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# Analyzing problems involving related rates

Related rates problems are word problems where we reason about the rate of change of a quantity by using information we have about the rate of change of another quantity that's related to it. Let's get acquainted with this sort of problem.
Related rates problems are applied problems where we find the rate at which one quantity is changing by relating it to other quantities whose rates are known.

## Worked example of solving a related rates problem

Imagine we are given the following problem:
The radius $r\left(t\right)$ of a circle is increasing at a rate of $3$ centimeters per second. At a certain instant ${t}_{0}$, the radius is $8$ centimeters.
What is the rate of change of the area $A\left(t\right)$ of the circle at that instant?

#### Making sense of the quantities and their rates

In general, we are dealing here with a circle whose size is changing over time. There are two quantities referenced in the problem:
$r\left(t\right)$ is the radius of the circle after $t$ seconds. It is measured in centimeters.
$A\left(t\right)$ is the area of the circle after $t$ seconds. It is measured in square centimeters.
The problem also refers to the rates of those quantities. The rate of change of each quantity is given by its derivative:
${r}^{\prime }\left(t\right)$ is the instantaneous rate at which the radius changes at time $t$. It is measured in centimeters per second.
${A}^{\prime }\left(t\right)$ is the instantaneous rate at which the area changes at time $t$. It is measured in square centimeters per second.

#### Making sense of the given information

We are given that the radius is increasing at a rate of $3$ centimeters per second. This means that ${r}^{\prime }\left(t\right)=3$ for any value of $t$.
We are also given that at a certain instant ${t}_{0}$, the radius is $8$ centimeters. This means that $r\left({t}_{0}\right)=8$. Notice that this is only the case for ${t}_{0}$, and not for just any value of $t$.
Finally, we are asked to find the rate of change of $A\left(t\right)$ at the instant ${t}_{0}$. Mathematically, we are looking for ${A}^{\prime }\left({t}_{0}\right)$.

#### Relating the area and the radius

After we've made sense of the relevant quantities, we should look for an equation, or a formula, that relates them. The quantities in our case are the area and the radius of a circle. These quantities are related using the formula for the area of a circle:
$A=\pi {r}^{2}$

#### Differentiating

To find ${A}^{\prime }\left({t}_{0}\right)$ we need to take the derivative of each side of the equation. Once we do that, we will be able to relate ${A}^{\prime }\left({t}_{0}\right)$ with other known values, like ${r}^{\prime }\left({t}_{0}\right)$, which will allow us to solve for ${A}^{\prime }\left({t}_{0}\right)$.
Since we don't have the explicit formulas for $A\left(t\right)$ and $r\left(t\right)$, we will use implicit differentiation:
$\begin{array}{rl}A\left(t\right)& =\pi \left[r\left(t\right){\right]}^{2}\\ \\ \frac{d}{dt}\left[A\left(t\right)\right]& =\frac{d}{dt}\left[\pi \left[r\left(t\right){\right]}^{2}\right]\\ \\ {A}^{\prime }\left(t\right)& =2\pi r\left(t\right){r}^{\prime }\left(t\right)\end{array}$
This is the core of our solution: by relating the quantities (i.e. $A$ and $r$) we were able to relate their rates (i.e. ${A}^{\prime }$ and ${r}^{\prime }$) through differentiation. This is why these problems are called "related rates"!

#### Solving

Note that the equation we got is true for any value of $t$ and specifically for ${t}_{0}$. We can substitute $r\left({t}_{0}\right)=8$ and ${r}^{\prime }\left({t}_{0}\right)=3$ into that equation:
$\begin{array}{rl}{A}^{\prime }\left({t}_{0}\right)& =2\pi r\left({t}_{0}\right){r}^{\prime }\left({t}_{0}\right)\\ \\ & =2\pi \left(8\right)\left(3\right)\\ \\ & =48\pi \end{array}$
In conclusion, we found that at ${t}_{0}$, the area is increasing at a rate of $48\pi$ square centimeters per second.
Problem 1.A
Problem set 1 will walk you through the steps of analyzing the following problem:
The base $b\left(t\right)$ of a triangle is decreasing at a rate of $13$ $\text{m/h}$ and the height $h\left(t\right)$ of the triangle is increasing at a rate of $6$ $\text{m/h}$. At a certain instant ${t}_{0}$, the base is $5$ $\text{m}$ and the height is $1$ $\text{m}$. What is the rate of change of the area $A\left(t\right)$ of the triangle at that instant?
Match each expression with its units.
$\text{m}$‍ ${b}^{\prime }\left(t\right)$‍ $A\left({t}_{0}\right)$‍ $h\left({t}_{0}\right)$‍ $\frac{dA}{dt}$‍

