## RELATED RATES – Sphere Volume Problem

The radius of a sphere is increasing at a rate of 4 $\frac{mm}{s}$. How fast is the volume increasing when the diameter is 80 mm?

This question states pretty clearly that we will be working with a sphere here. Since it gives information about how it is changing and asks us to find how quickly another value is changing, we know it’s a related rates problem.

As with all of the other related rates problems I’ve worked through, we are going to be following the same four step process. I will go through them one step at a time, but you can also find where I introduce the steps here.

## 1. Draw a sketch

The first thing we need to do is draw a sketch of the scene being described. Obviously we are dealing with a sphere, but we are really only told two things:

• how quickly the radius is changing, and
• what the diameter is at the specific moment we’re concerned with.

Since that’s all we know, it will be pretty simple to put that into a drawing. It would look something like this:

So we have a sphere whose radius is increasing at a rate of 4 $\frac{mm}{s}$ and we need to consider the moment when its diameter is 80 mm.

## 2. Come up with your equation

As with any related rates problem, we need to create our equation once we have created our drawing. To do this we need to consider the information we have been given and how it relates to the piece of information we’re looking for.

#### What are we looking for?

The question asks us to find how fast the volume is increasing when the diameter is 80 mm. Asking about how fast something is changing refers to its rate of change. Therefore, we can tell that this question is asking us about the rate of change of the volume.

Since we will later be taking the derivative of the equation we are currently building, we only need to make sure to include the volume. Once we take the derivative of the equation, this will introduce the rate of change of the volume.

#### What do we know about?

This question didn’t provide a lot of information and it’s fairly straight forward.

• Rate of change of the radius.
• The diameter of the figure at this moment.

#### Putting it into an equation

Up to this point we know that we need to include the sphere’s volume in this equation. We have also figured out that we know some information about the diameter, which can easily be used to find the radius. And we also know about the rate of change of the radius. So we basically know everything about the radius that we might need.

What is the first equation or formula you think of that relates the volume of a sphere to its radius?

I think a good place to start is with the formula for the volume of a sphere.

$$V=\frac{4}{3} \pi r^3$$

We can see that this equation only contains V (volume) and r (radius). As a result, I’d say this is as good of a place to start as any. Let’s proceed with this equation.

## 3. Implicit differentiation

Now that we have come up with our equation, we need to take its derivative with respect to time. This will allow us to introduce and work with the rates of change of our measurements.

Since we will be taking the derivative with respect to time, we will need to treat V and r as functions of time rather than variables. In order to do this we will need to use the chain rule. So, taking the derivative of our equation gives us:

$$\frac{d}{dt} \big[ V \big] = \frac{d}{dt} \bigg[ \frac{4}{3} \pi r^3 \bigg]$$

$$\frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt}$$

$$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}.$$

## 4. Solve for the desired rate of change

Finally, all we need to do now is solve for the rate of change the question is asking us to find. The problem asked us to find how fast the volume is changing at this moment. This is exactly what $\frac{dV}{dt}$ represents. Since this is already isolated, all we need to do is plug in the other values we know about.

The other values that are needed are r and $\frac{dr}{dt}$, which represent the radius and its rate of change respectively.

The radius of our figure wasn’t given directly, but we do know that its diameter is 80 mm at this instant. Since the radius of a sphere is always half of the diameter, this tells us that the radius is 40 mm, or

$$r=40.$$

We were given that the figure’s radius is increasing at a rate of 4 $\frac{mm}{s}$. Therefore, we know

$$\frac{dr}{dt} = 4.$$

#### Plugging it all in

Now we simply need to plug these values into the differentiated equation we found in step three.

$$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$$

$$\frac{dV}{dt} = 4 \pi (40)^2 (4)$$

$$\frac{dV}{dt} = 25,600 \pi$$

$$\frac{dV}{dt} \approx 80,424.772$$

So this tells us that the volume of the sphere is increasing at a rate of 25,600$\mathbf{\pi \ \frac{mm^3}{s}}$, or about 80,424.772 $\mathbf{\frac{mm^3}{s}}$ when its diameter is 80 mm.

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 practice 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!

I also encourage you to join my email list! Just enter your name and email below and I’ll send you a copy of my calculus 1 study guide to help you get through your homework faster and the exams with better scores!

## RELATED RATES – Sphere Surface Area Problem

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

By looking at the given statement, we can gather a few important fact quickly.

• The object in question is a snowball. This means that we will be dealing with a sphere here, so we will likely be needing some formula relating different dimensions and measurements of a sphere.
• The problem gives information about the rate of change of a specific measurement of the snowball. When a problem gives information like this, it’s a strong hint that we have a related rates problem.

So we know that we’re dealing with a related rates problem. Therefore, we are going to follow the four steps that these will all follow. If you want to look back at these steps, I discussed them in my related rates lesson.

