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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:
- Read the problem carefully and write down all the given information.
- Sketch and label a graph or diagram, if applicable.
- 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.
- Simplify using appropriate substitutions, so that chosen equation has only two variables (known and unknown).
- Differentiate the equation implicitly with respect to time.
- 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)
Ladder Sliding Down Wall
- 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 – 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.
- 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
- 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
- For a certain rectangle the length of one side is always three times the length of the other side.
- If the shorter side is decreasing at a rate of 2 inches/minute at what rate is the longer side decreasing?
- 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?
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.
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
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
Then substituting the expression for r into our relation for V:
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
3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
&= \frac<4> <75>\pi \frac
&= \frac<4> <75>\pi \left(3h^2 \frac
&= \frac<4> <25>\pi h^2 \frac
While the derivative of $h^3$ with respect to h is $\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,
&= \frac<4><75>\pi\,3[h(t)]^2 \frac
&= \frac<4><25>\pi [h(t)]^2 \frac
[Recall that we’re looking for $\dfrac
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
4. Solve for the quantity you’re after.
At this point we’re just substituting values. We have $\dfrac
Starting from our last expression above:
&= \frac<25> <4\pi (10)^2>(-15) \\ \\
&= \frac<25> <4\pi (100)>(-15) \\ \\
&= -\frac<15> <16\pi>\text < cm/s>\quad \cmark
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
- Rate ladder’s top falls as ladder slides away from house
- 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.
For more example problems with complete solutions, please visit our free Related Rates page!
We’d love your comments:
- 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:
I derive the formula: $\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?
2 Answers 2
Hint: How fast the slope is changing is $\frac
Added: The question does not ask for the slope, it asks for the rate of change of the slope. We have $\frac
For the rate of change of this, differentiate $4-3x^2$ with respect to $t$. We get that $$\frac
A slightly different explanation that some may follow more easily:
The trick here is reading the problem closely. It asks for the rate of change of the slope (which is itself a rate of change). The derivative of the function will give you the slope. The second derivative will give you the rate of change of the slope (the slope of the slope).
Since the question is asking about the change in the slope of the equation $y = 4x-x^3$, the first order of business would be to get the equation for the slope (which is the derivative, $y^<'>$ or $\frac
This is the equation for the slope of the function. So we don’t let confusing notation get in the way, let’s just call this equation for slope $s$. So, $s = 4 – 3x^2$.
Now we want to know how the slope is changing with time. We know that $x$ is increasing at a rate of $1 \over 3$ units/sec and we want to know how the slope changes with small changes in $x$. In other words, we were given $\frac
Substituting known values, $$\frac
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.
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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.
Solve for your unknown variable.
Inflating and deflating balloon related rates problems
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Finding rate of change of the radius, given rate of change of volume
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
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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Unformatted text preview: Aim: How do we solve related rates(part 3)? Get Ready: A cylindrical fish tank is filling with water at a m3 72 rate of min . The radius of the tank is 6m and the height is 10 m. How fast is the level rising? a. What is the question asking? b. What is given? Are they changing or unchanging? c. Write an equation relating the variables: d. Differentiate Implicitly: e. Plug in “snapshot” values: II. Extended Practice 1. A rectangular well is 6 feet long, 4 feet wide, and 8 feet deep. If water is running into the well ft 3 3 at the rate of sec , find how fast the water is rising? 2. A spherical hot air balloon is being inflated. If ft 3 2 air is blown into the balloon at the rate of sec , a. find how fast the radius of the balloon is changing when the radius is 3 feet. b. find how fast the surface area is increasing at the same time. 3. A child flies a kite which is 120 ft. directly above the child’s hand. If the wind carries the kite horizontally at the rate of 30 ft./min, at what rate is the string being pulled out when the length of the string is 150 ft.? Aim: How do we solve related rates(part 3)? 4. Ryan is flying a kite which is 100 feet above the ground. If the string is pulled out at the rate of 10 ft/sec (the wind carries the kite horizontally), what is the rate of change of the angle the kite makes with the vertical height when the angle is 30° . 5. A plane is flying west at 500 ft./sec at an altitude of 4000 ft. The plane is being tracked by a searchlight on the ground. If the light is to be fixed on the plane, find the rate of change in the angle of elevation of the searchlight after the plane has traveled a horizontal distance of 2000 ft. 6. How fast does the radius of a spherical soap bubble change when you blow into it at the rate cm 3 10 of sec at the time when the radius is 2 cm? 7. Water runs into a conical tank at the rate of ft 3 9 min . The tank has a height of 10 feet and a radius of 5 feet(radius is half the height). How fast is the water level rising when the water is 6 feet deep? Aim: How do we solve related rates(part 3)? 8. Two trucks leave a factory at the same time. Truck A travels east at 40 mph and Truck B travels north at 30 mph. How fast is the distance between the trucks changing in 30 minutes. 9. Two jets are both flying at 520 mph towards an airport. Plane A is flying south and is 50 miles from the airport while plane B is flying west and is 120 miles from the airport. How fast is the distance between the two planes changing at this time? 10. A spherical lollipop is losing volume at a 1 in 3 steady rate of 4 min . How fast will the radius be decreasing when the lollipop has a radius of 3/8 inches. 11. A rowboat is pulled toward a dock through a ring on the dock 6 feet above the boat. If the rope is hauled in at 2 ft./sec, how fast is the boat approaching the dock when there is 10 feet of rope connecting the boat. Aim: How do we solve related rates(part 3)? Similar Triangles 12. A. A 5 foot girl is walking toward a 20 foot lamppost at a rate of 6 ft./sec. How fast is the tip of her shadow (cast by the lamp) moving? What is being asked(goal)? dy dt What is given? What do I know about the triangles? “Similar” y= x+s s y y−x y = = s = y − x 5 20 5 20 3y = 4x 3y = 4x Differentiate Implicitly Plug and Solve. B. How fast is the length of the girl’s shadow changing when she is 12 feet from the lamppost. ds Goal: dt Given: Same as previous Snapshot: x=12 s y = Formula: 5 20 Differentiate: Solve: 13. Ryan and Franny leave home for work in the morning. Ryan walks north at a rate of 8 ft./sec while Franny walks east at a rate of 6 ft./sec. How fast is the distance between Ryan and Franny changing when Ryan is 20 ft. and Franny is 15 ft. from home? 14. Ohm’s law for certain electrical circuits is given by V = I * R where V is the voltage, I is the current in amperes, and R is the resistance in ohms. Suppose V is increasing at the rate of one volt per second while I is decreasing at the rate of one third ampere per second. Let t be time in seconds. Find the rate at which the resistance is changing when there are 12 volts passing through the circuit and current is 2 amperes. Hint: R is implied. .
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