Related Rates - Online Tutor, Practice Problems & Exam Prep

Intro To Related Rates

In recent videos, we were introduced to the idea of implicit differentiation. Recall that when doing these types of problems, we were able to take the derivative of one variable with respect to another variable using the chain rule. This idea is going to be very important to understand when solving another problem type you'll see in this course called related rates. Now, related rates are problems that students often find very tricky when they first learn them, but don't sweat it. Because in this video and over the course of the next few videos, I'm going to be walking you through these related rates problems you'll see in the course, including some of the more complicated ones that you'll see.

And hopefully, you're going to find this to be very straightforward because a lot of this is just related to implicit differentiation, which we've already talked about. So let's get right into things. Now, whenever you have a situation where 2 or more variables are related to each other, chances are you're dealing with a related rates problem. And the first thing you want to do with any related rates problem is to see how the variables are changing with time by taking time derivatives to see how one variable affects another. Now to understand this, let's just get into an example here where we are given this equation, \( y = x^3 \).

Now what we're asked to do here is find the rate of change, \( \frac{dy}{dt} \). And notice what's also given to us in this problem is the rate of change for \( x \), \( \frac{dx}{dt} \), and we're given \( x \) at one certain instance, which is equal to 4. Now this might seem like a lot of information, and we're not really sure where to start. But like I mentioned before, a good step for any related rates problem is to take the time derivative on both sides of the equation. So what I'm going to do is take the time derivative of this entire equation up here.

Now if I do this, we're literally just taking this derivative of every term. So we'll have \( \frac{d}{dt} \) for \( y \), which is the derivative of \( y \), then we'll have the time derivative of \( x^3 \). Now the way that we can find these derivatives is by recalling what we did with implicit differentiation. And remember, when we took the derivative of one variable with respect to another, we got this kind of rate. And we know that we can use the chain rule when doing this as well, so that's what we're gonna do over here.

So I know the derivative of \( y \) with respect to \( t \) is just gonna give me \( \frac{dy}{dt} \) on the left side. We learned that from implicit differentiation. And I know that for taking the derivative of \( x^3 \), I can use the power rule. Doing that is going to give me \( 3x^2 \). Now you might think that we can just finish right about here, but this is actually not the final result after taking the derivative because notice we're taking a derivative of \( x \) with respect to \( t \).

In the past when we've done this, we took the derivative of \( x \) with respect to \( x \), which would give us \( \frac{dx}{dx} \), which naturally we don't even have to write. But since we're taking the derivative of \( x \) with respect to \( t \), we need to use the chain rule and multiply this by \( \frac{dx}{dt} \). So this right here would truly be the answer for our derivative, and that's how you can differentiate both sides with respect to time, \( t \) being the time variable here. So this would be step 1 right here, which is taking the time derivative on both sides and using implicit differentiation. Now step 2 is going to be to isolate the target rate of change.

I can see clearly in this problem the target rate of change is going to be this \( dy/dt \). But notice something, \( dy/dt \) is already isolated on one side of the equal sign. So we don't even need to worry about step 2. We already have this rate isolated. So really, we can just move to step 3, which tells us to plug in the known values or rates and solve.

So to do this, well, let's just see what we're given in the problem. Now I can see for this equation here, we have 3 and then we have \( x^2 \). \( x \) we're told is 4 at this instance, so that's going to be \( 4^2 \). That's going to be multiplied by \( dx/dt \), which I can see right here is 2. Now I trust at this point you can do this math in your head, \( 3 \times 4^2 \times 2 \), and all of this should come out to 96.

So this right here is going to be the rate \( dy/dt \), and that would be the solution to this problem. So this is how you can solve these types of related rates problems. And something that I'm going to mention here is this was actually a bit more of a simple related rates problem. Because notice in this example, we were given the rate \( dx/dt \). We were also given the equation that we need to start with, and we were asked to find the missing rate here after being directly given the values.

That's not always going to happen in related rates, and that's why students often find this to be a bit trickier. But make sure to check out the next couple of videos because I'm going to walk you through some of those more complicated examples, and hopefully, you're going to see that this process is the same. See you there.

Given the equation below, find $dy/dt$ when $dx/dt=12$ and $x=\frac{15}{2}$.

