta) =100a^{-1}$$

3. Implicit differentiation

As with any related rates problem, we now need to take the derivative of both sides of the equation with respect to time.  Since \(\theta\) and \(a\) are both functions of time, we will need to use chain rule for both sides of this equation.  We know they are functions of time because they are both going to be dependent on the position of the kite as time progresses.  We don’t have an explicit formula for either of these functions, but we know their values are dependent on time.

$$\frac{d}{dt}tan(\theta) =\frac{d}{dt}100a^{-1}$$

$$\frac{d}{dt}\frac{sin(\theta)}{cos(\theta)} =\frac{d}{dt}100a^{-1}$$

To find the derivative of the left side of this equation you will need to use the quotient rule and the chain rule.  I’m not going to show all the steps of how to do this but if you want a refresher, you can read about the quotient rule here and the chain rule here.  Using Wolfram Alpha, you can see that

$$\frac{d}{dx}tan(x)=\frac{1}{cos^2x}$$

Therefore, we can say that

$$\frac{d}{dt}tan(\theta)=\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt}$$

Plugging this back into the left side of our equation, we get

$$\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt} =\frac{d}{dt}100a^{-1}$$

$$\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt}$$

4. Solve for desired rate of change

The last step of any related rates problem is to solve for the desired rate of change.  Now remember the thing we need to find is the rate of change of our angle \(\theta\).  This is exactly what \(\frac{d\theta}{dt}\) represents.  So now we just need to solve for \(\frac{d\theta}{dt}\).

$$\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt}$$

$$\frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt} \cdot cos^2 \theta$$

Now we just need to plug in the values for \(a\), \(\frac{da}{dt}\), and \(\theta\) and we will have our answer.  We don’t know all of these values but we can find them.

Finding a

As I mentioned before, we can find \(a\) by using Pythagorean Theorem.  Looking back at our drawing, we have a right triangle with side lengths of 100 ft, 200 ft, and \(a\).  We know that

$$100^2 + a^2 = 200^2$$

$$10,000 + a^2 = 40,000$$

$$a^2 = 30,000$$

$$a = \sqrt{30,000}$$

$$a = 100\sqrt{3}$$

Finding  \(\mathbf{\frac{da}{dt}}\)

This was actually given.  We know that \(a\) is the horizontal distance the kite is away from the person flying the kite.  We know that the kite is moving horizontally at a speed of 8 \(\frac{ft}{s}\).  Because of this we know that this is also the rate at which \(a\) is changing.  Since \(\frac{da}{dt}\) is the rate of change of \(a\), we know

$$\frac{da}{dt} = 8$$

Finding \(\mathbf{\theta}\)

To find \(\theta\) we will need to go back to the original equation we came up with before the implicit differentiation step:

$$tan(\theta) = \frac{100}{a}$$

Since we know \(a\), we can plug it in here and solve for \(\theta\).

$$tan(\theta) = \frac{100}{100\sqrt{3}}$$

$$tan(\theta) = \frac{1}{\sqrt{3}}$$

This angle is actually on the unit circle and by using this we know:

$$\theta = \frac{\pi}{6}$$

Note that \(\theta\) will be in radians.

Now we can plug all of these into our equation for \(\frac{d\theta}{dt}\).

$$\frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt} \cdot cos^2 \theta$$

$$\frac{d\theta}{dt} =-100 \big(100\sqrt{3} \big)^{-2} \cdot 8 \cdot cos^2 \Bigg( \frac{\pi}{6} \Bigg)$$

$$\frac{d\theta}{dt} =-\frac{1}{300} \cdot 8 \cdot \Bigg( \frac{\sqrt{3}}{2} \Bigg)^2$$

$$\frac{d\theta}{dt} =-\frac{1}{300} \cdot 8 \cdot \frac{3}{4}$$

$$\frac{d\theta}{dt} =-\frac{1}{50}$$

So we can say that the angle between the string and the horizontal is decreasing at a rate of \(\frac{1}{50} \ \frac{radians}{s}\) when 200 ft of string has been let out.

And that’s the answer to the question!  Hopefully that wasn’t too bad, but if you have any questions I’d love to hear them.  I know related rates problems can be challenging so you can email me any questions or suggestions at jakesmathlessons@gmail.com.  If you have any other problems you’d like to see worked out go ahead and send me an email.

Enter your name and email below to get your FREE copy of my calculus 1 study guide to help you boost your test scores!

If you feel you need some more practice with related rates, you can check out the lesson where I discussed related rates for more examples.

Also, if you want to check out some other problems and get some practice with derivatives, go check out my derivatives page.  You can see what other topics I’ve already covered and problems I’ve worked through.  If you can’t find your problem there just let me know and I may post the solution to your problem.

