The top of a ladder slides down a vertical wall at a rate of 0.15 m/s. At the moment when the bottom of the ladder is 3 m from the wall, it slides away from the wall at a rate of 0.2 m/s. How long is the ladder?

This is a fairly common example of a related rates problem and a common application of derivatives and implicit differentiation. I’m sure you may have come across a related rates ladder problem like this. If I can offer one piece of advice for this type of problem it’d be this: don’t use this ladder, it always falls…

Alright, bad jokes aside, this is going to follow the same 4 steps as all the other related rates problems I’ve done. If you’d rather watch a video, then check out my video below. But otherwise, let’s jump into it with the usual process!

## 1. Draw a sketch

As always, we’ll start by drawing a quick sketch of all of the information that is being described in the problem. To do this we should first think about what information we have. First of all, we need to think about the shape that’s being formed with the ladder.

Since the ladder is standing on the ground and leaning up against a vertical wall, we can say that a triangle would be formed by the 3 objects in the problem. More specifically we know that the vertical wall forms a 90 degree angle with the ground. Therefore, the triangle formed by the ground, the wall, and the ladder would be a right triangle.

On top of this, the problem also gives us a few pieces of information about the dimensions of the triangle and how they are changing. It actually tells us about how fast the ladder is moving, but since the ladder is what forms the triangle, we can deduce how the dimensions of the triangle are changing.

We are given 3 pieces of information about the position of the ladder as well as how the ladder is moving at the specific instant we are looking at.

• Bottom of the ladder is 3 m away from the wall.
• Top of the ladder is moving down the wall at a rate of 0.15 m/s.
• Bottom of the ladder is moving away from the wall at a rate of 0.2 m/s.

Adding these labels to our drawing from above would give us something like this:

This sketch gives us a pretty good idea of what is going on in this problem. Not only that, but we will be able to use this to get an idea of what kind of equation we will need to come up with.

## 2. Come up with your equation

Before we come up with our equation we want to sort through the information we are given and asked to find. This is important because we need to decide what measurements and variables we want in the equation.

#### What are we looking for?

The question asks us to find the length of the ladder. Therefore, we will need to find the length of the hypotenuse of the triangle in our drawing. Because of this we want to be sure to include the hypotenuse of our triangle in our equation.

#### What do we know about?

Looking back up at our labeled drawing, you can see that we really only have information about the bottom and side of our triangle. We know the length of the bottom side of the triangle and the rate of change of this side.

And we were also given information about the rate of change of the left side of the triangle. But remember that our equation in this step cannot include rates of change. Instead, the fact that we know this rate of change tells us that we can use the left side length of the triangle in our equation, not its rate of change.

We aren’t given any information about the angles in the triangle other than the fact that it’s a right triangle. As a result, we probably don’t want our equation to involve the angles of this triangle.

Since we know that our equation either needs to include, or can include the lengths of the sides of the triangle we should label them. Let’s go back to our drawing and label the side of the triangle. You can label them whatever you’d like, but I’ll go with x, y, and z.

#### Putting it into an equation

At this point we’ve figured out that we need an equation that related the sides of a right triangle.

What do you know about the relationship between the sides of a right triangle, but neither of the other two angles?

You’re probably thinking Pythagorean Theorem. If you are, you’re right! Pythagorean Theorem tells us that the square of the hypotenuse is equal to the sum of the squares of the other two sides of a right triangle. Remember this can only be applied to the sides of a right triangle, so noticing that is actually very important. In other words we know $$z^2=x^2+y^2.$$

## 3. Implicit differentiation

Now that we have come up with our equation, we need to apply implicit differentiation to take the derivative of both sides of our equation with respect to time.

$$\frac{d}{dt} \Big[ z^2 = x^2 + y^2 \Big]$$

Before we do this though I want to point something out. Let’s look at each of the letters in this equation and consider how we need to treat them when we differentiate with respect to time.

