# How to solve related rates in calculus

// Last Updated: February 22, 2021 – Watch Video //

How do you solve related rates?

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

And that’s just what you’re going to learn how to do in today’s calculus lesson.

## What Are Related Rates (Real Life Examples)

Have you ever watched a baseball player who is rounding third and heading for home and wondered if they had enough speed to make it before getting tagged out by the thrower?

Or have you ever watched a basketball player shoot a free-throw and speculate if the ball has enough height and distance?

Or perhaps you’ve listened to a guitar solo and contemplated the number of vibrations per second needed to make the guitar strings hum at the perfect pitch?

If you have, and even if you haven’t, all of these queries have something in common — something is changing with respect to time.

The baseball player’s distance to the home plate is changing with respect to the runner’s speed per second. The success of a free-throw is related to the ball’s projectile motion and the instantaneous rate of change of the height and distance traveled. And when a guitar string is plucked, the rate of the guitar string’s vibration (frequency) produces high or low pitches, which make the music we hear sound pleasing.

## What Does It Mean If Two Rates Are Related

Let’s make sense of things using the image to the right.

Imagine a person is outside looking up into the sky and they spot an airplane that is flying at an altitude of 6 miles above the ground. Now, as the plane continues on its flight path, several things are changing with respect to each other.

The angle of elevation (theta), the line of sight (hypotenuse), as well as the horizontal distance are all changing as the plane flies overhead and with respect to time.

And that’s what it means for two or more rates to be related — as one rate changes, so does the other.

Calculate the Speed of an Airplane

## How To Solve Related Rates Problems

We use the principles of problem-solving when solving related rates. The steps are as follows:

1. Read the problem carefully and write down all the given information.
2. Sketch and label a graph or diagram, if applicable.
3. Find an equation that relates the unknown variable and known variable(s) by looking for geometric shapes, known formulas, ratios such as the Pythagorean theorem, area and volume formulas, or trig identities.
4. Simplify using appropriate substitutions, so that chosen equation has only two variables (known and unknown).
5. Differentiate the equation implicitly with respect to time.
6. Substitute all known values into the derivative and solve for the final answer.

### Ex) Cone Filling With Water

Alright, so now let’s put these problem-solving steps into practice by looking at a question that frequently appears in AP calculus, college, and university classes — the cone problem.

If water is being pumped into the tank at a rate of 3 cubic feet per second, find the rate at which the water level is rising when the water is 4ft deep.

Related Rates – How Things Change Over Time

#### Steps 1 & 2: Read and Sketch

First, we will sketch and label a cone, vertex down, and identify all pertinent information.

Related Rates — Cone Problem

#### Step 3: Find An Equation That Relates The Unknown Variables

Because we were given the rate of change of the volume as well as the height of the cone, the equation that relates both V and h is the formula for the volume of a cone.

But here’s where it can get tricky. Our equation has three variables (V, r, and h), but we only have two derivatives, dh, and dV.

Hmmm, that means we have to reduce the number of variables so that the number of variables equals the number of derivatives.

How does that work?

#### Step 4: Simplify To Get Known & Unknown Variables

We use an incredibly useful ratio found by the similar triangles created from the cone above (HINT: this ratio will be used quite often when solving related rate problems). And in so doing, we will also create a proportion, using the conical tank’s original dimensions, and solve for r. This way, we can eliminate the r in volume formula.

#### Step 5: Implicit Differentiation

We will now use implicit differentiation on both sides with respect to t.

#### Step 6: Substitute Back In

And lastly, we will substitute our given information and solve the unknown rate, dh/dt.

See, all we have to do is follow the steps and arrive at our answer!

And don’t worry, we will do two more cone problems in the video below, so you’ll become a master at these questions in no time!

Let’s get after it!