$y=\sqrt{2x+1}$

• A. 6
• B. 1.5
• C. 36
• D. 3

Given the equation below, find $dz/dt$ when $dx/dt=3$, $dy/dt=2$, $x=2$, $y=4,$ and $z=1$.

$x^2+y^2-3z^2=125$

• A. $\frac{14}{3}$
• B. $\frac{16}{3}$
• C. $-\frac{14}{3}$
• D. $-\frac{16}{3}$

Changing Geometries

In the previous video, we were introduced to this idea of related rates. And recall that when doing these types of problems, what we did was we took time derivatives on both sides of the equation and then used implicit differentiation to find a missing rate with respect to time. Now this is the process that you're going to need to use for any type of related rates problem. Only in this video, we're going to learn that some of the problems are not as straightforward as simply being given an equation and having to solve it. But don't sweat it, because in this video, I'm going to walk you through how to solve some of these more complicated types of problems.

So let's just get right into things. Now a common problem type that you'll see is a changing geometries problem, and one of those problems would look something like this. Here, we're told in this example that a cube grows at a rate of 2 cubic centimeters per second. And we're asked how fast the side of the cube is growing when it has a length of 0.8 centimeters? Now you'll notice right off the bat, this is not a situation where we are being given an equation nor are we given any of the rates really that are happening in a straightforward manner.

So in this problem, we're going to have to do a couple of extra steps to figure out what all the rates are that we're dealing with and what equation we need to use. Now a good first step would be to draw and label a picture of the scenario. Now we already have a picture drawn for us right here of a cube. Now we're told specifically this cube is growing at 2 cubic centimeters per second. So what that means is that the volume of this cube is expanding, and the rate at which this is expanding is given to us in this problem as 2.

Now something that I'll mention is this is a very common problem type where you have a shape that will either grow or shrink over time. Now in situations where you are dealing with a shape that is growing, what you want to do is make your rate positive. And since I can see here that this shape is growing, then we have a positive 2 cubic centimeters per second as our rate. If this cube were shrinking, we would need to make this value negative instead. But since it's growing, we'll keep it positive.

Now we're also told that at one instance, we're told that it has a side length of 0.8 centimeters. So we're asked for this moment of time where this is a 0.8 centimeter side length, and we want to know at this moment how fast one side of the cube here is growing. So this is what we're trying to figure out. Now this still might seem a bit confusing, but what I can do is extract the variables that we have in this problem. Since I can see here that we have cubic centimeters per second as our rate, that means we have the changing volume with respect to time.

We're told here that it's 2 cubic centimeters per second. Now another variable that we have is the side length at one instant. It's going to be 0.8 centimeters. So these are two variables that we can find from our problem, and what we're looking for is how fast the side of the cube is growing. So this rate would be the side length with respect to time, which we could say is dxdt.

And this is what we're ultimately looking for. Now looking at all my variables here, I've gone ahead and identified and labeled the diagram. So my next step would be to figure out the equations that I need that's going to relate all of the variables together. And the equation that I'm looking for, well, I see here that I have a changing volume with respect to time. I have a side length that I have a changing side length with respect to time.

So I think the equation that's going to do the trick is the volume of a cube. Because I know that the volume of a cube is just x3 where x is one side length of the cube. And what I can do to figure this out is all the steps that we've done already when solving related rates. Because we know from here, we just need to take the time derivative on both sides and then use implicit differentiation. And so let's go ahead and do that step, which we talked about in the previous video.

So I'm going to go ahead and take the time derivative on both sides of the equation, and that's literally just taking the time derivative of every variable that we see. So we're going to have the time derivative of volume on the left side of the equation, and then we'll have the time derivative of x3 on the right side of the equation. Now taking the derivative of volume with respect to time, that's just going to give you 1. Then we need to multiply this using the chain rule by the derivative of v with respect to t, which is going to give you dvdt. Now dvdt multiplied by 1 is just going to be this rate.

So I'm just going to write dvdt. And then on the right side, the derivative of x3, well, that's going to be 3x2 using the power rule here. But again, we need to use the chain rule since we're taking the derivative of x with respect to t, and doing that will give you dxdt. So this is going to be the other rate that we get out of this equation. Now notice something here.

