A cylindrical tank with radius *5 m* is being ﬁlled with water at a rate of *3* \(\frac{m^3}{min}\). How fast is the height of the water increasing?

This is an interesting example because at first glance it doesn’t seem like we have been given enough information to solve this problem. If you compare this to the related rates cone problem we did, you can notice a few things that were given in that example but not this one.

- We don’t know the
**height of the cylindrical tank**. - We don’t know the
**height of the water**at the instant we need to find its rate of change.

These things were both given in the cone problem we did.

You may want to check out this video of this problem. I solved it in a slightly different way in the video than I did in this post. Check out both so you can see it a couple different ways and decide which one makes more sense to you!

*How do we deal with this missing information?*

We actually don’t need it! Because the surface of the water in this tank will always be a circle with radius *5 m* as this tank fills up, the height of the water or the tank won’t impact our answer. So there’s two possible ways to deal with this:

- Create an
**equation that doesn’t include the height**of the tank or the water, - or plug in
**some random value for the height**and see what we get. If this one works we should be able to plug in any height and get the same answer for all of them.

One of these options actually won’t work, but we will explore them further later in this example. Let’s not get ahead of ourselves. Since we have a related rates problem here, we will want to follow the **same four steps as all other related rates problems**.

## 1. Draw a sketch

As with any related rates problem, the first thing we need to do is draw the situation being described to us. This is a relatively simple situation being described, so we can go ahead and draw it.

Looking at the above drawing, you can see that water is being poured into a cylindrical tank at a rate of *3* \(\frac{m^3}{min}\). The height of the tank is unknown, but we know the radius is *5 m*. We can also see that the water level is rising, but we don’t know the height of the water at this instant.

## 2. Come up with your equation

Now that we have a drawing of the situation being described, we need to come up with our equation. To do this, we need to sort out what information we know and what we are looking for.

*What are we looking for?*

The question is asking us to find **how fast the height of the water is increasing**. Once we take the derivative of our equation, that will introduce the variable representing “how fast things are changing.” Therefore, we know that we will need the **height of the water** to be in our equation.

Remember in the last section I said we may want to create an equation that didn’t contain the height of either cylinder because we have no information about them? Well now we know that’s **not **really an option. The question is asking us to find the rate of change of the height of one of the cylinders. To find this, we need the height of the cylinder to be in the equation.

*What do we know about?*

This question didn’t give us a lot of information but we can figure out a little extra. First let’s think about what was directly given. In order to do this, we should consider that this example essentially has two cylinders in it. The **large cylinder is the tank**, and the **small cylinder is the water** in the tank.

- We know that water is flowing into the tank at a rate of
*3*\(\frac{m^3}{min}\). This means that the**volume of the small cone is increasing at a rate of**\(\mathbf{\frac{m^3}{min}}\).**3** - The problem also says that
**the tank has a radius of**.**5 m**

And this is all the information that is explicitly given in the problem. However, there is one other fact we can infer. As the tank is filled with water, the liquid takes on a cylindrical shape as well. The cylinder made up of the water would be shorter than the tank, but it would have the same width. This leads to our last fact.

- We can conclude that
**the water also has a radius of**.*5 m*

*Putting it into an equation*

So far in this section, we have figured out that we will need to include the height of the water. We also determined that the information we know is about the radius of both cylinders and the rate of change of the volume of the small cylinder.

So we need our equation to relate the **height, radius, and volume of the liquid**. Because of this, I think a good place to start would be with the equation for the volume of a cylinder:

$$V=\pi r^2 h$$

In this equation *V* is the volume, *r* is the radius, and *h* is the height of the liquid in the tank.

One thing worth pointing out is that once we take the derivative of this equation, we will still have an *h*. But we don’t know anything about the height of the liquid in the tank at this instant. It seems like we wouldn’t have enough information to successfully use this equation, but **it’s possible that the height at a given moment won’t actually impact the rate of change of the height**. If this is true, we should be able to use this equation anyway. So let’s proceed and see what happens.

## 3. Implicit differentiation

As with any related rates problem, once we create our equation we need to take its derivative. Since we are taking the **derivative with respect to time, we need to treat ****V****, ****r****, and ****h**** as functions of time** rather than variables. Therefore, we need to use the chain rule. In this case, since we have *r* and *h* being multiplied, we will also have to use the product rule.

$$\frac{d}{dt}[V]=\frac{d}{dt} \big[\pi r^2 h \big]$$

*Using the product rule*

To find the derivative of the right side of this equation we need to start by using the product rule. So we are trying to find

$$\frac{d}{dt} \big[\pi r^2 h \big].$$

Since \(\pi\) is a constant being multiplied by the rest of the function we are taking the derivative of, we can pull it out of the derivative and deal with it later. Therefore, we can think of the right side of our equation as

$$\pi \frac{d}{dt} \big[r^2 h \big].$$

So we need to find the derivative of \(r^2h\). Just as I did in the product rule lesson, we will start by deciding which part of this equation we will call *f* and *g*.