## Video Tutorial w/ Full Lesson & Detailed Examples (Video)

• Overview of Related Rates + Tips to Solve Them
• 00:02:58 – Increasing Area of a Circle
• 00:12:30 – Expanding Volume of a Sphere
• 00:21:15 – Expanding Volume of a Cube
• 00:26:32 – Calculate the Speed of an Airplane
• 00:39:13 – Conical Sand Pile
• 00:51:19 – Conical Water Tank
• 00:59:59 – Boat & Winch
• 01:09:13 – Ladder Sliding Down A Wall

## Related Rates

Related Rates – 2 Examples

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Related rates problems require us to find the rate of change of one value, given the rate of change of a related value. We must find an equation that associates the two values and apply the chain rule to differentiate each side of the equation with respect to time.. Suppose we have a variable v. The derivative, dv/dt would be the rate of change of v.

When solving related rates problems, we should follow the steps listed below.

1) Draw a diagram. This is the most helpful step in related rates problems. It allows us to visualize the problem.

2) Assign variables to each quantitiy in the problem that is a function of time. Each of these values will have some rate of change over time.

3) List all information that is given in the problem and the rate of change that we are trying to find.

4) Write an equation that associates the variables with one another. If there are variables for which we are not given the rates of change (except for the rate of change that we are trying to determine), we must find some relation from the nature of the question that allows us to write these variables in terms of variables for which the rates of change are given. We must then substitute these relations into the main equation.

Note: For an example of this situation, see example #3 below.

5) Using the chain rule, differentiate each side of the equation with respect to time.

6) Substitute all given information into the equation and solve for the required rate of change.

Note: It is important to wait until the equation has been differentiated to substitute information into the equation. If values are substituted too early, it can lead to an incorrect answer.

A function f has an absolute maximum at c if f(c) ³ f(x), for all x in the domain of f. At x=c, the graph reaches its highest point. The number f(c) is called the maximum value of f.

A function f has an absolute minimum at c if f(c) £ f(x), for all x in the domain of f. At x=c, the graph reaches its lowest point. The number f(c) is called the minimum value of f.

Together, the maximum and minimum values are called the extreme values of the function f.

A function f(x), defined on the open interval (a, b) has a local maximum at a point c in (a, b), if f(c) ³ f(x) for all x near c. This means that there is an interval around c (possibly very small), such that f(c) ³ f(x) for all x in the interval.

A function f(x), defined on the open interval (a, b) has a local minimum at a point c in (a, b), if f(c) £ f(x) for all x near c. This means that there is an interval around c (possibly very small), such that f(c) £ f(x) for all x in the interval.

Suppose we have a function defined on a closed interval [c, d]. A local maximum or minimum can not occur at the endpoints of this interval because the definition requires that the point is contained in some open interval (a, b). Since the function is not defined for some open interval around either c or d, a local maximum or local minimum cannot occur at this point. An absolute maximum or minimum can occur, however, because the definition requires that the point simply be in the domain of the function.

If f has a local maximum or minimum at c and f'(c) exists, then f'(c) = 0.

The converse of this theorem is not true. If f'(c) = 0, then c is not necessarily a maxiumum or minimum value. There may also be a maximum or minimum value when f'(c) does not exist.

A critical number of a function f is a number c in the domain of f, such that f'(c)=0 or f'(c) does not exist.

Using the definition of a critical number, we can rephrase Fermat’s Theorem as : If f has a local maximum or minimum at c, then c is a critical number of f.

To find the absolute maximum or minumum values of a continuous function f on a closed interval [a, b], we first find the critical numbers of the function in (a, b) and calculate the value of the function at each critical number. Next we find the values of the function at the endpoints of the interval. We then compare all of these values. The largest value is the absolute maximum of the function on the interval [a, b], while the smallest value is the absolute minimum.

The concept of maximum and minimum values allows us to solve optimization problems. These problems are one of the most practical applications of differential calculus. They allow us to find the optimal way to perform some task.

To solve optimization problems, we follow the steps listed below

1) Draw a diagram, if necessary, to help visualize the problem.