Consider z first. z represents the length of the hypotenuse of the triangle. This is the side that is formed by the ladder. As time changes what happens to the length of the ladder? Nothing. It doesn’t change at all. It’s constant. Therefore we can treat z like a constant. If z is a constant and never changes, then $$z^2$$ would be constant too. It doesn’t change as time changes.

So when we take the derivative of z with respect to time, the derivative will be 0. The derivative of any constant is 0.

Unfortunately x and y won’t be as convenient. Looking back at our drawings you can see that the sides labelled x and y are changing over time. As the ladder slides down and away from the wall, these two sides of the triangle change in length. Therefore, when we take the derivative of $$x^2$$ and $$y^2$$ we will need to treat x and y as functions of time. Doing this means that we will need to use the chain rule, where x and y are the inside function and they are being plugged into another function that squares them.

#### Back to the derivative

Knowing how to treat each letter in our equation, let’s go ahead and take the derivative of both sides with respect to time.

$$\frac{d}{dt} \Big[ z^2 = x^2 + y^2 \Big]$$ $$0 = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$$

## 4. Solve for the desired rate of change

Now all we need to do is plug in all of the information we have and solve for the right variable. However, this one is a little weird. The reason I say this is that we are actually not looking for a rate of change.

Remember the question told us to find the length of the ladder. Which means we need to find the value of z. The differential equation we just ended up with doesn’t even have a z in it, so how can we use it to find z?

Well, we’re actually going to need to go back to our original equation. $$z^2=x^2+y^2$$ We know the value of x based on information we were given, but we don’t know the value of y yet. If we could figure out what y was, then we could use this equation, plug in the value for x and y, then solve for z.

#### How do we find y?

This is what we will use our differential equation from the previous step for. That equation has a y in it, and we know the value of all the other variables.

• We are told that the moment we are considering is when the bottom of the ladder is 3 m from the wall. Since that corresponds to the side of our triangle labelled x, we know $$\mathbf{x=3}$$.
• We are also told that the ladder is moving away from the wall at a rate of 0.2 m/s. Therefore, x must be getting longer, or increasing, at that rate. So $$\mathbf{\frac{dx}{dt}=0.2}$$.
• And finally, the ladder is sliding down the wall at a rate of 0.15 m/s. So y must be getting shorter, or decreasing, at that rate. This means $$\mathbf{\frac{dy}{dt}=-0.15}$$.

#### Plugging it all into our equation

Knowing all of the values in our equation aside from y, we can plug these in and solve for y.

$$0 = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$$ $$0 = 2(3)(0.2) + 2y(-0.15)$$ $$0=1.2-0.3y$$ $$0.3y=1.2$$ $$y=4$$

Now that we know x and y, we can plug them back into our original equation and solve for z.

$$z^2=x^2+y^2$$ $$z^2=(3)^2+(4)^2$$ $$z^2=25$$ $$z=5$$

So the ladder must be 5 m long!

## RELATED RATES – Square Problem

Each side of a square is increasing at a rate of 6 $$\frac{cm}{s}$$. At what rate is the area of the square increasing when the area of the square is 16 $$cm^2$$?

If you haven’t already, you should check out my related rates lesson. I go through the steps that can be used to solve any related rates problem. I will use these steps here, but it might be useful for you to see a more detailed explanation of why we use these steps.

## 1. Draw a sketch

The first thing we will always want to do is draw a sketch of the situation described by the problem. This problem is relatively simple to draw out. Here we have a square whose sides are growing at a constant rate. We know the sides are growing at a rate of 6 $$\frac{cm}{s}$$ and we want to consider the moment when the area of the square is 16 $$cm^2$$.

## 2. Come up with your equation

Now that we have our drawing laid out, we need to create an equation whose derivative we will need to find.

#### What are we looking for?

This question is asking us to find the rate at which the area of the square is increasing. So it tells us that we need to have some variable that represents the rate of change of the area at some point.

Of course, we will need to take the derivative of our equation soon. Doing this will introduce the ‘rate of change’ part. Therefore, we need to make sure that our equation we make includes the area of the square. As long as we do this, we will end up with the rate of change of the area once we take its derivative.