Want more practice? Try this exercise.

### Common mistake: Confusing which expressions are variables and which are constants

As you've seen, related rates problems involve multiple expressions. Some represent quantities and some represent their rates. Some are changing, some are constants.
It's important to make sure you understand the meaning of all expressions and are able to assign their appropriate values (when given).
We recommend performing an analysis similar to those shown in the example and in Problem set 1: what are all the relevant quantities? What are their rates? What are their units? What are their values?
Problem 2
Consider this problem:
Two cars are driving towards an intersection from perpendicular directions. The first car's velocity is and the second car's velocity is . At a certain instant ${t}_{0}$, the first car is a distance $x\left({t}_{0}\right)$ of from the intersection and the second car is a distance $y\left({t}_{0}\right)$ of from the intersection. What is the rate of change of the distance $d\left(t\right)$ between the cars at that instant?
Which equation should be used to solve the problem?

### Common mistake: Selecting an equation that misrepresents the given problem

As you've seen, the equation that relates all the quantities plays a crucial role in the solution of the problem. It's usually helpful to have some kind of diagram that describes the situation with all the relevant quantities. Let's take Problem 2 for example. The problem describes a right triangle.
The diagram makes it clearer that the equation we're looking for relates all three sides of the triangle, which can be done using the Pythagoream theorem:
$\left[d\left(t\right){\right]}^{2}=\left[x\left(t\right){\right]}^{2}+\left[y\left(t\right){\right]}^{2}$
Without the diagram, we might accidentally treat $d\left(t\right)$ as if it's the triangle's area...
$d\left(t\right)=\frac{x\left(t\right)\cdot y\left(t\right)}{2}$
...or treat $x\left(t\right)$, $y\left(t\right)$, and $d\left(t\right)$ as if they are the three angles of the triangle...
$d\left(t\right)+x\left(t\right)+y\left(t\right)=180$
...or maybe treat $d\left(t\right)$ as if it's an angle and form some trig equation
$\mathrm{tan}\left[d\left(t\right)\right]=\frac{y\left(t\right)}{x\left(t\right)}$.
All of these equations might be useful in other related rates problems, but not in the one from Problem 2.
Problem 3
Consider this problem:
A $20$-meter ladder is leaning against a wall. The distance $x\left(t\right)$ between the bottom of the ladder and the wall is increasing at a rate of $3$ meters per minute. At a certain instant ${t}_{0}$, the top of the ladder is a distance $y\left({t}_{0}\right)$ of $15$ meters from the ground. What is the rate of change of the angle $\theta \left(t\right)$ between the ground and the ladder at that instant?
Which equation should be used to solve the problem?

Want more practice? Try this exercise.

## Want to join the conversation?

• For Problems 2 and 3: Correct me if I'm wrong, but what you're really asking is, "Which type of equation (below) should be used as part of the process of solving the problem?"

The problem had me confused for several minutes. The first time I saw it, I was perplexed and thought, "None of these equations is the right one because there are no derivatives in any of them."

After all, you can plug into the equation the numbers you've been given but you're still several steps short of having the solution.

However, what you're asking is which of the equations provided forms the proper structure by which you can calculate the correct answer. Am I understanding that right? • A 20-meter ladder is leaning against a wall. The distance x(t), between the bottom of the ladder and the wall is increasing at a rate of 3 meters per minute. At a certain instant t0 the top of the ladder is y0, 15m from the ground. What is rate of change of the angle between ground and ladder.