2) Assign variables to the quantity to be optimized and all other unknown quantities given in the question.

3) Write an equation that associates the optimal quantity to the other variables. If the optimal quantity is expressed in terms of more than one variable, we must eliminate the extra variables. We use the nature of the question to find some relation between the variables and substitute these relations into the equation for the optimal quantity. The optimal quantity equation should be in terms of only one variable so that it has the form f(x).

4) Find the absolute maximum or minimum of f(x), depending on the question. If the domain of f is closed, use the closed interval method.

For more practice with the concepts covered in this tutorial, visit the Related Rates and Optimization Problems page at the link below. The solutions to the problems will be posted after these chapters are covered in your calculus course.

To test your knowledge of these application problems, try taking the general related rates and optimization test on the iLrn website or the advanced related rates and optimization test at the link below.

Please forward any questions, comments, or problems you have experienced with this website to Alex Karassev.

1. In the following assume that $$x$$ and $$y$$ are both functions of $$t$$. Given $$x = – 2$$, $$y = 1$$ and $$x’ = – 4$$ determine $$y’$$ for the following equation. $6 + = 2 – <<\bf>^<4 - 4y>>$ Solution
2. In the following assume that $$x$$, $$y$$ and $$z$$ are all functions of $$t$$. Given $$x = 4$$, $$y = – 2$$, $$z = 1$$, $$x’ = 9$$ and $$y’ = – 3$$ determine $$z’$$ for the following equation. $x\left( <1 - y>\right) + 5 = + – 3$ Solution
3. For a certain rectangle the length of one side is always three times the length of the other side.

1. If the shorter side is decreasing at a rate of 2 inches/minute at what rate is the longer side decreasing?
2. At what rate is the enclosed area decreasing when the shorter side is 6 inches long and is decreasing at a rate of 2 inches/minute?

Solution

• A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of the sheet is decreasing at a rate of 0.5 m 2 /sec at what rate is the radius decreasing when the area of the sheet is 12 m 2 ? Solution
• A person is standing 350 feet away from a model rocket that is fired straight up into the air at a rate of 15 ft/sec. At what rate is the distance between the person and the rocket increasing (a) 20 seconds after liftoff? (b) 1 minute after liftoff? Solution
• A plane is 750 meters in the air flying parallel to the ground at a speed of 100 m/s and is initially 2.5 kilometers away from a radar station. At what rate is the distance between the plane and the radar station changing (a) initially and (b) 30 seconds after it passes over the radar station?

## 4 Steps to Solve Any Related Rates Problem – Part 2

In our last post, we developed four steps to solve any related rates problem.

Most frequently (> 80% of the time) you will use the Pythagorean theorem or similar triangles.

• Take the derivative with respect to time of both sides of your equation. Remember the Chain Rule.
• Solve for the quantity you’re after.
• We introduced three examples to illustrate the basic ideas, and solved two of them there.
As promised, we’ll solve the third here.

### Water Leaving a Cone Example

Here’s the problem statement, now with some additional details about the cone itself and the moment we’re interested in:

Water in a Cone Example. Given: An inverted cone is 20 cm tall, has an opening radius of 8 cm, and was initially full of water. It is now being drained of water at the constant rate of 15 cm$^3$ each second. The water’s surface level falls as a result. Question: At what rate is the water level falling when the water is halfway down the cone? (Note: The volume of a cone is $\dfrac<1><3>\pi r^<2>h$. You may leave $\pi$ in your answer; do not use a calculator to find a decimal answer.)

Let’s use our Problem Solving Strategy to answer the question.

1. Draw a picture of the physical situation.
See the figure.

We are given that the volume of water in the cup is decreasing at the rate of 15 cm$^3$/s, so $\dfrac = -15\, \tfrac<\text^3><\text>$. Remember that we have to insert that negative sign “by hand” since the water’s volume is decreasing.