#### What do we know about?

There was only two pieces of information that the question directly told us.

• Each side of a square is increasing at a rate of 6 $$\frac{cm}{s}$$.
• The area of the square in this instant is 16 $$cm^2$$.

#### Putting it into an equation

So at this point, we have figured out that we need our equation to include the square’s area, and we know something about the square’s area and the rate of change of its sides.

Although, we don’t know anything about the actual side lengths at the given instant, we can figure that out if we have to. Therefore, it’s fine if our equation includes the length of the square’s sides.

To summarize, the only measurements of this square we have discussed so far are its side lengths and its area. So our equation should relate its area and its side lengths. Keep in mind, we don’t want to put anything that represents a rate of change into our equation. These will come when we take the derivative of our equation.

The simplest equation that relates the area of a square with its side lengths would likely be the formula for the area of a square that you are already familiar with. $$A=l^2$$ Where A is the area of the square, and l is the length of its sides.

## 3. Implicit differentiation

Once we have created our equation like we did above, we need to go ahead and take its derivative with respect to time. This will be done using implicit differentiation.

Since we will be taking the derivative with respect to time, we will need to treat the A and the l in our equation as functions of time. This will require using the chain rule to find their derivatives.

$$\frac{d}{dt} \big[ A \big] = \frac{d}{dt} \Big[ l^2 \Big]$$ $$\frac{dA}{dt} = 2l \cdot \frac{dl}{dt}$$

## 4. Solve for the desired rate of change

Now all we have to do is solve for the rate of change the question is asking about. Just like I said earlier, we need to find the rate of change of the square’s area. Since this is exactly what $$\frac{dA}{dt}$$ represents and we have already isolated this, we don’t have much else to do. All we need to do is plug in the values we have for everything else in our equation.

The only other variables we need to plug in are l and $$\frac{dl}{dt}$$. The tricky thing here is that we don’t directly know what l is at the moment in this problem. Instead we know that the area of the square is
16 $$cm^2$$. Since we know that the relationship between the area of a square and its side lengths is $$A=l^2$$ we can find l in this instant. $$16=l^2$$ $$4=l$$ So we know that in this moment the length of the square’s sides is 4 cm.

Remember, the problem also told us that the side lengths are increasing at a rate of 6 $$\frac{cm}{s}$$. This directly tells us the rate of change of the sides lengths. So we also know that $$\frac{dl}{dt}=6$$

Putting all of this into our equation will give us: $$\frac{dA}{dt} = 2l \cdot \frac{dl}{dt}$$ $$\frac{dA}{dt} = 2(4)(6)$$ $$\frac{dA}{dt} = 48$$

So the area of this square is increasing at a rate of 48 $$\mathbf{\frac{cm^2}{s}}$$ when the area is 16 $$\mathbf{cm^2}$$.

If you’re still having some trouble with related rates problems or just want some more practice you should check out my related rates lesson. At the bottom of this lesson there is a list of related rates problems that I have posted along with their solutions. I also have several other lessons and problems on the derivatives page you can check out. If you can’t find the topic or question you’re looking for just let me know by emailing me at jakesmathlessons@gmail.com!

You can also enter your name and email using the form below and I will send you my calculus 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?

### 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.

### 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.

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

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.

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.

## RELATED RATES – 4 Simple Steps

Related rates problems are one of the most common types of problems that are built around implicit differentiation and derivatives.  Typically when you’re dealing with a related rates problem, it will be a word problem describing some real world situation.

Typically related rates problems will follow a similar pattern.  They can usually be broken down into the following four related rates steps:

• The first thing you will usually want to do after reading the problem is to draw a sketch of the situation being described.
• Then you will need to come up with some equation that relates the different quantities described to you, which may be volumes, areas, or distances.
• Once you have this equation, you’ll perform implicit differentiation on both sides of the equation, usually with respect to time.
• Then you just need to solve for the desired rate of change that the question is asking about.