*Choosing f and g*

It doesn’t really make a difference which piece of this function we call *f *and which we call *g*. As long as we make a decision now and stick with it throughout our solution, it will work out in the end. We will say

$$f=r^2,$$

$$g=h.$$

*Finding f’ and g’*

Now that we have figured out *f* and *g*, the next step of the product rule is to find each of their derivatives. Keep in mind that **we are taking the derivative with respect to time** in this example.

To find *f’*, we will need to use the chain rule as well since *r* is a function of time. doing this tells us that

$$f’=2r \frac{dr}{dt}$$

Lastly, we need to find *g’* by taking the derivative of *g* with respect to time.

$$g’=\frac{dh}{dt}$$

*Putting the pieces of the product rule together*

Now we have figured out all four of the pieces of the product rule, so we can just plug them into the product rule formula to find the derivative we’re looking for.

$$\frac{d}{dt} \big[r^2 h \big] = 2r \frac{dr}{dt} \cdot h + \frac{dh}{dt} \cdot r^2$$

And lastly, we just need to bring the \(\pi\) back into the equation.

$$\frac{d}{dt} \big[\pi r^2 h \big] = \pi \bigg[ 2r \frac{dr}{dt} \cdot h + \frac{dh}{dt} \cdot r^2 \bigg]$$

### Putting it all together

Now that we have the derivative of the right side of our equation, we can go back and figure out the left side.

$$\frac{dV}{dt}= \pi \bigg[ 2r \frac{dr}{dt}h + \frac{dh}{dt}r^2 \bigg]$$

## 4. Solve for desired rate of change

The last step of any related rates problem is to solve for the rate of change we need to find. The question asks us to find how fast the height of the water is increasing. This is exactly what \(\frac{dh}{dt}\) represents, so we need to isolate that variable.

$$\frac{dV}{dt}= \pi \bigg[ 2r \frac{dr}{dt}h + \frac{dh}{dt}r^2 \bigg]$$

$$\frac{1}{\pi}\frac{dV}{dt}= 2r \frac{dr}{dt}h + \frac{dh}{dt}r^2$$

$$\frac{1}{\pi}\frac{dV}{dt} – 2r \frac{dr}{dt}h = \frac{dh}{dt}r^2$$

$$\frac{1}{r^2} \bigg[ \frac{1}{\pi}\frac{dV}{dt} – 2r \frac{dr}{dt}h \bigg] = \frac{dh}{dt}$$

Now that we have isolated the term we need to solve for, we just need to plug in the values for all of the other variables. Before we just into this, I want to focus on just one of these variables.

*What do we do with h?*

Remember back in the beginning of this solution I said we may need to make up some value for *h* and just go with it? Let’s consider this for a moment.

Notice that *h *only appears in our equation in one place and it is being multiplied by two other variables, *r* and \(\frac{dr}{dt}\). Think about what \(\frac{dr}{dt}\) represents. It is the rate of change of the radius of the water. But **our water has a constant radius**, it’s always *5 m*.

Since the radius of the cylinder is never changing, **its rate of change must always be zero!** Therefore, we know that

$$\frac{dr}{dt} = 0.$$

Since *h* is being multiplied by another term that is always *zero*, it’s not going to matter what *h* is. We can essentially use any number for *h *and it won’t matter because we are going to multiply it by *zero *anyways. We can also just leave it as *h* and not plug anything in for it.

*Back to plugging in our values*

We have already figured out that \(\frac{dr}{dt}=0\) and luckily the other variables’ values were given.

Looking back at the original sketch, you can see that the radius of our cylinder is *5 m*. So

$$r=5.$$

Also, you can see that water is flowing into the tank at a rate of *3* \(\frac{m^3}{min}\). Since this is the only factor that will be impacting the volume of water in the tank, this must be the exact **rate that the volume of the water is increasing**.

$$\frac{dV}{dt}=3$$

Now we can plug all of these into the equation then simplifying to get our answer!

$$\frac{1}{r^2} \bigg[ \frac{1}{\pi}\frac{dV}{dt} – 2r \frac{dr}{dt}h \bigg] = \frac{dh}{dt}$$

$$\frac{1}{(5)^2} \bigg[ \frac{1}{\pi}(3) – 2(5)(0)h \bigg] = \frac{dh}{dt}$$

$$\frac{1}{25} \bigg[ \frac{3}{\pi} – 0 \bigg] = \frac{dh}{dt}$$

$$\frac{1}{25} \cdot \frac{3}{\pi} = \frac{dh}{dt}$$

$$\frac{3}{25 \pi} = \frac{dh}{dt}$$

So we now know that the rate of change of the height of the water is \(\frac{3}{25 \pi}\) no matter what the height of the cylinder is at that moment. In other words, we can say that** the height of the water is increasing at a rate of **

$$\mathbf{\frac{3}{25 \pi} \ \frac{m}{min}}.$$

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 a solution of. 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**!

**Also, if you want a copy of my calculus 1 study guide, just enter your name and email below and I’ll send you a copy** and add you to my email list so you can see as soon as I have more knowledge to share!