2. Write an equation that relates the quantities of interest.
A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
The height of the water changes as time passes, so we’re going to keep that height as a variable, h.

B. To develop your equation, you will probably use . . . similar triangles.
We have a relation between the volume of water in the cup at any moment, and the water’s current height, h:

Notice that this relation expresses the water’s volume as the function of two variables, r and h. We can only take the derivative with respect to one variable, so we need to eliminate one of those two. Since the question asks us to find the rate at which the water is falling when its at a particular height, let’s keep h and eliminate r as a variable using similar triangles.

Begin subproblem to eliminate r as a variable.

The figure is the same as in Step 1, but with the rest of the cone removed for clarity. Note that there are two triangles, a small one inside a larger one. Because these are similar triangles, the ratio of the base of the small triangle to that of the big triangle $\left(\dfrac <8>\right)$ must equal the ratio of the height of the small triangle to that of the big triangle $\left(\dfrac <20>\right)$:
\begin
\frac <8>&= \frac <20>\\[8px] r &= \frac<8> <20>h \\[8px] &= \frac<2> <5>h
\end
End subproblem.

Then substituting the expression for r into our relation for V:
\begin
V &= \frac<1> <3>\pi \left(\frac<2> <5>h \right)^2h \\ \\
&= \frac<1><3>\frac<4> <25>\pi h^3 \\ \\
&= \frac<4> <75>\pi h^3
\end
3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
\begin
\frac

&= \frac

\left(\frac<4> <75>\pi h^3 \right) \\ \\
&= \frac<4> <75>\pi \frac

\left(h^3 \right) \\ \\
&= \frac<4> <75>\pi \left(3h^2 \frac

\right) \\ \\
&= \frac<4> <25>\pi h^2 \frac

\end

While the derivative of $h^3$ with respect to h is $\dfrach^3 = 3h^2$, the derivative of $h^3$ with respect to time t is $\dfrac h^3 = 3h^2\dfrac$.

Remember that h is a function of time t: the water’s height decreases as time passes. We could have captured this time-dependence explicitly by always writing the water’s height as h(t), and then explicitly showing the water’s volume in the cone as a function of time as
$$V(t) = \frac<4> <75>\pi [h(t)]^3$$
Then when we take the derivative,
\begin
\frac

&= \frac<4> <75>\pi \frac

[h(t)]^3 \\ \\
&= \frac<4><75>\pi\,3[h(t)]^2 \frac

h(t)\\ \\
&= \frac<4><25>\pi [h(t)]^2 \frac

\end

[Recall that we’re looking for $\dfrac$ in this problem.]

Most people find that writing the explicit time-dependence V(t) and h(t) annoying, and so just write V and h instead. Regardless, you must remember that h depends on t, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac$ term.

4. Solve for the quantity you’re after.
At this point we’re just substituting values. We have $\dfrac = -15 \, \tfrac<\text^3><\text>$, and want to find $\dfrac$ at the instant when h = 10 cm.
Starting from our last expression above:

\begin
\frac

&= \frac<4> <25>\pi h^2 \frac

\\ \\
\frac

&= \frac<25> <4\pi h^2>\frac

\\ \\
&= \frac<25> <4\pi (10)^2>(-15) \\ \\
&= \frac<25> <4\pi (100)>(-15) \\ \\
&= -\frac<15> <16\pi>\text < cm/s>\quad \cmark
\end
The negative value indicates that the water’s height h is decreasing, which is correct.

Notice how our “Four Steps to Solve Any Related Rates Problem” led us straightforwardly to the solution. This is the strategy we use time and again; you can too.

