### Introduction

One of the most common related rates problems is the ladder problem, which looks at how a ladder slides down a wall, assuming that the ladder always makes a right triangle with the wall.

Let’s see how to solve these sorts of problems by working through a simple example.

### Example

Let’s consider a ladder that is $10$ feet long. At a given moment the top of the ladder is $8$ feet above the ground and the foot of the ladder is sliding away from the wall at a rate of $3$ feet per second. We’ll try to find the speed the top of the ladder is sliding down the wall at this moment.

At the moment we are considering, the ladder makes a right triangle with the wall, as shown in the diagram, so we can use the Pythagorean Theorem to relate the ladder and the sides of the wall. We’ll call the distance between the corner of the wall and the top of the ladder in feet $a$ and the distance between the corner of the wall and the foot of the ladder in feet $b$. Since the length of the ladder is unchanging despite $a$ and $b$, we can relate them with $a^2+b^2=100$.

Now, given the information about the ladder we have, we can find more related information. We know that $a=8$ and that the bottom of the ladder is sliding to the left at $3$ feet per second. We can easily find $b$ using the formula we created and compute that $b=6$. We’ll now look at the derivative of the formula we just created with respect to time. We need to derive with respect to time because we are looking at how the ladder slides down the wall as time increases, so time or $t$ can be treated as a separate variable. $$\dfrac{d}{dt}a^2+b^2= \dfrac{d}{dt}100$$ Because $a$ and $b$ are functions of time, we have to use the chain rule on them:

$$2a\cdot\dfrac{da}{dt}+2b\cdot\dfrac{db}{dt}=0.$$

Looking at the information we already know, we know that the ladder is sliding left at 3 feet per second, or in other words, $b$ is increasing at $3$ feet per second, so $\dfrac{db}{dt}=3$. We can then substitute the information we know.$$2\cdot8\cdot\dfrac{da}{dt}+2\cdot6\cdot3=0$$ So $\dfrac{da}{dt}$ equals $-\dfrac{9}{4}$, which means the distance between the top of the latter and the corner is decreasing by $\dfrac{9}{4}$ feet per second, which is our answer.

### Conclusion

You should now be familiar with the basic form of the ladder problem and how to use the Pythagorean Theorem and its derivative to solve for variables. The ladder problem may seem daunting at first, however using basic derivative rules, it can be very approachable.