Implicit Differentiation

Before getting into implicit differentiation, I would like to take some time to discuss variables, functions, and constants.  The reason for this is that when you do an implicit differentiation problem, you will likely be dealing with equations containing multiple letters.

Up to this point, most of the functions you have taken the derivative of usually contain one variable, usually \(x\), and any other letters in the function would be constants.  But with implicit differentiation, you will also need to deal with having another letter that represents another unknown function.

Before you start implicitly differentiating a problem I recommend determining whether each letter represents a function, or if it’s a variable or a constant.  This is because each one will be treated differently when you take its derivative.  Since implicit differentiation is essentially just taking the derivative of an equation that contains functions, variables, and sometimes constants, it is important to know which letters are functions, variables, and constants, so you can take their derivative properly.

In many cases, the problem will tell you if a letter represents a constant.  If a letter is a constant, that means you would treat it like it’s a number.  If this is a little confusing, just imagine what would happen if you were to actually put some number in for the constant and think about what would happen to that number when you take the function’s derivative.  Then revert back to having the letter in the equation and treat the constant the same way you treated the number.

Let’s jump into an example and I will explain the process along the way.

Example 1

Find the derivative of \(f(x)=cx^2+d\) where \(c\) and \(d\) are constants.

What to do with constants?

Like I said before, since \(c\) and \(d\) are constants, we can treat them as if they are just some number and take the derivative of the remaining function with \(x\) being the variable.

Let’s imagine \(c\) and \(d\) have been replaced with \(2\) and \(4\) respectively, and see what happens.

$$f(x)=2x^2+4$$

This is a case where we can just use the power rule to find:

$$f'(x)=2(2x)=4x$$

Now, we can revert back to having \(c\) and \(d\) in our function and we would see that:

$$f(x)=cx^2+d$$

$$f'(x)=c(2x)$$

$$f'(x)=2cx$$

Notice, the \(d\) disappeared because the derivative of a constant is just \(0\).

What about functions and variables?

Now that we have discussed some methods for identifying constants and how to deal with them when taking a derivative, I will discuss indicators for classifying letters that represent functions and those that are variables.

Frequently, when doing an implicit differentiation problem, you will simply be asked to find \(\frac{dy}{dx}\).  Then you will be shown some equation that contains at least one \(y\) and at least one \(x\).  Although this seems like you haven’t been given much direction, the \(\frac{dy}{dx}\) is actually an indicator that gives us all the information we need.  This notation is one way to write:

The derivative of \(y\) with respect to \(x\).

Or in other words, it’s telling you that you are trying to find the derivative of your function, \(y\), with respect to the variable, \(x\).  Which tells you that \(y\) is a function of \(x\).

In fact, this notation will always give you those two pieces of information.  For example, \(\frac{dh}{dt}\) is a symbol that represents “the derivative of \(h\) with respect to \(t\).”  Therefore, \(h\) must be a function and \(t\) must be its variable.

Example 2

Find \(\frac{dy}{dx}\) if \(y^2=4x^5-e^x\).

In a problem like this, since we know we need to find \(\frac{dy}{dx}\) and we are given an equation which relates \(y\) and \(x\), we need to find the derivative of both sides of the equation.

$$y^2=4x^5-e^x$$

Now we can take the derivative of both sides.  Remember, as long as we do the same thing to both sides of an equation, the results will be equal to each other also.

$$\frac{d}{dx}\big[y^2\big]=\frac{d}{dx}\big[4x^5-e^x\big]$$

The Left Side of the Equation:

The \(\frac{d}{dx}\) just means that you need to take the derivative of whatever follows, treating \(x\) as the variable.  The left side of this equation is the tricky part.  Since the question told us to find \(\frac{dy}{dx}\), we know that \(y\) is a function of \(x\).  The fact that \(y\) is a function tells us that we can’t just use the power rule to find the derivative of the left side of the equation.  We will actually need to use the chain rule.

We need to do the chain rule because \(y\) is not a variable here.  Since \(y\) is a function of \(x\) and we are taking the derivative with respect to \(x\), we cannot say that the derivative of \(y^2\) is \(2y\)!  Now that we have determined that we need to use the chain rule, we need to determine our inside and outside functions.  Remember, we need to figure out some \(f(x)\) and a \(g(x)\) so that \(f\big(g(x)\big)=y^2\) (if you need a refresher on the chain rule, click here).

Typically, when we have a letter that represents a function and we take its derivative with respect to a different variable, we can call our inside function just the single letter which represents a function.  Therefore, we can say our inside function is \(g(x)=y\).