I always think the best way to learn a new concept is practice, practice, practice.  So let’s jump into an example.  If you want to skip ahead, there is a list of other examples at the bottom of this page with a link to their solutions.

## Example 1

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?

### 1. Draw a sketch

Like I said before, the best thing to do first is draw a picture of what is being described.  The problem tells us where these ships are, and the direction and speed at which they’re moving at noon, so I think we should start by drawing our scene at noon, seen in Figure 2.1.

Now let’s think about what will be happening between noon and 4:00 PM.  We know the speed and direction both of these ships are traveling.  We also know that they will be moving for 4 hours before we consider their position again.

Ship A:  This ship is moving 35 km/h for 4 hours.  Therefore, between noon and 4:00 PM we know it will move east

$$35\frac{km}{h}*4h=140km.$$

Ship B:  This ship is moving 25 km/h for 4 hours.  Therefore, between noon and 4:00 PM we know it will move north

$$25\frac{km}{h}*4h=100km.$$

Considering both of these facts, at 4:00 PM, our scene would look something like this:

### 2. Come up with your equation

Now, the next thing we need to do is come up with an equation that relates the different distances given based on where the boats are at 4:00 PM.  We also want to consider what the question is asking before we come up with this equation.  The question is asking us to find how fast the distance between the boats is changing.  We won’t be able to find an equation for this right away, but this tells us that we need an equation that involves the distance between the two boats.  From there, we can figure out how fast that distance is changing.

To do this, we will want to think of the drawing in Figure 2.2 as a triangle.  The three vertices of the triangle would be ship A, ship B, and the point in the water where boat B was at noon, which is shown in the drawing above.  Doing this gives us something like this:

It is important to remember that both ships are still moving, they don’t stop at 4:00.  As a result of this, the sides of our triangle are not constants.  Instead, they would be variables, so the triangle we may want to consider is this one:

From this, we can create our equation.  What we want to think about is what variable has to do with the value the question is asking us to find and what variables do we know something about.

What are we looking for?

The question asks us to find how fast the distance between the ships is changing.  By comparing Figure 2.4 to the prior drawings, we can see that side $$z$$ is the one that represents the distance between the two ships.  Therefore, we will need to include $$z$$ in our equation which we will eventually differentiate.

This is where we want to consider the actual numerical values we have figured out and how they relate to the different variables in our drawing.  At 4:00 PM we know the values of $$x$$ and $$y$$.  We also know the speed at which the ships are moving in relation to a fixed point, which is the vertex of the triangle that makes up the right angle.  Therefore, we not only know the values of $$x$$ and $$y$$, but we also know their rates of change.  These will simply be the speeds of the ships.

Putting it into an equation.

We need an equation that relates the thing we are looking for with the things we already know.  Since we’re looking for some information about $$z$$ (the rate of change of $$z$$), and we know everything about $$x$$, $$y$$, and both of their rates of change at 4:00, we need an equation relating $$x$$, $$y$$, and $$z$$.

Since we know the relationship between these variables is that they are the side lengths of a right triangle, the simplest equation we can use is Pythagorean Theorem.  Based on this we know that

$$z^2=x^2+y^2.$$

### 3. Implicit differentiation

Now that we have our equation, we need to take its derivative.  This is where the implicit differentiation comes in.  Before we do this, let’s think about what we want to differentiate with respect to.

I mentioned in the beginning of this article that we will usually differentiate with respect to time.  The reason for this is that $$x$$, $$y$$, and $$z$$ are all changing as time passes.  If fact, since we know the ships’ initial positions and their velocities, we could actually write $$x$$, $$y$$, and $$z$$ as functions of time.  Therefore, when we differentiate both sides of our equation, we will treat $$x$$, $$y$$, and $$z$$ as functions, and time (represented by $$t$$) will be the variable.  If you want a bit more explanation on the next few step, I explained this a bit more here.

$$z^2=x^2+y^2$$

$$\frac{d}{dt}\big[ z^2\big]=\frac{d}{dt}\big[ x^2+y^2\big]$$

$$2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$$

Now we can divide both sides by $$2$$ to simplify.

