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
from the wall.*3 m*away - 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 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 and 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**. - 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**. - 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 .

#### 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!