### Quick links: Related Rates Problems & Solutions

• Snowball melts
• Rate angle changes as a ladder slides away from a house
• Water level falls as it drains from a cone
• Lamp post casts a shadow of a man walking
• Given an equation with x and y, find a rate

### Time to practice

Of course just reading our solution, or watching someone else solve problems, won’t really help you get better at solving calculus problems. Instead you need to practice for yourself, pencil in your hand, so you can get stuck and make mistakes and do all the other things people do when they’re learning something new. (And ideally do all those things before you’re taking an exam!) We have lots of problems for you to use, each with a complete step-by-step solution.

• What tips do you have to share about solving Related Rates problems?
• Or what questions do you have?
• Or how can we make posts such as this one more useful to you?

I am trying to solve a related rates problem. The problem states:

If $y = 4x -x^3$ and the x-coordinate is increasing at the rate of 1/3 unit/sec. How fast is the slope of the graph changing at the instant when $x = 2$?

I have done this:

Let $\frac = \frac <1><3>$ unit/sec

I derive the formula: $\frac = 4 \frac – 3x^2 \frac$

I substitute to solve for $dy \over dt$.

and then I solve for the slope as (dy/dt)/(dx/dt) = (-8/3)/3 which is -8 unit/sec

But the answer is supposed to be -4 units/sec

What am I doing wrong?

Hint: How fast the slope is changing is $\frac$ where $w=\frac$. So you will be differentiating again.

Added: The question does not ask for the slope, it asks for the rate of change of the slope. We have $\frac=4-3x^2$, so the slope is $4-3x^2$.

For the rate of change of this, differentiate $4-3x^2$ with respect to $t$. We get that $$\frac \left(\frac\right)=-6x\frac .$$ Since $\frac =\frac<1><3>$, when $x=2$ we have $$\frac \left(\frac\right)=(-6)(2)(1/3)=-4.$$

A slightly different explanation that some may follow more easily:

The trick here is reading the problem closely. It asks for the rate of change of the slope (which is itself a rate of change). The derivative of the function will give you the slope. The second derivative will give you the rate of change of the slope (the slope of the slope).

Since the question is asking about the change in the slope of the equation $y = 4x-x^3$, the first order of business would be to get the equation for the slope (which is the derivative, $y^<'>$ or $\frac$): $$y = 4x-x^3$$ $$\frac = 4 – 3x^2$$

This is the equation for the slope of the function. So we don’t let confusing notation get in the way, let’s just call this equation for slope $s$. So, $s = 4 – 3x^2$.

Now we want to know how the slope is changing with time. We know that $x$ is increasing at a rate of $1 \over 3$ units/sec and we want to know how the slope changes with small changes in $x$. In other words, we were given $\frac$ and we want to find $\frac$. So, we take the derivative of $s$ with respect to time:

Substituting known values, $$\frac = -6 \cdot 2 \cdot \frac <1> <3>= -4$$

## Steps we use to solve a related rates problem

Related Rates are an application of implicit differentiation, and are usually easy to spot.

They ask you to find how quickly one variable is changing when you know how quickly another variable is changing.

To solve a related rates problem, complete the following steps:

Construct an equation containing all the relevant variables.

Differentiate the entire equation with respect to (time), before plugging in any of the values you know.

Plug in all the values you know, leaving only the one you’re solving for.

## Finding rate of change of the radius, given rate of change of volume

Example

How fast is the radius of a spherical balloon increasing when the radius is . 100. cm, if air is being pumped into it at . 400. cm. ^3. /s?

In this example, we’re asked to find the rate of change of the radius, given the rate of change of the volume.

The formula that relates the volume and radius of a sphere to one another is simply the formula for the volume of a sphere.

Before doing anything else, we use implicit differentiation to differentiate both sides with respect to time . t.

They ask you to find how quickly one variable is changing when you know how quickly another variable is changing.

Now we plug in everything that we know. Keep in mind that . dV/dt. is the rate at which the volume is changing, . dr/dt. is the rate at which the radius is changing, and . r. is the length of the radius at a specific moment.

Our problem tells us that the rate of change of the volume is . 400. and that the length of the radius at the specific moment we’re interested in is . 100.