Now, to find our outside function, we can look at the entire function and replace the inside function with a single \(x\).  We are replacing it with an \(x\) because that is the variable we are differentiating with respect to.  So if we take our function (\(y^2\)) and replace the inside function (\(y\)) with a single \(x\), we are left with our outside function \(f(x)=x^2\).  At this point we have figured out:

$$f(x)=x^2,$$

$$g(x)=y.$$

The next thing we need to do is find the derivative of both our inside and outside functions.  Finding \(f'(x)\) can be found simply using the power rule:

$$f'(x)=2x.$$

Now we need to find \(g'(x)\).  This is a little more tricky.  The key thing, which I will continue to remind you of, is that we are taking the derivative of \(y\) with respect to \(x\).  Therefore, we cannot say that the derivative of \(y\) is \(1\).

In fact, we do not know the derivative of \(y\).  Since \(y\) is some function of \(x\) that we actually don’t know, we can’t explicitly write its derivative either.  But luckily, we don’t need to be able to do this.  All we need to say is that the derivative of \(y\) is the symbol I mentioned earlier which represents “the derivative of \(y\) with respect to \(x\).”  We can simply use \(\frac{dy}{dx}\) to represent this.  Therefore, we know that:

$$g'(x)=\frac{dy}{dx}.$$

Now we can just use these pieces and plug them into the chain rule formula.

$$\frac{d}{dx}\big[y^2\big]=f’\big(g(x)\big)\cdot g'(x)$$

$$\frac{d}{dx}\big[y^2\big]=2(y)\cdot \frac{dy}{dx}$$

$$\frac{d}{dx}\big[y^2\big]=2y\frac{dy}{dx}$$

The Right Side of the Equation:

Finding \(\frac{d}{dx}\big[4x^5-e^x\big]\) is quite a bit easier than the left side of the equation.  Since this side contains no other letters besides \(x\), which is the variable we are differentiating with respect to, this will be like any other derivative we have taken up to this point.

$$\frac{d}{dx}\big[4x^5-e^x\big]=4(5x^4)-e^x$$

$$\frac{d}{dx}\big[4x^5-e^x\big]=20x^4-e^x$$

Putting It All Together:

Back to our original equation, we had:

$$\frac{d}{dx}\big[y^2\big]=\frac{d}{dx}\big[4x^5-e^x\big].$$

And as we just showed above, this means:

$$2y\frac{dy}{dx}=20x^4-e^x.$$

Now, once we have taken the derivative of both sides, you can see that our equation contains a \(\frac{dy}{dx}\).  Since our goal here is to find \(\frac{dy}{dx}\), now that we have an equation that contains it, all we have to do is solve for \(\frac{dy}{dx}\).

All we have to do is divide both sides by \(2y\).

$$\frac{2y\frac{dy}{dx}}{2y}=\frac{20x^4-e^x}{2y}$$

Once we simplify the left side we are left with

$$\frac{dy}{dx}=\frac{20x^4-e^x}{2y}.$$

Although this looks a little strange, since our equation for \(\frac{dy}{dx}\) contains both \(x\) and \(y\), this is sometimes the best we can do.  Implicit differentiation is most useful in the cases where we can’t get an explicit equation for \(y\), making it difficult or impossible to get an explicit equation for \(\frac{dy}{dx}\) that only contains \(x\).  Therefore, we have our answer!

I would like to point out that this example is actually a case where we could have solved for \(y\) in terms of \(x\) before taking the derivative.  Doing this would have meant that we could have used other derivative tricks and avoided implicit differentiation, but the way I solved it shows the process of implicit differentiation which is applicable in cases where it is absolutely necessary.

In those cases the general idea and process is the same: we have some function that relates \(y\) and \(x\) and we need to take the derivative of both sides, then use algebra to solve for \(\frac{dy}{dx}\).  This may not always be as simple as the above example, but the process will be extremely similar.

Example 3

Find \(\frac{dy}{dx}\) if \(y=x^x\).

This problem is going to be a bit more tricky than the first two examples.  Click here to see the full solution.

More Examples

\(\mathbf{1. \ \ ycos(x) = x^2 + y^2} \) | Solution

\(\mathbf{2. \ \ xy=x-y} \) | Solution

\(\mathbf{3. \ \ x^2-4xy+y^2=4} \) | Solution

\(\mathbf{4. \ \ \sqrt{x+y}=x^4+y^4} \) | Solution

\(\mathbf{5. \ \ e^{x^2y}=x+y} \) | Solution

As always, don’t forget to let me know if you have any questions on this lesson or if you have any suggestions for other lessons you want to see in the future.  Go check out my derivatives page to see what other material I’ve covered.  I want to know what you want to see on this site, so any and all suggestions and questions are welcome if you can’t find an answer to your question in another lesson.  Just go ahead and leave a comment on this post or email me at jakesmathlessons@gmail.com!