Notice when I went ahead and took this time derivative here, notice we got all the variables that we need. I see that we have our dvdt. I see that we have our dxdt, and then I see that we have the x that was the one side length right here. So we know that we used the right equation, so we're in a good position here. Now our next step, since we know we did use the right equation, would be to isolate the target rate of change.

Now the target rate of change in this example, well, that's going to be this dxdt that we're trying to find right here. So if I want to isolate that on one side, well, what I'm going to go ahead and do is divide this 3x2 on both sides of the equation. So I'm just going to use some quick algebra here, and that's going to get the 3x2 to cancel on the right side. So doing this, I'm going to end up with just my rate dxdt on one side of the equal sign. And then on the other side, we're going to have this whole equation right here, which I'm going to write like this, well dvdt on top and that whole thing is going to be divided by 3x2.

So this is the equation. Now at this point, what I need to do for my last step is plug in all of the known variables and solve. Now I can see we have the dxdt solved for on one side of the equation. So I can see we have dvdt, which is given to us as 2 cubic centimeters per second. We already figured this out in the problem.

And I can divide that by 3x2, which in this case, x is 0.8 centimeters. So we have 2 divided by 3 times 0.8 squared. And if you do this on a calculator, all of this should come out to about 1.04. Now on your calculator, it's probably going to read as 1.0416666, and this is just going to keep repeating here. Let's go ahead and round this off.

I'm just going to say it's about 1.04. And since it's a length with respect to time, we're going to have centimeters per second. So this right here is going to be the missing rate and that is the solution to this problem. So as you can see, this process really is the same idea as what we did in the previous video. The main difference when dealing with these types of problems is notice in this situation, we had to first identify all the missing variables, and then we had to find our equation.

In the previous video, we had all of this information up here given to us, but here we need to extract it from the problem. But if you're able to draw out a diagram and figure out what exactly is happening, then doing this should be a pretty straightforward process to find exactly the equation that you need. So hope you found this video helpful, and let's go ahead and try getting some more practice.

A sphere is growing at a rate of $50\frac{\operatorname{\mathrm{cm}}^3}{s}$. At what rate is the radius of the sphere increasing when the radius is $5\operatorname{\mathrm{cm}}$?

• A. $1.57\operatorname{\frac{\mathrm{cm}}{s}}$
• B. $0.637\operatorname{\frac{\mathrm{cm}}{s}}$
• C. $0.159\frac{\operatorname{\mathrm{cm}}}{s}$
• D. $0.318\operatorname{\frac{cm}{s}}$

A right tringle has a base of $10\operatorname{\mathrm{cm}}$ and a height of $12\operatorname{\mathrm{cm}}$. The height of the right triangle is decreasing at a rate of $0.4\operatorname{\frac{\mathrm{cm}}{s}}$, at what rate is the area of the triangle decreasing?

• A. $-2\operatorname{.4\frac{cm^2}{s}}^{}$
• B. $-1\operatorname{\frac{cm^2}{s}}^{}$
• C. $-2\operatorname{\frac{cm^2}{s}}^{}$
• D. $-4\operatorname{\frac{cm^2}{s}}^{}$

The perimeter of a rectangle is fixed at $30\operatorname{\mathrm{cm}}$. If the length $L$ is increasing at a rate of $2\operatorname{\frac{\mathrm{cm}}{s}}$, for what value of $L$ does the area start to decrease? Hint: the rectangle's area starts to decrease when the rate of change for the area is less than 0.

• A. $L>15\operatorname{\mathrm{cm}}$
• B. $L>7.5\operatorname{\mathrm{cm}}$
• C. $L>0\operatorname{\mathrm{cm}}$
• D. $L>10\operatorname{cm}$

Real World Application

We've recently been spending a lot of time focusing on related rates problems. Recall for these problems that we see, we're dealing with situations where we have an equation or a shape that is changing over time. It turns out that a lot of these related rates problems that you're going to see are going to be real-world application problems, which, in my opinion, are some of the trickiest problems to solve. But don't sweat it because in this video, I'm going to walk you through an example of a real-world application problem and show you the process for solving these types of problems. I think you're going to find it's very similar to the process we've used for dealing with related rates up to this point.