Solving for . dr/dt. gives

Therefore, we know that the radius of the balloon is increasing at a rate of . 1/100\pi. cm per second.

The derivative can be used to determine the rate of change of one variable with respect to another. A few examples are population growth rates, production rates, water flow rates, velocity, and acceleration.

Rates of change with respect to some other quantity in our daily life are given below :

1) Slope is the rate of change in vertical length with respect to horizontal length.

2) Velocity is the rate of displacement with respect to time.

3) Acceleration is the rate of change in velocity with respect to time.

4) The steepness of a hillside is the rate of change in its elevation with respect to linear distance

The average rate of change in an interval [a, b] is

whereas, the instantaneous rate of change at a point x is fвЂІ(x) for the given function

## Examples

A particles moves along a line according to the law s(t) = 2t 3 вЂ“ 9t 2 +12t вЂ“ 4, where t 0.

(i) At what time the particle changes direction?

(ii) Find the total distance travelled by the particle in the first 4 seconds.

(iii) Find the particleвЂ™s acceleration each time the velocity is zero.

(i) The particle changes direction when velocity changes its sign.

Now, v(t) = = 6t 2 вЂ“ 18t + 12 = 0 gives t = 1 and t = 2.

If 1 2 then both (t – 1) 0 and (6t – 12) 0, then

If t > 2 then both (t – 1) and (6t – 12) > 0, then

So, the particle changes direction when t = 1 and t = 2.

(ii) Given s(t) = 2t 3 вЂ“ 9t 2 +12t вЂ“ 4 —(1),

s(0) = 2(0) 3 – 9(0) 2 + 12(0) вЂ“ 4 = -4

s(1) = 2(1) 3 – 9(1) 2 + 12(1) вЂ“ 4 = 1

s(2) = 2(2) 3 – 9(2) 2 + 12(2) вЂ“ 4 = 0

s(3) = 2(3) 3 – 9(3) 2 + 12(3) вЂ“ 4 = 5

s(4) = 2(4) 3 – 9(4) 2 + 12(4) вЂ“ 4 = 28

The total distance traveled by the particle in the first 4 seconds,

|s(1) – s(0)| + |s(2) – s(1)| + |s(3) – s(2)| + |s(4) – s(3)|

= |1 – (-4)| + |0 – 1| + |5 – 0| + |28 – 5|

So, the total distance traveled by the particle in the first 4 seconds is 34 m

(iii)The acceleration is given by a(t) = 12t вЂ“ 18.

Velocity is zero, when t = 1 and t = 2.

Therefore, the acceleration when t = 1 sec is -6 m/sec 2 and the acceleration when t = 2 sec is 6 m/sec 2 .

If the volume of a cube of side length x is v = x 3 . Find the rate of change of the volume with respect to x when x = 5 units.

The rate of change of the volume with respect to length x is 3x 2 .

The rate of change of the volume with respect to x, when x = 5 units is

If the mass m(x) (in kilogram) of a thin rod of length x (in meters) is given by, m(x) = в€љ(3x) then what is the rate of change of mass with respect to the length when it is x=3 and x = 27 meters.

The rate of change of mass with respect to the length is dm/dx = в€љ3/ в€љ(2x)

When x = 3, dm/dx = (1/2) kg/m.

When x = 27, dm/dx = (1/6) kg/m.

A stone is dropped into a pond causing ripples in the form of concentric circles. The radius r of the outer ripple is increasing at a constant rate at 2 cm per second. When the radius is 5 cm find the rate of changing of the total area of the disturbed water?

Given = 2 cm/sec. Total area of a circle is a = ПЂ r 2 . The rate of change of the total area = 2 ПЂ r ((dr/dt).

The rate of changing of the total area when the radius is 5 cm

So, the rate of changing of the total area of the disturbed water when the radius is 5 cm is 20 ПЂ sq.cm/sec.

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