## 1. Draw a sketch

Looking back at the problem, you can see that there isn’t a lot of information that has been described to us. This will end up being a very simple sketch, but that’s all it takes sometimes.

So we have a sphere whose surface area is decreasing at a rate of 1 $\frac{cm^2}{min}$ and we are looking at the instant when its diameter is 10 cm.

## 2. Come up with your equation

Now that we have our situation drawn out, we need our equation. To create this equation, we need to incorporate any relevant information that we were given and we need to consider what the question asks us to find.

#### What are we looking for?

The question is asking us to find “the rate at which the diameter decreases” at the instant when the diameter is 10 cm. This is the instant that we captured in our drawing. So the important thing here is that we are looking for the rate the diameter is decreasing. Another way to put this is the rate of change of the diameter.

Since we will eventually need to take the derivative, which will provide the rate of change part, we just need to make sure that our equation contains the diameter.

#### What do we know about?

This question didn’t provide a lot of information to us. We really only know two things:

• The rate of change of the sphere’s surface area.
• The diameter of the sphere at this instant.

#### Putting it into an equation

Up to this point we have figured out that we need to include the sphere’s diameter in our equation. We also know that we have some information relating to the snowball’s surface area. So we need to come up with an equation that relates a sphere’s surface area and diameter.

A good place to start may be the equation for the surface area of a sphere.

$$A=4 \pi r^2$$

In this equation A represents the surface area of the sphere and r is the radius.

Since we need an equation relating the surface area and the diameter, we will need to make an adjustment. The only thing we need to consider here is that a sphere’s diameter is always double the radius. Or in other words

$$d=2r.$$

But our equation contains the radius. So we’ll want to solve for r so we can plug that into our equation. Dividing both sides by 2 will accomplish this.

$$r=\frac{d}{2}$$

Now we can plug this in for r in our equation for the surface area of a sphere.

$$A=4 \pi \bigg( \frac{d}{2} \bigg)^2$$

So we know have an equation that contains only the surface area and the diameter of our sphere, exactly what we needed. To make things a little easier later, we will simplify this equation a bit.

$$A=4 \pi \bigg( \frac{d^2}{4} \bigg)$$

$$A=\pi d^2$$

## 3. Implicit differentiation

Now that we have created our equation we need to take its derivative. This will bring in the rates of change we discussed earlier. The most important one being the rate of change of the snowball’s diameter which is what we need to find.

Keep in mind that we will be taking the derivative with respect to time. This means we will need to treat A and d as functions of time, not variables. Therefore, we will need to apply the chain rule here.

$$\frac{d}{dt} \big[ A \big]= \frac{d}{dt} \big[ \pi d^2 \big]$$

$$\frac{dA}{dt} = 2 \pi d \cdot \frac{dd}{dt}$$

## 4. Solve for the desired rate of change

The fourth and final step of a problem like this is to isolate the rate of change we need and find its value. The question wants us to find the rate at which the diameter is decreasing when the diameter is 10 cm. This tells us we need to solve for the rate of change of the diameter, which is represented by $\frac{dd}{dt}$.

$$\frac{dA}{dt} = 2 \pi d \cdot \frac{dd}{dt}$$

$$\frac{1}{2 \pi d} \frac{dA}{dt} = \frac{dd}{dt}$$

So now that we have isolated the variable that we need to find we can simply plug in all of the values we have for the other variables.

#### What values do we know?

There are only two variables we need to plug in a value for: d and $\frac{dA}{dt}$. The question specifically asked us to find $\frac{dd}{dt}$ when the diameter of the snowball is 10 cm. Therefore, we know that we will have

$$d=10.$$

The question also told us that the surface area of the snowball is decreasing at a rate of 1 $\mathbf{\frac{cm^2}{min}}$. Since the surface area is decreasing, this tells us that

$$\frac{dA}{dt}=-1.$$

#### Plugging it all in

Now we just need to go back to our equation and plug in all of these other values.

$$\frac{dd}{dt} = \frac{1}{2 \pi d} \frac{dA}{dt}$$

$$\frac{dd}{dt} = \frac{1}{2 \pi (10)} \cdot (-1)$$

$$\frac{dd}{dt} = \frac{-1}{20 \pi}$$

So this tells us that the diameter of the snowball is changing at a rate of $\frac{-1}{20 \pi} \ \frac{cm}{min}$. Therefore, we can say that the diameter of the snowball is decreasing at a rate of $\mathbf{\frac{1}{20 \pi} \ \frac{cm}{min}}$ or about 0.0159 $\mathbf{\frac{cm}{min}}$.

Note that the question asked for the rate at which the diameter is decreasing. As a result of this, our answer will actually be a positive number despite the fact that $\mathbf{\frac{dd}{dt}}$ was negative.

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 these types of 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!

Enter your name and email below and I’ll send you a FREE copy of my calculus 1 study guide as well! It’s packed full of helpful formulas and shortcuts to help you get through calculus easier!

## 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.