So let's just go ahead and get right into things. Now in this example down here, this would be an example of a real work problem. We're told as an ice cube melts, each side changes at negative 3 centimeters per minute. Find the rate of change of the ice cube's volume when each side is 0.9 centimeters. The way that you can tell you're dealing with a real-world situation here is because we have this ice cube melting, and we're not really given an explicit shape or equation that we need to solve for, and we have to extract all that information ourselves.

For real-world problems, you need to determine what shapes are formed and how they change over time. The way that I'm going to do this is just by walking through the same steps that we've seen in the past with related rates, which is to first start by drawing and labeling a picture of the scenario. All I really know about this situation is that I'm dealing with an ice cube that is melting. But what I can figure out is, based on the sentence here, what information and variables I actually have. We're told each side changes at negative 3 centimeters per minute.

Now I want you to pay close attention to something. Notice how each side is changing at negative 3 centimeters per minute. We were told in previous videos when dealing with a positive sign, that means your shape is growing. When dealing with a a negative sign, your shape is shrinking. Since we have a negative sign here, that means that this entire ice cube is shrinking, which makes sense since it's a melting ice cube. I'm going to focus on one side of the ice cube even though all sides are the same. Just going to focus on this side for now since I know that all the sides are going to be melting uniformly.

This side is shrinking. It's becoming smaller at a rate of negative 3 centimeters for every minute. So that's one of the rates that we have. I'm also told that at one instant, this ice cube has each side equal to 0.9 centimeters. So I'll focus on another side of the ice cube here. Again, they're all the same. But I'll focus on another side here, which I'll say is 0.9 centimeters at one instant. Now what I'm trying to find is the rate at which the ice cube's volume changes.

And the way that I can do that is by recognizing since this shape is shrinking, the ice cube is melting, this entire volume is going to be shrinking, and that's the rate we're ultimately looking for. So what variables do we have in this situation? Well, I can see that we have the rate at which one side of the cube is changing. So that's going to be dxdt, which is negative 3 centimeters per minute. That's a changing side with respect to time. I can also see here that we have one instance where the side of the cube is 0.9 centimeters. So each side is going to be this length at one instant. And what we're ultimately trying to find is the rate at which the volume is changing the whole cube, which is dvdt. So that's what we don't have.

Now based on these variables here, I've gone ahead and done step 1, which is labeling the situation and finding these variables. So my second step is going to be to identify all the equations or identify all the equations we need that relate these variables. And if I'm thinking about the equations that are going to work, the one equation that comes to mind is the volume of a cube. So we can see that we have an ice cube. All of these have the same length, and it's uniformly melting. I can see we have the rate for one side changing, and we're looking for the rate at which the volume changes. So I'm pretty sure the volume of a cube, which I know is just going to be one side x3, I'm pretty sure this is going to give me what I'm looking for. But let's see if it does by doing this third step here, which is taking the derivative on both sides and using implicit differentiation.

So I'm going to take the derivative with respect to time of this entire equation. So that means I'll have the derivative with respect to time of the volume, and that's going to be equal to the derivative with respect to time of this term right here, x3. Now I can find the derivative of volume with respect to time, which we know using the chain rule is just going to come out to dvdt. And for finding the derivative of x3, well, using the power rule, I know that that's going to be 3x2. Then I need to multiply this by the derivative of x itself using the chain rule since x is the inside term, and that's just going to give us dxdt multiplied on the outside. So this right here is the equation that we need. An

A 15-foot plank leans against a vertical pole. The top of the plank begins to slide down the pole at a steady speed of 2 inches per second. How fast is the bottom of the plank moving away from the pole when it is 8 feet away from the base of the pole (in inches per second)?

• A. $dx/dt=1.58\frac{in}{s}$
• B. $dx/dt=3.46\frac{in}{s}$
• C. $dx/dt=5.04\frac{in}{s}$
• D. $dx/dt=3.17\frac{in}{s}$

Two cars leave the same intersection and drive in perpendicular directions. Car A travels east at a speed of $60\frac{mi}{hr}$, Car B travels north at a speed of $40\frac{mi}{hr}$. Car A leaves the intersection at $2pm$, while Car B leaves at $2:30pm$. Determine the rate at which the distance between the two cars is changing at $3pm$.

• A. $69.62\frac{mi}{hr}$
• B. $1.58\frac{mi}{hr}$
• C. $18.99\frac{mi}{hr}$
• D. $50.41\frac{mi}{hr}$