$$z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$$

### 4. Solve for desired rate of change

Now remember, the thing we are trying to find in this problem is the rate of change of $$z$$.  This is exactly what $$\frac{dz}{dt}$$ represents, so we will solve for this by dividing both sides by $$z$$.

$$\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}$$

All we have left to do now is plug in all the pieces on the right side of the equation and that would give us our answer.  We are looking for the value of $$\frac{dz}{dt}$$ at 4:00 PM, so we need to use the values of all the other variables based on what they are at 4:00 PM also.  We can gather most of the information we need from Figure 2.2, shown above.  Here it is again:

Clearly, we can see that $$x=10km$$ and $$y=100km$$ at 4:00.  We also know the value of $$\frac{dx}{dt}$$ and $$\frac{dy}{dt}$$. The reason for this is that these two things represent the rate of change of $$x$$ and $$y$$.  Since $$x$$ and $$y$$ represent the distance between each of the ships and a fixed point, $$\frac{dx}{dt}$$ and $$\frac{dy}{dt}$$ would be given by the speed of each ship.  It is also important to point out that is works because the ships are moving directly toward the fixed point or directly away from the fixed point.  The length of $$x$$ is shrinking by $$35\frac{km}{h}$$ because ship A is moving at that speed.  Therefore, we know

$$\frac{dx}{dt}=-35\frac{km}{h}.$$

Notice this value is negative.  This is simply because $$x$$ is getting smaller at 4:00.  Similarly, we know that $$y$$ is getting larger by $$25\frac{km}{h}$$ at 4:00 and therefore we can say that

$$\frac{dy}{dt}=25\frac{km}{h}.$$

Now the only thing we still need to figure out is the value of $$z$$ at 4:00 PM.  To do this, let’s go back to our Pythagorean Theorem equation we looked at earlier.  We know

$$z^2=x^2+y^2.$$

Since we already know the values of $$x$$ and $$y$$ at 4:00, we can just plug them into the Pythagorean Theorem equation to get the value of $$z$$ at 4:00.

$$z^2=10^2+100^2$$

$$z^2=100+10,000$$

$$z^2=10,100$$

$$z=\sqrt{10,100}$$

Now that we have all the pieces figured out, we can go back and plug them into our equation for $$\frac{dz}{dt}$$.

$$\frac{dz}{dt}=\frac{10*(-35)+100*25}{\sqrt{10,100}}$$

$$\frac{dz}{dt}=\frac{2,150}{\sqrt{10,100}}\approx 21.393\frac{km}{h}$$

So we can say that the distance between the two ships is changing at about $$21.393\frac{km}{h}$$ at 4:00 PM.

## Other Examples

Like I said before, the best way to gain an understanding of related rates problems is practice.  Here are some more complete solutions of other fun related rates problems.  Just click on the problem to see the full solution.

#### Triangles

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 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?

The top of a ladder slides down a vertical wall at a rate of 0.15 m/s. At the moment when the bottom of the ladder is 3 m from the wall, it slides away from the wall at a rate of 0.2 m/s. How long is the ladder?

#### Squares

Each side of a square is increasing at a rate of 6 cm/s. At what rate is the area of the square increasing when the area of the square is 16 cm^2?

#### Cones

Water is leaking out of an inverted conical tank at a rate of 10,000 cm^3/min at the same time water is being pumped into the tank at a constant rate. The tank has a height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.

#### Spheres

If a snowball melts so that its surface area decreases at a rate of 1 cm^2/min, find the rate at which the diameter decreases when the diameter is 10 cm.

The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when the diameter is 80 mm?

#### Cylinders

A cylindrical tank with radius 5 m is being ﬁlled with water at a rate of 3 m^3/min. How fast is the height of the water increasing?

As always, I’d love to see your questions!  You can leave a comment below or email your questions to me at jakesmathlessons@gmail.com.  Whether you have questions about this lesson or want me to post a solution for another problem you’re stuck on, just send me a message.

## 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!