Whouldnt you use Sin instead of Cos? You have the opposite - 15m, you have the hyp -20m, you need to find the angle, which should be opp/ hyp? • If rate of change of the radius over time is true for every value of time. If radius changes to 17, then does the new radius affect the rate? • Hello, can you help me with this question, when we relate the rate of change of radius of sphere to its rate of change of volume, why is the rate of volume change not constant but the rate of change of radius is...? Is it because they aren’t proportional to each other ? Kinda urgent ..thanks • It's because rate of volume change doesn't depend only on rate of change of radius, it also depends on the instantaneous radius of the sphere. We know that volume of a sphere is (4/3)(pi)(r)^3.
Now if we differentiate volume with respect to time, we will get the rate of change of volume. So,
dV/dt = d/dt((4/3)(pi)(r)^3)
applying chain rule(i.e. differentiating first wrt r and then wrt t. Remember constant can be taken out.
dV/dt = 4/3*pi(d/dt(r^3))
= 4/3*pi*(3r^2)(dr/dt)
Therefore rate of change of volume of a sphere with constant rate of change of radius(dr/dt) (let's call it C for constant) is
dV/dt = (4piC)(r)^2 where r is constantly changing and hence volume rate isn't constant.
• Heello, for the implicit differentation of A(t)'=d/dt[pi(r(t)^2)]. I undertsand why the result was 2piR but where did you get the dr/dt come from, thank you. • The dr/dt part comes from the chain rule. 2pi*r was the result of differentiating the right side with respect to r. But we need to differentiate both sides with respect to t (not r). Think of it as essentially we are multiplying both sides of the equation by d/dt

So to do that for the right side, you'd need to first take the derivative with respect to r d/dr (which you did to get 2pi*r, but then that result has to be multiplied by dr/dt so that the END result would be you finding out what d/dt is (differentiating that side with respect to time).

If you looked at just the derivative operators with no numbers or symbols there for the right side, it may be more clear. d/dr * dr/dt... The dr terms cancel out, and you're left with d/dt. And both sides of the equation have to be multiplied by the same thing (which was d/dt). So it ends up being equal to what you did on the left side to get dA/dt.

Hope that helps.
• Is there a more intuitive way to determine which formula to use? • for the 2nd problem, you could also use the following equation, d(t)=sqrt ((x^2)+(y^2)), and take the derivate of both sides to solve the problem.

it then seems like the unit of dd(t)/dt (the rate of change of the distance) is km^3/h? how do we make sense of this unit of the rate of change of the distance? Thanks!
(1 vote) • True, but here, we aren't concerned about how to solve it. We're only seeing the setup. But yeah, that's how you'd solve it.

(Also, a small tip from my side. When you have a square on one side, don't take it to the other side and make it a square root. Sometimes, keeping it as a square makes stuff much better, both in terms of solving and the amount of writing. I've shown it later on where I take the derivative of s^2 instead of just s)

For the other question, I'm not quite sure how you ended up with those units. I've done a dimensional analysis below (I'll just take d(t) = s to avoid confusion between distance and the differential operator)

So, we have s^2 = x^2 + y^2

Differentiating both sides w.r.t. time, we have

d(s^2)/dt = d(x^2)/dt + d(y^2)/dt

2s(ds/dt) = 2x(dx/dt) + 2y(dy/dt)

Now, dx/dt and dy/dt are instantaneous rates of change of position in the x and y directions respectively. So, I'll call them v_x and v_y. So,

2s(ds/dt) = 2x(v_x) + 2y(v_y)

Simplifying a bit, we have

s(ds/dt) = x(v_x) + y(v_y)

Solving for ds/dt, we have

(ds/dt) = (x(v_x) + y(v_y))/s

Now, s = sqrt(x^2 + y^2). Using this substitution, we have,

(ds/dt) = (x(v_x) + y(v_y))/(sqrt(x^2 + y^2))

Now, let's see the dimensions for this. Both x(v_x) and y(v_y) have the dimensions [L^2T^(-1)]. And the denominator has the dimensions of [L]. So, the dimensions of ds/dt would be [L^2T^(-1)]/[L] = [LT^(-1)]. As the length is in km and time in hours, the units would be km/hr.   